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Dual Nature Of Matter And Radiation Quantum Theory Introduction

Quantum theory or quantum physics Is the mainstay of modern physics. This theory is primarily applicable to the microscopic world, i.e., the physics of the atomic domain. The study of tills brunch started almost at the beginning of the twentieth century.

In quantum theory, we come across those phenomena or facts that are beyond our common experience and are non-realistic. Scientists who established the main foundation of this theory were also surprised to see the inferences obtained horn theoretical analysis of the theory. However, nil experiments conducted so far have strengthened the base of the theory.

Quantized quantity and quantum

Some quantities, obtained in daily life, can have only chosen values. These values are obtained, generally, by multiplying a primary value by an integer. Such quandaries are called quantized qiiantldcs and the primary value is called a quantum of the respective quantity.

As, the currency is quantized and previously, in Indian currency, 1 paisa was its quantum. It was possible to pay 1 rupee 6 paise or 106 paise but payment of 106.5 paise was not possible.

Read and Learn More Class 12 Physics Notes

Scientist Max Planck propounded his quantum theory in 1900 AD. In spite of the astounding success of the wave theory of light, this theory cannot explain phenomena like black body radiations, photoelectric effect, atomic spectra, etc. To explain black body radiation spectrum, Max Planck introduced quantum theory. Later, the concept of photon particles, introduced by Einstein, established the theory more firmly.

Properties of an electron

1. Charge: Electron is negatively charged. The magnitude of charge of an electron is,

e = 1.6 ×10^-19 C in SI

= 4.8 × 10^-10 esu of charge in the CGS system

From different experiments, we learned that the charge of a body is a quantized quantity and the quantum of charge is the charge of an electron (e). So values of charges can be +2e, -5e, 1000, etc. but values like 1.5e, -2.be are nonrealistic.

2. Rest mass: Rest mass of an electron

m₀ = 9.1 × 10^-31 kg = 9.1 × 10^-28 g

If the speed of an electron In much lens than the speed of light, there In no difference between its rest mass (m) and effective mass (m). Thus men of electron, m = m₀

3. Kinetic energy of an electron electronvolt:

The velocity of electron Incrcane on being attracted by a positive potential and hence Its kinetic energy also Increases, On the other hand, when repelled by a negative potential, the velocity of the electron decreases. The kinetic energy of an electron Is usually expressed In the electronvolt (eV) unit.

Electronvolt Definition:

The change In kinetic energy of an unbound electron, as it travels across a potential difference of IV, is called 1eV.

1eV= charge of an electron × IV

= 1.6 × 10^-19C × IV  = 1.6 × 10^-19 J

∴ C.V = J

= 1.6 × 10^-19×10⁷ erg = 1.6 × 10^-12 erg

Electronvolt Is a very small unit compared to erg or joule. Hence, It is mainly used In nuclear or atomic physics only.

1keV = 10³eV; 1 MeV = 10⁶eV

Quarks

The discovery of quarks by Gell-Mann In 1964 has destroyed the myth that nucleons (protons and neutrons) are the fundamental particles of matter that are incapable of further division and that the charge on the electron was the smallest possible, charge existing In nature.

Quarks have been identified as the fundamental charged particles constituting baryons and mesons. So far, six quarks with their corresponding antiquarks \((\bar{u} \bar{d} \bar{c} \bar{s} \bar{t} \bar{b}\)) have been detected: up (u), down (d), charm (c), strange(s), top (f), bottom (b), with electric charge +\(\frac{2}{3}\)e, and – \(\frac{1}{3}\) e [to be taken alternately in that order], e being the electronic charge.

Thus, for example, a proton is composed of u, u, d, while a neutron is composed of u, d, d, held by mediator particles called gluons. Mesons are composed of quark-antiquark pairs. Incidentally, in the current view.

All matter consists of three kinds of particles: Leptons, quarks, and mediators, (details are beyond the scope of the present discussion).

Dual Nature Of Matter And Radiation Quantum Theory Numerical Examples

1. What is the energy of a photoelectron, in electronvolt, moving with a velocity of 2 × 10⁷ m .s^-1? (Given, mass of electron = 9.1 × 10^-28 g )
Solution:

Mass of electron = 9.1 × 10^-28 g = 9.1 × 10^-31 kg

The kinetic energy of the electron

= ½mv² = ½ × ( 9.1 × 10^-31)×( 2 × 10⁷)²J

= ½ × \(\left\{\frac{9.1 \times 10^{-31} \times\left(2 \times 10^7\right)^2}{1.6 \times 10^{-19}}\right\}\)eV

= 1137.5 eV

Dual Nature Of Matter And Radiation Photoelectric Effect

Photoelectric emission Definition:

The emission of electrons from matter (metals and non-metallic solids) as a consequence of the absorption of energy from electromagnetic radiation of very short wavelengths (such as visible and ultraviolet radiation), is called photoelectric emission.

Observation of Hertz and Contemporary Scientists

In 1887 German scientist Hertz observed that when ultraviolet rays fell on the negative electrode of a discharge tube, electric discharge occurred easily.

Subsequently in Hallwach’s experiment two zinc plates were placed in an evacuated quartz bulb and when ultraviolet rays fell on the plate connected to the negative terminal of the battery, immediately a current was found to flow in the circuit. But when the ultraviolet rays fell on the positive plate, there was no flow of current. He also noticed that, as soon as the ultraviolet rays were stopped, the current also stopped. Hallwachs however could not explain this phenomenon

In 1900 Lenard proved that when ultraviolet rays fell on a metallic plate, electrons were emitted from the plate and current was constituted due to the flow of electrons. Since in this case, the flow of current is due to light, it is called the photoelectric effect (photo = light). For this phenomenon, light of short wavelength or high frequency is more effective than light of long wavelength or low frequency. Alkali metals.

For example: Lithium, Sodium, Potassium, etc., exhibit a photoelectric effect even in ordinary visible light

Lenard’s experiment

G is an evacuated glass bulb with a quartz window Q on its lower face C is a metal plate, kept at potential -V. Another plate A, having a hole at its center is kept at zero potential by earthing. Hence the potential difference between anode and cathode = 0- (-V) = V.

Now, the cathode is illuminated by a monochromatic beam of light, entering through the window. Here ultraviolet rays or visible light of small wavelengths are used according to the nature of cathode plate metal.

Suppose, due to the incidence of light, cathode C emits a beam of negatively charged particles having charge -q. These charged particles are attracted towards the positive plate A. So, the kinetic energy of each charged particle, just before reaching the plate is,

⇒ \(\frac{1}{2} m v^2=q V \quad \text { or, } \frac{q}{m}=\frac{v^2}{2 V}\) ………………………………… (1)

Where, m = mass of each charged particle and v = velocity of the charged particle

A beam of these particles, passing through the hole of the anode is incident on the plate P, which is connected to an electrometer to detect the current. Now, between A and P, a magnetic field B is applied perpendicularly upward concerning the plane of the paper.

Due to this field, the charged particles are forced to move in circular paths. By controlling the magnetic field B, the particles are made incident on the plate D where they follow a circular path of radius of curvature R. The electrometer, connected with the plate D shows the current. Here, the magnetic force, acting on each particle

⇒ \(\vec{F}=-q \vec{v} \times \vec{B}\)

As \(\vec{v} \text { and } \vec{B}\) both are perpendicularÿ to each other, the magnitude of this force,

F =  \(\vec{v}\) = qvBsin 90°

= qvB

This force acts, as a centripetal force for the revolving particle.

⇒ \(q v B=\frac{m v^2}{R} \quad \text { or, } \frac{q}{m}=\frac{v}{B R}\)

Or, \(\left(\frac{q}{m}\right)^2=\frac{v^2}{B^2 R^2}\)………………………………….(2)

Dividing the equation (2) by equation (1) we get,

⇒ \(\frac{q}{m}=\frac{2 V}{B^2 R^2}\) ……………………………………. (3)

The value of \(\) can be evaluated by putting the values of V, R, and B in equation (3). From the experimental result thus obtained, Lenard, had shown that the value of \(\frac{q}{m}\) is the same as the specific charge of the electron,\(\frac{e}{m}\)  (= 1.76 × 10¹¹C . kg^-1) previously known. From this result, it could be concluded that the emitted negatively charged particles from the cathode are electrons

Electrons emitted in this manner, are called photoelectrons. With proper arrangements, the motion of photoelectrons can be made unidirectional. The stream of unidirectional photoelectrons thus produced, develops a current, namely, photoelectric current

Work function

The minimum energy required to remove an electron from the surface of a particular substance to a point just outside the surface is called the work function of that substance.

Here the final position of an electron is far from the surface on the atomic scale but still close to the substance on a macroscopic scale.

Work function depends only on the nature of the metal and is independent of the method of acquiring energy by the electron. Work function is measured in electronvolt. Alkali metals like sodium and potassium have work functions lower than that of other metals but, nowadays, for photoelectric emission, suitable alloys are mostly used.

Demonstrative Experiment

An evacuated glass bulb G with a quartz windowing is used. Through the window, a monochromatic beam of light is incident a plate T that can emit electrons. Plate T is generally coated. With an alkali metal (like sodium or potassium). When plate C is kept at a positive potential concerning T, it attracts photoelectrons emitted from T. Hence a current is set up in the circuit monochromatic.

This force acts, as a centripetal force for the revolving particle. qvB-Sg or, which can be recorded by the galvanometer G’. T is called photocathode and C is called anode. Rheostat Rh, in series with battery B, can be used to increase or decrease the potential difference V. Using the commutator C’, C can also be kept at negative potential concerning T.

Ampere-Volt Characteristics Stopping Potential

Stopping Potential Definition:

The minimum negative potential of anode concerning photocathode, for which photoelectric current becomes zero, is called stopping potential

Keeping the frequency of light constant

Graphs are drawn showing the dependence of I on V; where I = photoelectric current and V = potential difference between anode and cathode. Here a monochromatic light is used so that the frequency (f), of the incident light remains constant. 1 and 2 in this graph, represent I-V characteristics for different intensities of incident light.

Detailed study of this graph reveals the following facts:

1. Saturation current: Characteristic curves become horizontal for higher values of V. This shows that saturation current has been achieved. So, all the electrons, emitted by the photocathode, have been collected by the anode.
2. Effect of intensity of incident light: At constant V, with a decrease in intensity i.e., the brightness of the incident, monochromatic light, the photoelectric current also decreases. Photoelectric current is directly proportional to the intensity of the incident light.
3. Stopping potential: A When a negative potential is applied to anode concerning the photocathode, photoelectric Current docs do not show hut decreases gradually with an Increase In negative potential on C.

This Indicates that photoelectrons possess some Initial kinetic energy due to which they can reach the anode, overcoming the repulsive force of negative potential, With an Increase In the negative potential of the anode, the photoelectric current becomes zero ultimately. The negative potential at this stage Is called the stopping potential, cut-off potential, or cut-off voltage, VQ

The value of stopping potential depends on two factors:

1.  Nature of the surface of the photocathode and
2. Frequency of the incident lightAnalysing graphs 1 and 2, we see that stopping potential VQ does not depend on the intensity of incident light.

An increase in the intensity of incident light only increases the t£p value of the saturation current

Keeping the intensity of light constant

Graphs are drawn showing the dependence of I on V; where = photoelectric current and V = potential difference between anode and cathode. Lights of different frequencies but of the same intensity are used as incident light on the cathode. In this figure, graphs 1 and 2 represent the I-V characteristic curves for different frequencies

In this case, as the frequency of incident light on the photocathode increases the y-value of stopping potential also increases and vice versa. But SEflyÿtion current is independent of the frequency of light

Relation between kinetic energy of photoelectrons and stopping potential:

When, the anode potential becomes equal to the stopping potential V₀, photoelectrons with even the highest kinetic energy, cannot reach the anode. Hence, the maximum kinetic energy of photoelectron (E_max) = loss of energy of electron for overcoming negative potential V₀.

When an electron of charge e overcomes a negative potential VQ, loss of energy of electron = eV₀. Thus,

E_max = eV₀ …………………….. (1)

Also, if the maximum initial velocity of the electron is v_max, then

⇒ \(E_{\max }=\frac{1}{2} m v_{\max }^2\)

m = Mass of electron

⇒  \(\frac{1}{2} m v_{\max }^2=e V_0\)

⇒  \(v_{\max }=\sqrt{\frac{2 e V_0}{m}}\)

The maximum kinetic energy of an electron Is independent of the Intensity of light

Dual Nature Of Matter And Radiation Photoelectric Effect Numerical Examples

Example 1. The stopping potential for monochromatic light of a metal surface is 4V. What is the maximum kinetic energy of photoelectrons?
Solution:

From the relation, £max = eV₀ we can say that, when the stopping potential is 4 V, maximum kinetic energy, E_max = 4eV.

Example 2. For a metal surface, the ratio of the stopping potentials for two different frequencies of incident light is 1: 4. What is the ratio of the maximum velocities in the two cases? max 
Solution:

Stopping potential oc maximum kinetic energy. Again maximum kinetic energy ∝ (maximum velocity)². Hence, stopping potential ∝ . (maximum velocity)².  If v₁ and v₂ are the maximum velocities in the two given cases, respectively then

⇒ \(\frac{1}{4}=\left(\frac{v_1}{v_2}\right)^2\)

i.e., v₁= v₂

= 1:2

Threshold Frequency or Cut-off Frequency

Threshold Frequency Definition:

The minimum frequency of incident radiation which can eject photoelectrons from the surface of a sub¬stance, is called the threshold frequency for that substance.

Frequency versus stopping potential graph:

In photoelectricity, neither the stopping potential nor maximum energy of photoelectrons is a constant quantity. Magnitudes of both V₀ and E_max depend on

1. The frequency of the incident light and
2. The nature of the surface of the substance used.

The relation between frequency and stopping potential for different metals is shown graphically

Characteristics of the graph:

For each substance, there is a certain frequency f₀ of incident light, for which the maximum energy of photoelectrons becomes zero.

In other words, there is no emission of photoelectrons. Hence, whatever may be the intensity of incident radiation, no electron can leave the metal, surface for the light incident with a frequency equal to or less than f₀ i.e., photoelectric emission stops. This f₀ is the threshold frequency.

The maximum wavelength corresponding to the minimum frequency f₀ is called the threshold wavelength. It is given by,

⇒ \(\lambda_0=\frac{c}{f_0}\) c= speed of light

Discussions:

1. From what we see, the stopping potential or maximum kinetic energy increases as the frequency of incident radiation increases. Hence in practice, almost in all cases, ultraviolet rays are used, as the frequency of ultraviolet rays is much more than that of visible violet ray
2. Alkali metals (For example,  sodium, potassium, cesium, etc.) emit photoelectrons even for comparatively low-frequency light.

Characteristics of Photoelectric Effect

1. Photoelectric current is directly proportional to the intensity of the incident light.
2. The maximum velocity or kinetic energy of the photoelectron is independent of the intensity of incident light. On the other hand, maximum velocity or kinetic energy increases with an increase in the frequency of the incident light.
3. For a given material, there exists a certain minimum frequency (threshold frequency, f₀) of incident light bel which no photoelectrohs are emitted. Photoelectric effect is usually prominent in the range of frequencies of yellow, to ultraviolet radiation.
4. Threshold frequency is different for different materials. Photoelectrons emitted from the surface of a substance have any velocity between zero and maximum velocity.
5. The emission of a photoelectron is an instantaneous process, which means that photoelectrons are emitted as soon as light falls on the metal surface. There is practically no time gap between these two incidents.
6. The emission of photoelectrons makes the rest of the surface very slightly positively charged (this principle is followed in making photovoltaic cells). of electrons in, the photoelectric effect does not depend on the temperature of the surface

Photoelectric Cell

Photoelectric Cell Definition:

Cells, designed to convert light energy to electri¬ cal energy, based on the principle of the photoelectric effect, are photoelectric cells.

Photo-voltaic cell: 

In this cell, an emf is developed directly from the photoelectric effect. This cell is an electric cell as it works as a source of EMF without any aid of an auxiliary cell.

Description:

A copper plate is taken and on one surface, a layer of cuprous oxide (Cu₂O) is deposited by the method of oxidation. Over this layer of Cu₂O, using the evaporation technique, a very thin coating of gold or silver is applied. This coating is so thin that light, especially ultraviolet rays, can easily penetrate it and reach the layer of Cu₂O.

Working principle:

When light Tails on Cu₂O, photoelectrons are emitted. These electrons instantaneously spread over the gold or silver coating. As a result, this coating achieves a negative potential concerning the copper plate, i.e., an effective emf is developed between the copper plate and the gold or silver coating. This emf sets up a current in the load resistance (R_L) used in the external circuit. This current is directly proportional to the intensity of incident light.

Use of photoelectric cells:

1. Automatic switch: In fire alarms, railway signals, streetlights, etc., automatic switches are made using photoelectric cells. QFJ Sound recording and reproduction: Photoelectric cells are used in sound recording and its reproduction, on the soundtrack of cinema and television.

2. Solar cell: A photo-voltaic cell is used to construct a solar battery. This battery is indispensable in spaceships and artificial satellites-

3. Television camera: In this device, the photoelectric cell is j used to convert optical images into video signals for television broadcasts.

4. Automatic counting device: A photoelectric cell is used to make an automatic counting device. The number of viewers entering or emerging from a hall can be counted with this device.

5. Automatic camera: A photoelectric cell is used in automatic cameras where the controlling of picture quality depends on the intensity of light.

Failure of Wave Theory of Light

The photoelectric effect cannot be explained by the concept of the wave theory of light. The following observations are in contradiction with the wave nature of light:

1. Maximum kinetic energy of photoelectrons: According to wave theory, energy carried by light waves increases with an increase in the intensity of light. Hence, when the material surface is illuminated with highly intense light, the kinetic energy of emitted photoelectrons will be very high and it will not have any upper limit. But from the value of stopping potential V₀, we see that the kinetic energy of the photoelectron cannot be more than e V₀.
2. Threshold frequency: A highly intense light wave, even of a frequency less than threshold frequency f₀, carries a large amount of energy. This energy is sufficient to cause photoelectric emission. But no photoelectron is emitted in such cases. On the other hand, even a very low-intensity light beam of frequency greater than f₀ can start photoelectric emission.
3. The photoelectric effect is instantaneous: The energy carried by light waves incident on the surface of a substance needs a little time to be centralized in the limited space of an electron. Hence, there should be a time gap between the incidence of light waves on the surface and the emission of photoelectrons from the surface. But the photoelectric effect is instantaneous; that means there is no time gap between incidence and emission

Dual Nature Of Matter And Radiation Quantum Theory of Radiation

Photon:

It has already been stated that the photoelectric effect cannot be explained in terms of the wave theory of light. In 1905, Einstein used Planck’s quantum theory and introduced the concept of photon particles. Thus, he could explain the photoelectric effect. The particle concept of radiation is the basis of quantum theory.

The basic point of the theory is that electromagnetic radiation is not a wave by nature but consists of a stream of particles called photons.

Photons Properties:

The main properties of photons are:

1. Photons are electrically neutral.
2. Photons travel with the speed of light, which does change under any circumstances, (velocity of light, c = 3 × 10⁸ m . s^-8 )
3. The energy carried by a photon, E = hf; where f = frequency of radiation and h = Planck’s constant. The energy radiated increases with the increased number of photons in its stream and hence the intensity of radiation also increases
4. According to Einstein’s theory of relativity, the vast mass of a particle is zero, if it trawls at the speed of light Hence, the vast mass of each photon is zero.
5. According to the theory of relativity, if the rest mass of a particle is m₀(1 and its momentum is p, the energy of the particle,

E = \(\sqrt{p^2 c^2+m_0^2 c^4}\),  In case of a photon, m₀ = 0 ,

Hence E = pc, or p = \(\frac{E}{C}\)  = hf/c. Thus, despite the photon being a massless particle, it has a definite momentum.

Planck’s constant

It is a universal constant

Unit of h in SI = \(\frac{\text { unit of } E}{\text { unit of } f}=\frac{\mathrm{I}}{\mathrm{s}^{-1}}\) = j- s

Unit of h in CGS system = erg .s.

This unit is the same as the unit of angular momentum. Thus h is a measure of angular momentum

Value of h = 6.625 × 10^-34J . s =  6.625 × 10^-27J . s

Relation between the wavelength of radiation and the photon energy

Energy of a photon, E = hf = \(\)

1 eV= 1.6. × 10^-19J . s and 1 A° = 10^-10m

Hence, expressing E in the eV unit and λ in the A° unit

λ A° = λ  × 10^-10 m EeV = E × (1.6 ×10^-19) J

Hence, E × (1.6 ×10^-19) J =   \(\frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\lambda \times 10^{-10}}\)

Or,  E = \(\frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\left(1.6 \times 10^{-19}\right) \times \lambda \times 10^{-10}} \approx \frac{12422}{\lambda}\)

Usually, the number on the right-hand side is taken as 12400.

Hence

E = 12400/ λ(inÅ) eV

Dual Nature Of Matter And Radiation Quantum Theory of Radiation Numerical Examples

Find the energy of a photon of wavelength 4950 A In eV ( h = 6.62 × 10^-12  erg .s ). What is the momentum of this photon?
Solution:

A = 4950 A = 4950 ×10^-8  cm

Hence energy of a photon

E = hf \(\frac{h c}{\lambda}=\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{4950 \times 10^{-8}}\)

= 4.012 × 10^-12  erg

= \(\frac{4.012 \times 10^{-12}}{1.6 \times 10^{-12}}\)

= 2.5 eV

Momentum of photon

p = \(\frac{E}{c}=\frac{4.012 \times 10^{-12}}{3 \times 10^{10}}\)

= 1.34 × 10^-22 dyn. s

Example 2. Find the photons emitted per second by a source of power 25W. Assume, the wavelength of emitted light = 6000A°. h= 6.62 × 10^-34 J.s.
Solution:

λ = 6000 A° = 6000m × 10^-10 ,c = 3 × 10⁸ m.s^-1

Number of photons per shroud emitted by a source of power P Is,

n = \(\frac{p}{h f}=\frac{p}{h c / \lambda}=\frac{P \lambda}{h c}\)

n= \(\frac{25 \times 6000 \times 10^{-10}}{6,62 \times 10^{-34} \times 3 \times 10^8}\)

= 7.55 × 10¹⁹

Example 3. The wavelength of ultraviolet light Is 3 × 10^-5cm, What will be the energy of a photon of this light, in eV?  (C = 3 ×10^-10 m.s^-1 )
Solution:

Planck’s constant, h = 6.625 × 10^-27  erg .s

Wavelength, λ = 3 × 10^-5 cm

The energy of a photon

E = \(\frac{h c}{\lambda}=\frac{\left(6.625 \times 10^{-27}\right) \times\left(3 \times 10^{10}\right)}{3 \times 10^{-5}} \mathrm{erg}\)

= \(\frac{\left(6.625 \times 10^{-27}\right) \times\left(3 \times 10^{10}\right)}{\left(3 \times 10^{-5}\right) \times\left(1.6 \times 10^{12}\right)} \mathrm{eV}\)

= 4.14 eV

Example 4. Work functions of three metals A, H, and C arc 1.92 eV, 2.0 eV, and 5.0 eV respectively. Which metal will emit photoelectrons when a light of wavelength 4100A is Incident on the metal surfaces?
Solution:

Energy of incident photon

E = hf

= \(h \cdot \frac{c}{\lambda}=\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{4100 \times 10^{-8}} \mathrm{erg}\)

= \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{1.6 \times 10^{-12} \times 4100 \times 10^{-8}} \mathrm{eV}\) ≈ 3.03 eV

Hence, this photon will be able to emit photoelectrons from metals A and B but not from C

Example 5. In a microwave oven, electromagnetic waves are generated having wavelengths of the order of 1cm. Find the energy of the microwave photon. (h = 6.33 × 10^-34 J.s).
Solution:

= 1 cm = 10^-2 m; c = 3 × 10⁸ m .s^-1

E = hf

= \(\frac{h c}{\lambda}=\frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{10^{-2}}\)

= 1.989 × 10^-23 J

= \(\frac{1.989 \times 10^{-23}}{1.6 \times 10^{-19}}\) eV

= 1.24 × 10^-4 10 eV

Dual Nature Of Matter And Radiation Einstein’s Photoelectric Equation

Einstein made the following assumptions to explain the photoelectric effect.

Einstein’s postulates

IQ A beam of light is incident on a metal surface as streams of photon particles. The energy of each photon having frequency is, E = hf (h = Planck’s constant).

Incident photons collide with electrons of metal. The collision may produce either of the two effects: The o photon gets reflected with its full energy hf or the Q photon transfers its entire energy hf to the electron.

Einstein used the quantum theory of radiation to explain the photoelectric effect.

The entire energy hf of die incident photon, when transferred to an electron of the metal, is spent in two ways:

1. A part of the energy is spent to release the electron from the metal whose minimum value is equal to the work function WQ of the metal. But, due to the interaction of positive and negative charges inside a metal, most of the electrons need more energy than WQ for release.
2. Rest of the energy, changes to the kinetic energy of released electrons. These moving electrons are photoelectrons that can set up photoelectric current. If energy absorbed by the electron to leave from the metal surface is the least i.e., WQ, the emitted electron attains maximum kinetic energy

Hence, hf = W₀+ E_max

E_max= hf – W₀ ………………………..(1)

If the mass of an electron is m and the maximum velocity of a photoelectron is equation (1) we get,

½mv²_max =  hf -W₀ ……………………..(2)

Also, if V₀ is the stopping potential for the incident light of frequency/, then Emax = eV0 (e = charge of an electron) Hence, from equation (1),

eV0 = hf-W₀

Equations (1), (2), and (3) are practically the same. Each of these is called Einstein’s photoelectric equation. In most collisions of photons with electrons, there is no energy transfer and the photons are reflected with their full energy hf. Hence the probability of photoelectric emission from the metal surface is low and the strength of photoelectric current never becomes very high

Explanation of Photoelectric Effect by Quantum Theory

Einstein’s photoelectric equation is based on the quantum theory of radiation.

This equation correctly explains the following observations in the photoelectric effect:

1. The maximum kinetic energy of photoelectrons:

Work function W₀ is a constant for a fixed material surface; also the frequency of monochromatic light, f is a constant. Hence, E_max = hf- W₀, is also a constant. Thus, for fixed wavelength or fixed frequency of incident light, whatever may be the intensity, emitted photoelectrons cannot attain kinetic energy more than E_max

2. Threshold frequency:

The work function, JV0 is also a constant for a fixed material surface. If the frequency of incident light is decreased then as evident from equation E_max. = hf- W₀, the value of £max will come down to zero for a certain value of f = f₀ (say).

0= hf₀– W₀ Or, hf₀ = W₀

Or,  f₀ = \(=\frac{W_0}{h}\) …………………………………(1)

If the value of f happens to be below f₀, the energy of the photoelectron turns out to be negative and there is no photoelectron emission. Hence, f0 is the threshold frequency. Putting W₀ = hf₀ in Einstien’s photoelectric equation, E_max  = hf – W₀ ,we get,

E_max  = hf – hf₀ = h( f- f E_max  = h (f -f₀ ) ………………………………(2)

Also, if λ and λ₀ are the wavelength of the incident light and”threshold wavelength for the. metal surface respectively, then

f = c/λ and f₀= c/λ₀

c = Speed of light

Putting these values in equation (2), we get

E_max=  \(h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\) …………………………..(3)

Equations (2)equation. and (3) are the other forms of the arc of Einstein’s photoelectric equation.

3. Photoelectric emission is instantaneous:

Energy transfer takes place between a photon of energy hf and an electron In the metal due to their elastic collision. Hence, there is no delay in photoelectron emission after the incidence of light.

4. Dependence of photoelectric current on the intensity of incident light:

An increase in the intensity of incident light of a constant frequency increases the number of photons incident on the surface of the material.

Hence, the number of collisions between the photons and electrons increases. So, more electrons are emitted which increases photoelectric current This agrees with the results obtained experimentally

Graphical representations of Einstein’s equation

1. Frequency (f) versus stopping potential ( V₀) graph:

From equation (3) we have,

eV₀ = hf- W₀

Or, V₀ =  \(\frac{h}{e} f-\frac{W_0}{e}\)

The graph obtained by plotting V₀ against f is a straight line of the type, y = mx + c Knowing the charge of an electron e,  Planck’s constant h can be calculated from the slope of the graph.

Work function W₀ can be obtained from the Intercept \(\) in the y-axis. Also, the intercept with the x-axis gives the threshold frequency f₀.

It is important to note that the gradient of the straight Line Is \(\) for all substances but Intercepts from y- and x-axes, and f₀, respectively are different for different substances.

2. Frequency (f) versus maximum kinetic energy (E_max) graph:

From equation (1)

E_0max = hf – W₀

The graph obtained by plotting E_max against f is a straight line. Comparing the above relation with y = mx + c, we note

The slope of the graph is h, x -the axis Intercept is f₀ and the y-axis intercept is – W₀.

Dual Nature Of Matter And Radiation Einstein’s Photoelectric Equation Numerical Examples

Example 1. The work function for zinc Is 3.6 eV. If the threshold frequency for zinc is 9 × 10¹⁴ cps, determine the value of Planck’s constant. (1eV = 1.6× 10^-12 erg).
Solution:

Work function, W₀ = 3.6eV = 3.6 ×  1.6 × 10^-12 erg

Threshold frequency, f₀ = 9 × 10¹⁴ cps = 9 × 10¹⁴ Hz

As, W₀ = hf₀

So, h = \(\frac{W_0}{f_0}=\frac{3.6 \times 1.6 \times 10^{-12}}{9 \times 10^{14}}\)

= 6.4 x  10^-27  erg.s

Example 2. The maximum kinetic energy of the released photoelectrons emitted from metallic sodium, when a light Is an Incident on It, is 0,73 eV. If the work function of sodium Is 1.82 eV, find the energy of the Incident photon In eV. Find the wavelength of incident light. (h = 6.63 × 10^-27erg s, eV = 1.6 × 10^-12 erg )
Solution:

From Einstein’s photoelectric equation, Emax hf – W₀, we get the energy of the incident photon as,

E = hf = E_max + W₀ = 0.73 + 1.82 = 2.55 eV

Hence wavelength of incident light, \(\)

∴ λ = \(\frac{h c}{E}=\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{2.55 \times 1.6 \times 10^{-12}}\) cm

λ  = \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10} \times 10^8}{2.55 \times 1.6 \times 10^{-12}}\)A°

λ = 4875 A°

Example 3. Light of wavelength 6000A° is Incident on. a metal. To. release an electron from the metal surface, 1.77 eV of energy Is needed. Find the kinetic energy of the fastest photoelectron. What is the threshold frequency of the metal (h = 6.63 × 10^-27erg s, eV = 1.6 × 10^-12 erg ).
Solution:

The energy of a photon,

hf = \(h \frac{c}{\lambda}=\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{6000 \times 10^{-16}} \mathrm{erg}\)

= \(\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{6000 \times 10^{-8} \times 1.6 \times 10^{-12}} \mathrm{eV}\)

= 2.07 eV

As per Einstein’s photoelectric equation,

E_max  = hf – W₀= 2. 07- 1. 77 = 0. 3eV

Threshold frequency,

⇒ \(\frac{W_0}{h}=\frac{1.77 \times 1.6 \times 10^{-12}}{6.62 \times 10^{-27}}\)

= 4.28 × 10¹⁴ Hz

Example 4. The photoelectric threshold wavelength for a metal is  3800A°. Find the maximum kinetic energy of the emitted photoelectron, when ultraviolet radiation of length 2000A° is incident on the metal surface. Planck’s constant, h = 6.62 × 10^–34 J s
Solution:

Maximum kinetic energy of photoelectron,

E_max = hf-W₀ = hf-hf₀ = \(\frac{h c}{\lambda}-\frac{h c}{\lambda_0}\)

= hc \(h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)=h c \frac{\lambda_0-\lambda}{\lambda \lambda_0}\)

In this case, the wavelength of the incident light,

λ = 2000 A° = 2000 × 10^-10 m = 2 × 10^-7 m

Threshold wavelength

λ₀ = \(3800 \times 10^{-10} \mathrm{~m}=3.8 \times 10^{-7} \mathrm{~m}\)

Hence, E_max  = \(\left(6.62 \times 10^{-34}\right) \times\left(3 \times 10^8\right) \times \frac{(3.8-2) \times 10^{-7}}{3.8 \times 2 \times 10^{-14}} \mathrm{~J}\)

= \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8 \times 1.8}{3.8 \times 2 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV}\)

= 2.94 eV

Example 5. The threshold wavelength for photoelectric emission from a metal surface is 3800 A. Ultraviolet light of wavelength 2600A° is incident on the metal surface, 

1. Find the work function of the metal and
2. Maximum kinetic energy of emitted photoelectron. ( h= 6.63  × 10^–27 erg.s )

Solution:

Threshold wavelength

λ₀ = 3800 A° =  3800 × 10^–8  cm

∴ Work function,

W₀ – hf₀ = \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{3800 \times 10^{-8}} \mathrm{erg}\)

= \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{3800 \times 10^{-8} \times 1.6 \times 10^{-12}} \mathrm{eV}\)

= 3.27 eV

As Per Einstein’s photoelectric equation, the maximum kinetic energy of photoelectron, E_max = hf-W₀

hf = kinetic energy of the incident photon

= hf₀  \(h f_0 \times \frac{f}{f_0}=h f_0 \frac{c / \lambda}{c / \lambda_0}=h f_0 \frac{\lambda_0}{\lambda}\)

= \(3.27 \times \frac{3800}{2600}\)

= 4.78

E_max = 4.78 – 3.27 = 1.51 eV

Example 6. When radiation of wavelength 4940 A° is incident on a metal surface photoelectricity is generated. For a potential difference of 0.6 V between the cathode and anode, photocurrent stops. For another incident radiation, the stopping potential changes to 1.1V. Find the work function of the metal and wavelength of the second radiation. ( h= 6.63  × 10^–27 erg.s , e = 1.6 × 10^–19 C)
Solution:

For the first radiation, stopping potential V₀ = 0.6V

∴ Maximum kinetic energy of photoelectron,

E_max  = eV₀ = 0. 6eV

Wavelength, A = 4940 A° = 4940 × 10^–8 cm

∴ The energy of an incident photon

= hf = \(h \frac{c}{\lambda}=\frac{6.6 \times 10^{-27} \times 3 \times 10^{10}}{4940 \times 10^{-8}}\)

= \(\frac{6.6 \times 10^{-27} \times 3 \times 10^{10}}{4940 \times 10^{-8} \times 1.6 \times 10^{-12}}\)

= 2.5 eV

If the work function of the metal is WQ, from Einstein’s equation

E_max= hf – W₀

Or,  W₀ = hf – E_max

= 2.5 -0.6

= 1.9 eV

For the second radiation, V’₀ = 1.1V

Hence, E’_max = 1.1 eV

∴ E’_max = hf- W₀

Or, hf’ = E’_max + W₀ = 1.1 + 1.9

= 3.0 eV

Hence, \(\frac{h f}{h f^{\prime}}=\frac{2.5}{3.0} \text { or, } \frac{f}{f^{\prime}}=\frac{5}{6}\)

Or, \(\frac{c / \lambda}{c / \lambda^{\prime}}=\frac{5}{6} \quad \text { or, } \frac{\lambda^{\prime}}{\lambda}=\frac{5}{6}\)

Or, λ’ = \(\lambda \times \frac{5}{6}=4940 \times \frac{5}{6}\)

= 4117 A°(approx)

Example 7. A stream of photons of energy 10.6 eV and intensity 2.0 W.m^–2 is incident on a platinum surface. The area of the surface is 1.0 × 10^–4 m² and its work function is 5.6 eV. 0.53% of incident photons emit photoelectrons. Find the number of photoelectrons emitted per second and the maximum and minimum energies of the emitted photoelectrons in eV. ( 1eV = 1.6 × 10^–19 J)
Solution:

If the intensity of incident light is I, the energy incident on a surface area A is IA. Hence, number of photons incident per second n = \(\frac{I A}{h f}\)

If x% of photons help to emit photoelectrons, the number of photoelectrons emitted per second,

N = \(n \times \frac{x}{100}=\frac{I A x}{h f \cdot 100}\)

Given, I = 2.0 W m^–2 , A = 1.0 × 10^–4 m^–2

hf = 10.6 eV = 10.6 × 1.6 × 10^–19 J  and x = .0.53

N = \(\frac{2.0 \times 1.0 \times 10^{-4} \times 0.53}{10.6 \times 1.6 \times 10^{-19} \times 100}\)

= 6.25 × 10¹¹

Minimum kinetic energy of emitted photoelectron = 0

Maximum kinetic energy, E_max = hf- W₀ = 10.6 – 5.6 = 5 eV

Example 8. At what temperature would the kinetic energy of a gas molecule be equal to the energy of a photon of wavelength 6000A°? Given, Boltzmann’s constant,  k = 1.38 × 10^–23J K^–1, Plank’s constant , h = 6.625 × 10^–34 J. s
Solution:

Let the required temperature be TK. We know, the kinetic energy of a gas molecule

= \(\frac{3}{2} k T=\frac{3}{2} \times 1.38 \times 10^{-23} \times T\)

Again, the kinetic energy of the photon

= \(h f=\frac{h c}{\lambda}=\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}}\)

Hence,

= \(\frac{3}{2} \times 1.38 \times 10^{-23} \times T\)

= \(\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}}\)

T = \(\frac{2}{3} \times \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{1.38 \times 10^{-23} \times 6000 \times 10^{-10}}\)

= 1.6 × 10⁴ k

Example 9.  The ratio of the work functions of two metal surfaces is 1: 2. If the threshold wavelength of the photoelectric effect for the 1st metal is 6000 A°, what is the corresponding value for the 2nd metal surface?
Solution:

If the work function of 1st and 2nd metals be W₀ and W’₀ , respectively then

⇒ \(\frac{W_0}{W_0^{\prime}}=\frac{h f_0}{h f_0^{\prime}}=\frac{h c / \lambda_0}{h c / \lambda_0^{\prime}}=\frac{\lambda_0^{\prime}}{\lambda_0}\)

Or, \(\lambda_0^{\prime}=\lambda_0 \times \frac{W_0}{W_0^{\prime}}\)

= \(6000 \times \frac{1}{2}\)

= 3000 A°

Example 10. The work function of a metal surface is 2 eV. The maximum kinetic energy of photoelectrons emitted from the surface for Incidence of light of wavelength 4140 A is 1 eV. What is the threshold wavelength of radiation for that surface
Solution:

From Einstein’s photoelectric equation,

E_max = hf-W₀

Or hf = E_max  + W₀= 1+2 = 3 eV

Now, \(\frac{h f}{W_0}=\frac{h f}{h f_0}=\frac{f}{f_0}=\frac{c / \lambda}{c / \lambda_0}=\frac{\lambda_0}{\lambda}\)

∴ λ₀ =  \(\lambda \frac{h f}{W_0}\)

= 4140 ×\(\frac{3}{2}\)

= 6210A°

Example 11. The work function of a metal is 4.0 eV. Find the maximum value of the wavelength of radiation that can emit photoelectrons from the metal
Solution:

Let, threshold frequency = f₀ , threshold wavelength λ₀ and work function = W₀

W₀ =\(h f_0=\frac{h c}{\lambda_0} \text { or, } \lambda_0=\frac{h c}{W_0}\)

Given W₀ = 4.0 eV = 4 × 1.6 × 10^–12 erg

We know, h = 6.60 × 10^–27 ergs. And

And c = 3 × 10¹⁰ cm .s^–1

λ₀ = \(=\frac{6.60 \times 10^{-27} \times 3 \times 10^{10}}{1.6 \times 10^{-12} \times 4} \mathrm{~cm}\)

λ₀ = 3.09375 × 10^–5 cm≈ 3094 A°

Example 12. The maximum energies of photoelectrons emitted by a metal are E₁ and E₂ when the incident radiation has frequencies f₁ and f₂ respectively. Show that the Planks comment h and the work function W0 of the metal are \(\)
Solution:

According to Einstein’s photoelectric equation we here,

E= hf₁ – W₀ …………………………………(1)

E= hf₂-W₀ ……………………………..(2)

E₁– E₂ = h(f₁-f₂)

∴ h = \(\frac{E_1-E_2}{f_1-f_2}\) ………………………………….(3)

For E₁ > E₂

Again multiplying equation (1) by f2 and equation (2) by f1 we obtain.

E₁f₂= hf₁f₂ – f₂W₀ …………………………………..(4)

E₂f₁= hf₁f₂ – f₁W₀ ………………………………..(5)

Subtracting equation (5) from equation (4), we get

E₁f₂ – Ff = f₁W₀-f₂W₀= W₂(f₁-f₂)

∴ W₀ = \(\frac{E_1 f_2-E_2 f_1}{f_1-f_2}\)

Example 13. A photoelectric source is illuminated successively by monochromatic light of wavelength λ and λ/2 calculates the work function of the material of the source. If the maximum kinetic energy of the emitted photoelectric in the second case is 3 times that in the first case.
Solution:

We know the kinetic energy of emitted photoelectrons,

k = hf-W₀ = \(\frac{h c}{\lambda}\) – W₀

In the first case,

K₁ = \(\frac{h c}{\lambda}\) – W₀ = \(\frac{h c}{\lambda}\) – W₀ 

In the second case,

K₂ = \(\) – W₀ = \(\) – W₀ 

It is given that, K₂ = 3K₁

Or, 2hc/λ = W₀ 

= 3(hc/λ – W₀ )

2W₀  = hc/λ

∴ W₀  = hc/2λ

Dual Nature Of Matter And Radiation Nature Of Radiation: Wave-Particle Duality

Electromagnetic radiations, if assumed to be streams of photon particles, can explain phenomena like photoelectric emission, blackbody radiation, atomic spectra, etc. However the theory fails to explain other optical phenomena like Interference, diffraction, polarisation, etc.

On die other hand, the wave theory of radiation can interpret these phenomena successfully. Hence, depending on the type of experiment, radiation sometimes behaves like waves and sometimes like a stream of particles. Thus, theory and particle theory are not contradictory but complementary to each other. ‘Ibis Is railed wave-particle duality

Dual Nature Of Matter And Radiation Matter Wave

It has already been stated that radiation shows both wave nature and particle nature. But the fact that matter can show wave nature was unimaginable till 1924 when French physical Louis de Broglie put forward the theory that a stream of material particles may behave as a wave.

Most probably the following reasons led him to such a conclusion:

1. Nature prefers symmetry. Hence, two physical entities, matter and energy must co-exist In symmetry,
2. If radiation can have both particle and wave nature would also possess particle and wave nature,
3. We know that a beam of light, which Is a wave, can transfer energy and momentum at different points of a substance, ‘similarly, a stream of particles can also transfer energy and momentum at different points of a substance. Therefore, this stream of particles may be a matter wave.

de Broglie’s hypothesis:

Matter also consists of waves. For a radiation of frequency f, the energy of a photon

E =hf Or, E = hc/λ

∴ c = fλ

Or, λ = \(\frac{h}{E / c}=\frac{h}{p}\) ………………………………… (1)

Where p is the momentum of the photon.

As per de Broglie’s hypothesis, equation (1) is also applicable to an electron or any other particle. In this case, A gives the wavelength of the electron (or particle) of momentum p and is known as the de Broglie wavelength.

Thus, substituting the values of Planck’s constant h and momentum of the particle p in equation (1), we get the de Broglie wavelength of the wave associated with the moving particle.

Hence a stream of any particle behaves like a beam of light, i.e., like a wave. The wave is known as the matter wave. The wavelength of this matter wave,

λ = \(\frac{h}{p}=\frac{h}{m \nu}\) ……………………………………. (2)

Where, m = mass of the particle, v = velocity of the particle, p = mv = momentum of the particle.

We can make the following inferences from the above relation connecting wavelength (a characteristic of the wave) And

Momentum (a characteristic of the particle) :

1. If v = 0, then λ = ∞, it means the waves are associated with moving material particles only.
2. The de Broglie wavelength does not depend on whether the moving particle is charged or uncharged. It means that matter waves are not electromagnetic waves because electromagnetic waves are produced from accelerated charged particles.
3. If the mass m and the velocity v of the particle are large, the associated de Broglie wavelength becomes very small. If the momentum of the particle increases, the wavelength decreases.
4. The wave nature and particle nature of any physical entity (matter or radiation) are mutually exclusive, i.e., if we consider the particle nature of radiation at any instant, the wave nature of radiation is to be excluded at that instant.

In 1927 C J Davisson and H. Germer of Bell Telephone, laboratories and George P Thomson of the University of Aberdeen, Scot¬ land, were able to show diffraction of electron streams and hence established experimentally the existence of matter waves

Hence, any moving stream of particle or matter exhibits interference, diffraction, and polarisation phenomena which can only be explained with wave theory.

The interference pattern, obtained by using a double slit type of experiment, using about 70000 moving electrons is shown in

Dual Nature Of Matter And Radiation de Broglie Wavelength Of Moving Electron

Let an electron of mass m move with velocity v. Its de Broglie wavelength is given by

λ = h/mv

If the kinetic energy of the electron is K, then

K = ½ mv² Or, v = \(\sqrt{\frac{2 K}{m}}\)

So, de Broglie wavelength is

λ = \(\frac{h}{m \sqrt{\frac{2 K}{m}}}=\frac{h}{\sqrt{2 m K}}\) ………………………………. (1)

Equation (1) is the expression for the de Broglie wavelength ofa moving particle in terms of its kinetic energy.

Now suppose an electron is at rest. It is accelerated through a potential difference V. The kinetic energy acquired by the electron is K = eV; e = charge of the electron.

On substituting the value of K in equation (1), the de Broglie wavelength associated with the electron is given by

λ = \(\frac{h}{\sqrt{2 m e V}}\) ……………………… (2)

Putting the values of h,e,m in ‘equation (2) we have,

λ = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}}\)

= \(\frac{12.27 \times 10^{-10}}{\sqrt{V}} \mathrm{~m}\)

= \(\frac{12.27}{\sqrt{V}}\)

Wavelength of matter-wave

Let the velocity of an electron (mass = 9.1 × 10¹⁰ cm .s^–1), v = 10⁷ m. s^–1. de Broglie wavelength of the electron

λ = \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{\left(9.1 \times 10^{-31}\right) \times 10^7}\)

= \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{\left(9.1 \times 10^{-31}\right) \times 10^7}\)

= 7.3 × 10^-11 cm .s^–1

= 0.73 A°

This wavelength is equivalent to the wavelength of X-rays.

1. Let the mass of a moving marble, m = 10 g = 0.01 kg, and its velocity, v = 10 m s-1. Then de Broglie wavelength of the marble

⇒ \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{0.01 \times 10}\)

= 6.63 × 10^-33 cm .s^–1m

The value of this wavelength is too small to be measured or to be observed by any known experiment Existence of such small wavelengths in electromagnetic radiation or any other real waves is still unknown to us

From the above discussions, we can infer that de Broglie’s hypothesis is of no use in the case of the macroscopic objects that we encounter in our daily lives.

The concept of matter waves is only Important In the case of particles of atomic dimensions.

Dual Nature Of Matter And Radiation de Broglie Wavelength Of Moving Electron Numerical Examples

Example 1. What is the de-Broglie- wavelength-related to an electron of energy 100 eV? (Given, the mass of the electron, m =  9.1 x 10^-31 kg, e = 6.63 × 10^-34  J. s)
Solution:

Let the velocity of the electron be v. Its kinetic energy,

½ mv² = E

Or, m²v² = 2mE

Or, mv = \(\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}\)

de Broglie wavelength related to the electron,

λ = \(\frac{h}{m \nu}=\frac{h}{\sqrt{2 m E}}\)

= \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}}\)

= 1.23 × 10^-10 m = 1.23A°

Example 2. Calculate the momentum of a photon of frequency 5 × 10¹³ Hz. Given h = 6.6 × 10^-34 J.s and c= 3 × 10⁸ m.s-1
Solution:

de Broglie wavelength \(\frac{h}{m v}=\frac{h}{p}\)

Momentum p = h/λ

Therefore, the momentum of the photon.

p = \(\frac{h}{\lambda}=\frac{h f}{c}=\frac{\left(6.6 \times 10^{-34}\right) \times\left(5 \times 10^{13}\right)}{3 \times 10^3}\)

= 1.1 × 10^-28 kg.m.s^-1

Example 3. The wavelength A of a photon and the de Broglie wavelength of an electron have the same value. Show that the energy of the photon \(\) kinetic energy of the electron. Here m, their usual meaning
Solution:

The energy of the photon. F. = hf = hc/λ

The kinetic energy of the electron

E’ = \(\frac{1}{2} m v^2=\frac{p^2}{2 m}\)

Where , p= mv= momentum

As p = \(\frac{h}{\lambda}\) So, E’ = \(\frac{h^2}{2 m \lambda^2}\)

E’ = \(\frac{h^2}{2 m \lambda^2}\)

Or, E =  \(\frac{E}{E^{\prime}}=\frac{h c}{\lambda} \cdot \frac{2 m \lambda^2}{h^2}=\frac{2 \lambda m c}{h}\)

E = \(\frac{2 \lambda m c}{h}\) E’

Example 4. An electron and a photon have the same de Hrogilr wave¬ length λ = I0^-10m. Compare the kinetic energy of the electron with the total energy of the photon
Solution:

The kinetic energy of an electron having mass m and velocity v,  K =  ½ mv²

Wavelength λ  = h/mv

Or, v = h/m λ

∴ K =½ m. \(\frac{h^2}{m^2 \lambda^2}=\frac{h^2}{2 m \lambda^2}\)

The energy of the photon of wavelength λ, E \(\frac{h c}{\lambda}\)

∴  \(\frac{K}{E}=\frac{h^2}{2 m \lambda^2} \cdot \frac{\lambda}{h c}=\frac{h}{2 m \lambda c}\)

= \(\frac{6.6 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 10^{-10} \times 3 \times 10^8}\)

= 0.012<1

Hence, the kinetic energy- of the electron is lower than the total energy of the photon

Example 5. Calculate the de Broglie wavelength of an electron of kinetic energy 500 eV.
Solution:

The kinetic energy of an electron 500 eV means the electron is accelerated by the potential 500 V So, the de Broglie wavelength associated with the electron

λ = \(\frac{12.27}{\sqrt{500}}\)

= 0.55 A°

Dual Nature Of Matter And Radiation Experimental Study Of Matter-Wave

Davisson and Germer Experiment:

We physicists C J Davisson and L H Germer first established the reality of matter waves. The experiment conducted by them in the year 1927 is described below.

1. Davisson and Germer experiment:

The electrons evaporated from the heated filament (F) after coming out of the electron gun are passed through a potential difference V. Thus, they acquire a high velocity and hence increased kinetic energy.

These electrons are then directed through a narrow hole and are thus collimated into a unidirectional beam. The kinetic energy of each single electron of this beam is eV.

Arrangements are such that this beam impinges normally on a specially hewn nickel single crystal. Because of this impact, the electrons may be scattered in different directions through different angles ranging from 10° to 90°. The incident and scattered beams are generally referred to as the incident ray and scattered ray.

A collector D capable of rotating around the crystal measures the intensity of the rays scattered in different directions. What is measured in the process is the number of electrons collected per second

2. Davisson and Germer Observation:

The intensity I obtained for the different values of the angle of scattering θ of the scattered ray is expressed using a polar graph where the scattered intensity is plotted as a function of the scattering angle.

The convention in this respect is this:

1. The point of incidence on the crystal is taken as the origin O of the graph
2. The direction OP opposite the direction of the incident ray PO is taken as the standard axis of the graph.

Suppose, for a stream of particles with definite energy, which is incident on the p crystal, the scattering angles are θ₁, θ_2, and the corresponding intensities  I₁, I_2…….Hence, the line joining the points is represented by the polar coordinates (I₁, θ₁ ), (I₂, θ₂ ), …, etc.

Will be the polar Q graph indicating the result of the experiment. The polar coordinates of the points A and B are (I₁, θ₁ ) and (I₂, θ₂ ); that is, I₁ = OA, I₂ = OB, and θ₁ = ∠AOP, θ₂= ∠BOP. In this figure, the line OABQ joining the points A, B… is the polar graph.

In this experiment, if the kinetic energy of the incident electrons is considerably low [as for the 40 eV electrons,  then it is observed that the intensity I corresponding to changes in the angle of scattering from 10° to 90° keeps decreasing continuously, which does not indicate any characteristic property of the electron beam.

After this, Davisson and Germer noticed that when the kinetic energy of the incident electrons is gradually increased, a distinct hump appears at 44 eV at a scattering angle of about 60°. The hump becomes most prominent at 54 eV and a scattering angle of 50°. At still higher potentials, the hump decreases until it disappears completely at 68.

If the nickel crystal is compared with a diffraction grating, it is observed that if an X-ray of wavelength 1.65A is incident normally on the nickel crystal instead of the electron beam, then also, a peak representing maximum intensity will be obtained at 50°

The angle of scattering. On the other hand, when the de Broglie formula for matter wave is applied, the de Broglie wavelength of 54 eV electrons is λ =  \(\frac{12.27}{\sqrt{54}}\) = 1.67A°

This striking similarity of the experimental value to the theoretical value according to de Broglie’s relation demonstrates the behavior of the electron stream as a matter wave. The experiment is termed an electron diffraction experiment as well

1. The atoms of pure solids are arranged in crystal lattices of definite shapes. If an infinitely large number, say, 1020 or. more, if such crystals are aligned in a three-dimensional array to form a large piece of matter, then the specimen is called a single crystal. For example, the cubical grains of common salt thus formed are each a single crystal.
2. Davisson and Germer used it for their experiment comparatively. slow-moving electrons with energy varying between 30 eV and 60 eV. Later on, G.P. Thomson successfully conducted a diffraction experiment with even high-energy electron beams; electrons of energy as high as 10 keV to 50 keV. In the subsequent years, the wave nature of other elementary particles, like the proton and the neutron, has also been established experimentally

The nature of matter-wave

According to de Broglie’s hypothesis, every moving particle can be represented as a matter wave, which, obviously, therefore, has to obey certain conditions

1. Corresponding to a moving particle, a matter wave must also be a moving wave or progressive wave.
2. The wave velocity must be equal to the particle velocity.
3. Because the particle has a definite position at any time, the matter wave must be such that it can indicate the position of the particle at that time.

A pure sine curve cannot represent a matter wave, which means that no proper idea can be formed about the true nature of the matter wave from such curves. Let us suppose that a progressive wave moving in the positive  x-direction is given by

Ψ = a sin (ωt – kx + δ) where a = amplitude,  ω = angular velocity, fc= propagation constant; and  δ = phase difference

1. This wave extends from x = ∞ to -∞  with no dissipation or damping anywhere along the path. J fence, It can never Indicate the Instantaneous position of the moving partly, OH If it is assumed however that the matter wave Js analogs, to the pure sine wave, then it can be shown.
2. That the velocity ofthe de Broglie wave turns out to be greater than the speed of light in a vacuum, which is Impossible contradicting Einstein’s special theory of relativity,

Moreover, a sine wave does not exist In nature. No real wave can extend from -∞ to +∞ without any damping. This implies that what we observe in reality is a wave group. For a practical example, consider the shape of waves formed when a stone is dropped in a pond. A wave group as shown comprising just a couple of wave crests and wave troughs is formed in the water and proceeds in circles over the water’s surface.

Wave group or wave packet:

If there is a superposition of two or more sine waves of different frequencies, then the waveform changes. If any such sine waves of continuously varying frequencies are superposed, the resultant wave that is formed has a general form like the one This is what is called wave group or wave packet.

Characteristics of wave packets are

This too is a progressive wave moving in a definite direction, in this case, along the + x-axis.

It is a localized wave, which means that it is limited within a rather small interval. Hence, such a wave group can indicate the possible instantaneous location of a moving particle.

It can be shown analytically that the velocity of such wave groups, \(\).  It is called group velocity. Rigorous calculation shows that vg = v; that is, the group velocity is equal to the particle velocity. Thus properly constituted wave group or wave packet alone can correctly represent a matter wave.

Wave function

It has to be noted with particular care that a matter wave is neither an elastic wave like a sound wave nor an electromagnetic wave like a light wave. This is because an elastic wave is associated with the vibration of particles in an elastic medium, whereas an electromagnetic wave involves the simultaneous vibrations of the electric field and the magnetic field vectors.

In analogy, u matter wave is associated with a quantity known as wave function – Ψ. The origin and propagation ofthe matter wave are perfectly consistent with the vibrations of this Ψ – function concerning position and time.

It can be noticed that while the moving particle exists at a particular point at a particular moment of its motion, the correspond¬ ing matter wave occupies an extent of space at that moment, rather than be limited to a point.

It can be inferred from quantum mechanics that this property is an inherent property of matter waves; a wave packet is never concentrated at a single point. The probability of the particle existing at a particular point at a particular time is given by Ψ², the modulus squared off,

Photon wave

We have seen, even before discussing wave-particle duality, that electromagnetic radiation has a dual nature too; radiation is sometimes represented by waves, at other times in terms of photon beams. It is possible also to represent moving photons by a corresponding waveform as is done with a moving particle.

A single photon will naturally be represented by a wave group. The only difference here is that this (photon) wave packet must be an electromagnetic wave packet with group velocity in vacuum or air equal to the velocity of light.

Dual Nature Of Matter And Radiation Experimental Study Of Matter-Wave Numerical Examples

Example 1. Find de Broglie wavelength of the neutron at 127°C. Given, K = 1.38 × 10^-23 ; h = J. mol^–1.K^–1. Plank’s constant, h = 6.626 × 10^-34J.s. mass of neutron, m = 1.66 × 10^-27 kg.

Kinetic energy of neutron, E = \(\frac{3}{2}\) kT

We know E = ½ mv²

Or, 2Em = m²v²

Or, mv = \(\sqrt{2 E m}=\sqrt{2 m \times \frac{3}{2} k T}=\sqrt{3 m k T}\)

de Broglie wavelength

λ = \(\frac{h}{m v}=\frac{h}{\sqrt{3 m k T}}\)

= \(\frac{6.626 \times 10^{-34}}{\sqrt{3 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 400}}\)

= 1.264 ×  10^-10m

Example 2. Under what potential difference should an electron be accelerated to obtain electron waves of A = 0.6 A f Given, the mass of the electron, m = 9.1x 10^-31  kg; Planck’s constant, h = 6.62 x 10^-34 J. s
Solution:

We know λ = ½mv²  = eV

∴ mv= \(\sqrt{2 m e V} \text { or, } \lambda=\frac{h}{\sqrt{2 m e V}}\)

Or, λ² =  \(\frac{h^2}{2 m e V} \text { or, } V=\left(\frac{h}{\lambda}\right)^2 \cdot \frac{1}{2 m e}\)

V =  \(\left(\frac{6.62 \times 10^{-34}}{0.6 \times 10^{-10}}\right)^2 \times \frac{1}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}}\)

= 418.04 V

Example 3. An a -particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of Broglie wavelengths associated with them.
Solution:

The kinetic energy of a particle of mass m.

E = \(\frac{p^2}{2 m}\)

Where p = \(\sqrt{2 m E}\)

So, de Broglie wavelength, λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

Now, if a particle of charge q is accelerated by applying potential difference V, then

E = qV

λ = \(\frac{h}{\sqrt{2 m q V}}\)

∴ λ ∝ \(\frac{1}{\sqrt{m q}}\) when V is constant

Hence \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{m_2 \cdot q_2}{m_1 \cdot q_1}}\)

For proton and a -particle \(\frac{m_a}{m_p}=4, \frac{q_a}{q_p}\) = 2

∴ \(\frac{\lambda_p}{\lambda_a}=\sqrt{\frac{m_a}{m_p} \cdot \frac{q_a}{q_p}}\)

= \(=\sqrt{4 \times 2}\)

= 2 \(\sqrt{2}\)

Hence, the required ratio 2\(\sqrt{2}\):1

Example 4. For what kinetic energy of neutron will the associated de Broglie wavelength be 1.40 x 10^-10 m? The mass of a neutron is 1.675 x 10^-27 kg and h = 6.63 x 10^-34 J
Solution:

If m is the mass and K is the kinetic energy of the neutron, the de Broglie wavelength associated with it is given by,

λ = \(\frac{h}{\sqrt{2 m K}}\)

Or, K = \(\frac{h^2}{2 m \lambda^2}=\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times 1.675 \times 10^{-27} \times\left(1.40 \times 10^{-10}\right)^2}\)

= 6.69 × 10^-21 J

Example 5. Find the wavelength of an electron having kinetic energy 10eV.(h =  6.33 × 10^-34J m_e = 9 × 10^-31 kg
Solution:

The kinetic energy of the electron,

E = ½ mv² = 10 eV = 10 × (1.6 x 10^-19)J

m²v² = 2mE, or, momentum p = mv = \(\sqrt{2 m E }\)

The de Broglie wavelength of the electron,

λ = \(\frac{h}{\sqrt{2 m E}}=\frac{6.63 \times 10^{-34}}{\left[2 \times\left(9 \times 10^{-31}\right) \times\left(10 \times 1.6 \times 10^{-19}\right)\right]^{1 / 2}}\)

= 3.9 ×  10^-10 m

= 3.9 A°

Example 6. An α – particle moves In a circular path of radius  0.83 cm in the presence of a magnetic field of 0.25 Wb/m². What is the de Broglie wavelength associated with the particle? ,
Solution:

The radius of a charged particle rotating in a circular path in a magnetic field

R = \(\frac{m v}{B q}\) or, mv = RBq

The de Broglie wavelength associated with the particle,

λ = \(\frac{h}{m v}=\frac{h}{R B q}\)

Here, R = 0.83 cm = 0.83 × 10^-2 m, B = 0.25 Wb. m^–²

q = 2e = 2 ×1.6 × 10^-19C

[Since α – particle]

λ = \(\frac{6.6 \times 10^{-34}}{0.83 \times 10^{-2} \times 0.25 \times 2 \times 1.6 \times 10^{-19}}\)

= 0.01 A°

Dual Nature Of Matter And Radiation Very Short Questions And Answers

Question 1. Below a minimum frequency of light, photoelectric emis¬ sion does not occur Is the statement true or false?
Answer: True

Question 2. Above the threshold wavelength for a metal surface, even a light of low intensity can emit photoelectrons. Is the statement true?
Answer: No

Question 3. What will be the effect on the velocity of emitted photo¬ electrons if the wavelength of incident light is gradually
Answer: The velocity of electrons will increase

Question 4. Why are alkali metals highly photo-sensitive?
Answer: Work functions of alkali metals are very low

Question 5. Express in eV the amount of kinetic energy gained by an electron when it is passed through a potential difference of
Answer: 100 eV

Question 6. The photoelectric threshold wavelength for a metal is 2100 A. If the wavelength of incident radiation is 1800 A, will there be any emission of photoelectrons?
Answer: Yes

Question 7. Which property of photoelectric particles was discovered from Hertz’s experiment?
Answer: Property of negative charge

Question 8. Which property of photoelectric particles was measured
Answer: Specific charge

Question 9. What is the relation between the stopping potential VQ and the maximum velocity v_max of photoelectrons?
Answer:

⇒ \(\left[v_{\max }=\sqrt{\frac{2 e V_0}{m}}\right]\)

Question 10. How does the kinetic energy of photoelectrons change due to an increase in the intensity of incident light?
Answer: No change

Question 11. Give an example of the production of photons by electrons.
Answer: X-ray emission

Question 12. Give an example of the production of electrons by photons.
Answer: Photoelectric effect

Question 13. If the intensity of incident radiation on a metal surface is doubled what happens to the kinetic energy of the electrons emitted?
Answer: No change in KE

Question 14. Write down the relation between threshold frequency and photoelectric work function for a metal.
Answer: \(\left[f_0=\frac{W}{h}\right]\)

Question 15. Which property of light is used to explain the characteristics of the photoelectric effect?
Answer: Particle (photon) nature

Question 16. What is the rest mass of a photon?
Answer: Zero

Question 17. The wavelength of electromagnetic radiation is X . What is the energy of a photon of this radiation?
Answer: E =  hc/λ

Question 18. The light coming from a hydrogen-filled discharge tube falls on sodium metal. The work function of sodium is 1.82 eV and the kinetic energy of the fastest photoelectron is 0.73 eV. Fine, the energy of incident light.
Answer: 2.55 eV

Question 19. In the case of electromagnetic radiation of frequency f, what is the momentum of the associated photon?
Answer: hf/c

Question 20. If photons of energy 6 eV are incident on a metallic sur¬ face, the kinetic energy of the fastest electrons becomes 4eV. What is the value of the stopping potential?
Answer: 4V

Question 21. The threshold wavelength of a metal having work function W is X . What will be the threshold wavelength of a metal having work function 2W?
Answer: λ/2

Question 22. The work function of a metal surface for electron emission is W. What will be the threshold frequency of incident radiation for photoelectric emission?
Answer: W/h

Question 23. What is the effect on the velocity of the emitted photoelectrons if the wavelength of the incident light is decreased?
Answer: Velocity will increase

Question 24. Two metals A and B have work functions 4eV and 10 eV respectively. Which metal has a higher threshold wavelength
Answer: Metal A

Question 25. Ultraviolet light is incident on two photosensitive materials having work functions W₁ and W₂(W₁ > W₂). In which case will the kinetic energy of the emitted electrons be greater? Why
Answer: For the material of work function W₂

Question 26. Is it possible to bring about the interference of electrons?
Answer: Yes

Question 27. What type of wave is suitable to represent the wave associated with a moving particle?
Answer: Wave Packet

Question 28. The wavelength of a stream of charged particles accelerated by a voltage Vis A. What will be the wavelength if the voltage is increased to 4 V?
Answer: λ /2

Question 29. The de Broglie wavelength of an electron and the wave¬ length of a photon are equal and its value is A = 10^-10m. Which one has higher kinetic energy?
Answer: Photon

Question 32. An electron and a proton have the same kinetic energy. Identify the particle whose de Broglie wavelength would be
Answer: Electron

Question 36. The de Broglie wavelength of a particle of kinetic energy K is A. What would be the wavelength of the particle if its kinetic energy were K/4?
Answer: 2 λ

Question 37. The de Broglie wavelength associated with an electron accelerated through a potential difference V is A. What will be its wavelength when the accelerating potential is increased to 4 V?
Answer: λ /2

Question 38. What will be the kinetic energy of emitted photoelectrons if light of threshold frequency falls on a metal?
Answer: Zero

Question 39. What conclusion Is drawn from the Davisson-Germer experiment?
Answer:

The Davisson-Germer experiment proves the existence of matter waves. It can be concluded from the experiment that any stream of particles behaves as waves

Question 40. Name the phenomenon which shows the quantum nature of electromagnetic radiation.
Answer:

The phenomenon which shows the quantum nature of electromagnetic radiation is the photoelectric effect

Dual Nature Of Matter And Radiation Fill In The Blanks

Question 1. Maximum kinetic energy of photoelectrons depends on __________________ of light used but does not depend on the of light
Answer: Frequency, intensity

Question 2. In the case of photoelectric emission, the maximum kinetic energy of photoelectrons increases with an increase in the _______________ of incident light
Answer: Frequency

Question 3. Lenard concluded from his experiment that the particles emitted in the photoelectric effect are _________________
Answer: Electrons

Question 4. In the case of photoelectric emission, the maximum kinetic energy of emitted electrons depends linearly on the _________________ of incident radiation
Answer: Frequency

Question 5. Davison and Germer experimentally demonstrated the existence of _____________________ waves
Answer: Matter

Dual Nature Of Matter And Radiation Assertion Type

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements,

1. Statement 1 Is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.
2. Statement 1 Is true, statement 2 Is true; statement 2 Is not a correct explanation for statement 1.
3. Statement 1 Is true, statement 2 Is false.
4. Statement 1 Is false, and statement 3 is true.

Question 1.

Statement 1: Any light wave having a frequency less than 4.8 × 10¹⁴ Hz cannot emit photoelectrons from a metal surface having work function 2.0 eV.

Statement 2:  If the work function of a metal is W₀ (in eV), then the maximum wavelength (in A°) of the light capable of initiating a photoelectric effect in the metal is \(\lambda_{\max }=\frac{12400}{W_0} \)

Answer: 1. Statement 1 Is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.

Question 2.

Statement 1: The energy of the associated photon max becomes half when the wavelength of the electromagnetic wave is doubled.

Statement 2: Momentum of a photon

= \(\frac{\text { energy of the photon }}{\text { velocity of light }}\)

Answer: 2. Statement 1 is true, and statement 2 Is true; statement 2 Is not a correct explanation for statement 1.

Question 3.

Statement 1: In the photoelectric effect the value of stopping potential is not at all dependent on the wavelength of the light

Statement 2: The maximum kinetic energy of the emitted photoelectron and stopping potential are proportional to each other.

Answer: 4. Statement 1 is false, and statement 3 is true.

Question 4.

Statement 1: The maximum surface velocity does not of increase the photo electron even if the intensity of the incident electromagnetic wave is increased.

Statement 2: Einstein’s photoelectric equation

½ mv²_max = hf – W₀

Where the symbols have their usual meaning

Answer: 1. 1. Statement 1 is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.

Question 5.

Statement 1: The stopping potential becomes double when the frequency of the incident radiation is doubled.

Statement 2: The work function of the metal and the threshold frequency of the photoelectric effect are proportional to each other.

Answer: 4. Statement 1 is false, and statement 3 is true.

Question 6.

Statement 1: If the kinetic energy of particles with different masses is the same, then the de Broglie wavelength ofthe particles is inversely proportional to their mass.

Statement 2: The momentum of moving particles is inversely proportional to their de Broglie wavelengths.

Answer: 4. Statement 1 is false, and statement 3 is true.

Question 7.

Statement 1: If a stationary electron is accelerated with a potential difference of IV, its de Broglie wavelength becomes 12.27 Å approximately.

Statement 2: The relation between the de Broglie length A and the accelerating potential V of an electron is given by λ  = \(\frac{12.27}{V}\) Å

Answer: 3. Statement 1 Is true, and statement 2 Is false.

Question 8.

Statement 1: A moving particle is represented by a progressive wave group.

Statement 2: A pure sinusoidal wave cannot represent the instantaneous velocity or position of a moving particle.

Answer: 1. Statement 1 Is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.

Question 9.

Statement 1: The wavelength of 100 eV photon is 124 Å.

Statement 2: The energy of a photon of wavelength λ in Å is

E = \(\frac{12400}{\lambda}\) eV

Answer: 1. Statement 1 Is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.

Question 10.

Statement 1: A proton, a neutron, and an a -particle are accelerated by the same potential difference. Their velocities will be in the ratio of 1: 1: √2.

Statement 2: Kinetic energy, E = qV =½mv²

Answer: 4. Statement 1 Is false, and statement 3 is true.

Dual Nature Of Matter And Radiation Match The Columns

Question 1. Light of fixed intensity is incident on a metal surface.’ Match the columns in case of the resulting photoelectric effect

Answer: 1-B, 2 – C, 3-D, 4-A

Question 2.

Answer: 1-A, 2-C, 3-D, 4-B

Question 3. Several stationary charged particles are accelerated with appropriate potential differences so that their de Broglie wavelengths are the same

Answer: 1-B, 2-D, 3-A, 4-C

Question 4. The work function and threshold frequency of photoelecj trie emission of a metal surface are WQ and f respectively. Light of frequency/ is incident on the surface. The mass and charge of the electron are m and e respectively

Answer: 1-B, 2-D, 3-C, 4-A

Optics

Light Wave And Interference Of Light Introduction

Light propagates through the vacuum or any medium in the form of waves l light wave is a transverse electromagnetic wave. Optical phenomena observed in daily life, as produced by slits or orifices, obstacles, reflecting planes, and refracting planes, have sizes many times more than the wavelength of light.

For example, the wavelength of visible light ranges from 4 × 10^-5 cm but even a very small slit in the path of light generally lias a diameter not less than I mm ie, 0.1 hence, the diameter of the orifice (slit) Is 1000 times more than the wavelength of visible light. The wavelength of light is taken as zero. with respect to large orifices, slits, obstacles, reflecting surfaces or refracting surfaces, i.e., light is not considered as waves

However In cases, when the size of an orifice or obstacle, in the path of light is too small and is comparable to the wavelength of light, the wavelength of visible light can no more be taken as zero, light has to be considered waves

Read and Learn More Class 12 Physics Notes

Optics: This branch of physics deals with the properties of light

Optics is divided into two parts

1. Geometrical optics and
2. Physical optics.

Geometrical optics: In this branch of optics, it is assumed that the wavelength of light is negligible in comparison with the sizes of instruments used in experiments (like an orifice ot an obstacle).

Physical optics:  it is assumed that the sizes of Instruments used In the experiment (litre size of an orifice ot an obstacle) are i comparable with the wavelength of light

Light Wave And Interference Of Light Wavefront

Concept of wavefront

When a Stone is chopped in a reservoir, waves are set up on its still-wet surface. I tear the centre of disturbance, i.e., where the stone is dropped, waves spread out in all directions at constant velocity. Water particles ‘ however, do not spread out horizontally with the wave.

Instead, they vibrate vertically let us that at .in instant of time the displacement of a water p.irtirle. at some distance from the centre of the disturbance. Is m.minimum At the it moment, displacements of all water particles situated on the circumference of the circle of radius same as the distance of the point in reference, are maximum.

This implies that water particles on the circumference of all circles- concentric with the circle described above, should all circle- concentric with the circle described above, should be medium in the form of circles Hence. where spreading of a wave in a dimensional medium.

If a sphere is imagined with the outer of disturb nice a; its centre, all points lying on the surface of this inference sphere, ate In the same phase. In this case spherical wavefront is obtained. If the radius of the spherical spherical wavefront is obtained. If the radius of the spherical wavefront is large enough, a small part of the wavefront can be taken as a plane wavefront

Definition: As a wave generated from a source spreads in all diet nuns through the vacuum or a medium, the locus (line or surface) points it the path of the wave which is in equal phase at any moment Is called a wavefront

Different types of wavefronts

1. Spherical wavefront:

The waveforms formed during the propagation of light coming hum a point source are considered spherical in shape the locus of the points in the same phase is spherical I In, which is known as a spherical wavefront

2. Cylindrical wavefront:

The wavefront produced during the propagation of light coming from a line source (for example, a single slit) is considered cylindrical in shape

3. Plane wavefront: The rays coming from infinity are parallel and the wavefront thus associated can be considered a plane wavefront

Properties of wavefronts

1. A perpendicular drawn at any point on a wavefront shows the direction of velocity of the wave at that point
2. The velocity of a wave actually denotes the velocity of the wavefront. If v is the wave velocity, the velocity of each wavefront the perpendicular distance between two consecutive wavefronts in the same phase is called the wavelength of that wave.
3. The phase difference between any two points located on the 1 same wavefront is zero

Ray: Perpendicular drawn to the wavefront is called Energy of a wave is transmitted from one part to another in a vacuum or in a medium along these rays. Direction of rays by yellow arrowhead.

Light Wave And Interference Of Light Huygens’ Principle Of Wave Propagation

Dutch scientist Christian Huygens introduced a geometrical method suggesting the propagation of wavefront which is very useful in interpreting reflection, refraction, interference, diffraction etc,, of waves.From the position and shape at any subsequent time Huygens principle.

Huygens’ principle

Each point on a wavefront acts as a new source called a secondary source. From these secondary sources, secondary waves or wavelets generate and propagate in all directions at the same speed of the wave. The new wavefront at a later stage is simply the common envelope or tangential plane to these wavelets

Let S be a point source of light from which waves spread in all directions. At some instant of time, AB is a spherical wavefront

According to Huygens’ principle, points P₁,  P₂, and P₃ wavefront AB act as new sources, and wavelets from these sources spread out at the same speed. Small spheres each of radius ct (c = velocity of light) with centres P₁,  P₂, P₃ respectively are imagined.

Clearly, wavelets produced by the secondary sources reach P₁,  P₂, P₃ surfaces of these imaginary spheres after a time interval. Following Huygens’ Principle, the common tangential plane AJBJ, to the front surface of small spheres should be the new wavefront In the forward direction, after a time t. In this context, it is pointed out that A₁B₁ is also a spherical surface with a centre at S.

Wavefronts, close to a source of light are spherical. But when a wavefront is far enough from the source of light, the Wavefront is taken to be a plane wavefront

From Huygens’ method of tracing wavefronts, like wavefront, A₁B₁ another wavefront A₂B₂ is also obtained behind AB, which can be called back wavefront. But in Huygens’ opinion, there is no existence of such a back wavefront. Later this opinion was supported theoretically and also experimentally.

Verification of the Laws of Reflection

Using Huygens’ principle, laws of reflection of light can be proved. Reflection of a plane wavefront from a plane reflecting surface

Let AC, part of a plane wavefront, perpendicular to the plane of paper, be incident on the surface of a plane reflector XY. Note that the plane wavefront and the plane of paper intersect each other along line AC. The plane of the paper and the plane of the reflector are also perpendicular to each other. According to Huygens! each point on wavefront AC acts as a source of secondary wavelet.

Let at time t = 0, one end of the wavefront touches the reflecting surface at A. At the same moment, wavelets form from each point on the wavefront AC. These wavelets gradually reach the reflector. After a time t say, wavelet from C reaches point F of the reflector. In accordance with Huygens’ principle, all points from A to F on the reflecting plane in turn generate wavelets. Hence by the time.

The wavelet from C reaches F, wavelet generated at A reaches D in the same medium. Centring point A, an arc of radius CF is drawn. Assuming c to be the speed of light in the medium under consideration, we have CF = ct. Tangent FD is drawn on the arc. FD is the reflected wavefront after time t.

The perpendiculars M₁A and M₂F drawn on the incident wavefront AC are the incident rays and the perpendiculars AM’₁[ and FM’₂ drawn on the reflected wavefront DF are the corresponding reflected rays. NA and N’F are the normals drawn at the points of incidence on the reflector.

.-. ∠M₁AN = Angle of incidence

And ∠M’₂FN’ = Angle of reflection (r)

According to

∠CAF = 90°- ∠NAC= ∠M₁AC-∠NAC

= ∠M₁AN =i

And ∠AFD = 90°-∠N’FD= ∠M₂FD-∠N’FD

= ∠M’₂FN’ = r

Now from ΔACF and ΔAFD, ∠AGF = ∠ADF = 90°,

CF = AD = ct and AF is the common side.

Hence the triangles are congruent.

∠CAF = ∠AFD

∴  Angle of incidence (i) = Angle of reflection (r)

Thus one of the laws of reflection is proved.

AC, FD and AF are the lines of intersection of the incident wavefront, reflected wavefront and plane of the reflector with the plane of paper respectively, which means AC, FD and AF lie on the plane of paper. Hence the perpendiculars to these lines, that is, incident ray (M₁A or M₂F), reflected ray (AM_1′or FM’₂) and the perpendicular to the reflector (AN or FN’) at the point of incidence lie on the same plane. Thus, the other law of reflection is also proved.

∴ sini = sin r -∠CAF = \(\frac{C F}{A F}\)

And in r = sin ∠AFD = \(\frac{A D}{A F}\)

Verification of the Laws of Refraction

Laws of refraction can also be proved using Huygens’ theory of wave propagation. Refraction of a plane wavefront in a plane refracting surface is shown in

Let XY be a plane refracting surface which is the surface of separation of medium 1 and medium 2 . The speed of light in medium 1 is cx and in medium 2 is c₂. Refractive indices of media 1 and 2 are μ₁ and μ₂  respectively.

There fore μ₁ = \(\frac{c}{c_1}\) and μ₂ = \(=\frac{c}{c_2}\) where c is speed of light in vacuum

The refractive index of the medium 2 with respect to that of medium

μ₂ = \(\frac{\mu_2}{\mu_1}=\frac{c / c_2}{c / c_1}=\frac{c_1}{c_2}\) ………………. (1)

Assume that, part of a plane wavefront normal to the plane of paper is incident on the surface of separation of two refractive media. The wavefront and the plane of paper intersect each other along the line AC. The plane of refraction XY is also at right angles to the plane of the paper.

According to Huygens’ principle, all the points on the wavefront AC act as sources of secondary wavelets. Suppose, at t = 0, one end of the wavefront touches the plane of the refracting medium at A.

At the same moment, wavelets form from each point on the die wavefront AC. These wavelets gradually reach the plane of the refracting surface. After a time t, say, the other end C of the wavefront touches F on the surface of separation.

Following Huygens’ principle, all points from A to F on the plane of separation will in turn generate wavelets. Hence, by the time the wavelet from C reaches F, wavelets are generated at A toward medium 2.

Centring point A, nt FD is drawn to the arc from point A to F on the time the wavelet from C reaches F, wavelets generated at A advance toward medium 2. Centring point A, an are of radius C₂ t is drawn. Tangent FD is drawn to the arc from point F. FD is the refracted wavefront after time t.

Naturally , CF = c₁t …………. (2)

And = AD = c₂ t …………. (3)

The perpendiculars M₁A and M₂F wavefront AC are the incident rays and the perpendiculars AM₁ and FM₂ drawn on the refracted wavefront DF are the corresponding refracted rays. NA and NfF are the normals drawn at the points of evidence on the surface of separation

∠M₁AN = angle of incidence (i)

and ∠M₂FN’ = angle of refraction (r)

According to

∠CAF = 90°- ∠NAC= ∠M₁AC-∠NAC

= ∠M₁AN = i And ∠AFD = 90°-∠N’FD= ∠M’₂FD- ∠N’FD

= ∠M₂‘FN’ = r

∴ sini = sin ∠CAF = \(\frac{A D}{A F}\)

∴ \(\frac{\sin i}{\sin r}=\frac{\frac{C F}{A F}}{\frac{A D}{A F}}\)

⇒ \(\frac{C F}{A D}=\frac{c_1 t}{c_2 t}=\frac{c_1}{c_2}\)

Or, \(1 \mu_2=\frac{\mu_2}{\mu_1}\)

Hence Snell’s law of refraction is proved.

AC, FD and AF are the lines of intersection of the incident. wavefront, refracted wavefront, and the plane of refraction with ” the plane of paper respectively which means AC, FD, and AF lie on the plane of the paper.

Hence the perpendiculars on them, that is, incident ray (M₁A or M₂F), refracted ray (AM₁‘ or FM_1′) and perpendicular to the refracting surface (AN or FN’) I lie on the same plane. Thus the other law of refraction is also proved.

Relating to the above discussion live has been drawn for the refraction from a rarer to a denser medium. The entire proof can be done for the refraction from a denser to a rarer medium as well in the same manner.

The velocity of light changes while passing from one medium to another keeping its frequency unchanged.

From equation (1) we get

⇒ \(_1 \mu_2=\frac{c_1}{c_2}=\frac{n \lambda_1}{n \lambda_2}=\frac{\lambda_1}{\lambda_2}\) …………………….. (4)

∴ c = nλ; n – frequency of light

λ₁ and λ₂ are the wavelengths of light in media (1) and (2) respectively. It means, the wavelength of light changes due to refraction.

LightWave And Interference Of Light Huygens’ Principle Of Wave Propagation Numerical Examples

Example 1. A plane wavefront, after being Incident on a plane reflector at an angle of Incidence, reflects from It. Show that the Incident wavefront and the reflected wavefront are Inclined at an angle (180° -2i) with each other.
Solution:

AB and CD are the incident and reflected wavefronts respectively. On extending the wavefronts, they meet at E and the angle between them is 0. From ∠BAC =l and ∠DCA = r.

From ΔACE, 6 + 1+r = 180°

Or,  θ = 180- (i+r) Or, θ= 180 – 2i

∴ i = r

Example 2. The wavelength of a light ray in vacuum is 5896A What will be Its velocity and wavelength when It passes through glass? Given, the refractive index of glass = 1.5 and the velocity of light In vacuum = 3 × 10⁸ m. s^-1
Solution:

Here, X = 5896A°, c = 3 × 10⁸ m .s^-1 μ_g = 1.5

We know , \(\mu_g=\frac{c}{c_g}\)

Or, \(c_g=\frac{c}{\mu_g}\)

⇒ \(\frac{3 \times 10^8}{1.5}\)

= 2 × 10⁸ m. s^-1

Again, nλ =c (in vacuum) nλ_g = c_g (in glass)

⇒ \(\frac{n \lambda_g}{n \lambda}=\frac{c_g}{c}\)

= \(\lambda^c \frac{c_g}{c}=\frac{\lambda}{\mu_g}\)

= \(\frac{5896}{1.5}\)

= 3931 A°

Example 3. Refractive indices of glass with respect to water and air were 1.13 and 1.51 respectively. If the velocity of light In air is 3 × 10⁸ m. s^-1 what will be The velocity in water
Solution:

We know, \({ }_a \mu_g=\frac{c_a}{c_g}\)

∴ \(1.51=\frac{3 \times 10^8}{c_g}\)

c_g = Velocity of light in the glass

Or, \(c_g=\frac{3 \times 10^8}{1.51}\)

= 2 × 10⁸ m. s^-1

Again , \({ }_{w^{\prime}}{ }_g=\frac{c_w}{c_g}\) = velocity of light in water

Or, 1.13 = \(\frac{c_w}{2 \times 10^8}\)

Or, c_w = 2.26 × 10⁸ m. s^-1

Light Wave And Interference Of Light Principle Of Superposition Of Waves

Simultaneous propagation of a number of waves through the same space In a medium is called superposition During superposition, while one wave superposes on another, individual properties remain unaltered

Let us consider the situation where at any point in a medium a number of waves are incident at the same time. The displacement of the point would have been different if the waves passed through it individually. But as all the waves are incident at the same time, the point undergoes a resultant displacement (since displacement is a vector quantity). Clearly, resultant displacement is the vector sum of the displacements produced by each wave. This is the principle of superposition of waves. The principle can be expressed as follows

At any instant, the resultant displacement of appointing a medium due to the influence of a number of waves is equal to the vector sum of displacements produced by each individual wave at that point at that instant.

Light Wave And Interference Of Light Interference Of Light

Interference Of Light Definition :

When two light waves of the same frequency and amplitude (or nearly equal amplitude) superpose in a certain region of a medium, the intensity of the resultant light wave increases at certain points and decreases at some other points in that region.

This phenomenon is known as interference of light An increase in intensity of light is due to constructive interference and a decrease in intensity is due to destructive interference. The increase or decrease of intensity in the resultant light wave depends on the phase difference of the superposing light waves at that point.

The principle of superposition of waves is applicable to all types of | waves, that is for sound waves, light waves, and other electromagnetic waves, as well. An example of the superposition of waves is interference which was first demonstrated experimentally by Youngin in 1801.

Young’s Double Slit experiment

Young’s experimental arrangement is as follows. Two narrow slits A and B are made in close proximity to each other on an obstacle 0 placed in front of a source of a monochromatic light M. Being placed symmetrically about source M, slits A and B act as a pair of coherent sources when illuminated by the source If the laboratory is sufficiently dark,

Alternate bright and dark lines can be seen on screen S placed behind O. These alternate dark and bright lines are called interference fringes

Constructive and destructive interferences

Assume that the amplitude of each of the two light waves of the same frequency is A. While propagating in the same direction through a medium they superpose at a point in the medium. The resultant amplitude at the point is equal to the vector sum of the amplitude of the original waves (by the principle of superposition of waves). If superposition takes place in the same phase, then the resultant amplitude = A + A = 2A.

As intensity is directly proportional to the square of amplitude, the resultant light will be four times as intense as the individual wave. This is called constructive interference. On the other hand, if superposition takes place in opposite phases, the resultant amplitude = A-A = 0 and the intensity of light is also zero. This is called destructive interference

If the phase difference or phase relation between two waves remains the same then the interference pattern at every point of the medium remains the same.

It may be noted that destructive interference does not imply the destruction of energy. No loss of energy takes place, only the energy of the dark points is transferred to that of the bright points so that the total energy of the incident waves remains constant. In other words, there is only redistribution of light energy over the region of superposition.

Analytical treatment of interference:

Let c and D be two sources of monochromatic light. The amplitude of each wave = A, wavelength = λ and speed = c.

Two light waves A moving in the same direction superpose at the point E. The A resultant displacement of point E due to superposition is the 1 algebraic sum of two individual displacements produced by the J two waves.

Distances of the point E from the two sources are x and (x+ δ), respectively. So the path difference of the waves at that point is δ . If y₁ and y₂ are the displacements of point E due to the waves produced from sources C and D In time t

⇒ \(y_1=A \sin \frac{2 \pi}{\lambda}(c t-x)\)

⇒  \(y_2=A \sin \frac{2 \pi}{\lambda}[c t-(x+\delta)]\)

The resultant displacement of point E

y = y₁+ y₂

⇒ \(A\left[\sin \frac{2 \pi}{\lambda}(c t-x)+\frac{\sin 2 \pi}{\lambda}\{c t-(x+\delta)\}\right]\)

⇒ \(2 A \sin \frac{2 \pi}{\lambda}\left\{c t-\left(x+\frac{\delta}{2}\right)\right\} \cdot \cos \frac{\pi \delta}{\lambda}\)

⇒  \(\sin C+\sin D=2 \sin \frac{C+D}{2} \cdot \cos \frac{C-D}{2}\)

Or, \(y=A^{\prime} \sin \frac{2 \pi}{\lambda}\left\{c t-\left(x+\frac{\delta}{2}\right)\right\}\) …………………. (2)

Where A’ = 2A \(\cos \frac{\pi \delta}{\lambda}\) …………………. (3)

The sine function in equation (1) suggests that the resultant wave is also a wave of velocity c and wavelength A.

Again equation (2) shows, that the amplitude A’ of the resultant wave is not equal to the amplitude of the two superposing waves but modified by their path difference δ.

1. Condition for destructive interference:

If \(\delta=\frac{\lambda}{2}, \frac{3 \lambda}{2}\), \(\frac{5 \lambda}{2}\) ….. = \((2 n+1) \frac{\lambda}{2}\) (where n = 0 or any integer), \(\cos \frac{\pi \delta}{\lambda}\) = 0 and hence A’ = 0 and hence A’ = 0. Amplitude being zero

Intensity is also zero. Thus at points where the path difference between the waves are odd multiples of, the intensity of light is zero. These points are dark points. The two waves produce destructive interference at such points. This phase difference between the two waves at points where destructive interference takes place is

⇒ \(\Delta \phi=\frac{2 \pi}{\lambda}(2 n+1) \frac{\lambda}{2}=(2 n+1) \pi\)

∴ [since phase difference, path difference ]

2. Condition for constructive interference:

If = 0 , \(\frac{2 \lambda}{2}\), \(\frac{4 \lambda}{2}, \ldots=2 n \frac{\lambda}{2}\)(where n = 0 or any integer) that is A’ = 2A . Amplitude is the highest, and intensity is also the maximum.

Thus at points where the path difference between the waves is an even multiple of y the intensity of light is maximum. These points are bright points. Two waves produce constructive interference at these points

Thus the difference between the two waves at points where conNtrnctlvc Interference takes place Is

⇒ \(\Delta \phi=\frac{2 \pi}{\lambda} \cdot \frac{2 n \lambda}{2}\) = 2n

It Is to be noted that, point E is not a single point that satisfies equation (1). That Is, either of the conditions of destructive and constructive interference is obeyed not only at a single point In space but for a set of points.

The locus of the point E is, In general, hyperbolic. But when E is at a large distance from the sources C and D, compared to their mutual distance CD, the locus of E is effectively a straight line.

As a result, dark and bright straight lines are obtained instead of dark and bright points, due to destruc¬ tive and constructive interference, respectively. These are called dark and bright interference fringes

Intensity:

Clearly, interference of two waves results in a variation of intensity of light at different points. The amplitude of the resultant wave, A’ = 2A cos \(\cos \frac{\pi \delta}{\lambda}\) and therefore A’ can vary from 0 to ±2A.

As intensity is directly proportional to the square of amplitude, the value of intensity increases from 0 to 4A². That is maximum intensity can be four times the intensity of a single wave. IfI is the intensity at a point on the screen then,

I∝ A’²

⇒ \(4 A^2 \cos ^2 \frac{\pi \delta}{\lambda}\)

Or, I = \(k 4 A^2 \cos ^2 \frac{\pi \delta}{\lambda}\)

Or, \(I_0 \cos ^2 \phi\)

I_o= 4A²k, (is the maximum intensity) and \(\frac{\pi \delta}{\lambda}\)

Hence the intensity in the region of superposition follows the cosine-squared rule. Variation of intensity with phase difference is shown below

1. Let waves coming from two coherent sources of equal frequency be of amplitudes A₁ and A₂, intensities I₁ and 1₂ with phase difference Φ.

1. On the superposition of two coherent waves, the expression for resultant amplitude is

⇒ \(A^{\prime 2}=A_1^2+A_2^2+2 A_1 A_2 \cos \phi\)

2. As intensity is directly proportional to the square of amplitude, the resultant intensity is

I = \(I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi\)

⇒  \(I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)

And \(I_{\min }=\dot{I}_1+I_2-2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)

3. Moreover, as the amplitude of bright fringe = A₁+ A₂ and amplitude of dark fringe =|A₁A₂|

⇒  \(\frac{I_{\max }}{I_{\min }}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}\)

2. If the sources are not coherent, the resultant intensity at any point will just be the sum of the intensities of the individual waves

Conditions of sustained interference:

• Two light sources must be monochromatic and should emit waves of the same wavelength.
• The amplitude of the waves should be equal or nearly equal.
• There must always be a constant phase difference between the two waves. With the change in phase of any wave there: should be a simultaneous change in the other to the same extent.
• Such pair of sources are called coherent sources

Formation of Coherent Sources

It has already been stated that for sustained interference, a pair of coherent sources is necessary. When sources are not coherent, the intensity at any point changes so rapidly that no interference fringe is observed in practice, and all points appear equally bright

Two similar but separate sources do not form coherent sources. Even two very close points on the same source are not coherent.

Hence there are special methods of creating coherent sources. Some are described below: 

• In Young’s double slit experiment, two slits S₁ and S₂, kept at a fixed distance from light source S, act as coherent sources
• In Lloyd’s single mirror experiment, a thin illuminated slit S and its virtual image, S’, formed due to reflection from a plane mirror, serve as a pair of coherent sources
• In the experiment with Fresnel’s biprism, two virtual images and S₂ of a source S, produced by refraction through the biprism 1, act as coherent sources

Explanation of Formation of Interference Fringes

The formation of interference fringes can be explained by the principle of superposition of waves. In Young’s experiment, slits A and B are equidistant from source M. In that case, the secondary sources will also be coherent and hence in phase. Wavefronts from these two sources, propagate one after another

Through the space between the obstacle and the screen. Hence they superpose in solid line arcs to show wavefronts in the same phase. Broken line arcs, in between represent wavefronts in the opposite phase.

Clearly, waves coming from two sources proceeding towards the points a, c and e, superpose in the same phase. Thus amplitude of the resultant wave becomes maximum and constructive interference takes place along the lines leading to the points a, c and e on the screen.

On the other hand, waves are directed towards points b or d, superposing the opposite phase. Thus amplitude of the resultant wave is zero and destructive interference takes place. Therefore, interference fringes consisting of bright and dark lines are displayed on screen S.

The wavelength of monochromatic light be A, the path difference between two waves along the dark lines happens to be odd multiples \(\frac{\lambda}{2}\) of and that along bright lines, even multiples of \(\frac{\lambda}{2}\).

1. If white light is taken instead of monochromatic light in Young’s double slit experiment, the central bright fringe will be white and bright-coloured fringes will be observed on either side of the central fringe. This is because wavelengths of the colours, forming white are different and therefore each one produces its characteristic interference fringe pattern. Thus fringes of different colours are produced.

As the central line of the interference pattern is equidistant from each of the two coherent sources; the light of all component colours reaches phase and a bright white light is formed at that point.

2. The fringes disappear when one of the slits A or B is covered by an opaque plate and the screen gets illuminated with a uniform intensity.

3. If one of the slits is covered with translucent paper, fringes of the same width will be formed. But the bright band will look less bright and the dark band will look less dark.

Width of interference fringes

In A and B are two coherent sources of monochromatic light, S is a screen, 2d = distance between A and B, O is the mid-point of AB, and D = perpendicular distance of AB from the screen. As AC = BC, waves starting from two of the sources in phase reach point C in phase. Hence, the resultant amplitude as well as intensity at C is maximum.

All points on the line perpendicular to the plane of the paper and passing through C will be equidistant from the I sources A and B. Thus a central bright fringe is formed which is a straight line.

Now a point P is taken at a distance x from C i.e., CP = x; AP and BP are joined. As the lengths of two paths are not equal there will be a phase difference between the waves reaching P from A and B. Thus whether constructive or

Destructive interference will take place at point P depending on the path difference between the two incoming waves

From the

BP² = D² + (x+ d)² and AP² = D² + (x-d)²

BP² – AP² = (x+ d)²- (x-d)² = 4xd

Or, Bp- Ap = \(\frac{4 x d}{B P+A P}=\frac{4 x d}{2 D}=\frac{2 x d}{D}\) ……………… (1)

[Since D»d hence BP = AP ~D]

This path difference between the two waves in reaching P

Bp -Ap = \(\frac{2 x d}{D}\)

By the condition of interference, for n -th bright fringe at P, path difference

⇒ \(\delta=\frac{2 x_n d}{D}=2 n \cdot \frac{\lambda}{2}\) [n = 0, 1, 2……]

Or, \(\frac{D n \lambda}{2 d}\) ……………. (2)

x_n is the distance of n -th bright fringe on either side of C] Taking n = 1, 2, 3 …, the positions of 1st, 2nd, 3rd etc. bright fringe, on either side of the central bright fringe, can be found out

For (n + 1) -th bright fringe

⇒ \(x_{n+1}=\frac{D(n+1) \lambda}{2 d}\) …………… (3)

So, the distance between two consecutive bright fringes

= \(x_{n+1}-x_n=\frac{D(n+1) \lambda}{2 d}-\frac{D n \lambda}{2 d}=\frac{D}{2 d} \cdot \lambda\)

By the condition for the formation of the n -th dark fringe at P is

⇒  \(\delta=\frac{2 x_n d}{d}=(2 n+1) \frac{\lambda}{2} \text { or, } x_n=\frac{D}{2 d}(2 n+1) \frac{\lambda}{2}\)

Taking n = 0, 1, 2 etc, positions of 1st, 2nd, 3rd, etc., dark fringes on either side of the central bright fringe can be found out

Hence for (n + 1) -th dark fringe

⇒ \(x_{n+1}=\frac{D}{2 d}\{2(n+1)+1\} \frac{\lambda}{2}\) ………………………….. (6)

Hence the distance between two consecutive dark fringes

= \(x_{n+1}-x_n=\frac{D}{2 d}\{2(n+1)+1\} \frac{\lambda}{2}-\frac{D}{2 d}(2 n+1) \frac{\lambda}{2}\)

= \(\frac{D}{2 d} \cdot \lambda\) ………………………….. (7)

Thus the distance between two consecutive bright or dark fringes is the same. This distance is called fringe width. Therefore if y is the fringe width then,

= \(\)………………………….. (8)

It is clear from equation (8) that

1. Since there is no ‘n’ in the expression of y, it can be said that fringe width does not depend on the order of the fringe. All hinges are of the same width.
2. Fringe width is directly proportional to the wavelength of light used. For a greater wavelength, fringe width increases, i.e., fringes will be wider. Similarly, for shorter wavelengths, fringes will be thinner.
3. If the value of D is large, the fringe width will Increase.
4. If the value of d is small, the fringe width will Increase.

Also If the total experiment is conducted In any other medium, wavelength decreases (\(\lambda^{\prime}=\frac{\lambda}{\mu}0\)  ). Hence, fringe width decreases.

Further, in whichever position the screen is placed In front of sources A and D, the interference pattern is always observed. As interference patterns are not restricted to a fixed place, these are called non-localised fringes.

Angular fringe width

If the angular position of the n-th fringe on the screen be Q_n then,

⇒ \(\theta_n=\frac{x_n}{D}=\frac{\frac{D n \lambda}{2 d}}{D}=\frac{n \lambda}{2 d}\)

For (n + 1) -th fringe , \(\theta_{n+1}=\frac{(n+1) \lambda}{2 d}\)

So, the angular separation between two successive fringes i.e., the angular width of the fringe

⇒ \(\theta=\theta_{n+1}-\theta_n=\frac{(n+1) \lambda}{2 d}-\frac{n \lambda}{2 d}=\frac{\lambda}{2 d}\) …………………………………………… (9)

The angular width of the fringe does position of the screen

Angular width decreases with the Increase of the separation between the coherent sources and vice versa. 0 = R)

If the entire experiment Is performed Inside any liquid (refractive Index for say), the angular width decreases

Optical path

The equivalent optical path of the distance covered by a certain monochromatic light through a certain medium In a certain interval of time Is the path covered by light In the same Interval of time through a vacuum.

Suppose, the refractive Index of a medium for a certain monochromatic ray Is fi. The velocity of the ray In that medium Is v.

⇒ \(\frac{c}{v}\) Or, c= v ………………………………….. (10)

Where, c – velocity of light In vacuum]

Now If the said monochromatic light takes time t to travel through the said medium then,

v = \(\frac{x}{t}\)

And If light can travel distances in the same Interval of time through a vacuum then

c = \(\frac{l}{t}\)

Hence from equation (10), we get

⇒ \(\frac{l}{t}=\mu \cdot \frac{x}{t}\)

Or, l = μ. x

Here l,  according to the definition of optical path, Is the equivalent optical path of x.

For more than one medium, we can write optical path

l = \(\Sigma \mu_i x_i\)

Displacement of fringes due to the introduction of a thin plate:

A and H are two monochromatic, coherent sources of light, S’ It the screen, the 2d distance between A and U, 0 is the midpoint of AB, and Dr I perpendicular distance of All front of the screen.

Waves coining from two coherent sources produce an Interference pattern on the screen. Generally, point C Is the position of the central bright fringe as AC = BC. A point P Is considered on the screen. Now a glass plate is introduced perpendicularly in the path AP. Let its thickness be t and the refractive index of glass he mm. it results in the shifting of the entire fringe along with its central fringe.

This is caused by the difference in the velocity of light in air and glass

While moving from A to P, light travels a path (AP-t) through air (ft = 1) and a path t through glass

Hence optical path from A to P= (AP-t). 1+μ. 1

= Ap + t (-1)

Differences in optical paths from the points A and B

δ = Bp- Ap- t(μ-1)

= (Bp- Ap)- t(μ-1)

= \(\frac{2 x d}{D}-t(\mu-1)\)

If P is at the centre of n -th bright fringe

= \(\delta=2 n \cdot \frac{\lambda}{2}=n \lambda\) , where n = 0,1,2……..

= \(\frac{2 x d}{D}-t(\mu-1)=n \lambda\)

Or, x = \(\frac{D}{2 d}\{n \lambda+t(\mu-1)\}\)

Hence, because of the presence of a glass plate, a distance of n -th bright fringe from C, x = \(\frac{D}{2 d}\{n \lambda+t(\mu-1)\}\). Putting n = 0 in this equation we get the displacement of the central bright fringe due to the insertion of the plate

Let the displacement be x₀

⇒ \(x_0=\frac{D}{2 d}(\mu-1)\)t

The refractive index of glass p is greater than 1, hence xQ is positive, i.e., the central bright fringe will move further from C towards the glass plate.

Putting n = 1, displacement of the 1st bright fringe l.e., xl is obtained.

∴  \(\frac{D}{2 d}\{n \lambda+t(\mu-1)\}\)

∴ Fringe width

y = \(x_1-x_0=\frac{D}{2 d}\{\lambda+(\mu-1) t\}-\frac{D}{2 d}(\mu-1) t=\frac{D}{2 d} \cdot \lambda\)

It proves that though a displacement occurs In fringe pattern due to introduction of a glass plate, fringe width remains unaltered

⇒ \(\frac{y}{\lambda}=\frac{2 D}{d}\) …………. (12)

From equations (11) and (12),

⇒ \(x_0=\frac{y}{\lambda}(\mu-1) t\) …………. (13)

If x₀,y, t and A are known, the refractive index p of the material of the plate can be found out from equation (13).

Also if p is known, t can be calculated out.

Due to the introduction of the glass plate, if the central bright fringe shifts a distance equal to the length of m bright fringes, then

⇒ \(x_0=m y \quad \text { or, } \quad \frac{y}{\lambda}(\mu-1) t=m y\)

Or, \((\mu-1) t=m \lambda\)

Or, m = \(\frac{(\mu-1) t}{\lambda}\)

This equation gives the equivalent number of fringe width by which a displacement of total fringe pattern occurs.

Light Wave And Interference Of Light Interference Of Light Numerical examples

Example 1. A screen is placed at a distance of 5 cm from a point source. A 5 mm thick piece of glass with a refractive index of 1.5 is placed between them. What is the length of the call path between the source and the screen
Solution:

Length of air path between the source and screen = 5 – 0.5 = 4.5 cm

Airpath, equivalent to path through the glass plate

= Refractive index x thickness = 1.5 × 0.5 = 0.75 cm

length of optical path = 4.5 + 0.75 = 5.25 cm

Example 2. Two straight and parallel slits, 0.4 ap-ft sraOhiminated by a source of monochromatic light. Interference pattern of fringe width 0.5 mm I* produced 40 cm away from the sUts. Find the length of the light used.
Solution:

In this case 2d = 0.4 mm = 0.04 cm , y= 0.5 mm = 0.05cm, D = 40 cm

We know, the fringe width,

y = \(\frac{\lambda D}{2 d}\)

⇒ \(\lambda=\frac{2 d \cdot y}{D}=\frac{0.04 \times 0.05}{40}\)

5 × 10^-5 cm

= 5000 × 10^-8cm = 5000 A°

Example 3.  In Young’s double-slit experiment on interference, the distance between two vertical slits was 0.5 mm and the distance of the screen from the plane of slits was 100 cm. It was observed that the 4th bright band was 2.945 mm away from the second dark band. Find the wavelength of light used.
Solution:

From it is observed that the distance between the second dark band to 4th bright band = 2.5 x bandwidth. If bandwidth or fringe width is y then,

2.5 × y = 2.945

⇒ \(\frac{2.945}{2.5}\)

= 1.178 mm

As, \(=\frac{D \lambda}{2 d}\)

So, \(\lambda=\frac{2 d \cdot y}{D}=\frac{0.05 \times 0.1178}{100}\)

= 5890 × 10^-8 cm = 5890 A°

Example 4. Monochromatic light of wavelength 6000A° was used to set up interference fringes. On the path of one of the interfering waves, a mica sheet 12 × 10^-5 cm thick was placed when the central bright fringe was found to be displaced by a distance equal to the width of a bright fringe. What is the refractive index of mica?
Solution:

On placing the mica sheet, if the central bright fringe shifts by the distance of m number of bright fringes then

(μ-1)t = mλ

Here, m = 1, λ = 6000 A° = 6000 × 10^-8 cm

t = 12× 10^-5 cm

((μ-1) ×12× 10^-5

= 1× 6000 × 10^-8

Or, (μ-1) = \(\frac{1}{2}\)

Or, mm = 1.5

Example 5. A ray of light of wavelength 6 × 10^-5cm, after passing through two narrow slits, 1mm apart, forms interference fringes on a screen placed lm away. Find the list between two successive bright bands Of the fringes.
Solution:

The distance between two successive bright bands is the fringe width

Width of the fringe \(, \lambda=6 \times 10^{-5}\)

D = 1m, λ = 6 × 10^-5 cm

0.06cm

Example 6. In a double-slit experiment using monochromatic a screen at a light, interference fringes are formed at a particular distance from the slits. If the screen is moved towards the slits by 5 × 10^-2 m, then the fringe width changes by  3 × 10^-2m. If the distance between the two slit is 10^-3m, then determine the wavelength of the light used
Solution:

Fringe width , y = \(\frac{\lambda D}{2 d}\)

Here, A is the wavelength of light used, D is the distance of the screen from the slits, 2d is the separation between the slits. Here screen distance changes by ΔD and fringe width changes by Δy

\(\Delta y=\frac{\lambda \Delta D}{2 d}\)

Or, \(\lambda=\frac{\Delta y 2 d}{\Delta D}\)

Now , \(\Delta y=3 \times 10^{-3} \mathrm{~m}, 2 d=10^{-3} \mathrm{~m}, \Delta D=5 \times 10^{-2} \mathrm{~m}\)

λ = \(\frac{3 \times 10^{-5} \times 10^{-3}}{5 \times 10^{-2}}\)

= 6 × 10^-2

= 6000 A°

Example 7. In Young’s double slit experiment, the width of the fringe is 2.0 mm. Find the distance between the 9th bright band and the 2nd dark band. 
Solution:

It is seen from the distance of the 9th bright band from the second dark band

= 7.5 × width of fringe = 7.5  ×  0.2 cm = 1.5 cm

Example 8. Using light of wavelength 600 nm in Young’s double slit experiment 12 bands are found on one part of the screen. If the wavelength of light is changed to 400 nm, then what will be the number of bands on that part of the screen?
Solution:

In wavelength, λ₁ = 600 nm = 600 × 10^-9 m; the number of fringes, n₁ = 12, fringe width y₁ In the second case, wavelength, λ₂ = 400 nm = 400 × 10^-9 m’ number of fringes n₂ and fringe width y₂ first case, sucLet the distance between the slits and the screen be D, separation between the slits be 2d and length of the entire fringe pattern be x.

Now \(\frac{\lambda_1 D}{2 d}=\frac{600 \times 10^{-9} D}{2 d}\)

And n₁ = \(\frac{x}{y_1}\)

Or, = \(12=\frac{x \times 2 d}{600 \times 10^{-9} D}\)

or x = \(\frac{12 \times 600 \times 10^{-9} D}{2 d}\)

Also, y= \(\frac{400 \times 10^{-9} D}{2 d}\)

n₂  = \(\frac{x}{y_2}=\frac{12 \times 600 \times 10^{-9} D \times 2 d}{2 D \times 400 \times 10^{-9} D}\)

= 18

Example 9.  The green light of wavelength 5100 A° from- a narrow slit is incident on a double slit. The overall separation of 10 fringes on a screen 200 cm away is 2 cm, find the separation between the slits
Solution:

Width of 10 fringes, x = \(\frac{D}{2 d} \lambda\)

= 2 cm

Here, D = distance of this screen = 200 cm, 2d = separation between the two slits,  λ= wavelength of light = 5100A°

= 5100 × 10^-8cm

2d = \(10 \frac{D}{x} \lambda=\frac{10 \times 200 \times 5100 \times 10^{-8}}{2}\)

= 51 × 10^-3 = 0.051 cm

Example 10. The ratio of the intensities between two coherent light sources used in youngs double slit experiment, is n. Find the ratio of the intensities of the principle maximum and minimum of the band.

Let the intensities of the sources be I₁ and I₂ According to question, I₁ = nI₂

Now, if I_max and I_min are the intensities of central maximum and minima respectively then

⇒ \(\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\frac{\left(\sqrt{n} \cdot \sqrt{I_2}+\sqrt{I_2}\right)^2}{\left(\sqrt{n} \cdot \sqrt{I_2}-\sqrt{I_2}\right)^2}\)

= \(\frac{(\sqrt{n}+1)^2}{(\sqrt{n}-1)^2}=\frac{n+1+2 \sqrt{n}}{n+1-2 \sqrt{n}}\)

Example 11. In Young’s experiment, with a monochromatic light of wavelength 5890 A°, the seats are separated by a distance of 1mm. Find the angular width between two sucessive interference fringes
Solution:

Here, ,1 = 5890 A° = 5890 × 10^-8cm and 2d = 1mm

= 0.1 cm

Angular width between two successive bright or dark bands

= \(\frac{\lambda}{2 d}=\frac{5890 \times 10^{-8}}{0.1}\) radian

= \(5890 \times 10^{-7} \times \frac{180}{\pi}\)

= 0.03 degree

Example 12. In Young’s double slit experiment, the angular width of a fringe formed on a distant screen is 0.1 0. The wavelength of the light used in the experiment is 6000A. What is the distance between the two slits?
Solution:

We know, the angular width of a fringe,

⇒ \(\theta=\frac{\lambda}{2 d}\)

= \(\frac{\lambda}{\theta}=\frac{6000 \times 10^{-10}}{\frac{\pi}{180} \times 0.1}=\frac{6000 \times 180 \times 10^{-10}}{\frac{22}{7} \times 0.1}\)

= 3.44 × 10^-3 m

∴ Distance between the slits = 3.44 × 10^-4m

Example 13. In the experiment, the path difference between two Interfering waves at a point on the screen is 167.5 times the wavelength of the monochromatic light used. Is the point dark or bright? If the path difference is 0.101 mm, find the wavelength of light used.
Solution: Path difference

167.5 = \(335 \times \frac{\lambda}{2}\)

Wavelength of light

As the number 335 is odd So the point in Question is darkpoint

Here, \(\frac{28}{335}=\frac{2 \times 0.101}{335}\)

⇒ \(6.03 \times 10^{-4}\)

= \(6.03 \times 10^{-4}\) x 107 A = 6030 A

Example 14. The optical path traversed by a monochromatic ray of light are same while the ray either passes through a distance of 4 cm in glass or a distance of 4.5 cm in water. What is the refractive index of water if that of glass is 1. 53?
Solution:

As the optical path is the same in both cases.

So, μ_gx_g= μ_wx_w

Here, x_g = distance travelled through glass and xw = distance travelled through water.

Or, \(\mu_w=\mu_g \cdot \frac{x_g}{x_w}\)

= \(1.53 \times \frac{4.0}{4.5}\)

= 1.36

Refractive index of water = 1.36

Light Wave And Interference Of Light Very Short Question And Answers

Waves and Wavefronts

Question 1. At what angle does a ray of light remain inclined to the wavefront?
Answer: \(\left[\frac{\pi}{2}\right]\)

Question 2. What will be the nature of the wavefront of light emitted from a line source?
Answer: Cylindrical

Question 3. Can two wavefronts of same wave cut each other?
Answer: No

Question 4. A plane wavefront is incident on a prism. What will be the nature of emergent wavefront?
Answer: Plane

Question 5. What is the relationship between the intensity and amplitude of a wave
Answer: Intensity is the square of amplitude

Question 6. If the path difference is A, what will be the phase difference?
Answer: 2 π

Question 7. Do the two electric bulbs connected to the same electric supply line behave as coherent sources?
Answer: No

Question 8. What is the path difference between two waves for constructive interference?
Answer: 2n\(\) n = 0,1,2…..

Question 9. What is the path difference between two waves for destructive interference?
Answer: \((2 n+1) \frac{\lambda}{2}\):, n – 0,1,2…..

Question 10. What is the most important condition for interference of light
Answer: The two light sources must be coherent

Question 11. If Young’s double slit experiment is performed by using a source of white light, then only white and dark fringe patterns is seen. Is the statement true or false?
Answer: False

Question 12. What change will you observe if the whole arrangement used in youngs double slit expeiment is immersed in water
Answer: Fringe lines become narrower

Question 13. For which colour oflight, in Young’s double slit experiment, will the fringe width be minimum?
Answer: For violet light

Question 14. In Young’s double slit experiment, if the distance between the two sources is increased, how will the fringe width be changed?
Answer: The width of wringer would be reduced

Question 15. Does interference of light give any information about the nature of light waves? (whether it is longitudinal or tranverse)
Answer: No

Question 16. What is the phase difference between two points situated on a wavefront
Answer: Zero

Light Wave And Interference Of Light Fill In The Blanks

Question 1. The source of a spherical wavefront is____________________
Answer: A point source

Question 2. If a spherical wavefront is propagated up to infinite distance, then a part of that wavefront is called _________________  front
Answer: Plane

Question 3. Two coherent monochromatic light sources produce constructive interference when the phase difference between them is_____________
Answer: 2nπ

Question 4. In case of interference oflight, when the path difference is odd multiplies of half wavelength, then_______ interference takes place _____________
Answer: Destructive

Question 5. In case of interference of light ____________ remains conserved
Answer: Energy

Question 6. If the distance between the two slits in Young’s double slit experiment is halved and the distance between the slit and the screen is doubled then the fringe width will be_ times the previous fringe width
Answer: Four

Question 7. If the light of a smaller wavelength is used in Young’s double slit experiment, then fringe width will be
Answer: Decreased

Question 8 . In Young’s double slit experiment, if a glass plate is placed perpendicular to the direction of propagation oflight, there will be a _________________ of interference fringe, and there would be in the fringe width
Answer: Displacement, No Change

Question 9.  In Young’s double slit experiment, if a glass plate is placed perpendicular to the direction of propagation oflight, there will be a ______________________of interference fringe, and there would be in the fringe width
Answer: Number of fringes

Light Wave And Interference Of Light Assertion Reason Type

Direction: These questions have statement 1 and statement 2. of the four choices given in below, choose the one best describes the two statements

1. Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true, and statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true

Question 1. 

Statement 1: A ray of light entering from glass to air suffers from a change in frequency.

Statement 2: The velocity of light in glass is less than that in

Answer:  4. Statement 1 is false, statement 2 is true

Question 2.

Statement 1: If the phase difference between the tight waves passing through the slits in the Young’s experiment is n -radian, the central fringe will be dark.

Statement 2: Phase difference is equal to \(\frac{2 \pi}{\lambda}\) times the ath difference.

Answer:  2. Statement 1 is true, statement 2 is true, and statement 2 is not a correct explanation for statement 1.

Question 3.

Statement 1: Interference obeys the law of conservation of energy.

Statement 2: The energy is redistributed in case of interference.

Answer:  1. Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.

Question 4.

Statement 1: When the apparatus in Young’s double slit experiment is immersed in a liquid the fringe width will increase

Statement 2: The wavelength oflight lh a liquid is a lesser titan than in air. That is \(\lambda^{\prime}=\frac{\lambda}{\mu}\)

Answer:  4. Statement 1 is false, statement 2 is true

Question 5. 

Statement 1: Interference pattern is obtained on a screen due to two identical coherent sources of monochromatic light. The intensity at the central part of the screen becomes one-fourth if one of the sources is blocked.

Statement 2: The resultant intensity is the sum of the intensities due to two sources.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 6.

Statement 1: If the two interfering waves have Intensities in the ratio 9: 4, the ratio of maximum to minimum amplitudes becomes 3:2

Statement 2: Maximum amplitude = A₁+A₂ amplitude = A₁-A₂

Also \(\frac{I_1}{I_2}=\frac{\left(A_1\right)^2}{\left(A_2\right)^2}\)

Answer:  4. Statement 1 is false, statement 2 is true

Light Wave And Interference Of Light Match The Column

Question 1. Match the wavefronts in column 2 with their sources in column 1

Answer:  1- C, 2-A, 3-B

Question 2.

Answer: 1- C, 2- D, 3-B, 4-A

Question 3.

Answer:  1- B, 2- C, 3- A,D,5-C

Question 4. In Young’s double slit experiment the path difference, Ax =(S₂P~ S₁P). Now a glass slab is placed in front of the slit S₂.

Now from the above information match the columns.

Answer:  1-A, 2-C, 3-E, 4-C

5. Column I shows four situations of Young’s double slit experimental arrangement with the screen placed far away from the slits S₁ and S₂ . In each of these cases S₁P₀ = S₂P₀

S₁P₁ – S₂P₂   \(\frac{\lambda}{4}\)

S₁P₁ – S₂P₂   \(\frac{\lambda}{3}\)

Where A is the wavelength of the light used. In cases 2, 3 and 4 a transparent sheet of refractive index p and thickness t is pasted on slit S₂: The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by S(P) and the intensity by I(P). Match each situation given in column I with the statement(s) in column valid for that situation

Answer: 1- A,D, 2- B, 3- E, 4- C,D,E

Light Wave And Interference Of Light Conclusion

1. Light wave is a transverse electromagnetic wave, As a wave generated from a source spreads hr all directions through vacuum or any medium, the locus (line or surface) of the points in the path of the wave in the same phase at any moment constitute a wavefront.

2. According tolUrygens’ principle, each point on a Wavefront acts as a secondary source oflight. It means that from each of these secondary sources secondary wavelets are produced and they spread all around with the same speed, Tire new wavefront at a later stage is simply the common envelope or tangential plane of these wavelets.

3. The principle of superposition of waves is that at any moment the resultant displacement at any point in a medium due to the influence of a number of waves Is equal to the vector sum of the displacements of the component waves at that point.

4. If two light waves having the same amplitude and

Frequency is superposed at any region In a medium the Intensity of the resultant light wave Increases at some points In that region and decreases at some other points. This phenomenon Is known as Interference of light.

The phenomena or Increase hr light Intensity Is called constructive Interference and the decrease In light Intensity Is called destructive interference.

5. To get sustained Interference fringes two coherent sources of light are necessary, 4 Scientist Thomas Young first experimentally demonstrated the Interference of light.

6. In Young’s double-slit experiment, It Is seen that the separation of two consecutive bright or dark hands is equal. This separation Is called fringe width. Fringe width is directly proportional to the wavelength of light

7. Let the two waves

y₁ = \(A \sin \frac{2 \pi}{\lambda}(c t-x)\)

y₂ = \(\lambda \sin \frac{2 \pi}{\lambda}\{c t-(x+\delta)\}\)

Superpose on each other to form interference. Then the resultant displacement produced at a point Is

y = \(=y_1+y_2=A^{\prime} \sin \frac{2 \pi}{\lambda}\left\{c t-\left(x+\frac{\delta}{2}\right)\right\}\)

Where A’ =\(2 A \cos \frac{\pi \delta}{\lambda}\) and δ – path difference between the two waves

8. For constructive Interference, \(2 n \cdot \frac{\lambda}{2}\)

Phase difference = 2nπ

And for destructive interference, δ = (2n+1) \(\frac{\lambda}{2}\)

Phase difference = (2n+1)π

Where n = 0 or any positive integers

In case of young double slit experiment the fringe width y = \(=\frac{D}{2 d} \cdot \lambda\)

Where D = Perpendicular distance of the screen from the

Plane of two monochromatic sources (slits), 2d = distance between two coherent sources, A = wavelength oflight used. Displacement of fringes due to the introduction of a thin plate on the path of one of the waves,

x = \(\frac{y}{\lambda}(\mu-1) t\)

Where y = fringe width, A = wavelength of light, t = thickness of the plate

Due to the inclusion of a glass plate if the central bright band is shifted through a distance of the previous m -bright bands, then (μ-1)t = mλ

Optics

Diffraction And Polarisation Of Light

Diffraction And Polarisation Of Light Definition:

Light rays, while passing round the edges of an A obstacle or aperture, instead of traveling in a straight line, bend to some extent. This phenomenon is known as the diffraction of light.

We know from our daily experience that sound wave bends while passing round the edges of an obstacle or spreads in all directions while passing through a slit or aperture. The same thing happens with light waves. A thin tin sheet placed in sunlight casts its shadow on a wall. Sunrays can be treated as parallel rays and according to geometrical optics, they travel in a straight line. So, a sharp shadow of the thin tin sheet should be observed on the wall.

But if the shadow is examined carefully, it will be seen that the edges are not very distinct. The direction of the light wave changes while passing through the edges of an obstacle or through an aperture. This is called ‘diffraction’ oflight

Read and Learn More Class 12 Physics Notes

In point of In a blade, formed by diffraction monochromatic pattern, mm formed just outside the shadow of the blade, is shown in an enlarged form. The shadow of the side of the blade is not very distinct. In diffraction due to slit or aperture, the deviation of the propagation of the wave depends on its wavelength and on the size of the slit or aperture

Comparison of diffraction Of light with diffraction of sound

The wavelength of an audible sound is sufficiently long (from 1.6 cm to 16 m). Even if there is a big hole in the line of propagation of the wave, the wave deviates considerably while passing through it. On the other hand, the wavelength of visible light is very small, 4000 A° -8000 A°.

Even a very fine slit, like the eye of a needle, is large enough in comparison to the wavelength oflight For a light wave, while passing through a slit large enough in comparison to its wavelength, there is no noticeable change in the direction of light, i.e., diffraction oflight isin distinguishable

Now for the same wavelength oflight, as the aperture is gradually made finer, the diffraction of light becomes more distinct. On the other hand, a distinct diffraction can also be made to occur by increasing the wavelength oflight used, so that the slit can now be comparable in size with the wavelength of light .

Some Special Conclusions: Observing the phenomenon of diffraction of light

The following conclusions can be drawn:

• Like other waves, light also spreads like a wave
• If the size of the apertures is much larger than the wavelength of light, diffraction of light is not easily detectable. In
• In that case,  can be said that light travels in a straight line. In the case of a very fine aperture, when light hends from its straight path, we come to know of the limitations of geometrical optics. That is why, the rectilinear behavior of light according to geometrical optics is actually an approximate behavior.
• When the edges of the obstacle or aperture are sharp, diffraction is more distinctly detectable.
• Diffraction validates the wave theory of light, but it does not give any information about the nature of light waves (whether it is longitudinal or transverse).
• As the wavelengths are long, sound waves and radio waves are diffracted more prominently than other kind of waves.

Optics

Diffraction And Polarisation Of Light Comparison Between Interference And Diffraction Of Light

Similarity: Both interference and diffraction of light take place due to the superposition of waves. Diffraction fringes are formed mainly due to the interference of waves.

Dissimilarity: There are some basic differences between interference and diffraction of light. The differences are as follows

Difference Between Interference And Diffraction

Optics

Diffraction And Polarisation Of Light Classification Of Diffraction

The phenomena of diffraction oflight can be classified mainly into two classes

1. Fresnel diffraction and
2. Fraunhofer diffraction.

1. Fresnel diffraction

• The diffraction, where both the source of light and the screen, are at finite distances from the obstacle or the aperture is called Fresnel diffraction.
• In this diffraction, wavefronts incident on screen are either spherical or cylindrical.
• Obstacles with sharp edges, narrow slits, thin wires, small circular obstacles or holes etc. can produce Fresnel diffraction.

2. Fraunhofer diffraction

• The diffraction, where the source of light and the screen are virtually at an infinite distance, is called Fraunhofer diffraction.
• In this case, the incident wavefront is a plane. Single slit, double slit, diffraction grating etc. produce Fraunhofer diffraction

Optics

Diffraction And Polarisation Of Light Fraunhofer Diffraction By Single Slit

Experimental arrangement

In this experiment of light from a monochromatic light source, Is made to a convex lens L₁, through a narrow silt S. The slit S is held at e focus of the lens Lx. Hence, rays retracted from lens L₁ parallel.

This parallel beam of monochromatic light is incident normally on the slit AB placed perpendicular to the plane of the paper. The ray is now focussed by a convex lens I₁ on a screen MN, where we observe diffraction fringes.  Instead of the screen, if the diffraction pattern is observed by an objective, the tire pattern will be observed in its focal plane

 Fraunhofer Diffraction By Single Slit Explanation

According to geometrical optics, light rays merging out from the slit AB, if focussed by the lens L₂, should produce a sharp image of the slit, at point O of the screen. But actually, this does not happen.

This is because light is passing through AB, and does not propagate in straight lines. Getting diffracted by AB, the light rays spread upwards of point A id downwards of point B. So with the formation of a sharp Image of the slit O, alternate bright ml dark diffraction fringes are produced on both sides of O

Central of principle maximum

Is the midpoint of the silt AB. CO Is the principal axis, the livery point of the plane wavefront, which Is an Incident on the slit, and Is of tho name phase. All wavelets originating from those points and proceeding parallel to CO are focussed by L₂ at O. Since those wavelets have no path difference, they are In the same phase. So, they make constructive Interference, and point O appears very bright. O is called principal or control maximum. Simplified form of

Conditions for the formation of minima and secondary maxima

Suppose, some wavelets after being diffracted through an angle  are focussed at Ox by the lens L₂

Now, the condition for the formation of constructive or destructive interference at the point Ox depends on the path difference of the wavelets originating from A and B. From A, a perpendicular AP is drawn on BOx.

So the path difference between the wavelets emergent from points A and B =BP. Now, BP =AB sin ∠BAP = a sin θ

[where, AB = a- width of the slit]

1. For minima:

To obtain the condition for minima being formed at O₁, slit AB is notionally divided into halves, AC and CB, i.e., AC = CB = \(\frac{a}{2}\)

Let the wavelength of incident monochromatic light = λ. If the path difference between two wavelets, originating from points A and C be  \(\frac{\lambda}{2}\)  they would cause destructive interference.

S°, the condition of formation of first minima on both sides of O for diffracting angle θ₁ is,

⇒ \(\frac{a}{2} \sin \theta_1=\frac{\lambda}{2} \quad \text { or, } a \sin \theta_1=\lambda\)

Or, \(\sin \theta_1=\frac{\lambda}{a}\) …………………………… (1)

In general, the condition for the formation of n th minimum on both sides of O, for diffracting angle θ_n is, a side = nλ …………………………… (2)

Putting n = ±1, ±2, ±3,………….  in equation (2), we would get simultaneously, first, second, third etc. minima on either side of the principal maximum. Here, ± sign is used to indicate diffractions on either side of the central line.

2. For secondary maxima:

If the path difference of the wavelets emitted from A and B, BP = \(\frac{3 \lambda}{2}, \frac{5 \lambda}{2}, \cdots(2 n+1) \frac{\lambda}{2}\) then at points O₂, O₄, etc. they would produce first, second, etc. maxima. At these points, the two waves superpose in the same phase. These are called secondary maxima. If ‘ is the corresponding angle of diffraction for the nth secondary maximum, then

⇒ \(a \sin \theta_n^{\prime}=(2 n+1) \frac{\lambda}{2}\) ………………..(3)

[where, n = ±1, ±2, ±3……. etc.]

It is to be noted that the intensity of the secondary maxima gradually decreases

The linear distance of the nth minimum from the central maximum

Generally, the wavelength of visible light (for example, A = 5 × 10^-7m) is much lower than the width of the slit (for example, a = 10^-4m ). For such values of θ, sin θ ≈θ. With this approximation, equation (2) becomes

⇒ \(\theta_n=\frac{n \lambda}{a}\) …………………………(4)

Let the distance from principal maximum point 0 to n th minimum point On, OOn = xn and distance from screen to slit =D

As the value of θ_n is very small,  \(\theta_n=\frac{x_n}{D}\)

Putting the value of θ_n in equation (4) we get

⇒ \(a \cdot \frac{x_n}{D}=n \lambda \quad \text { or, } x_n=\frac{n \lambda D}{a}\) ………. (5)

Putting n = ±1,±2, ±3-” etc. in equation (5), linear distances of various minima from central maximum are obtained

Width of central maximum

The angle between the first minima on either side of the central maximum is called the angular width of the central maximum.

According to equation (4), the angular spread of the central maximum on either side is

⇒ \(\theta=\frac{\lambda}{a}\)

Angular width of central maximum,

⇒ \(2 \theta=\frac{2 \lambda}{a}\) ……………………. (6)

Therefore, linear width of central maximum \(=D \cdot 2 \theta=\frac{2 D \lambda}{a}\) where D = distance of slit from the screen. If lens L₂ is located very close to the slit AB, or if the screen MN is located far away from lens L₂ > then D ≈the focal length of the lens (f).

In that case, linear width central maximum point In general, the condition for the formation of n th minimum on = \(\frac{2 f \lambda}{a}\)

Optics

Diffraction And Polarisation Of Light Resolving Power Of Optical Instruments

Two types of resolving power are relevant for different optical instruments: O Spatial resolving power and Q Spectral resolving power.

Spatial or angular resolving power

Our eye is an optical instrument. If two point objects (or their images) are very close to each other, our eyes may not see them as separate objects. They seem to be the same object or the same image. It can be verified by a simple experiment.

Let a white paper be fixed on a wall in front of us. On the paper black parallel lines are drawn at 2 mm distance apart. When we stand very close to the wall, we can see all the parallel lines. When we gradually move away from the wall, the angle formed by any two lines at our eye gets diminished and at one point it seems that die lines have merged with each other i.e., the lines can no longer be identified separately.

It can be inferred that whether two objects placed side by side can be differentiated, depends on the angle formed by the two objects at our eyes. It has been established through experiments that if the angle becomes less than 1 minute or \(\frac{1}{60}\) degree, eyes will not be able to see two objects separately.

This angle is called the angular limit of the resolution of our eyes. This means our eyes, as well as optical instrument, has their own limit of

Resolving images of two different objects located very near to each other:

1. Limit of Resolution: The limit of Resolution is the smallest linear distance or the angular separation between two objects that can be directly seen through an optical instrument, is called the spatial limit of resolution of that instrument.
2. Resolving power: The power or ability of an instrument to produce distinct separate images of two close objects, is called the spatial resolving power of the instrument.

Spatial resolving power is measured by the reciprocal of the limit of resolution. If Δx or is the linear or angular limit, then the resolving power would be \(\frac{1}{\Delta x} \text { or } \frac{1}{\Delta \theta}\) respectively

Spectral resolving power

Instruments like prism and diffraction grating are used to separate spectral lines of different wavelengths. For example, the D, and D₂ lines of sodium spe trim have a separation of 6 A of wavelength between them. Usually, a prism cannot separate them, but a diffraction grating can. So we say that the limit of resolution of a grating is 6 A° or less, whereas that of a prism is greater than 6 A°.

If an optical instrument just resolves two spectral lines of wavelengths λ and λ + Δλ, then its limit of resolution is defined as Δλ and its spectral resolving power as \(\frac{\lambda}{\Delta \lambda}\)

Rayleigh criterion:

This defines the spectral limit of resolution of an optical instrument. Its statement is:

Two images are said to be just resolved when the central maxi¬ mum in the diffraction pattern due to one of them is situated at the first minimum in the diffraction pattern due to the other.

It is to be noted that, spatial resolving power is intimately related to spectral resolving power; because, to observe the spatial 1. on stars separation between two objects, we often have to use Instruments working on the principle of wavelength separation, phenomenon of diffraction, etc.

Resolving power of Microscope

If a microscope is able lo show the image, of two point objects, lying close lo each other, separably, then the reciprocal of the distance between these two objects is the resolving power of that microscope.

This power depends on the wavelength (λ) of light used, the refractive index (μ) of the medium between two objects and the objective of the microscope, and the cone angle (θ) formed by the radius of the objective on any one of the objects

If the internal distance between two objects is And, then the resolving power of the microscope’

R = \(\frac{1}{\Delta d}=\frac{2 \mu \sin \theta}{\lambda}\)

To increase the resolving power of a microscope, the objects and the objective of the instrument are immersed in oil. Hence, as the value of fj increases, the resolving power, R also increases.

The expression μ sinθ is called the numerical aperture of a microscope. It is a special characteristic of a microscope. It is mentioned in some microscopes

Resolving power of Astronomical Telescope

When a telescope is able to analyze two separate distant objects lying closely, then the reciprocal of the angle subtended by the two objects at its objective is called the resolving power of that telescope.

If the angle subtended, by the two objects at the objective, be Δθ, then the resolving power of the telescope,

R = \(\frac{1}{\Delta \theta}=\frac{a}{1.22 \lambda}\)

[where a = diameter of the objective of the telescope]

Hence, if the diameter of the objective of the telescope is increased, its resolving power increases. Again if the wavelength of the incident light decreases, The resolving power increases.

1. The angular spread Ad of a telescope depends solely on its objective. If the objective of a telescope is unable to analyze two stars located extremely far away, then these cannot be analyzed by the telescope even by increasing the magnification of its eye piece.
2. To see different astronomical objects in the sky, telescopes widi an objective having a diameter 1mm or more are used

Optics

Diffraction And Polarisation Of Light Polarisation Of Light

Polarisation of light Definition:

The phenomenon of restricting the vibrations of the electric vector of a light wave along a particular axis in a plane perpendicular to the direction of the light wave is called polarisation of light.

The phenomena of interference and diffraction demonstrate that light propagates in the form of waves. Butitis is not understandable from interference and diffraction, whether the light waves are transverseorlongitudinalinnaturebecausebothlongitudinaland transverse waves exhibit interference and diffraction.

The topic of discussion of this section is the polarisation of light. This phenomenon of light distinctly proves that light waves are transverse in nature and not longitudinal like sound waves.

Polarisation of mechanical waves

Two narrow slits A and B are cut in the middle portions of two cardboards C₁ and C₂. A thin long string OE, tied at one end E to a rigid support is passed through the slits A and B

Now holding the end O of the string, it is made to vibrate perpendicularly along the direction to result,is a taken transverse long wave OE, the advanced particles of the string will vibrate perpendicular to the x-axis, i.e., In the y: plane. Holding the end O, the string can be made to vibrate randomly i.e., In any direction in the yz plane. In that case, each particle of the string, in the intermediate portion of the string between O and A, will have two vertical components of transverse vibration along the y and z-axes

At first tire cardboards, C₁ and C₁ are so placed that both the slits A and B are parallel to y-axis. Clearly, the component of vibration will be obstructed by the slit A, but the y -component will pass through A without any obstruction and reach the section AB.

So in spite of random vibrations of the string in portion OA, the vibration of the string in section AB will be confined only along the y-axis. This phenomenon of converting the random vibrations of a transverse wave to unidirectional vibration is called polarization.

In this case, the transverse wave in section OA is unpolarised, but it turns into a polarised wave in die section AB by the slit A because the wave of this section (section AB ) vibrates only along the y-axis.

Since slit B is parallel to the y-axis, so the vibration of the wave along y-axis in the section AB, will pass through slit B also without any obstruction. Thus, the transverse wave will propagate up to point E.

Now if the slit B is rotated through 90° with respect to OE, then the slit becomes parallel to the z-axis. Clearly, the vibrations of the string of section AB along the y-axis, get completely obstructed by the slit B. So, no vibration exists in the section BE of the string i.e., the transverse wave cannot propagate along BE.

From the above discussion, it is clear that if the vibration is longitudinal, that is, parallel to the x-axis, then, they are not at all obstructed by slits A and B in any of their orientations. Thus the longitudinal wave can propagate up to point E. Hence, it can be said that polarisation is a phenomenon that is not exhibited by longitudinal waves. For example, sound wave is a longitudinal wave, hence sound wave is not polarised.

Unpolarised Light

In the usual sources of light like the sun, candle, electric lamp, etc., electrons, ions, or other charged particles vibrate randomly. Hence the transverse vibrations of the waves emitted from these sources have no definite direction.

This type of light is called ordinary light or unpolarised light.  In this case, the transverse vibration may be referred to as the sum of the two perpendicular components of equal amplitude.

Light is an electromagnetic wave. The electric field £ and the magnetic field B of this wave always vibrate perpendicular to the direction of the wave. The vibration is confined to a certain plane. The wave propagates in a direction perpendicular to the plane

The polarization of light can be easily explained by an experiment with tourmaline crystals.

Experiment with a tourmaline crystal

Tourmaline is a hexagonal crystal. The crystal cut in the form of a thin plate behaves almost like a transparent substance. The longest diagonal of the hexagonal crystal is called the crystallographic axis or optical axis C₁ and C₂ are two thin tourma¬ line crystal sheets and M₁N₁ and M₂N₂ are their optical axes respectively

O is an ordinary source of light. Keeping the eye fixed at position E, one is looking towards O. Here x-axis is taken along OE. At first, the crystal C₁ is placed on the way of the ray OE at location A in such a way that its optical axis M₁N₁ lies perpendicular to x -x-axis.

If the crystal is so placed, the intensity of light is found to be a little diminished. If the crystal is made to rotate about OE as the axis of rotation, the intensity of the transmitted light remains unchanged.

Now the crystal C₂ is also placed at B on the way of the ray OE in such a way that the optic axes of both C₁ and C₂ are parallel to y -axis. It is found that light comes out undiminished in intensity in spite of C₂ being placed.

But as the crystal C₂ is rotated slowly about point B with OE as the axis of rotation, it is found that the intensity of light decreases. When the axis of C₂ makes an angle 90° with the axis of C₁i.e., crystallographic axis M₂N₂ becomes parallel to the z-axis, no light from the source reaches the eye. When C₂ is rotated further, the intensity of the light gradually increases. When C₂ is rotated through another 90°, i.e.,it is rotated through 180° from its initial position, light reappears with its earlier intensity

Explanation of the result of the experiment

The above experiment can be explained if we consider light waves as transverse nature and the crystallographic axes of the tourma¬ line crystals as narrow slits. The transverse vibrations of the light waves emitted from the source O are random in nature.

So, two perpendicular components of vibration along y and z-axes exist in each point of the section OA of the light ray. Since the axis M₁N₁ of the crystal C₁ is placed parallel to y -y-axis, so the y -y-component of the transverse vibration of the light wave passes through the crystal, but the z -z-component is completely absorbed. Since one component is absorbed completely, the intensity of the transmitted light becomes half. Only y -the component of the transverse vibration of the light wave has been shown in section AB

Now if the crystallographic axis M₂N₂ of the crystal C₂ also becomes parallel to y -the axis, the y -y-component of the transverse vibration passes through the crystal and reaches the eyes But by rotating the optical axis M₂N₂ through 90°, if it is placed parallel to z-axis, the crystal C₂ absorbs the y component of the vibration completely.

So, no vibration exists in the section BE, lightwave is absent here. Hence no light reaches the eye  When crystal C₂ is rotated through another 90° i.e., when total rotation is 180°, the optical axis M₂N₂ becomes parallel to the y-axis again, as a result, the y -component of the transverse vibration can pass through the crystal C₂

Polarised Light Conclusion:

When an ordinary light wave passes through a tourmaline crystal or a similar medium, its random transverse vibrations are converted to a unidirectional transverse vibration.

This phenomenon is called polarisation of light and the light is called polarised light. In the, light of the section AB is called polarised light.

Any transverse wave, like a light wave can be polarised.

Polariser:

The instrument by which unpolarised light is made polarised is called a polariser. The tourmaline crystal C₁ is called the polariser and the crystallographic axis M₁N₁ is called polarising axis.

Analyzer:

The instrument examines whether light is polarised or not and the type of polarization. Is called analyser. The tourmaline crystal C₂ is called an analyzer because it examines whether the light is polarised or not and what type of polarization has been produced by the crystal C₁.

When the crystallographic axes of the crystals C₁, and C₂ arc parallel, It is called the parallel position of polarizer and analyzer. When their crystallographic axes are perpendicular to each other, they are said to be in a crossed position.

Optics

Diffraction And Polarisation Of Light Plan Of Vibration And Plane Of Polarisation

Plane of vibration Definition: The plane in which the vibration of the polarised light remains confined is called the plane of vibration.

Plane of polarisation Definition: The plane containing the ray of light and perpendicular to the plane of vibration is called the plane of polarisation.

 Description:

The direction of propagation of the polarised light through the tourmaline crystal AB is shown. Light rays are advancing along x -the axis and the electric field of polarised light is vibrating along y -the y-axis, xy plane i.e., the plane ABCD is the plane of vibration of polarised light.

Unit 6 Optics Chapter 7 Diffraction And Polarisation Of Light Plane Polarised Or Line arly Polarised Light

 Plane-polarised or linearly polarised light Definition: If the vibration of polarised light remains confined In a plane and takes place along a straight line, then it is called plane-polarised or linearly polarised light.

The vibration of an electric field of a light wave on a plane perpendicular to the direction of propagation of an ordinary light or unpolarised light can take place In any direction from a point. The direction of vibration In a particular plane perpendicular to the direction of propagation of a ray of light, Is shown by the arrowheads In different directions in that plane

Convention of representation of unpolarized and polarised light:

In the plane of the paper, unpolarUed and polarised lights are represented according to the following conventions.

Ordinary light i.e., unpolarised light has vibration In all directions In a plane, perpendicular to the direction of propagation of light  It is supposed to be made up of two mutually perpendicular vibrations. Hence ordinary unpolarized light is shown with dots and lines with arrows in opposite directions at the same time

If the polarised light has vibrations in the plane of the paper, it is shown with lines haring arrows in opposite directions perpendicular to its direction of propagation

If the vibration of polarised light in a direction perpendicular to the plane of the paper

They are shown by dots on the line of propagation:

1. When unpolarised light is transmitted through an analyzer, its intensity is halved.
2. 2. If polarised light is incident on an analyser dien die intensity of the transmitted light is given by Malus’ law.

Malus’ law:

When a beam of completely plane polarised light is incident on an analyzer, the resultant intensity of light (I) transmitted from the analyzer varies directly as the square of the cosine of the angle (θ) between the plane of transmission of the analyzer and polariser i.e

⇒ \(I \propto \cos ^2 \theta \text { or, } I=I_0 \cos ^2 \theta\)

Where I₀ is the intensity of the light incident on the analyzer.

Diffraction And Polarisation Of Light Polarisation By Reflection

In 1808 French scientist E L Malus discovered that plane polar¬ ised light can be produced by reflection. He showed that when ordinary light i.e., unpolarised light is reflected from the surface of a transparent medium such as glass or water, the reflected light becomes partially plane polarised. The degree of polarisation depends upon the angle of incidence

Angle of polarisation Definition:

For a particular angle of incidence, the degree of polarisation by reflection is maximized. This angle of incidence is called the angle of polarisation or polarising angle

The magnitude of this angle depends on the nature of the reflecting surface and the wavelength of the Incident light

For glass, the polarising angle Is an hour, and for pine water, it Is about 53°

The experiment of polarisation by reflection

Let us sup. pose, a black glass plate is placed perpendicularly a sheet of paper. MM’ is the smooth upper surface of the plate This is the reflecting surface. As the glass is black, the possibility of more than one reflection of the refracted ray is less. An ordinary’ my of light AO is incident on the reflecting surface at an angle 5(5° and is reflected along OH. The reflected ray OB will be plane-polarised.

To examine, a tourmaline crystal ( T”) is placed OB and looked from point E located behind the crystal along BO. Now the crystal is rotated slowly about the reflected ray OB. It will be seen that at a particular position of the crystal, no light Is transmitted through the crystal. The crystal is again rotated from this position slowly and when the rotating angle becomes 90°, the Intensity of light transmitted by the crystal will be maximum. This proves that the reflected beam OB is polarised.

Polarisation by Reflection Explanation:

The transverse vibrations of ordinary incident light may be supposed to consist of two mutually perpendicular vibrations

• One component in the plane of the paper i.e., lies in the plane of incidence and
• The other perpendicular to the plane of paper i.e., lies parallel to the reflecting surface.

Whatever be the value of the angle of incidence on MM’s plane, the vibration of the second component will always be parallel to the reflecting surface. As a result, if the incident angle changes, the vibration of that component will be parallel with the reflecting plane MM’ but the vibrations of the first component will make varying angles with the reflecting plane. If light is incident at a polarising angle, the vibrations will be refracted from air to glass and get absorbed inside the glass i.e., these rays will not be reflected.

Only the second component will be reflected. Hence reflected ray OB is the plane polarised light. The plane perpendicular to the sheet of paper is the vibration plane of die-polarised light. So, it can be said that plane-polarised light can be produced by reflection

Optics

Diffraction And Polarisation Of Light Brewster S Law

When polarised light is incident at a polarising angle on the interface of two media of different refractive indices; a portion of that light is reflected and completely polarised and the other portion is refracted and partly polarised.

It has been found from the experiment that in the event of such an unreflective and refraction of an unpolarised light, incident polarising angle, the reflected ray and the refracted ray become mutually M perpendicular.It is to be mentioned here that this polarising angle is also called Brewster’s angle

In ∠PON = angle of polarisation (polarising angle)

= i_p and ∠QON’ is the corresponding angle of refraction = r.

Then , i_p + r= 90°

Or, r = 90° – i_p ………………………… (1)

Then, according to Snell’s law; \(\frac{\sin i_p}{\sin r}=\frac{\mu_2}{\mu_1}\)

[where μ ₁ = refractive index of the medium of incidence

μ ₂= refractive index of the medium refracting of incidence

⇒ \(\frac{\sin i_p}{\cos i_p}=\frac{\mu_2}{\mu_1}\)

Or,  \(\tan i_p=\frac{\mu_2}{\mu_1}\) ………………………… (2)

If both the incident ray and the reflected then px = 1

In that case, if refractive index of the refracting medium is taken as n, the equation (2) can be written as follows

tan i_p = μ

i.e., the tangent of a polarising angle is numerically equal to the refractive, the index refractive of the index refractive of the index reflecting of a medium. depends this is Brewster’son the wavelength depends of on light wavelength.It can be said that the polarising angle also

Optics

Diffraction And Polarisation Of Light Polarisation By Refraction Brewster S Law

When an ordinary (unpolarised) light is incident on the upper surface of a parallel-faced glass plate at the polarising angle, the reflected light is completely plane polarised but its intensity is very low.

A major portion (about 85%) of the incident ray is refracted and only a very small portion (15%) is reflected. The refracted ray is also partly polarised. The two planes of polarisation of completely polarised reflected ray and partly polarised refracted ray are at right angles to each other.

So it is not possible to get a strongly reflected beam of polarised light with the help of a single plate. To overcome this defect, a number of plane parallel glass plates are placed parallel to each other and an unpolarised light is allowed to fall on the first plate at the polarising angle.

Due to successive reflections, strong beams of polarised reflected light will be obtained. Ultimately two plane polarised light will be separated one reflected polarised light with vibrations perpendicular to the plane of the paper i.e., the plane of incidence, and another refracted polarised light with vibrations in the plane of the paper

Optics

Diffraction And Polarisation Of Light Double Refraction Or Birefringence

In 1669 Ramus Bartholin discovered that when an ordinary ray of light is incident perpendicularly on the surface of a calcite crystal, it splits into two rays due to refraction
one of the refracted rays obeys the laws of refraction and is called the ordinary ray or O -ray. The other refracted ray does not obey these laws hence it is called the extraordinary ray or E- ray Both of these rays are plane polarised in mutually perpendicular planes.

The phenomenon by virtue of which an unpolarised ordinary ray, on entering a crystalline substance,

Splits up into two rays:

1. Herapathite la an organic compound whose chemical name is iodoquine sulphate.It is a dichroic crystal used to polarise light rays.
2. Each polaroid hasaparticularplaneofpolarisation.Apolaroid allows only those incident unpolarised light rays to refract, whose vibration plane is parallel to the polarisation plane of the polaroid. The direction of this polarisation plane of the polaroid is called the transmission axis

• Uses of PolaroidWith the help of polaroids, plane polarised light can be produced and analyzed very easily at a low cost
• Polaroids are widely used as polarising sunglasses.
• Polaroids are used to eliminate the headlight glare in motor cars.
• Polaroids are used as glass windows in trains and airplanes to control the intensity of light coming from outside.
• In calculators and watches, letters and numbers are formed by liquid crystal display (LCD) through the polarisation of light.
• Polaroid films are used in making 3D cinema or picture

Diffraction And Polarisation Of Light Numerical Examples

Example 1.  For producing a Fraunhofer diffraction fringe, a screen, Is placed 2m away from a single narrow slit. If the width of the slit is 0.2 mm, It is found that the first minimum lies 5 mm on either side of the central maximum. Find the wavelength of the incident light
Solution:

We know, if the distance between n nth minimum and the Central maximum is, the

⇒ \(x_n=\frac{n D \lambda}{a}\)

Or, 0.5 = \(\frac{1 \times 200 \times \lambda}{0.02}\)

[Here , n= 1 xn = 5mm = 0.5 cm , D = 2m = 200 cm , a= 0.2 mm = 0.02 cm ]

⇒\(\frac{0.5 \times 0.02}{1 \times 200}\)

= 5000 × 10^-8  cm

= 5000 A°

Example 2. A single narrow slit of width 0.1 mm, with a parallel beam of light of wavelength 600 ×  10^-9 m. An interference fringe is formed on a screen 40 cm away from the slit. At what distance, will the third minimum band be formed from the central maximum band?
Solution:

We know, if the distance of the nth minimum from the central maximum is x_n, then

⇒ \(x_n=\frac{n D \lambda}{a}\)

[Here, n = 3, a = 0.1 mm = 0.01 cm D = 40 cm and A = 600 ×  10^-9 m=  600 ×  10^-7 cm ]

⇒ \(\frac{3 \times 40 \times 600 \times 10^{-7}}{0.01}\)

= 0.72 cm

Example 3. A Fraunhofer diffraction pattern is formed by light, of wavelength 600 nm, through a slit of width 1.2 μm. Find the angular position of the first minimum and the angular width of the central maximum.
Solution:

If the angular position of the first minimum with respect to the central maximum be θ, then

sin θ = \(\frac{\lambda}{a}=\frac{600 \times 10^{-9}}{1.2 \times 10^{-6}}\)

= 0.5 = \(\frac{1}{2}\)

= sin 30°

θ = 30° [ Here , = 600 nm = 600 ×  10^-9 m

a = 1.2 μm = 1.2 ×  10^-6 m

∴ Angular width of central maximum = 2θ = 2 × 30° = 60°

Example 4.  A Fraunhofer diffraction pattern is formed by a light wave of frequency 5 × 10^-4 Hz through a slit of width m. Find the angular width of the central maximum [ Velocity of light in vaccum] = 5 × 10^-8m.s^-1

The angular width on either side of the central maximum

⇒ \(\frac{\lambda}{a}=\frac{c}{\nu a}\)

= \(\frac{c}{\nu}\)

= \(\frac{3 \times 10^8}{5 \times 10^{14} \times 10^{-2}}\)

[Here, ν = 5 × 10¹⁴ Hz

a = 10^-2M , C = 3 × 10^-8m.s^-1

= 0.6 × 10^-4 rad

∴ Angular width of central maximum = 2 = 1.2 × 10^-4 rad

Example – 5. A single narrow slit of width It Illuminated by a monochromatic parallel ray of light of wavelength 700 nm. Find the value of an In each cate following the given conditions

First minimum for 30 diffraction angle and

First secondary maximum for 30° diffraction angle

From the conditions of formation of n nth minimum,

a sin = n λ

a sin 30° = 1 × 700 × 10^-9 m

Since θ = 30°, n= 1

And λ = 700 nm = 700 × 10^-9 m]

a = \(\frac{700 \times 10^{-9}}{\frac{1}{2}}=14 \times 10^{-7}\)

From the condition of formation of n nth secondary maximum

a sin = (2n +1) \(\frac{\lambda}{2}\)

Or, a sin = \(\frac{3}{2} \times 700 \times 10^{-9}\)

= 30° for 1st secondary maximum, n = 1 and λ = 700 nm e 700 × 10^-9  m

∴ a = 21 × 10^-7 m

Example 6. A Star Is observed through a telescope. The diameter of the objective of the telescope is 203.2 cm. The wave¬ length of the light, corning from the star to the tele¬ scope U 6600A. Find the resolving power of the telescope.
Solution:

Resolving power of a telescope

R = \(\frac{a}{1.22 \lambda}=\frac{203.2}{1.22 \times 6600 \times 10^{-8}}\)

= \(2.52 \times 10^6\)

= 2.52 × 10^-6

Here, the diameter of the objective of the telescope,

a =  203.2 cm

And wavelength of the Incident light, λ =  6600 A° = 6600× 10^-8cm

Example 7. Find the Brewster angle for *|r to glass transmission.(R.1. of glass = 1.5)
Solution:

If the refractive index of glass relative to air be p and the polarising angle he ip, then according to Brewster’s law we have,

μ = tan i_p Or, tan i_p = 1.5

i_p = tan^-1 (1.5) = 56. 3

Example 8. A single narrow till of width a U Illuminated by white light or whal value of a will the for minimum of rd light of wavelength 650 nm, lie al point PI For what wavelength of the Incident light will the first secondary maximum lie at point P?
Solution:

From the condition of formation of the first minimum at point P.

a sin θ = nλ

Here, n = 1 ,  θ= 30° and A 650 nm

a sin 30° = 1 × 650 or, a = 1300 nm

a sin θ = \((2 n+1) \frac{\lambda^{\prime}}{2}\)

Or, a sin θ = \(\frac{3}{2} \lambda^{\prime}\)

[Here, A’ – wavelength to be determined and n Or, \(\lambda^{\prime}=\frac{2}{3} a \sin \theta\)

= \(\frac{2}{3} \times 1300 \times \frac{1}{2}\)

= 433.33

The required wavelength a 433.33 nm

Example 9. The refractive Index of glass la 1.55. What is the polarising angle? Determine the angle of refraction for the polarising angle.
Solution:

If the refractive index of glass relative to air be μ and the polarising angle be i_p, then according to Brewster’s law we have.

μ = tan i_p  or, tan i_p = 1.55

i_p =  tan^-1(1.55) = 57. 17°

Again for Incidence at the polarising angle,

i_p  + r = 90° [here r =  angle of refraction]

Or, r = 90°- i_p = 90°- 57.17º = 32.83

Example 10. The critical angle of n transparent crystal is 30°, What is the polarising angle of the crystal?
Solution:

If μ is the refractive Index and θ_c. is the critical angle of the crystal, then

μ = \(\frac{1}{\sin \theta_c}=\frac{1}{\sin 30^{\circ}}\) = 2

From Brewster’s law, we have

tan i_p =μ

[Here i_p = polarising angle]

Or, tan i_p = 2 Or, i_p= tan^-1 (2) = 63. 43°

Example 11. Determine the polarising angle of the light ray moving from the water of a refractive index of 1.33 to a glass with a refractive index of 1.5
Solution:

From Brewster’s law, we have,

⇒ \(_w \mu_g=\tan i_p\)

[Here  i_p = Polarising angle]

Or, \(\frac{a^{\mu_g}}{{ }_a \mu_w}=\tan i_p\)

Or, \(\frac{1.5}{1.33}=\tan i_p\)

Or, tan i_p = 1.13

Or, i_p = tan^-1 (1.13) = 48. 5

Example 12. When sun rays is incident at an angle 37 on the water surface, the reflected ray gets completely plane polarised. Find the angle ofrefraction and refracttve Index of water
Solution:

According to the question, angle of incidence

= (90° -37°) = 53°

The angle of polarization, i_p = 53°

Now,  i_p + r = 90°

Or, r = 90° – i_p

Or, r = 90°- 53°= 37°

According to Brewster’s law, the refractive index of water

μ = tan i_p = tan 53 °= 1.327

Diffraction And Polarisation Of Light Synopsis

1. When a wave passes close to the edges of an obstacle or an aperture, the direction of motion of the wave gets changed. This is called the diffraction of the wave.

2. Light is a wave, so it has the property of diffraction. Light, while passing around the edges of an obstacle or an aperture, bends a little departing from straight-line propagation. This incident is called the diffraction of light.

3. As wavelength increases, the amount of bending i.e., dif¬ fraction also increases. Hence sound waves and radio waves are diffracted more as their wavelengths are large.

4. Fringes of diffraction are formed due to interference of the waves.

5. There are two types of diffraction phenomena. They are— O Fresnel’s diffraction and Fraunhofer’s diffraction.

6. In Fresnel’s diffraction, the source of light and the screen on which the diffraction pattern is observed are at finite distances from the obstacle or aperture.

7. On the other hand/ In Fraunhofer’s diffraction, the source of light and the screen are virtually at an Infinite distance from the obstacle or aperture. Single narrow silt produces Fraunhofer’s diffraction.

8. Single narrow slits produce diffraction fringes on the screen. On either side of the central maximum, minima and secondary maxima are formed. The intensity of the secondary maxima gradually decreases.

9. If a microscope is just able to sec two objects, lying close to each other, separately, then the reciprocal of the distance between the two objects Is the resolving power of that microscope.

10. When a telescope Is able to analyse distinctly two separate objects lying closely, then the reciprocal of the angle subtended by the two objects at the objective of the telescope, Is called its resolving power.

11. Light emitted from the usual sources like the sun, candle, electric lamp etc. are unpolarised light.

12. Hit phenomenon of restricting the vibrations of an electric vector of a light wave along a particular direction of axis, In a plane perpendicular to the direction of the light wave. Is called the polarisation of light

13. When an unpolarized light wave passes through a tourma¬ line crystal or similar medium, its random transverse vibrations are converted into a unidirectional transverse vibration.

14. Polarisation proves that a light wave is transverse Sound waves can not he polarised as is a longitudinal wave.

The plane in which the wave. vibration of the polarised light remains confined is called a plane of vibration. The plane drawn through the light rays which are perpendicular to the plane ol vibration Is called the plane of polarization

15. Plane-polarised light can be produced by reflection. For a particular angle of incidence, the degree of polarisation by the reflection is maximum. This angleofincidenceiscalledangle polarization or Brewster’s angle. The magnitude of this angle depends on the nature of the reflecting surface.

16. When a ray of light is incident at the interface of two media at a polarising angle, the reflected and the refracted rays become perpendicular to each other.

17. Brewster’s law: The tangent of the polarising angle is numerically equal to the refractive index of the refracting medium.

18. When an ordinary ray of light (unpolarised ray) is incident perpendicularly on the surface of a calcite crystal, it splits into two rays due to refraction. One of the refracted rays is called an ordinary ray or O -ray which obeys the common

19. Laws of refraction and the other is called extraordinary ray or E -ray which does not obey these laws. This Incident Is called double refraction or birefringence. The crystals In which double refraction takes place are called double-refracting crystals. Examples of such crystals are calcite, quartz, tourmaline, etc.

20. Double refracting crystals are classified Into two types

1. Negative crystal and positive crystal.
2. In negative crystals (tourmaline, calcite),

μ_o > μ_E ,> V_E>V_o

And in positive crystals (ice, quartz),

μ_o > μ_E ,> V_o>V_E

21. A polaroid is a polarising sheet or film by which polarised light can be produced.

In the single-slit experiment, the condition for the formation of the nth minimum

a sin n = n λ

Where n = ±1,±2,±3, , a = width of the slit

λ = wavelength of light used and 8n = diffraction angle]

22. Distance of nth minimum from the central maximum,

⇒ \(x_n=\frac{n \lambda D}{a}\)

Where D= distance of the screen from the slit

Condition for the formation of the nth secondary maximum

⇒ \(a \sin \theta_n^{\prime}=(2 n+1) \frac{\lambda}{2}\)

Where n = +1, ±2, etc

23. Angular width of central maximum = \(\frac{2 \lambda}{a}\)

24. Linear width of central maximum = \(\frac{2 \dot{D} \lambda}{a}\)

Malus’ law: I = I_o = cos²θ

Where  I_o = Intensity of the light incident on the analyzer.

25. Brewster’s law: pt = tan i_p

Where n – refractive index of the refracting medium with respect to air, ip = angle of polarization.

26. i_p + r = 90°; where r = angle of refraction

Diffraction And Polarisation Of Light Very Short Questions And Answers

Question 1. Do the sound waves have the property of diffraction?
Answer: Yes

Question 2. What idea does diffraction of light give about the nature of light light waves?
Answer: Gives no idea

Question 3. Why do we feel more diffraction in sound waves than in light waves?
Answer:  As the wavelength of sound is larger than the light wave

Question 4. While passing around the comer of an obstacle, the bending with the increase of the wavelength of light
Answer: Increases

Question 5. What should be the nature of the incident wavefront ir case of Fresnel’s diffraction? Answer: Spherical or cylindrical diffraction

Question 6. A single slit produces ‘ diffraction [Fill in the blank].
Answer: Fraunhofer

Question 7. In Fresnel’s diffraction, the source of light is located at a distance from the aperture (silt)
Answer: Finite

Question 8. If the wavelength of the incident light in a single slit is increased, the Fraunhofer’s diffraction bands will be
Answer: Wider

Question 9. In a single slit diffraction, the intensity of secondary maxima gradually
Answer: Decreases

Question 10. How does the angular width of the central maximum change when the slit width is increased?
Answer: Angular width will decrease

Question 11.. Instead of violet light if red light is used in the formation of
the diffraction pattern in a single slit, the diffraction band will be wider—is the statement correct
Answer: Yes

Question 12. In single-slit diffraction, what is the condition for the formation of the first minimum point?
Answer: [a sin θ= λ, where a = width of the slit

Question 13. In a single slit diffraction, what is the expression of the linear width of the central maximum?
Answer:

a sin = \(\frac{3}{2} \lambda\) λ , where a = width of the slit

Question 14. In a single slit diffraction, what is the expression of the linear width of the central maximum?
Answer:

Linear width = 2f \(\frac{d}{a}\)

Question 15. What is called the power of an optical instrument to produce distinctly separate images of two close objects?
Answer: Resolving power

Question 16. What does the polarisation of light prove about the nature of
Answer: Lightwave is transverse

Question 17. Why does ultrasonic waves not exhibit polarisation?
Answer: Because the ultrasonic wave is a longitudinal wave

Question 18. When light is polarised, how does its intensity change?
Answer: Intensityis reduced

Question 19. What is the angle between the plane of polarisation and the direction to propagation to polarised light?
Answer: 0

Question 20. Tourmaline is a hexagonal crystal. The longest diagonal of the crystal is known as
Answer: Optic axis

Question 21. If a beam of light has its vibrations restricted to one plane instead of different planes, itis called
Answer: Polarisation

Question 22. The plane containing the direction of propagation of light and perpendicular to the plane of vibration is called
Answer: Plane of polarization

Question 23. A ray of light incident on a medium at a polarising angle. What will be the angle between the reflected and the refracted rays?
Or
An unpolarised ray of light is incident on a rectangular glass block at Brewster’s angle. What will be the angle between the reflected and the refracted rays?
Answer: 90°

Question 24. If the polarising angle for the air-glass interface is 56°, what will be the angle of refraction in glass?
Answer: 34°

Question 25. What is the relation between the polarising angle ip and refractive index μ of the medium?
Answer:  μ tan i_p

Question 26. For a slab, the polarising angle is rad. What is the refractive index of the slab?
Answer: 1.732

Question 27. The angle of polarisation for glass is about
Answer: 56

Question 28. The particular angle of incidence, which is the refractive index of the slab
Answer: 1.732

Question 29. If Brewster’s angle be θ, then the magnitude of the critical angle is [Fill in the blank
Answer: sin^-r5 ?(cotθ)

Question 30. Give an example of a double-refracting crystal
Answer: Tourmaline

Question 31. The ordinary ray i.e., O-ray obeys the general laws of refraction of light—is the statement correct?
Answer: Yes

Diffraction And Polarisation Of Light Fill In The Blanks

Question 1. Between sound and light, _______________ bends more while passing around the comer of an obstacle
Answer: Sound

Question 2. While passing around the tall buildings, radio wave produces ____________ but ___________ does not
Answer:  Diffraction, Light waves

Question 3. Intensity of all fringes in diffraction pattern are____________
Answer: Not same

Question 4. Small spherical obstacle produces ________________ diffraction
Answer: Fresnel type

Question 5. What is the phenomenon of diffraction more pronounced in a single slit?
Answer: Fraunhofer

Question 6. Resolving power of a telescope ___________ with the increase of the diameter of its objective
Answer: Increase

Question 7. Sunray, sodium light, and light of an automobile __________________ which of these lights are polarised?
Answer: None of these is polarised.

Question 8. Quartz is a ________________crystal
Answer: Opposite

Question 9. If refractive indices of a positive crystal for O-ray and E-ray are mu and mu respectively, then mu E will be _____________ than mu o
Answer: Less

Question 10. Use of ___________ instead of glass in high-quality sunglasses is more pleasant for eyes
Answer: Polaroid

Question 11. In the case of negative crystals, the velocity of the E-ray is than that of the O-ray _______________________
Answer: Greater

Diffraction And Polarisation Of Light Assertion  Reason Type

Direction: These questions have statement 1 and statement 2 Of the four choices given below, choose. the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is the correct explanation for statement 1
2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. ’
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: To observe the diffraction of light, the size of the obstacle or aperture should be of the order of 10-7 m.

Statement 2: 10-7 m is the order of Wavelength of visible light.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is the correct explanation for statement 1

Question 2.

Statement 1: We cannot get a diffraction pattern from a wide slit illuminated by monochromatic light.

Statement 2:  In the diffraction pattern all the bright bands are not of the same intensity.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. ’

Question 3.

Statement 1: The revolving power of a telescope increases on decreasing the aperture ofits objective lens.

Statement 2: Resolving power of a telescope, R =

Answer: 4. Statement 1 is false, statement 2 is true.

Question 4.

Statement 1: In a single slit experiment, the greater is the wavelength of the light used, the greater is the width of the central maximum.

Statement 2: The width of the central maximum is directly proportional to the wavelength of light used.

Answer: 1. These questions have statement 1 and statement 2 Of the four choices given below, choose. the one that best describes the two statements.

Question 5.

Statement 1: The value of polarising angle is independent of the color of incident light.

Statement 2: The polarising angle depends on the refractive index of the medium.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 6.

Statement 1: The electromagnetic waves of all wavelengths can be polarised.

Statement 2: Polarisation is independent of the wavelengths of electromagnetic waves.

Answer: 1. These questions have statement 1 and statement 2 Of the four choices given below, choose. the one that best describes the two statements.

Question 7.

Statement 1: Diffraction can be seen clearly if the edge of the obstacle or slit is very sharp.

Statement 2: As the size of the slit is much larger than the wavelength, it is very difficult to capture the effect of diffraction by the naked eye.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1. ’

Diffraction And Polarisation Of Light Match The Columns

Question 1. In column 1, some optical incidents and in column 2 some .experiments are mentioned. The experiment which the experiment is verified is to be matched

Answer: 1-C, 2-D,3- B, 4- A

Question 2. Factors on which the width of central maximum, formed due to single slit Fraunhofer diffraction, depends are to be matched.

Answer: 1-B, 2-D,3- A, 4- A

Refraction Of Light

Refraction Of Light Definition:

When a ray of light travelling in one medium enters another medium obliquely, the ray changes its direction at the interface. This phenomenon is known as the refraction of light. CD be the plane of separation of the media—air and glass

It is called a refracting surface. The ray AO incident obliquely at 0, changes direction after refraction and goes along the line OB. NON’ is drawn perpendicular to the surface of separation. The perpendicular to CD, NON’ is called normal. AO is the incident ray and OB is the refracted ray. The angle between the incident ray and the normal to the surface of separation at the point of incidence is called the angle of incidence

The angle between the refracted ray and the normal to the surface of separation is called the angle of refraction (r).

Refraction from rarer to denser medium:

If the ray travels from an optically rarer medium to a denser medium, say from air to glass, the refracted ray bends towards the normal. Here i>r.

Refraction from denser to the rarer medium:

If the ray travels from denser to rarer medium, say from glass to air, the refracted ray bends away from the normal Here r>i.

Read and Learn More Class 12 Physics Notes

Refraction Of Light Laws Of Refraction

Refraction of light is governed by the following two laws, called
laws of refraction.

1. The incident ray, the refracted ray and the normal at the point of incidence lie on the same plane.
2. The ratio of the sine of the angle of Incidence to the sine of the angle of refraction is constant.
3. This constant depends on the nature of the two concerned media and the colour of the ray used.

This second law of refraction is known as Snell’s law, named after Willebrord Snellius (1580-1626)

Refractive Index

If i is the die angle of incidence and r is the angle of refraction, then according to the second law,

⇒ \(\frac{\sin i}{\sin r}=\mu\) (pronounced as ‘mu’) = constant.

This constant is called the refractive index of the second medium concerning the first medium.

The value of the refractive index depends on:

1. Nature of the two concerned media and
2. Colour of the Incident light.

Whatever may be the value of the angle of incidence, the value of the refractive index will remain constant if the colour of the incident light (i.e., frequency) and the two media remain unchanged

Μ has no uni

Normal Incidence:

If a ray of light is incident perpendicularly on a refracting surface, then / = 0. According to Snell’s law

π sinr = sin 0 = 0 or, r = 0

So, the ray suffers no deviation

Relative refractive index:

When light passes from medium a into another medium b, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is called the refractive index of medium b concerning medium a. It is denoted by and i.e.,

_aμ_b=  \(\frac{\sin i}{\sin r}\)

[i = angle of incidence, r = angle of refraction]

This refractive index is called the relative refractive index. According to the principle of reversibility of light—a ray of light will follow the same path if its direction of travel is reversed.

Following this principle, we can say that the ray BO in medium b when incidents at an angle r on the interface of the second medium a, refracts at an angle i along the path OA. Comparing this figure with it is the opposite phenomenon.

Thus, \(b^{\mu_a}=\frac{\sin r}{\sin i}\)

Here _bμ_a is the refractive index of medium a concerning medium b.

So, \(a_a \mu_b \times{ }_b \mu_a=\frac{\sin i}{\sin r} \times \frac{\sin r}{\sin i}\)

= 1

Or, _aμ_b = \(\frac{1}{b^\mu}\)

For example, if the refractive index of water concerning air is \(\frac{4}{3}\), then the refractive index of air concerning water is\(\frac{3}{4}\)

Absolute refractive index:

When light is refracted from a vacuum to another medium, the ratio of, the sine of the angle of incidence to the sine of the angle of refraction is called the absolute refractive index of the medium

If the angle of incidence is i and the angle of refraction is r, then the absolute refractive index of the medium,

⇒ \(\mu=\frac{\sin i}{\sin r}\)

Therefore, the relative refractive index of a medium concerning a vacuum is the absolute refractive index of that medium. The refractive index of a vacuum is 1.

In general, the refractive index of a medium relative to an air medium is considered as the refractive index of that medium. But it is not absolute, 1 refractive index of the medium

It is an experimental fact that the difference in the values of the refractive index of a medium concerning air and its absolute refractive index is very small. For example, at STP, the absolute refractive index of air is 1.0002918. So the refractive index of any medium concerning air may be considered as its absolute refractive index

For example, the refractive index of glass is 1.5 which means that the refractive index of glass concerning air is 1.5. At STP, the absolute refractive index of air is 1.0002918 and the refractive index of glass concerning air is 1.5. Thus, the absolute refractive index of glass = \(\frac{1.5}{1.0002918}\) = 49956 ≈1.5 The Absolute refractive index of a medium is denoted by μ. If there is more than one medium μ₁, μ₂, μ₃ then is used.

Optical Density of a Medium

If the absolute refractive index (μ₁) of any medium is greater than that of another medium (μ₂), then the first medium is called optically denser and the second medium is called optically rarer. So, μ₁>μ₂ if then medium 1 is optically denser concerning medium 2 i.e., medium 2, is optically rarer concerning medium 1.

The optical density of a medium has no relation with its physical density or specific gravity. For example, the specific gravity of turpentine oil is 0.87 and that of water is 1. But the refractive index of turpentine oil is 1.47 and that of water is 1.33.

The refractive index of a medium depends on the colour of the incident light. It is greater for blue or violet than for red. The refracted ray in the case of violet light bends more than in the case of red light. The refractive index for yellow light is midway between these two.

So unless otherwise stated, the refractive index of a medium refers to yellow light.

Refractive indices of a few substances:

Refractive Index and Related Terms

Relation between the velocity of light and refractive index:

According to the wave theory of light, the velocity of light is different in different media. If pt is the absolute refractive index of a medium, then

μ = \(\frac{\text { velocity of light in vacuum }(c)}{\text { velocity of light in that medium }(v)}\)

∴ Speed of light in free space = ft x speed of light in that medium

For any medium μ > 1, so speed of light in a vacuum is greater than that in any medium. Thus, the speed of light is maximum in a vacuum.

Accordingly, if and is the refractive index of the medium b concerning medium a, then

⇒ \(a^{\mu_b}=\frac{\text { velocity of light in medium } a}{\text { velocity of light in medium } b}\)

= \(\frac{v_a}{v_b}\)

Medium b is denser than medium a then _aμ_b > 1 and in this case v_a>v_b.

The velocity of light in a denser medium is less than the velocity of light in a rarer medium. The velocity of light in a medium decreases with the increase of its refractive index.

Thus if the velocity of light in medium b is lesser, i.e., medium b is optically denser, and _aμ_b > 1 i.e., sin i>sinr or i< r. Thus, the angle of refraction is less than the angle of incidence, so the refracted ray bends towards the normal.

From the above discussion, it is clear that refraction variation in the speed of light in different media

Relation between relative refractive index and absolute refractive index:

If μ _a and μ_b are absolute refractive indices of media a and b respectively

⇒ \(\mu_a=\frac{c}{v_a} \text { and } \mu_b=\frac{c}{v_b}\)

So relative refractive index of medium b concerning medium a,

⇒ \({ }_a=\frac{v_a}{v_b}=\frac{\frac{c}{v_b}}{\frac{c}{v_a}}=\frac{\mu_b}{\mu_a}\)

Relation of the wavelength of light with refractive index:

The relative refractive index between two media depends on the wavelength of light. Cauchy’s equation for the dependence of refractive index on the wavelength of light is

⇒ \(\mu=A+\frac{B}{\lambda^2}\)

Here A and B are two constants; their values are different in different media. The refractive index of a medium decreases if the wavelength of light increases. The graph shows the variation of n with μ  with λ of BK7 glass.

In a medium apart from free space different coloured light travels at different speeds. In a particular medium red travels the fastest and violet travels the slowest. Thus refractive index of any medium for the red colour is the lowest and that for the violet colour is the highest. This is the reason why white light is dispersed

Relation of temperature with refractive index:

Generally, the refractive index of a medium decreases if the temperature of the medium increases. For a solid medium this change is small, for a liquid it is moderate and for a gas it is remarkable

It is to be remembered that, velocity, intensity and wave¬ length of light change due to refraction but its frequency and phase remain unchanged

Generalised Form of Snell’s Law

Let AB be the surface of the separation of two media 1 and 2. Medium 2 is denser and medium 1 is rarer. PO is the incident ray at the point O on the surface of separation and OQ is A the refracted ray at the point O . Let angle of incidence = i₁ angle of refraction = i₂

By Snells law \(\frac{\sin l_1}{\sin i_2}\)

We known \({ }_1 \mu_2=\frac{\mu_2}{\mu_1}\)

∴ \(\begin{equation}\frac{\sin i_1}{\sin i_2}=\frac{\mu_2}{\mu_1} \quad \text { or, } \mu_1 \sin i_1=\mu_2 \sin i_2\end{equation}\) …………………………. (1)

So for n number of media, it can be written as

\(\mu_1 \sin i_1=\mu_2 \sin i_2=\cdots=\mu_n \sin i_n\) …………………. (2)

This equation is known as the generalised form ofSnell’s law

Refraction Of Light Laws Of Refraction Numerical Examples

Example 1. A ray of light is incident from water on the surface of separation of air and water at an angle of 30°. Calculate the angle of refraction in air mu \(\frac{4}{3}\)
Solution:

Let the angle of refraction of the ray of light in the air be r

Since the ray of light is refracted from water to air,

⇒ \(w^\mu{ }_a=\frac{\sin i}{\sin r}\)

Or, \(\frac{1}{a^{\mu_w}}=\frac{\sin 30^{\circ}}{\sin r}\)

Or, \(\frac{1}{\frac{4}{3}}=\frac{1}{2 \sin r}\)

Or, \(2 \sin r=\frac{4}{3}\)

Or, \(\sin r=\frac{2}{3}\)

= 0.666

= sin 41.8°

or = r= 41.8°

Example 2. A ray of light is incident on a block of glass in such a way that the angle between the refracted ray and the refracted ray is 90°. Determine the relation between the angle of incidence refractive index of
Solution:

Here angle of incidence = i

The angle of reflection = i, angle of refraction = r

The angle between the reflected ray and the refracted ray = 90°

According to the

i+90°+r= 180°

Or, r = 90°-i

The refractive index of glass,

μ = \(\mu=\frac{\sin i}{\sin r}=\frac{\sin i}{\sin \left(90^{\circ}-i\right)}\)

= \(\frac{\sin i}{\cos i}\)

= tan i

Example 3.  The refractive index of glass is 1.5 and the refractive index of water is 1.33. If the velocity of light in glass is 2 × 10⁸ m s^–1 what is the velocity of light In water?
Solution:

μ_g = \(\frac{\text { velocity of light in vacuum }}{\text { velocity of light in glass }\left(v_g\right)}\)

Or, velocity light in a vacuum

Again, \(\mu_w=\frac{\text { velocity of light in vacuum }}{\text { velocity of light in water }\left(v_w\right)}\)

Or, Velocity of light in vacuum = μ_w. v_w

μ_g v_g= μ_w v_w

Or, \(\frac{\mu_g v_g}{\mu_w}=\frac{1.5 \times 2 \times 10^8}{1.33}\)

= 2.26x × 10⁸ m s^–1

Example 4. A monochromatic -ray of light-, is refracted from the vacuum. to a medium of refractive index pi. Determine the relation of the wavelengths of light in vacuum and in glass
Solution:

μ = \(\frac{c}{v}=\frac{n \lambda_0}{n \lambda}=\frac{\lambda_0}{\lambda}\)

So, λ₀ = μ λ

[Here, c and v are the velocities of light in vacuum and the medium respectively; n = frequency of light, which remains unchanged on refraction λ₀, λ = wavelengths in vacuum and in the medium respectively.]

Example 5. If a ray of light is incident on a plate inside the water at an angle of 45°, what is the angle of refraction inside the plate? Given that the absolute refractive Indices of the plate and water are 1.88 and 1.33 respectively.
Solution:

Let the angle of refraction inside,n-rriDT.tile plate be r

Here , \(w^{\mu_g}=\frac{\sin i{ }^0}{\sin r} \text { or, } \frac{\mu_g}{\mu_w}=\frac{\sin i}{\sin r}\)

Or, \(\frac{1.88}{1.33}=\frac{\sin 45^{\circ}}{\sin r}\)

Or, in r = \(\frac{1}{\sqrt{2}} \times \frac{1.33}{1.88}\) = 0.5

= \(\frac{1}{2}\)= sin 30°

r = 30°

Example 6. How much time will sunlight take to pass through the glass window of thickness 4 mm ? μ of glass =1.5.
Solution:

Velocity of sunlight in a vacuum or air,

C = 3×10^-8m s^-1

Thus velocity in a medium of refractive index μ

u = \(v\frac{c}{\mu}\)

So, to cross a thickness d, the time taken by light,

t = \(\frac{d}{v}=\frac{d \mu}{c}\)

= \(\frac{\left(4 \times 10^{-3}\right) \times 1.5}{3 \times 10^8}\)

[here d= 4 mm = 4 ×10m]

= 2 × 10^-11 s

Example 7. Green light of wavelength 5460 A°Is incident on an airglass interface. If the refractive index of glass is 1.5 what will be the wavelength of light In glass?
Solution:

The wavelength of the light in air, A0 = 5460 A°

The refractive index of glass concerning air, μ = 1.5

If the wavelength of the light In glass is λ, then

= \(\frac{\lambda_0}{\lambda}\)

Or, λ = \(\frac{\lambda_0}{\mu}=\frac{5460}{1.5}\)

= 3640 A°

Refraction Of Light Deviation Of A Ray

During reflection or refraction, the change In the direction of the tight is called its deviation.

The angle between the refracted ray and the direction of the incident ray gives the measure of deviation.

Deviation of incident ray AO, which after refraction proceeds along OD instead of OC. So the deviation of ray, S = ∠BOC

Now, δ = ∠BOC = ∠N’OC- ∠N’OB

= ∠AON- ∠N’OB = i-r

We know. If the angle of Incidence Increases, the angle of refraction also Increases

For normal Incidence, i = 0 thus r = 0 and so δ = 0 (minimum).

For  i = 90°, δ Is maximum

For refraction to a rarer medium from a denser medium, the angle of refraction is greater than the angle of incidence, l.e., r> i.

Now, δ = ∠BOC = ∠N’OB- ∠N’OC

=∠N’OB-  ∠AON =r – i

Optics

Refraction Of Light Image Due To Refraction

Suppose, a beam of rays from a point object after refraction reaches our eyes in another medium. Now if dierefracted rays are produced backwards they are nice at a point It seems that the refracted rays are diverging from die second point. The second point Is the image of the lift’s first point. The image of any point object is formed in the same way. Thus a complete image of the object is formed.

If the object Is situated In a denser medium and Is viewed from a rarer medium, It appears (closer to the surface of separation. For example, If we look at a fish inside water in a pond it appeals nearer to the surface than the actual position.

If the object Is situated in a rarer medium and Is viewed from a denser medium, it appears to move away from the surface of separation, for example, our earth is surrounded by a thick atmospheric layer composed of different gases. Starlight comes to our eyes through this atmosphere. Hence we are the observers on Earth in a denser medium whereas the stars are in a vacuum i.e., in the rarer medium. So, the animal position of a star Is far behind its normal viewing position.

Object In denser medium end eyes In rarer medium:

Let the refractive indices of the two media a and b be μ₁ and μ₂, respectively and μ₁ >μ₁, An object P situated in a is viewed from b. The surface of separation of a and b is a plane surface. A ray of light from P incident perpendicularly at A proceeds along AB without changing its direction. Another oblique ray PC incident at C is refracted along CD.

The refracted rays A B and CP when produced backwards meet at Q . So when the two refracted rays reach the rays of the observer. It will appear as if tyre two rays are coming from Q . So Q h the virtual Image of p.  In this case, the image rises towards the surface of the separation

If the angle of incidence and the angle of refraction are l and r.

⇒ \(\frac{\mu_2}{\mu_1}=\frac{\sin r}{\sin i}=\frac{\tan r}{\tan i}=\frac{A C / A Q}{A C / A P}=\frac{A P}{A Q}\)

For near-normal viewing, points A and C are close enough. So sin0 *s tarif

Hence, \(\frac{\mu_2}{\mu_1}=\frac{\sin r}{\sin i}=\frac{\tan r}{\tan l}=\frac{A C / A Q}{A C / A P}=\frac{A P}{A Q}\)………………………….(1)

If the observer is In the air then μ₁  = 1

Putting μ₁ = 1 and μ₂ =μ  (say) In equation (1) and writing AP = u, AQ =  v

We have \(\mu=\frac{A P}{A Q}=\frac{u}{v}\)

If d is die real deeds of the object then

μ = \(\frac{d}{\text { apparent depth of the object }}\)

Or, Apparent depth of the object = \(\frac{d}{\mu}\)

Hence apparent displacement

x = PQ = AP-AQ = d- \(\frac{d}{\mu}\)

= \(d\left(1-\frac{1}{\mu}\right)\)

Therefore, the apparent displacement of an object depends on the real depth (d) of the object and the refractive index (/z) of the denser medium.

The refractive index of water concerning air is|. If an object immersed in water is observed vertically from above the water, then its apparent displacement.

x = \(d\left(1-\frac{1}{4 / 3}\right)=\frac{d}{4}\)

General case:

The apparent depth of an object when viewed from the air through successive media of thicknesses d₁ d₂, d₃ ‘ ……….. d_n having refractive indices μ₁,  μ₂, μ₃ ‘ ……….. μ_n  respectively is

= \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}+\frac{d_3}{\mu_3}+\cdots+\frac{d_n}{\mu_n}=\sum_{i=1}^n \frac{d_i}{\mu_i}\)

Its apparent displacement is

⇒ \(d_1\left(1-\frac{1}{\mu_1}\right)+d_2\left(1-\frac{1}{\mu_2}\right)+d_3\left(1-\frac{1}{\mu_3}\right)+\cdots+d_n\left(1-\frac{1}{\mu_n}\right)\)

= \(\sum_{i=1}^n d_i\left(1-\frac{1}{\mu_i}\right)\)

Object in rarer medium and eye in denser medium:

Let the refractive indices of the two media a and b be μ₂ and μ₂ respectively and μ₂ > μ₁ An object P is situated in the medium b and it is seen from the medium. The surface of separation of a and b is a plane surface.

A ray of light from P is incident perpendicularly at point A on the surface of separation and proceeds straight along AB through the medium a without changing its direction. Another oblique ray PC is incident at C and proceeds along CD after j refraction. The refracted rays AB and CD when produced backwards meet at Q. So when the two refracted rays reach the observer, they will.

Appears to him that the two rays are coming from Q. So Q is the virtual image of P. In this case, the image appears to move farther away from the surface of separation.

If the angle of incidence and the angle of refraction of the incident ray at C are i and r respectively, then according to Snell’s law

μ₁ sin i= μ₂sin r

∴ \(\frac{\mu_2}{\mu_1}=\frac{\sin i}{\sin r}=\frac{\sin \angle P C N_1}{\sin \angle N C D}\)

Since the two lines PAB and NjCiV are parallel

∠PCN₁ = ∠APC and ∠NCD = ∠AQC

⇒ \(\frac{\mu_2}{\mu_1}=\frac{\sin \angle A P C}{\sin \angle A Q C}=\frac{\frac{A C}{C P}}{\frac{A C}{C Q}}=\frac{C Q}{C P}\)

If the points A and C are very close to each other i.e., if the ray PC is not so oblique, then CP ≈ AP and CQ ≈ AQ

∴ \(\frac{\mu_2}{\mu_1}=\frac{A Q}{A P}\)…………. (3)

If the rarer medium in which the object is situated is air then = 1.

Putting μ₁ = 1 and μ₂= p (say) in equation (3) we get

⇒ \(\mu=\frac{A Q}{A P}\)

The refractive index of the denser medium concerning air apparent height of the object from

= \(\frac{\text { the surface of separation }}{\text { real height of the object from the surface of separation }}\)

If the real height of the object, AP = μd, then

= \(\mu\frac{\text { apparent height of the object }(A Q)}{d}\)

Or, AQ = μd ………………… (4)

So the apparent displacement of the object

=PQ = AQ – AP – μd-d = (μ-1) d

General case:

The apparent height of an object in the air when viewed from a medium of refractive index pn through successive media of thicknesses d₁ d₂, d₃ ‘ ……….. d_n having refractive indices, μ₁,  μ₂, μ₃ ‘ ……….. μ_n   respectively is

⇒ \(\mu_1 d_1+\mu_2 d_2+\cdots+\mu_n d_n=\sum_{i=1}^n \mu_i d_i\)

Its apparent displacement is

⇒ \(\left(\mu_1-1\right) d_1+\left(\mu_2-1\right) d_2+\cdots+\left(\mu_n-1\right) d_n\)

= \(\sum_{i=1}^n\left(\mu_i-1\right) d_i\)

Let an object in a medium of refractive index fly be viewed from a medium of refractive index  Then we have,

= \(\frac{\text { Apparent depth of the object }}{\text { real depth of the object }}=\frac{\mu_2}{\mu_1}\)

=  \({ }^1 \mu_2\)

If the object is situated in a comparatively denser medium, then μ₁>μ₂.

In that case, apparent depth < real depth. If the object is situated in a comparatively rarer medium then μ₁<μ₂ In that case, apparent depth > real depth

Image Formed by Oblique Incident Rays

In the last section, we talked of almost normal viewing. For more B oblique incidence, the apparent displacement is higher

A point object O in an optically denser medium (say, water) is viewed from a rarer medium (say, air), at different angles. For different positions of the observer, the locus of the different positions of the image is a curved line. This curved line is called a caustic curve. The curve has two parts. These two parts meet at a point O’, known as the cusp. The image of an object at O when viewed vertically downward, is formed, at O’.  shows how the images A’, B’, C’, etc. of different points A, B, C, etc. on the base of a vessel or tank containing water will appear to an observer located at a given position. It is evident

The image for normal incidence is at the lowest position. The other images go on rising as the oblique rays from the base produce the images. The image of the base of the vessel will be a curved surface indicated by A’B’C’. With the increase of the distance of the base of the vessel from the eye, the curved surface appears, to rise higher. So if an observer stands in a shallow pond having equal depth everywhere, it appears to him that the pond near his feet is the deepest

Image of an Object under a Parallel Slab

ABCD is a parallel glass slab. Its thickness is d and its refractive index μ,  P is a point object placed in air under the surface AB of the slab. A ray of light PX normally incident on AB goes undeviated along XY. Another ray PQ incident obliquely is refracted along QR. After that, the ray is further refracted along RS. Since the two faces AB and DC of the glass slab are parallel, the rays

PQ and RS will be parallel. So an observer from above the slab will see the image of P at P’. PP’ is the apparent displacement of the point object towards the observer. The normal NQN₁ at Q intersects P’R at M. The quadrilateral PP’MQ is a parallelogram

PP’ = QM

The Apparent Displacement of the point object

∴ \(d\left(1-\frac{1}{\mu}\right)\)

PQ and RS will be parallel. So an observer from above the slab will see the image of P at P’. PP’ is the apparent displacement of the point object towards the observer.

The normal NQN₁ at Q intersects P’R at M. The quadrilateral PP’MQ is a parallelogram.

∴ PP’ = QM

∴ The apparent displacement of the point object

= PP’ = QM

= d(1- \(\frac{1}{\mu}\))

So the apparent displacement of an object does not depend on the position of the object under the lower face of the glass slab. It only depends on the thickness of the slab (d) and the refrac¬ index of its material (μ).

 Optics

Refraction Of Light Some Examples Of Refraction

A coin immersed in water:

A coin is placed at the bottom of a pot such that the coin is just not visible. If eyes are set at the same position and the pot is now filled with water, the coin becomes visible. Because the rays from P are refracted from denser to rarer medium and are bent away from the normal, they reach our eyes. As a result, the refracted rays appear to diverge from P’, i.e., the virtual image of the coin is formed at P’, situated above the coin.

A rod partly immersed in water:

Let a straight rod be partly immersed in water. When the rod is held obliquely in water, the portion of the rod in water will appear to be bent upward [Fig. 2.20]. The reason is the same as above. The light rays coming from the portion immersed in water are refracted from denser to rarer medium and hence bent away from the normal. So point A of the rod appears to be raised at B. This happens for every point of the immersed portion of the rod

Object and medium having approximately equal refractive index:

An object becomes invisible when it is sur¬ rounded by a medium having a nearly equal refractive index. Since the refractive indices of both of them are almost the same, negligible refraction does occur from their surface of separation, and for refraction lending of light is negligible. i.e., it travels undefeated. As a result, the surface of separation is not visible. ff99Wt9rf . The refractive indices of glycerine and glass are almost equal. So when a glass rod is immersed in glycerine thiqÿrod is not visible

Multiple images in a thick glass mirror:

If an object is placed in front of a thick mirror with silvered glass at the mirror at the back if a surface object is viewed from a slanting direction, a series of images are formed.

When ray PA is incident on the front face at point A, a very small portion of the light is reflected along AK producing a faint image at P₁. The remaining larger portion of the light is refracted into the mirror along AB and is reflected along BC from the silvered surface. A large portion of this reflected ray within the glass comes out along CL producing the second image P₂ which is the brightest of all the images.

The remaining part of the ray CD is reflected from surface Y to the silvered surface where it is again reflected. The process continues and gradually fainter images are formed. The different images lie on the line drawn perpendicular to the surface X from the object P.

The ray BC, which is the first reflected ray from the silvered surface, is the brightest and it travels along CL. Consequently, image P₂ is the brightest. Hence the brightest second image is considered to be the image of the object. The more oblique the incident rays are, the more the amount of reflection from the front surface of the mirror and the brightness of the image P₁ will increase accordingly

Apparent thickness of thick glass mirror:

In the case of a water-filled bowl, the depth of the bowl appears to be less. Similarly, a thick glass mirror seems to be less thick than it is.

We have, \( \frac{\text { real thickness of mirror }}{\text { apparent thickness of mirror }}\)

= Refractive index of glass = \(\frac{3}{2}\)

Therefore, the apparent thickness of a mirror

= \(\frac{2}{3 }\) × Real thickness of the mirror

Optics

Refraction Of Light Some Examples Of Refraction Numerical Examples

Example 1. There is a mark at the bottom of the beaker. A liquid with a refractive index of 1.4 is poured into it. If the depth then determines how much the (liquid is 3.5 cm mark appears to rise when it is viewed from above.
Solution:

If the object is in a denser medium and the observer is in a rarer medium, the refractive index of the denser medium relative to the rarer medium

= \(\frac{\text { real depth of the object }}{\text { apparent depth of the object }}\)

Or, 1.4 = \(\frac{3.5}{\text { apparent depth of the object }}\)

Or, apparent depth of the object = \(\frac{3.5}{1.4}\)

= 2.5 cm

Apparent upward displacement of the mark

= 3.5 – 2.5 = 1cm

Example 2.  There is a black spot at the bottom of a rectangular glass slab of thickness d and refractive index μ. When the spot is viewed perpendicularly from above, the spot appears to be shifted through a distance \(\frac{(\mu-1) d}{\mu}\) towards the observer. Prove it.
Solution:

The real depth of the black spot from the upper surface of

Let the apparent depth of the black spot from the upper surface of the glass slab be d₂

Refractive index of glass \(\mu=\frac{d}{d_1}\)

Or, \(d_1=\frac{d}{\mu}\)

∴ The apparent displacement of the black spot towards the observed

= \(d-d_1=d-\frac{d}{u}\)

= \(d\left(1-\frac{1}{\mu}\right)=d\left(\frac{\mu-1}{\mu}\right)\)

Example 3.  In a beaker partly filled with water, the depth of water seems to be 9 cm. On pouring more water into it, the real depth of water is increased by 4cm. Now the apparent depth of water seems to be 12cm. Determine the refractive index of water and the initial depth of water in the beaker.
Solution:

Let the refractive index of water be ft and the initial| depth of water in the beaker be x.

μ = \(\frac{x}{9}\)

or, x = 9μ

When more water is poured into the beaker, the real depth of water becomes (x + 4) cm.

In the second case, μ = \(\frac{x+4}{12}\)

Or, 12 μ = x + 4 or; 12 μ = 9 μ + 4 Or, 3 μ= 4

∴ μ = \(\frac{4}{3}\)

μ = 1.33

Initial depth of water in the beaker.

x = 9 μ = 9 ×  \(\frac{4}{3}\)

= 12cm

Example 4. A small air—bubble exists inside a transparent cube of side 15 cm each. The apparent distance of the bubble observed from one face is 6 cm and from the opposite face its apparent distance becomes 4 cm. Determine the real distance of the bubble from the first face and the refractive index of the material of the cube
Solution:

Let the real distance of the bubble from the first face be x cm.

The real distance of the bubble from the opposite face = (15-x) cm

Let the refractive index of the material of the cube be ft.

We know if the object lies in a denser medium and eye in the rarer medium,

μ = \(\frac{\text { real distance }}{\text { apparent distance }}\)

In the first case , μ=  \(\frac{x}{6}\)

In the second case,  μ  = \(\frac{15-x}{4}\)

⇒ \(\frac{x}{6}=\frac{15-x}{4}\)

Or, 4x = 90- 6x

Or, 10x = 90

Or, x = 9cm and

mu = \(\frac{9}{6}\) = 1.5

Therefore, the real distance of the bubble from the first face is 9 cm and the refractive index of the material of the cube is 1.5

Example 5. A vessel is filled with two mutually immiscible liquids with refractive indices μ₁  and μ₂. The depths of the two liquids are d₁ and d₂ respectively. There is a mark at the bottom of the vessel. Show that the apparent depth of the mark when viewed normally is given by \(\left(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\right)\)
Solution:

The image of P is formed at Q due to refraction at the surface of separation B of the 1st and 2nd liquid. Another final image due to refraction in air from the second liquid is formed at R

For the first refraction:

⇒ \(\frac{\mu_1}{\mu_2}=\frac{B P}{B Q}\)

Or, \(B Q=\frac{\mu_2}{\mu_1} \cdot B P\)

⇒ \(\frac{\mu_2}{\mu_1} d_1\)

For the second refraction:

⇒ \(\frac{\mu_2}{1}=\frac{A Q}{A R}\)

Or, \(A R=\frac{A Q}{\mu_2}\)

= \(\frac{1}{\mu_2}(A B+B Q)\)

Or, \(A R=\frac{1}{\mu_2}\left(d_2+\frac{\mu_2}{\mu_1} d_1\right)\)

= \(\frac{d_2}{\mu_2}+\frac{d_1}{\mu_1}\)

The apparent depth of the mark P when viewed normally

= AR = \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}\)

Example 6. A rectangular slab of refractive index [i is placed on another slab of refractive index 3. Both the slabs are of ror. the displacement of the image? the same dimensions. There is a coin at the bottom of the lower slab. What should be the value of n such that when viewed normally from above, the coin appears to be at the surface of separation of the two slabs?
Solution:

Let the thickness of each slab be d. According to the question the apparent depth of the coin =d

d = \(\frac{d}{\mu}+\frac{d}{3}\)

Or, 1= \(\frac{1}{\mu}+\frac{1}{3}\)

Or, \(\frac{1}{\mu}=\frac{2}{3}\)

Or, \(\frac{3}{2}\)

= 1.5

Example 7. A tank contains ethyl alcohol of a refractive index of 1.35 The depth of alcohol is 308 cm. A plane mirror is placed horizontally at a depth of 154 cm in it. An; object is placed 254 mm above the mirror. Calculate the apparent depth of the image formed by the mirror
Solution:

Depth of the mirror =1.54 m; object distance from the mirror = 0.254 m

The image of the object is formed at a distance of 0.254 m behind the mirror

= \(\frac{\text { real depth }}{\mu}\)

=\(\frac{1.794}{1.35}\)

= 1.33m

Example 8. A 20mm thick layer of water \(\left(\mu=\frac{4}{3}\right)\) 35mm thick layer of another liquid \(\left(\mu=\frac{7}{5}\right)\) = ‘ in a tank. A small coin lies at the bottom of the tank. Determine the apparent depth of the coin when viewed normally from above the water
Solution:

Real depth of the coin d+d = 20+ 35 = 55m m

∴ A parent depth of the coin

= \(\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}=\frac{20}{\frac{4}{3}}+\frac{35}{\frac{7}{5}}\)

= 15+ 25

= 40 mm

Example 9. If a point source is placed at a distance of 18 cm from the pole of a concave mirror, its image is formed at a distance of 9 cm from the mirror. A glass slab of thickness 6 cm is placed between the point source and the mirror such that the parallel faces of the glass slab remain perpendicular to the principal axis of the mirror the refractive index of glass is 1.5 what will be the displacement of the image?
Solution:

Let P be the position of the object and in the absence of the glass slab, Q be the position of the image formed by the concave mirror

u = 18 cm,

OQ = v= 9 cm

According to \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) we get,

= \(\frac{1}{-9}+\frac{1}{-18}=\frac{1}{f}\)

Or, f= -6 cm

If the glass slab of thickness 6 cm is placed between the point I source and the concave mirror, apparent displacement of the I Point source will take place towards the mirror. The rays coming from P appear to come from P’ after refraction.

The apparent displacement of P

PP’ = \(d\left(1-\frac{1}{\mu}\right)\)

= 6\(\left(1-\frac{1}{1.5}\right)\)

= 2cm

So in the second case, object distance u = -(18- 2) = -16 cm ; f = -6 cm; v = ?

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) Or, \(\frac{1}{v}+\frac{1}{-16}=\frac{1}{-6}\)

Or, \(\frac{1}{v}=\frac{1}{-6}+\frac{1}{16}=\frac{-10}{96}\)

Or, v = \(-\frac{96}{10}\)

= – 9.6 cm

Displacement of the image

QQ’ = OQ’ OQ – OQ = 9.6- 9 = 0.6 cm

Example 10. A plane is made of glass having a thickness of 1.5 cm. Its back surface is coated with memory. A man is standing at a distance of 50 cm from the front face of the mirror. If he looks at the mirror normally, where can he find his image behind the front face of the mirror? The refractive index of glass  = 1.5 
Solution:

The real depth of the mercury-coated surface from the upper surface of the mirror = 1.5 cm

If the apparent depth of the mercury coated mercury  cm, then \(\frac{1.5}{x}\) = 1.5 or, x= 1 cm

So the mercury-coated surface appears to be at n distance of 1 cm from the front face of the mirror.

‘Therefore, the distance of the man from the appetent position of the mercury-coated surface a = 50 +1  cm

So the distance of the Image from the apparent position of the mercury-coated surface =  51 cm

The distance of the Image of the man from the front tuifnee of the mirror =51 + 1 = 52 an

Example 11.  An observer can see the topmost point of a narrow rod of height through a small hole  The rod Is placed Inside a beaker. The beaker’s height is 3h and its radius Is h. When the beaker Is filled up to 2 h of its height with a liquid the observer can see the entire rod. What Is the value of the refractive Index of the liquid?

Solution:

Let us assume that  PQ is a thin straight rod kept in a beaker. B Is a small hole in the wall of the beaker. When the beaker is filled with a liquid up to a height of 2h, then Q can be seen through hole B.

Here, QD ray propagates through the liquid and gets refracted along DB in the air

According to ABRP is a square midpoint of the diagonal PB.

Here DE= PE = h

Again, ∠BDP = 45°  [∴ ∠DPE ]

Since the object was placed in the denser medium, according to Snell’s law

⇒ \(\frac{1}{\mu}=\frac{\sin i}{\sin r}=\frac{\frac{Q G}{Q D}}{\sin 45}\)

Or, \(\frac{1}{\mu}=\frac{\frac{h}{\sqrt{5 h}}}{\frac{1}{\sqrt{2}}}\)

Since,  QD²=QG²+GD²=h²+(2 h)²= 5 h²

Or, \(\frac{1}{\mu}=\sqrt{\frac{2}{5}} \quad \text { of, } \mu=\sqrt{\frac{5}{2}}\)

Therefore, the required refractive Index of the liquid is \(\sqrt{\frac{5}{2}}\)

Example 12. A ray of light Incident at the Interface of glass and water at an angle of Incidence i. If the ray finally emerges parallel to the surface of water Then what will be the value of μ_g?

Solution:

When refraction occurs due to the propagation of light rays from glim to water, we may write from Snell’s law,

⇒ \(\mu_g \sin i=\mu_w \sin r\) …………..(1)

Again, when refraction occurs due to the propagation of light rays from water to air, we may write from Snail * law

⇒ \(\mu_g \sin i=\mu_w \sin r\) …………..(2)

Therefore, the required refractive Index of glass is \(\frac{1}{\sin i}\)

Optics

Refraction Of Light Critical Angle’s Total Internal Reflection

We know, in refraction from denser to a rarer medium light | bents away from the normal. As a result angle of refraction ’ becomes larger than the angle of incidence.

L line AB represents the surface of the separation of water and air.

Ray P₁ O travelling through water is incident at O on the surface- of separation. A part of the ray is reflected into the water along OR₁ and another part is refracted into the air along OQ₁:  The angle of refraction ∠Q₁ON is greater than the angle of incidence ∠P₁ ON₁. The greater the angle of incidence, the greater the angle of refraction and in each case, both reflection and refraction will take place.

For a particular value of the angle of incidence, the angle of refraction becomes 90°, so that the refracted ray grazes the surface of separation. This limiting angle of incidence in the denser medium is called the critical angle for the two given media. Thus, ∠P₂ON₁ = critical angle (θ_c). In this case, the angle of refraction ∠NOQ₂ = 90°. Here also a part of the incident ray is reflected to water along OR₂.

If the angle of incidence exceeds the critical angle i.e., if i  > θ_c [as in the case of the incident rayP₃O ] no part of the incident ray is refracted in the second medium. The ray is completely reflected along OR₃ into the first medium. This phenomenon is called total internal reflection. In this case, the surface of the separation of the two media behaves as a mirror.

Critical angle:

It is that particular angle of incidence of a ray of light for a given pair of media, passing from denser one to rarer one, for which the corresponding angle of refraction is equal to 90 0 and the refracted ray grazes along the surface of the interface separating the two media. The critical angle of a pair of media depends on the colour of the incident light and the nature of the two media.

For example, in the case of two particular media, the critical angle for red light is greater than that for violet light. Again, the critical angle of water to air is 49°, while that of glass to air is 42°. The statement, ‘critical angle of glass to air is 42° ‘ means that a . ray of light from glass being incident on the surface of separation of glass and water at an angle of 42°, should go along the surface of separation after refraction i.e., the refracted angle will be 90°.

Total internal reflection:

When a ray of light travelling from a denser medium to a rarer medium is incident at the surface of separation of the two media at an angle greater than the critical angle for the media, there is no refraction; rather the whole of the incident ray is reflected. This phenomenon is known as total internal reflection.

Condition of total internal reflection:

The conditions to be satisfied for total internal reflection are as follows

1. The light must travel from a denser to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the two media

Reason for using the term “total’:

In ordinary reflection, a part of the incident light is reflected from the surface of separation and the rest is refracted. But in the case of internal reflection, no part of the incident light is refracted, rather the entire portion of the incident light is reflected to the first medium from the surface of separation of the two media. So this reflection is called total reflection.

Relation between critical angle and refractive index of the denser medium:

Let ∠P₂ON₁ = θ_c = critical angle between the two media, water and air, which

Impliesair concerning the angle to water of refraction is _aμ_w, which is then 90°. If the refractive index

_aμ_w = \(\frac{\sin \theta_c}{\sin 90^{\circ}}\)

Or,  sin θ_c = \(\frac{1}{a^{\mu_w}}\)

So, the value of critical depends on the refractive index of one medium concerning another

If the medium is a and b the,

⇒ \(\sin \theta_c=\frac{1}{b^{\mu_a}}=\frac{1}{\begin{array}{r}
\text { refractive index of denser medium } \\
\text {concerning rarer medium }
\end{array}}\)

= \(\frac{\mu_b}{\mu_a}=\frac{\text { absolute refractive index of medium } b}{\text { absolute refractive index of medium } a}\)

Optics

Refraction Of Light Critical Angle’s Total Internal Reflection Numerical Examples

Example 1. If the absolute refractive index of a medium is \(\sqrt{2}\), calculate the critical angle of glass to the medium. Given
Solution:

If the refractive index of glass concerning air is _aμ_g, then

_aμ_g= \(\frac{1}{\sin \theta_c}=\frac{1}{\sin 30^{\circ}}\)

= 2

If the refractive index of glass concerning the medium is _mμ_g, then

_aμ_g= \(\frac{a^{\mu_g}}{{ }_a \mu_m}\)

= \(\frac{2}{\sqrt{2}}\)

= \(\sqrt{2}\)

If the critical angle of glass to the medium is #c, then

sin = \(\sin \theta_c=\frac{1}{m^\mu}\)

= \(\frac{1}{m^\mu}=\frac{1}{\sqrt{2}}\)

= sin 45

Or, θ_c = 45

Example 2.  The refractive index of carbon disulphide for red light is 1.634 and the difference in the values of the critical angle for red and blue light at the surface of separation of carbon disulphide and air is 0°56/. What is the value of the refractive index of carbon disulphide for blue light
Solution:

Let the refractive index of carbon disulphide for red light = and circle angle =

Now, sin θ_r = \(\frac{1}{\mu_r}=\frac{1}{1.634}\)

= 0.6119 = sin 37.73

θ_r = 37.73

Let the critical angle for blue light be θ_b. The refractive index increases as the wavelength of light decreases. So the critical angle decreases.

∴ θ_b<θ_r

According to the Question,

θ_b = 37.73°-0°.56′

= 37.73°-0.93°

= 36.8°

So refractive index of carbon disulphide for blue light,

μ_b = \(\frac{1}{\sin \theta_b}=\frac{1}{\sin 36^{\circ} 48^{\prime}}\)

= \(\frac{1}{\sin 36.8^{\circ}}=\frac{1}{0.599}\)

= 1.669

Example 3. The refractive index of-diamond is 2.42,-which proves that all the beams of rays having an angle of incidence of more than 25° will be reflected, [sin 24.41° = 0.4132]
Solution:

If the critical angle is θ_c then

Sin θ_c = \(\frac{1}{\mu}=\frac{1}{2.42}\)

= 0.4312° = sin 24.41°

θ_c  = 24.41°

We know that if the angle of incidence of a ray of light is greater than the critical angle, the ray will be reflected. Here the critical angle is 24.41°. So rays of light having an angle of incidence greater than 25° will be reflected

Example 4. A ray of light will go from diamond to glass. What should be the minimum angle of incidence at the surface of separation of the two media, diamond and glass, so that the ray of light cannot be refracted in glass? μ of glass =1.51 and μ of diamond = 2.47; sin37.69° = 0.61134
Solution:

If the light ray is incident at the surface of separation of the two media diamond and glass at a critical angle, the ray is grazingly refracted in the glass. If the critical angle is QQ then

sin θ_c = \(\frac{1}{g^{\mu_d}}=\frac{1}{\frac{\mu_d}{\mu_g}}\)

= \(\frac{\mu_g}{\mu_d}=\frac{1.51}{2.47}\)

= 0.61134

= sin 37.69°

∴ θ_c  = 37.69°

So if the angle of incidence of a light ray is greater than 37.69° it | cannot be refracted in glass.

∴  Required minimum angle of incidence =37.69°

Example 5. cube has a refractive Index μ₁. There is a plate of refractive Index μ₂(μ₂<μ₁) A ray travelling through the air is incident on the side face of the cube. The refracted ray Is an Incident on the upper face of the cube at the minimum angle for total internal reflection to occur. Finally, the reflected ray emerges from the opposite face. Show that if the angle of emergence Is Φ then sin = \(\sqrt{\mu_1^2-\mu_2^2}\)
Solution:

PQ = incident ray on the side face of the cube, QR = refracted ray inside 0′ the cube, S = reflected ray from the upper face of the cube, ST = emergent ray from the opposite face.

Let the critical angle for total reflection be θ_c

According to the question θ’_c ≈θ_c

⇒ \(\sin \theta_c^{\prime} \approx \sin \theta_c=\frac{1}{2^{\mu_1}}=\frac{1}{\frac{\mu_1}{\mu_2}}=\frac{\mu_2}{\mu_1}\)

Angle of incidence of the ray RS = l = 90° – θ_c and angle of refraction -tf>

⇒ \(\sin \theta_c^{\prime} \approx \sin \theta_c=\frac{1}{2^{\mu_1}}=\frac{1}{\frac{\mu_1}{\mu_2}}=\frac{\mu_2}{\mu_1}\)  = Refractive index of air with respect to the cube

= \(\frac{1}{\text { refractive index of the cube with respect to air }}\)

Or, \(\frac{\sin i}{\sin \phi}=\frac{1}{\mu_1}\)

Or, sin Φ = μ₁ sin i

= \(\mu_1 \sin \left(90^{\circ}-\theta_c\right)=\mu_1 \cos \theta_c\)

= \(\mu_1, \sqrt{1-\sin ^2 \theta_c}\)

= \(\mu_1 \sqrt{1-\frac{\mu_2^2}{\mu_1^2}}=\sqrt{\mu_1^2-\mu_2^2}\)

Example 6. A ray of light travelling through a denser medium is incident at an angle i in a rarer medium. If the angle between the reflected ray and the refracted ray is 90° show that the critical angle of the two media, \(\theta_c=\sin ^{-1}(\tan i)\) 
Solution:

Suppose the angle of refraction in the medium =r

From  we get

i+ 90° + r = 180°

Or, r = 90° – i

According to Snell’s law.

⇒\(\frac{\sin l}{\sin r}={ }_1 \mu_2=\frac{\mu_2}{\mu_1}\)

Or, \(\frac{\sin i}{\sin \left(90^{\circ}-i\right)}=\frac{\mu_2}{\mu_1}\)

Or, \(\frac{\sin i}{\cos i}=\frac{\mu_2}{\mu_1}\)

Or, \(\tan i=\frac{\mu_2}{\mu_1}\)

If the critical angle for the two media is θ_c, then

⇒ \(\sin \theta_c=\frac{1}{{ }_2 \mu_1}\)

= \(\frac{1}{\frac{\mu_1}{\mu_2}}=\frac{\mu_2}{\mu_1}\)

= tan i

Or, \(\theta_c=\sin ^{-1}(\tan i)\)

Example 7. A nail Is fixed up perpendicularly at the centre of a circular wooden plate. Keeping the nail at the bottom, the circular plate Is made to float In water. What should be the maximum ratio of the radius of the plate and the length of the nail so that the nail will be out of vision? Refractive index of water \(\frac{4}{3}\)
Solution:

AB Is the circular wooden plate and CD is the nail. Suppose, the radius of the plate =r and the length of the nail =h Since the nail Is not seen from the air, the angle of incidence of the ray DA will be greater than 0 and the ray will be reflected.

We know, \(\sin \theta_c=\frac{1}{a^\mu{ }_w}=\frac{3}{4}\)

⇒  \(\cos \theta_c=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\)

⇒ \(\tan \theta_c=\frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}}=\frac{3}{\sqrt{7}}\)

Or, \(\frac{r}{h}=\frac{3}{\sqrt{7}}\)

This is the required ratio

Example 8.  The Critical angle of glass relative to a liquid is 57° 20′. Calculate the velocity of light in the liquid. Given,  μ of glass = 1.58, velocity of light in vacuum = 3×10⁸ m s^-1 sin 57° 20′ = 0.8418
Solution:

Critical angle of glass relative to the liquid,

θ_c = 57°20′

If the refractive index of glass concerning the liquid is {fiR then.

sinθ_c =\(\frac{1}{\nu_g}=\frac{1}{\frac{\mu_g}{\mu_l}}=\frac{\mu_l}{\mu_g}\)

⇒ \(\frac{\mu_l}{\mu_g}=\sin 57^{\circ} 20^{\prime}\)

= 0.8418

Or, \(\mu_l=0.8418 \times \mu_g=0.8418 \times 1.58\)

Again , \(\mu_l=\frac{\text { velocity of light in vacuum }}{\text { velocity of light in the liquid }}\)

The velocity of the light in the liquid

= \(\frac{3 \times 10^8}{\mu_l}=\frac{3 \times 10^8}{0.8418 \times 1.58}\)

= \(2.255 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 9.  A transparent solid cylindrical rod has a refractive Index of. It Is surrounded by air. A light ray Is Incident at the midpoint of one end of the rod.  Determine the incident angle 8 for which the light ray grazes along the wall of the rod

Solution:

For refraction of light at point B, we can write by applying Snell’s law

1 × sin θ = μ sin r

[where μ is the refractive index of the solid material)

or, sin θ = \(\frac{2}{\sqrt{3}} \sin r\) ……………………… (1)

The light ray BC is incident on point C making critical angle θ_c and propagates along CD

Thus, from Snell’s law,

μ = \(\frac{1}{\sin \theta_c}\)

Or, \(\theta_c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)

=\(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

= 60°

r = 180°- (60° + 90°) = 30°

Hence from equation (1), we can write

θ_c = \(\frac{2}{\sqrt{3}} \sin 30^{\circ}=\frac{1}{\sqrt{3}}\)

Or, \(\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)

Examples of Total Internal Reflection

A test tube dipped in water: A glass test tube half filled with water is held obliquely in a beaker containing water

The empty portion of the Immersed test tube appears shining if It is seen from above. This happens due to the total internal reflection of light. For the empty portion of the tube, the light goes from a denser to a rarer medium. Rays which are incident at angles greater than the critical angle of glass and air (48.5°) arc are reflected. So this portion of the glass appears shining.

A portion of the tube filled with water does not glow because here light enters water In test tube from water in a beaker. Thus, total internal reflection does not occur here. In this discussion, we do not take into account the existence of the glass wall of the tube due to its negligible thickness.

A metal ball coated wHb lampblack Immersed In water:

If a metal ball coated will lampblack Is Immersed in water, the ball appears shining. Due to the coating of the lamp¬ black. a thin layer of air surrounds the surface of the ball. Rays incident at an angle greater than the critical angle of water and air, are reflected. The ball appears shining when the reflected rays reach the eyes of the observer.

Glass tumbler full of water:

A glass tumbler full of water Is held above eye level. If the upper surface of water In the tumbler is seen from any one side, the surface appears shining. Rays coming from the side of the tumbler are incident on the surface of the separation of water and air. Hence total internal reflection takes place at particular angles of slantness and the surface of the water appears shining.

Air bubbles:

The air bubbles rising through the water look shiny. Rays travelling through water are incident on the surface of the air bubbles. Those rays which are incident at angles greater than the critical angle are reflected. When these reflected rays reach the eyes of the observer, the hubbies appear shining. For the same reason, air bubbles existing In paper weights appear to be shining

Natural Examples Of Total Internal Reflection

Mirage: It Is an optical Illusion brought about by total internal reflection. There are two types of mirage, one observed in hot regions and the other observed In extremely cold regions.

1. Inferior mirage or mirage in the desert:

People travelling through the desert sometimes see water at a distant place which is an optical illusion, called an inferior mirage, or simply, a mirage.

During daytime, the lower regions of the atmosphere become hot¬ ter than the upper regions. So density of air in the lower regions is less than that in the higher regions. Let us consider the atmo¬ sphere to be made up of layers of air, one above the other. A ray of light starting from a distant tree (P) and travelling downward happens to be going from a denser to a rarer medium.

So its angle of incidence at consecutive layers goes on increasing gradually till it exceeds the critical value when it is reflected due to total internal reflection. The traveller sees an inverted virtual image (P’) of the tree. Secondly, due to continuous temperature changes, there exists a temperature gradient in the layers which undergo a continuous change of density and hence in the refractive index as well. So the path of the rays coming through the layers of air is also continuously changing. Hence to the traveller, the image of the tree appears to be swaying. This completes the illusion of a pond lined with trees.

 2. Superior mirage or mirage in cold countries:

In cold countries, the temperature of air In the lower regions is lower than that of the upper region. So the density of air in the lower region is greater than that of the upper region. A ray of light starting from an object (P) travelling upwards, finds itself going from denser to rarer medium

So its angle of incidence at consecutive layers of air gradually increases till it reaches the critical value. Then it is reflected due to total internal reflection. To an observer, the ray appears to come from a point above, thus giving the impression that an inverted object (P’) is floating in the air which is an optical illusion. This phenomenon is called a superior mirage

View of an observer inside water:

To the eye of an observer or a fish Inside water, all objects above water appear to exist in n cone of semi-vertical angle <19° which Is the critical angle of water and air. This happens due to total Internal reflec¬ tion of light.

If a ray of light travelling from a denser medium is Incident at the critical angle, the refracted ray grazes the surface of separation. Conversely, if a ray of light travelling from a rarer medium is incident at an angle of 90°, the angle of refraction In the denser medium becomes equal to the critical angle. The critical angle of water and air is 49°. So if a ray of light S₁A coming from the rising sun AS₁, along the surface of water reaches eye E along the direction AE, then the angle of refraction in water becomes 49°

As the eye cannot follow ray AS₁, an observer inside water will sec the rising sun along the line EAC and this line will make an angle of 49° with the line OE. Similarly, the setting sun S₂ will be seen along the line EBD and this line also will make an angle of 49° with the line OE. So all the objects above water appear to exist in a cone of angle 98° to the eye of a fish or observer underwater.

It Is to be noted that the sun describes an arc of 180° to earthbound observers but to the eyes of a fish it describes an arc of 98°.

1.  Surface of water to the eye of an observer inside water:

The diameter of the circular base of the cone AEB is AB. If an observer keeping his eye on E looks at the circular section of water, he can see any object lying above water. But If the observer looks at the rest of the portion of water other than the circular portion, then

1. He cannot see any object above water, rather
2. He can see the images of the objects inside the water.

Reason explaining 1st Incident:

Any ray of light coming from outside water can reach point E only through the circular section but cannot reach point E if it comes through die remaining portion.

Reason explaining 2nd Incident:

Suppose the ray of light emerging from the object situated In water, reaches die point E after reflection from the surface of water. This reflection will take place from the surface of the water excluding the circular portion.

This reflection will be a total reflection. For example, if a ray of light from the object P situated inside water, is incident on the surface of water, the angle of incidence exceeds 49°. So, the ray after total reflection from the surface of the water reaches the eye of the observer and he observes the dead object, at P’.

So to the observer situated inside water, the surface of the water appears as a mirror with a circular hole in it, because he sees the objects situated outside water through the circular section and sees the images of the objects inside water in die remaining action of the surface of the water. The radius of the circular hole is OA or OB.

2. Determination of the radius of the hole:

Let the radius of the hole =OA = OB = r and OE – h. If the critical angle is θ then ∠OEA = Qc

⇒ \(\tan \theta_c=\frac{O A}{O E}=\frac{r}{h}\)

Or, \(r=h \tan \theta_c=h \frac{\sin \theta_c}{\cos \theta_c}\)

= \(h \frac{\sin \theta_c}{\sqrt{1-\sin ^2 \theta_c}}\)

Since ( sin \(\theta_c=\frac{1}{\mu}\))

= \(h \frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}}\)

r = \(\frac{h}{\sqrt{\mu^2-1}}\)

Sparkling of diamond:

Diamond is notable for its sparkle and shine. This characteristic of a diamond is based on total internal reflection. The refractive index of a diamond is 2.42 and its critical angle relative to air is only 24.4°. This value is quite small as compared to other pairs of media.

Therefore, there is a high probability of total internal reflection in the case of diamond. If the diamond is cut properly, it will have a large number of faces. Ray of light entering through one face undergoes total internal reflection at several faces. As the rays of light enter through many faces and are confined inside they emerge together through only a few faces, these faces appear to sparkle and shine

Transmission of Light through Optical Fibre

Optical fibre:

A beam of light can be sent from one place to another through an optical fibre made of glass, quartz or optical-grade plastic, by following successive total internal reflections. As water can be sent from one place to another through a hollow pipe, a fibre can allow light to flow through it from one place to another. Hence, an optical fibre is often loosely called a light pipe.

Construction and principle of action:

An optical fibre is a long and very thin pipe. Its diameter is about 10 × 10^-6 metres. The internal section of the die pipe is called the core. It is a die core through which light travels from one point to another. Above the core, there is a coating of a substance having a refractive index less than that of the core. This coating is called cladding.

A ray of light entering die fibre through one face undergoes successive total internal reflections at the surface of separation of core and cladding and emerges through the other face [Fig. 2.38].: As total internal reflection of light takes place inside a fibre, the intensity of the light remains almost the same.

The image of a large object cannot be sent through a single fibre. In that case, bundles of fibres or cables of fibres are used. A cable contains about a thousand fibres. The image of an object is focused on one end of the bundle. If the order of die fibres is properly maintained, the image obtained at the other end will be an exact reproduction. In, the letter ‘T’ has been focussed at one end of the die bundle and an exact image of ‘T’ has been obtained. Light rays from the different portions of ‘T’ travel through the different fibres and form a die image at the other end

Application of optical fibre:

• Optical fibres are extensively used in medical science and the field of communication.
• These are used to study the interior parts of the body which are inaccessible to the bare eye, for example, lungs, tissues, intestines etc. It can be used to transmit high-intensity laser light inside the body for medical purposes.
• These are used for sending signals from one place to another. This signal is mainly digital. It is information that the signal carries.
• This information is used in telephone, television, fax, computer etc. It is to be noted that, different digital signals may be sent through the same fibre at the same time, without any chance of overlapping.
• So many times it is needed to collect samples inside from human body to identify disease. For this purpose, optical fibre is used. Besides, optical fibre is used for operation inside the human body. Thus, in most cases, no major excising is needed outer part of the body

Advantages of optical fibre over copper wire:

• Comparatively less power Is required to send a signal,
• The loss in energy Is much less
• The capacity of carrying information is approximately times.
• There exists no influence of any external electromagnetic wave signal.
• Electrical resistance is much more.
• It is very light.
• killVelocity of the signal is very fast (approximately equal to that of light in vacuum).
• The possibilities of the illegal usage of the signal are very low.

Optics

Refraction Of Light  Transmission of Light through Optical Fibre Numerical Examples

Example 1. A point source of light is placed at a depth of h below the calm surface of the water. From the source, light rays can only be transmitted to air through a definite circular section,

1. Draw the circular section of the surface of the water by ray diagram and mark its radius r.
2. Determine the angle of incidence of a ray of light incident at any point on the circumference of the circular plane. [Given: refractive index of water,\(\frac{4}{3}\) = 48°36′ = sin^-1 0.7501 ]
3. Show that r= \(\frac{3}{\sqrt{7}} h\)

Solution:

Let MN be the open surface of water. O is the source of light at a depth h below the surface of water. Light rays incident on the surface of the water from 0 at angles less than critical angle transmit in air after refraction. At points A and B the light rays are incident at angles equal to the critical angle (θ). So the refracted rays at these two

Points graze along the surface of separation. So the light rays will transmit outside water only through the circular section of radius r =  AP = PB.  If the rays are Incident on the surface of water excluding this circular section, the»7 will be reflected from (be surface of the water and will return to water,

Let the angle of Incidence be θ.

⇒ \(\sin \theta=\frac{1}{a^{\mu_w}}=\frac{1}{\mu}=\frac{3}{4}\)

= sin 48°36′ or, = 48° 36′

From the triangle AOP,

⇒ \(\tan \theta=\frac{A P}{O P}=\frac{r}{h} \quad \text { or, } \frac{\sin \theta}{\cos \theta}=\frac{r}{h}\)

Or,\(\frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}}=\frac{r}{h}\)

∴ \(\sin \theta=\frac{1}{\mu}\)

Or, \(r=\frac{h}{\sqrt{\mu^2-1}}=\frac{h}{\sqrt{\frac{16}{9}-1}}=\frac{3}{\sqrt{7}}\)

Example 2.  The water in a pond has a refractive Index| of light and is placed 4 m below the surface of the water. Calculate the minimum radius of an opaque disc that needs to be floated on water so that light does not come out.
Solution:

Minimum radius of the opaque disc,

r = \(\frac{h}{\sqrt{\mu^2-1}}=\frac{4}{\sqrt{\left(\frac{5}{3}\right)^2-1}}\)

= \(\frac{4}{\frac{4}{3}}\)

= 3m

Example 3. Shows a longitudinal cross-section of an optical fibre made of glass with a refractive index of 1.68. The pipe is coated with a material of a refractive index of 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflection inside the fibre can take place?

Solution:

The refractive index of the outer coating concerning the glass pipe

⇒ \(g_g \mu_c=\frac{a^{\mu_c}}{a^{\mu_g}}=\frac{1.44}{1.68}\)

If the critical angle for the total reflection is θ, then

sinθ = \(\frac{1}{c^{\mu_g}}\)

= \(\frac{1.44}{1.68}\)

= 0.857 or, 59°

Thus’ total reflection takes place when i’ > 59° or when r < r_maxwhere r_max= 90° – 59° = 31°

So if the maximum angle of incidence on the fibre is imax then,

sin i_max  = μ_gsir r _max= 1.68 sin 31° = 0.865

_Imax  = 60°

So, the range of the angles of incidence for total internal reflection inside the fibre is from 0° to 60°.

Example 4. In which direction will the sun appear to set if the observer is inside the water of a pond? Refractive index of water, μ – 1.33.
Solution:

For a setting sun, the incident rays graze along the surface of the water, i.e., angle of incidence = 90°

∴ According to Snell’s law

μ_w= \(\frac{\sin i}{\sin r} \)

Or,   1.33 \(=\frac{\sin 90^{\circ}}{\sin r}\)

sir r  = \( \frac{1}{1.33}\)

= 0.7518

= sin 48.75°

r = 48.75°

Therefore, to see the setting sun the observer in water should look at an angle of 48.75° with the normal.

Optics

Refraction Of Light Atmospheric Refraction

Apparent position Of a star: The whole atmosphere surrounding the earth may be supposed to be divided into different horizontal layers. As the height above the earth’s surface increases, the density of the air decreases. Due to this, the refractive index of air also decreases with the increase in altitude. For this reason, the ray from a star S (say) proceeding towards the earth’s surface cannot travel straight but continually bends towards the normal at the surface of separation due to refraction as it penetrates from rarer to denser layers

This ray after several refractions reaches the observer at O. But our vision cannot follow the curved path OS. A tangent OS’ is drawn on OS at O . So, the observer sees the star at S’. This phenomenon is called atmospheric refraction.

Visibility of the sun before sunrise and after sunset:

The diameter of the sun subtends an angle of 0.5° at the eye of an observer on Earth. This value is equal to the deviation of sun¬ light due to atmospheric refraction, when at the horizon. So, the sun appears to just touch the horizon during sunset and sunrise when it is actually below it. What we see therefore is the raised image of the sun, formed due to atmospheric refraction. As a result, we see the sun a few minutes after sunset or before sunrise. Now, the sun covers a distance equal to its diameter in 2 min.

So, the sun becomes visible another 2 min earlier at sunrise and also remains visible for another 2 min after the actual sunset. Consequently, 4 min are added to the length of a day. This value is valid for observations from the equatorial region. At higher latitudes, this time increases

The oval shape of the sun when It Is near the horizon:

During sunrise or sunset, the lower edge of the sun remains nearer to the horizon than its upper edge. So, the rays coming from the lower edge of the sun are incident on an atmospheric layer at a larger angle than that for the rays coming from its upper edge. As the refracting angle increases with the increase of the incident angle, the rays coming from the lower edge bend more than the others due to multiple refractions at different atmo¬ spheric layers. As a result, the vertical diameter of the Sun appears to be reduced whereas the horizontal diameter remains unaffected.

Twinkling of Stars:

Due to atmospheric refraction, we see the stars twinkle. Light rays from the stars situated far and far away from us come to our eyes passing through various layers of air. The temperature of the layers does not remain constant and changes continuously. So the density of the various layers also. changes.

Again the refractive index of the layers changes with the change of density. So, when the rays of light from a star come to our eyes, the direction of the path of the rays changes continuously. As a result, the amount of light reaching our eyes also changes continuously. It seems as if the brightness of the stars is changing. So the stars appear twinkling.

As the planets are nearer to us than the stars, more amount of light comes to us. Therefore, the change in brightness of the planets due to changes in the refractive index of various layers of air is negligible. We cannot detect it with our eyes. So it appears that the planets are emitting light steadily.

Optics

Refraction Of Light Thin Prism

Thin Prism Definition:

The prism, whose refracting angle is very small (not more than 10°), is called a thin prism.

Deviation produced by a thin prism:

ABC is a thin prism. A ray PQ is incident on the refracting face AB nearly normally. For nearly normal incidence, i₁ ≈ 0, i₂≈0. If n is the refractive index of the material of the prism, then

μ = \(\frac{\sin i_1}{\sin r_1}=\frac{i_1}{r_1} \quad \text { or, } i_1=\mu r_1\)

And \(\mu=\frac{\sin i_2}{\sin r_2}=\frac{i_2}{r_2} \quad \text { or, } i_2=\mu r_2\)

Or,

So, the deviation of the ray

⇒ \(\delta=i_1+i_2-A=\mu r_1+\mu r_2-A=\mu\left(r_1+r_2\right)-A\)

= μA – A

Since = r₁+r₂ = A

= (μ- 1)

Again, if the ray PQ is incident on the face AB normally, then

i₁= r₁ = 0. So, A = r₂
.
Therefore, the deviation of the ray,

δ = i₁ + i₂ – A = μr₂-A =μA-A = (μ -1 )A

So, for normal and nearly normal incidence, the deviation of the array in a thin prism, δ = (μ-1)A

Thus it is seen that for normal or nearly normal incidence, the deviation of a ray in a thin prism depends only on the refract¬ ing angle of the prism and the refractive index of its material but not on the angle of incidence. So if the angle of the incidence is small, the deviation of a ray in the case of a thin prism remains constant.

Optics

Refraction Of Light Thin Prism Numerical Examples

Example 1. A very thin prism deviates a ray of light through 5°. If the refractive index of the material of the prism is 1.5, what is the value of the angle of the prism?
Solution:

The angle of deviation for a thin prism,

δ = (μ-1)A

Here  δ = 5° and μ = 1.5

Therefore from equation (1) we get,

5° = (1.5 -1)A or, A = 10°

Example 2. A prism haying refracting angle 4°. Is placed in the air. Calculate the angle of deviation of a ray incident nor¬ mally or nearly normally on It. The refractive index of the material of the prism| = \(\frac{3}{2}\)
Solution:

The refracting angle of the prism, A = 4°. So it is a thin prism. We know that the deviation of a ray in a thin prism for. normal or nearly normal incidence is given by,

⇒  \(\delta=(\mu-1)\)A

⇒ \(\delta=\left(\frac{3}{2}-1\right) \times 4^{\circ}\)

= 2°

Example 3. A thin prism with a refracting angle of 5° and having refractive index of 1.6 is kept adjacent to another thin prism having a refractive index of 1.5 such that one is inverted concerning the other. An incident ray falling vertically on the first prism passes through the second prism without any deviation. Calculate the refracting angle of the second prism.
Solution:

According to the condition,

(μ₁– 1)A₂ = (μ₂– 1)A₂

(1.6-1) × 5° = (1.5-1)A₂

Or, A₂ =\(\frac{0.6 \times 5^{\circ}}{0.5}\)

So the refracting angle of the second prism = 6°

Limiting Angle of a Prism for No Emergent Ray

A ray of light incident on a refracting surface of a prism may not emerge from the second refracting surface. It depends on the refracting angle of the prism. Every prism has a limiting value of its refracting angle. Light can emerge from the prism if the angle of the prism is equal to or less than this critical value, otherwise, no light can emerge from the prism.

Let ABC be the principal section of a prism  PQRS is the path of a ray through the prism placed in the air where ray RS grazes along the second face AC.

Let the angles of incidence and refraction at the face AB be i₁ and r₁ respectively and the corresponding angles at the face AC be r₂ and i₂, where i₂ = 90°.

So, r₂ = θ_c, the critical angle between glass and air. If the refracting angle of prism A is equal to the limiting angle, then the ray incident at an angle of incidence fj to the face AB of the prism makes a grazing emergence along the second refracting surface AC

A= r₁+r₂ ……………….(1)

For refraction at Q

⇒ \(\sin i_1=\mu \sin r_1 \quad \text { or, } r_1=\sin ^{-1}\left(\frac{\sin i_1}{\mu}\right)\)

For refraction At R,

⇒ \(\sin 90^{\circ}=\mu \sin r_2 \quad \text { or, } r_2=\sin ^{-1}\left(\frac{1}{\mu}\right)\)

Or, \(\theta_c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)

From equation (1) we get, A = \(A=\sin ^{-1}\left(\frac{\sin i_1}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)\) …………. (2)

Special cases:

Limiting angle of the prism for normal incidence on the first face: When ray, PQ is incident on the face AB normally, then fj = 0. In this case, if the emergent ray grazes along the surface AC then from equation (2) we get,

A = \(\sin ^{-1}\left(\frac{\sin 0}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)=\sin ^{-1}\left(\frac{1}{\mu}\right)=\theta_c\)

Hence, the ray can emerge from the prism through its second surface till the refracting angle of the prism remains less than its critical angle. But as the refracting angle of the prism becomes greater than its critical angle, no ray emerges from the surface AC. Then the face AC acts as a total reflecting surface.

Limiting angle of the prism for grazing incidence on the first face:

For grazing incidence on the face AB, i₁ = 90° Then from equation (2) we get

A = \(=\sin ^{-1}\left(\frac{\sin 90^{\circ}}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)\)

= \(\sin ^{-1}\left(\frac{1}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)=2 \sin ^{-1}\left(\frac{1}{\mu}\right)=2 \theta_c\)

So, if the refracting angle of the prism is greater than 20C and if a ray is incident on the face AB grazing the surface, then it will be reflected from the face AC. It means the ray will not emerge in the air through the face AC. Hence, no emergent ray will be obtained.

Thus, from the above discussions we conclude that for any incidence no ray can emerge from the prism If the angle of the prism is greater than twice the critical angle for the material concerning the surrounding medium.

Optics

Refraction Of Light Limiting Angle Of Incidence For No Emergent Ray From A Given Prism

Just like a prism has a limiting refracting angle for no emergent ray, a prism with a definite refracting angle also possesses a limiting angle of incidence for no emergent ray from it. If the angle of incidence i₁ becomes less than this limiting incident angle, then there will be no corresponding emergent ray.

Let ABC be the principal section of a prism. The ray of light PQ is incident at Q on the face AB. After refraction through the prism, the emergent ray RS grazes the second face AC of the prism.

Let the angles of incidence and refraction at the face AB be i₁ and r₁ the corresponding angles at the face AC be r₂ and i₂ respectively. Here, i₂ = 90° .

Now, the angle of the prism, A = r₁ + r₂ – constant.

From, A = r₁ + r₂; we get, r₂ = A – r₁ , reduces with the decrease of it. Again, r₁ increases with the decrease of r₁

Now, if r₂ is greater than θ_c the ray QR is reflected from the face AC inside the prism and it does not emerge in the air.

So when r₁ = θ_c then i₁ = limiting angle of incidence.

If μ is the refractive index of the material of the prism then

⇒ \(\sin \theta_c=\frac{1}{\mu}\)

⇒ \(\cos \theta_c=\sqrt{1-\sin ^2 \theta_c}=\sqrt{1-\frac{1}{\mu^2}}=\frac{\sqrt{\mu^2-1}}{\mu}\)

Considering the refraction of the ray at Q we have,

⇒ \(\cos \theta_c=\sqrt{1-\sin ^2 \theta_c}=\sqrt{1-\frac{1}{\mu^2}}=\frac{\sqrt{\mu^2-1}}{\mu}\)

= \(\cos \theta_c=\sqrt{1-\sin ^2 \theta_c}=\sqrt{1-\frac{1}{\mu^2}}=\frac{\sqrt{\mu^2-1}}{\mu}\)

= \(\sin i_1=\mu \sin r_1=\mu \sin \left(A-r_2\right)\)

Since A = r₁+r₂

= \(\mu \sin \left(A-\theta_c\right)=\mu\left[\sin A \cos \theta_c-\cos A \sin \theta_c\right]\)

= \(\mu\left[\sin A \cdot \frac{\sqrt{\mu^2-1}}{\mu}-\cos A \cdot \frac{1}{\mu}\right]\)

= \(\mu\left[\sin A \cdot \frac{\sqrt{\mu^2-1}}{\mu}-\cos A \cdot \frac{1}{\mu}\right]\)

= \(\sin A \sqrt{\mu^2-1}-\cos A\)

Or, i₁ = \(i_1=\sin ^{-1}\left[\sin A \sqrt{\mu^2-1}-\cos A\right]\)

This is the limiting angle of incidence. If the angle of incidence is less than this limiting angle, no ray will emerge from the second face of the prism

Optics

Refraction Of Light Limiting Angle Of Incidence For No Emergent Ray From A Given Prism Numerical Examples

Example 1. To get an emergent ray from a right-angled prism its refractive index should not exceed \(\sqrt{2}\) —prove it.
Solution:

The condition of getting an emergent ray from a prism is that the refracting angle of the prism should be equal to or less than twice the value of the critical angle

A≤ 2θ_c Or, 90°≤ 2θ_c 

Or, θ_c ≥ 45°

∴ sin θ_c ≥ sin 45° ,Or, sin θ_c ≥ \(\frac{1}{\sqrt{2}}\)

∴  \(\sin \theta_c=\frac{1}{\mu}\)

∴ \(\frac{1}{\mu}\frac{1}{\sqrt{2}}\)

Or, \(\sqrt{2}\)

Example 2. The refractive index of a prism having a refracting angle of 75° is \(\sqrt{2}\). What should be the minimum angle of incidence on a refracting surface so that the ray will emerge from the other refracting surface of the prism?

Solution: According to the question, the emergent angle is i₂ = 90°.

So for refraction at the second face of the prism

μ = \(\frac{\sin i_2}{\sin r_2}=\frac{\sin 90^{\circ}}{\sin r_2}\)

Or, \(\sin r_2=\frac{1}{\mu}=\frac{1}{\sqrt{2}}\)

= sin 45°

Or, r₂ = 45°

We, know A= r₁ +r₂

75= r₁ + 45°

Or, r₁ = 30°

For refraction at the first face

μ = \(\frac{\sin i_1}{\sin r_1}\)

Or, \(\sqrt{2}=\frac{\sin i_1}{\sin 30^{\circ}}\)

Or, \(\sin i_1=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}\)

= sin 45°

Or, i₁ = 45°

The required angle of incidence = 45°

Example 3. Find the value of the limiting angle of incidence if the refractive index of the material of the prism is 1.333 and the angle of the prism is 60°
Solution:

Here, the refractive index of the material of the prism, mu = 1.333; the angle of the prism, A = 60°

⇒ \(i_L=\sin ^{-1}\left[\sin A \sqrt{\mu^2-1}-\cos A\right]\)

= \(\sin ^{-1}\left[\frac{\sqrt{3}}{2} \sqrt{(1.333)^2-1}-\frac{1}{2}\right]\)

= \(\sin ^{-1}(0.2633)\)

= 15.27°

Example 4. The refracting angle of the prism is 60° and its refractive J index is Jl. What should be the minimum angle of | incidence on the first refracting surface so that the ray | can emerge somehow from the second refracting our face?
Solution:

Let i be the limiting angle of incidence, then

sin i = \(\sqrt{\mu^2-1} \cdot \sin A-\cos A\)

= \(\sqrt{\frac{7}{3}-1} \cdot \sin 60^{\circ}-\cos 60^{\circ}\)

= \(\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2}\)

= \(1-\frac{1}{2}=\frac{1}{2}\)

i = 30

Example 5. The refractive index of the material of a prism is \(\)  and the refracting angle is 90°. Calculate the angle of minimum deviation and the corresponding angle of incidence. Show that the limiting angle of incidence for getting emergent ray is 45°
Solution:

Here, the refractive index of the material of the prism,

M = \(\sqrt{\frac{3}{2}}\)

The angle of prism A =  90°

μ = \(\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}\)

Or, \(\sqrt{\frac{3}{2}}=\frac{\sin \frac{A+\delta_m}{2}}{\sin 45^{\circ}}\)

⇒ \(\sin \frac{A+\delta_m}{2}=\sqrt{\frac{3}{2}} \times \frac{1}{\sqrt{2}}=\frac{\sqrt{3}}{2}\)  = sin 60°

⇒  \(\frac{A+\delta_m}{2}\)  = 60°

Or, A + δ_m = 120°

δ_m = 120° – 90° = 30°

For minimum deviation, i₁= i₂

δ_m = i₁+ i₂ -A

30° = 2i₁ – 90

Or, 2i₁ = 120°

Or, i₂= 60°

Or minimum deviation angle of indecency = 60

To obtain the emergent ray I be the limiting angle of incidence. Then

sini = \(\sqrt{x^2-1} \sin x-\cos 4\)

= \(\sqrt{\frac{3}{2}-1} \cdot \sin 90^{\circ}-\cos 90^{\circ}=\frac{1}{\sqrt{2}}\)

= sin 45°

= 45°

Example 6. The refractive index of a prism is \(\sqrt{2}\). A ray of light is incident on the prism grazing along one of its refracting surfaces. What should be the limiting angle of the prism for no emergent ray from the other face?
Solution:

If The limiting angle of the prism for no emergent ray is A, then

A = \(2 \sin ^{-1} \frac{1}{\mu}\)

= \(2 \sin ^{-1} \frac{1}{\sqrt{2}}=2 \times 45^{\circ}\)

= 90°

Optics

Refraction Of Light Conclusion

1. When a ray of light enters a medium from another medium through the interface of the two media, then the path of the rav changes its direction and this phenomenon is known as refraction of light.

2. Laws of refraction:

• The incident ray, the refracted ray and the normal to the refracting surface at the point of incidence lie on the same plane.
• The sine of the angle of incidence bears a constant ratio to the sine of the angle of refraction. The value of this constant depends on the nature of the pair of media concerned and the colour of the incident light.

3. Relative refractive index:

When a ray of light is refracted from a medium a to another medium b then the ratio of the sine of the angle of Incidence to the sine of the angle of refraction is called the refractive index of the medium b concerning the medium a. This refractive index is called the relative refractive index.

4. Absolute refractive index:

When a ray of light is refracted from the vacuum to any other medium then the ratio of the sine of the angle of incidence to the sine of the angle of refraction is called the absolute refractive index of that medium.

5. Critical angle:

When a ray of light is refracted from a denser medium to a rarer medium than for a particular angle of incidence in the denser medium, the angle of refraction in the rarer medium is 90° i.e., the refracted ray grazes along the interface of the .two media. Then that particular angle of incidence is called the critical angle for the pair of media.

6. Total internal reflection:

While passing from a denser medium to a rarer medium, if a ray of light is incident on the surface of separation between the angle greater than the critical angle for the two media involved, the ray of light is reflected to the denser medium without undergoing refraction in the rarer medium. This phenomenon is known as the total internal reflection of light.

Through optical fibres tight rays can be transmitted from one place to another in the straight or curved path by successive total internal reflections. f If a ray of light from a rarer medium is incident on a prism of denser medium then after refraction through the prism, the ray of tight bends towards the base of tyre prism. -f In the case of reflection or refraction, the change of direction of tight is called deviation.

4- Minimum deviation:

For any prism, there is a certain angle of incidence for which the angle of deviation becomes minimum or the least. This angle of deviation is called the angle of minimum deviation for the prism.

At minimum deviation, the angle of incidence becomes equal to the angle of emergence.

5. Thin prism:

If the refracting angle of a prism is very small (not more than 10° ) then it is called a thin prism.

6. Total reflecting prism:

If the principal section of a prior made of crown glass is a right-angled isosceles triangle then the prism makes a total Internal reflection of light for any incidence on it hence this type of prism is called a total reflecting prism.

7. During the refraction of light, the velocity, intensity and wavelength of light change but its frequency and phase remain constant.

8. If there is a parallel glass slab in the path of light then Ugh rays remain undefeated after refraction through it but lateral displacement of light rays will occur.

9.  \(\frac{\sin i}{\sin r}={ }_a \mu_b=\frac{\mu_b}{\mu_a}\)

μ_a = absolute refractive index of medium a, nb = absolute refractive index of medium b, afib = relative refractive index of medium b concerning the medium a.

10.  \(\frac{\sin i}{\sin r}={ }_a \mu_b=\frac{\mu_b}{\mu_a}\)

[ va = velocity of light in the medium a, vb= velocity of tight in the medium b ]

⇒  \(\mu=\frac{c}{v}\)

[n = absolute refractive index of a medium, v = velocity of light in that medium, c = velocity of tight in vacuum]

[v= – frequency of the wave, A = wavelength]

11. \(a^{\mu_b}=\frac{v_a}{v_b}\)

The relation between (i and A given by scientist Cauchy

12. \(\mu=\frac{c}{v}\)

Deviation of a ray of tight due to refraction,

= i~r [i = angle of incidence, r = angle of refraction]

13. Lateral displacement of a ray of tight after suffering refraction through a parallel plate glass slab

= \(t \sin i_1\left[1-\frac{\cos i_1}{\sqrt{\mu^2-\sin ^2 i_1}}\right]\)

[where, i- the angle of incidence on the front surface of the tin slab, t = thickness of the slab, (J- = refractive index of the material of the slab]

If the angle of incidence ij is very small then lateral displacement = \(t i_1\left(1-\frac{1}{\mu}\right)\)

14. General formula ofSnell’s law:

μ₁ sin₁  = μ₂sin₂ =  μ₃ sin₃ = ………………. = μ₃ sin i₃

15. The apparent position of an object due to refraction:

The refractive index of rarer medium =; refractive index c denser medium = μ₁

1. If the object is in a denser medium and the observer is in: a rarer medium:

The refractive index of the denser medium with respect to the rarer medium real depth of the object from
= \(=\frac{\text { the surface of separation }(d)}{\text { apparent depth of the object from the surface of separartion}}\)

= \(\frac{\mu_2}{\mu_1}\)

2. If the object Is In the rarer medium and the observer is in a denser medium: The refractive index of the denser medium concerning the rarer medium apparent height of the object from

= \(\frac{\text { the surface of separation }\left(d^{\prime}\right)}{\text { real height of the object from the surface of separation (d)}}\)

16. The refractive index of glass

=\(\frac{\text { real thickness of a thick mirror }}{\text { apparent thickness of a thick mirror }}\)

⇒ \(\frac{1}{b^{\mu_a}}=\frac{\mu_b}{\mu_a}\)

[where, A = r₁+r₂]

17. Deviation of a ray of light after refraction through a prism,

[i = angle of incidence on the first face of the prism, i’μ_a = angle of emergence from the second face to the prism, r₁

= angle of refraction In the first face, = angle of incidence in the second face, A = refracting angle of the prism]

18. In case of minimum deviation, i₁ = i₂ = i (say)

r₁ = r₂ = r (say)

19. In that case, the minimum deviation, δ

m = 2i-r and A = 2r.

20. Relation between refractive index (fi) and minimum deviation (8m ) :

μ\(\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}\)

21. For the normal and nearly normal incidence of a ray of light, deviation in the case of a thin prism,

⇒ δ = (μ-1)A

22. Limiting the angle of a prism for no emergent light from a prism

⇒ \( \sin ^{-1}\left(\frac{\sin i_1}{\mu}\right)+\sin ^{-1}\left(\frac{1}{\mu}\right)\)

When iy = 0 , then A>θ_c; when iy = 90°, then A >2θ_c.

Limiting value of the angle of incidence in case of definite prism for no emergent light from it,

⇒\(i_1=\sin ^{-1}\left[\sin A \sqrt{\mu^2-1}-\cos A\right]\)

23. A ray of light passes through n number of media of refractive indices, μ₁,  μ₂, μ₃, ……………….. μ_n respectively. The planes of the media are parallel. If the emergent ray from the n -th medium is parallel to the ray incident on the first medium, then μ₁ = μ₂

24. For a hollow prism, the angle of the prism is A ≠ 0 but the angle of deviation is δ = 0.

25. A container of depth 2d is half filled with a liquid of refractive index μ₁ and μ₂    the remaining half is filled with another liquid of refractive index When seen from the top, the apparent depth of the container is

⇒ \(d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)\)

26. An observer is situated at a depth h in a water body. The water surface appears as a porous circular mirror to

 27. If a parallel plane glass slab of thickness t is kept in the path of a beam of converging rays, then the displacement of the intersecting point of the rays,

⇒ \(O O^{\prime}=x=\left(1-\frac{1}{\mu}\right) t\)

To transmit through a glass slab of thickness t and refractive index μ, time taken by light is, where c is the velocity of light in vacuum \(\frac{\mu t}{c}\)

Optics

Refraction Of Light Very Short Questions

Question 1. What is the angle of deviation due to the refraction of a ray of light incident perpendicularly on a refracting surface?
Answer: Zero

Question 2. Arrange the following media according to the increasing optical density Air, diamond, glass, water, glycerine.
Answer:

Air < Water < Glycerine < Glass < Diamond]

Question 3. Can the value of absolute refractive index of a medium be
Answer: No

Question 4. State the relation of velocity of light with refractive index
Answer:

⇒ \(\left[\mu=\frac{c}{v}\right]\)

Question 5. The refractive index of a medium depends on temperature’—is the statement correct or wrong?
Answer: Correct

Question 6. Arrange the following media according increasing velocity of light through them: vacuum, diamond, water, air, glass
Answer:

Diamond < Glass < Water < Air < Vacuum

Question 7. Does the velocity of light in a vacuum depend on

1. Wavelength of light
2. Frequency of light
3. Intensity of light?

Answer: No

Question 8 If water is heated, how refractive index it will change?
Answer: Decrease

Question 9. The refractive index of glass is 1.5. What is the velocity of light
Answer: [2 × 10⁸m/s].

Question 10. Can the relative refractive index of a medium concerning another be less than unity?
Answer: Yes

Question 11. The refractive index of a medium is a physical quantity having no dimension and no unit—is the statement true or
Answer: True

Question 12. A light ray of wavelength 4500 A enters a glass slab of refractive index 1.5, from the vacuum. What is the wavelength of light inside the slab?
Answer: 30000 A°

Question 13. For which colour of light is the refractive index of glass the minimum?
Answer: Red

Question 14. If the frequency of light increases will there be any change in the refractive index of the medium?
Answer: No

Question 15. For which colour of light is the refractive index of glass a maximum?
Answer: Violet

Question 16. Does the refractive index of glass depend on the colour of light? If so, how?
Answer: Yes, the refractive index increases with the decrease of wavelength

Question 18. When a ray of light is incident on a plane normally, then what is the value of the angle of refraction?
Answer: Zero

Question 17. If for refractive index of water relative to air is \(\), what will be the refractive index of air relative to water
Answer:

⇒ \(\frac{3}{4}\)

Question 18. When light travels from air to glass, how does its wavelength change? less than 1
Answer: Wavelength decreases

Question 19. The absolute refractive index of water and glass are \(\frac{4}{3}\) and \(\frac{3}{2}\). What is the ratio of velocity of light in glass and water?
Answer: [8:9]

Question 20. what is the value of the product of the refractive index of relative of the first medium and that of the first-second medium relative to the second medium?
Answer: 1

Question 21. For what angle of incidence the lateral shift produced by a parallel-sided glass plate is zero?
Answer: For i = 0

Question 22. For what angle of incidence the lateral shift produced by a parallel-sided glass plate is maximum?
Answer: For i = 90°

Question 23. what is the maximum lateral shift produced by a parallel-sided glass plate of thickness t?
Answer: t

Question 24. The bird descends vertically downwards in the direction of a pond during its flight. To a fish which is underwater and directly below the bird, what will be the apparent position of the bird?
Answer:

The bird’s apparent position will be slightly above its actual position i.e., the bird will appear higher up than it is

Question 25. An object lying inside a pond is viewed by a man from above (air) along a horizontal plane. Now if the man moves away from the object keeping his eyes along the same horizontal plane, how will the apparent depth of the object change?
Answer: Decrease along the caustic curve

Question 26. A transparent cube of glass of refractive index n and thickness ‘t is placed on a spot of ink drawn on white paper. When the spot is viewed from above (air) normally, the spot appears to shift through a distance of At towards the observer. What is the value of Δt?
Answer:

⇒ \(\left[\left(l-\frac{1}{\mu}\right) t\right]\)

Question 27. If a straight rod is held obliquely in water how does the immersed portion of the rod appear?
Answer: Refraction of light

Question 28. Multiple images are formed in a thick mirror. Which Image looks brightest?
Answer: Second images

Question 29. If a completely transparent object has to be made Invisible In a vacuum, what should be the value of its relative Index?
Answer: 1

Question 30. What Is the critical angle of light when passing from water \(\left(\mu \text { of water }=\frac{4}{3}\right)\)
Answer: 49

Question 31. For which colour of light is the critical angle between glass and air minimum?
Answer: Violet

Question 32. What are the factors of light which are responsible for creating mirages?
Answer: Refraction and total internal reflection]

Question 33. In which direction do we have to look to see the setting sun if we are underwater? (μ_w = 1.33)
Answer: At an angle of 49° with the normal drawn at the surface of the water]

Question 34. From sunrise to sunset, the sun subtends an angle of 180’ to our eyes. What will be the value of this angle to an observer underwater?
Answer: 98°

Question 35. Light passes through an optical fibre following which physical phenomenon?
Answer: Total internal reflection

Question 36. What is the approximate value of the refractive index of a diamond
Answer: 2.42

Question 37. what kind of image is formed in a desert by the formation
Answer: Virtual

Does critical depend on the colour of life

Question 38. The angular altitude at which we see a star is not its actual angular altitude’—is the statement true or false?
Answer: True

Question 39. The sun appears to be elliptical during sunset. What is the reason behind it?
Answer: Refraction of light

Question 40. Under what condition the angle of deviation of a refracted ray through a prism will be minimal?
Answer: If the angle of incidence and the angle of emergence are equal

Question 41. If the medium surrounding the four faces of a prism is denser an the entire material of the prism, then in which direction will the light rays bend when emerging from the prism?
Answer: It will bend upwards, towards the prism apex

Question 42. How can an inverted image be made erect by using a total reflecting prism?
Answer: By total internal reflection on the hypotenuse of a right-angled

Question 43. ‘When light travels from a rarer medium to a prism which is a denser medium, a light ray bends towards the base of the prism. Is the statement true or false?
Answer: True

Question 44. At the position of minimum deviation, what is the nature of the path of a light ray through a prism?
Answer: Symmetrical

Question 45. For a thin prism on what factor does the magnitude of the angle of deviation of light rays depend?
Answer: Angle of incidence

Question 47. what will be the angle of deviation of a ray of light incident normally on any smaller side of a total reflecting prism?
Answer: 90°c

Optics

Refraction Of Light Fill In The Blanks

Question 1. Which of the following_________________wavelength, frequency or velocity does not change during the refraction of light?
Answer: Frequency

Question 2. For any particular medium, the refractive index is greater If the wavelength of light is____________________ 
Answer: Small

Question 3. If μ_a,  μ_b and μ_c are the absolute refractive indices of three media then _aμ_b × _bμ_c  _________________________
Answer: _aμ_c

Question 4. When a man stands inside a shallow pond the depth of water at that place appears___________________places appear comparatively___________________ and the depth of other 
Answer: Maximum, Less

Question 5. Critical angle of which pair of medium is lesser___________air and water or air and diamond
Answer: Air and Diamond

Question 6. We see the sun ___________________ a few minutes ___________________ sunset or sunrise
Answer: After, Before

Question 7. If a ray of light moves from a ___________________ medium to a ___________________, total internal reflection does not take place
Answer: Rarer, Denser

 Optics

Refraction Of Light Assertion Reason Type

Direction: These questions have statement 1 and statement 2 of the four characters given below, choose t=the one that best describes the two statements

1. Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true, and statement 2 is not a correct explanation for statement 1
3. Statement I is true, and statement 2 is false
4. Statement 1 is false statement 2 is true

Question 1. 

Statement 1: The Greater the refractive index of a medium or denser the medium, the lesser the velocity of light In that medium

Statement 2: Refractive index is inversely proportional to velocity.

Answer: 1.  Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: The critical angle of the light passing from glass to is minimal for violet colour

Statement 2: The wavelength of violet light is greater than the light of other colours

Answer: 3. Statement I is true, and statement 2 is false

Question 3.

Statement 1: The twinkling of the star is due to the reflection of light

Statement 2: The velocity of light changes while going from one medium to the other.

Answer: 4. Statement 1 is false statement 2 is true

Question 4.

Statement 1: The relative refractive index of a medium can be less than unity.

Statement 2: The angle of incidence is equal to the angle of refraction

Answer: 3. Statement I is true, and statement 2 is false

Question 5.

Starement 1: When a ra> of light enters glass from atr. Its frequency changes

Sutrmem 2: The velocity of light in glass is less than that in oil.

Answer: 4. Statement 1 is false statement 2 is true

Question 6.

Statement 1: The refractive index of a medium Is Inversely proportional to temperature

Statement 2: Refractive index U is directly proportional to the density of the medium.

Answer: 2. Statement 1 is true, statement 2 is true, and statement 2 is not a correct explanation for statement 1

Question 7. 

Statement 1: The velocity of light rays* of different colours is the same. But the velocity of the light rays is different for anv another medium.

Statement 2: If v = velocity of light in the respective

Answer: 1.  Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.

Question 8.

Statement 1: The linages formed the total lniern.il reflections are much brighter than those formed by lenses.

Statement 2: There Is no loss of Intensity during total Internal reflection

Answer: 1.  Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation for statement 1.

Optics

Refraction Of Light Match The Columns

Question 1. 

Answer:  1- A, C, 2-A, C, 3- B, 4- 4, 5- A, C

Question 2.

Answer: 1-C, D, 2- A, 3- B, E

Question 3.

Answer: 1- D, 2- A, 3- B, 4. C

μ₁, μ₂ and μ₃ are the refractive indices of the first, second and third medium respectively

Question 4.

Answer: 1- B, 2-1, 3 -C

Digital Electronics & Logic Gates Short Questions And Answers

Question 1. Convert the number (120)3 into decimal.
Answer:

(120)₂ = (1 × 3²) + (2 × 3¹) + (0 × 3⁰)

= 9 + 6 + 0 = (15)₁₀

Question 2. Assume that the first and the second digits of any binary system are 7 and 6, respectively. Convert (76)₁₀ into this binary system.
Answer:

(76)₁₀ = (1001100)₂

Here, we have to take 0 → 7 and 1→ 6.

Therefore, according to the given binary process,

(76)₁₀ = (6776677)₂

Question 3. Express 39 as a binary number
Answer:

∴ (39)₁₀ = (10011)₂

Question 4. Why is it called a universal gate?
Answer:

NOR gate can be used to obtain all the possible gates by using it as a basic building. This is why it is called the universal gate. It may be used to realize the basic logic functions OR, AND, and NOT

Question 5. What is the decimal equivalent of the binary number 100117
Answer:

(1 × 2⁴) + (0 × 2³)  (1 × 2²) (1 × 2¹) +(1 × 2⁰)

= (16 + 0 + 0 + 2+ 1)

= (19)₁₀

Question 6. If two Inputs of n NAND gate arc are joined, what type of gate is formed
Answer:

If the joining of two inputs implies that the inputs are short-circuited, then the two inputs are generally the same. If the input is A, then the output of the NAND gate

Y = \(\overline{A \cdot A}=\bar{A}\)

Hence, in this case, a NOT gate Is formed

Question 7. Write down the value of \((\bar{X}+X) \text { and }(X \cdot \bar{X})\) in Boolean algebra
Answer:

⇒ \(\bar{X}+X\)= 1

⇒ \(X \cdot \bar{X}\) = 0

Question 8.  The Input waveforms A and B to a logic gate. 

Answer:

Question 9.  The figure shows the Input waveforms A and B for the AND gate. 

Output waveform will ho as follows:

Reflection Of Light Light Introduction

Any luminous body is a source of light. The sources of light ‘ may be of two kinds

1. Self-luminous source and
2. Nonluminous source.

The sun, the stars, electric bulbs, burning candles, etc. are lumi¬ nous sources. Non-luminous sources themselves become visible when light from a luminous body falls on them.

The moon and the planets are non-luminous bodies. These are visible as light from the sun falls on them and is reflected.

Reflection of light:

When a ray of light passing through a medium is incident on the interface with another medium then, a portion of light returns to the first medium. This phenomenon Is called the Reflection of light

 Reflection, absorption, and refraction of light:

Light travels in a straight line in a homogeneous medium.

When it is traveling in one homogeneous medium and meets the surface of another homogeneous medium.

Read and Learn More Class 12 Physics Notes

The following effects may occur:

1. A portion of the light falling on the surface of separation returns to the first medium. This phenomenon is called the reflection of light.
2. A part of the incident light is absorbed by the second medium.
3.  If the second medium is transparent or translucent then a portion of the incident light penetrates into the second medium and undergoes a change of direction at the surface separating the two media and continues to travel along a straight line.
4. This phenomenon is called refraction of light

The surface from which the reflection of light takes place is called the reflector.

The amount of reflected light depends on the following two factors:

1. Direction of Incident light:  The more obliquely the Inol dent light falls on the reflector, the more the amount of reflected light.
2. Nature of the first and second medium: It Is found from the experiment that if light Is Incident from air to glass nor¬ mally, about 4% of incident light Is reflected, hot If light from air is incident on u plane mirror normally, about W0% of incident light is reflected.

Again if light Is incident from air to glass the amount of light reflected Is more than the amount of light reflected when light Is Incident from air to water. On the other hand, If the reflector is black, most part of the incident light is absorbed by the reflector. As a result, there is negligible reflection from a black body or surface

Reflection Of Light Some Definitions

In M_1, M₂ is a reflector. A ray AO is incident on the reflector at point 0 and is reflected along OB. AO is the incident ray and OB is the reflected ray. 0 is the point of incidence.

The angle of incidence: The angle that the incident ray makes with the normal to the reflector at the point of incidence is called the angle of incidence. ON is the normal drawn on M₁ M₂ at O . Hence, ∠AON is the angle of incidence.

The angle of reflection:

The angle that the reflected ray makes with the normal to the reflector at the point of incidence is called the angle of reflection. ∠BON is the angle of reflection.

Reflection Of Light – Laws Of Reflection

Reflection of light obeys the following two laws:

1. The Incident ray the reflected ray and the normal to the reflecting surface at the point of incidence, are all He on the same plane.
2. The angle of incidence is equal to the angle of reflection.
   ∠AON = ∠BON.

Normal Incidence:

The ray of light AO is incident normally on M₁ M₂ So the angle of incidence is zero. According to the law of reflection, the angle of reflection is also zero. Thus die ray retraces its path into the first medium.

 Reflection Of Light – Types Of Reflection

Reflection can be of two types depending on the nature of the surface of the reflector. O regular reflection and 0 diffused reflection

1. Regular Reflection:

If a parallel beam of rays is incident on a smooth plane reflector, then it is reflected wholly as a parallel beam This type of reflection is called regular reflection. Such reflectors are plane mirrors, upper surfaces of undisturbed water, polished metal surfaces, etc.

In the case of a smooth plane surface, the normals drawn at the points of incidence of the beam of rays are parallel to each other. So in the case of regular reflection, if the incident rays are parallel, the reflected rays are also parallel to each other.

The image of an object is formed due to regular reflection [vide section So, we see our image In a plane mirror

It will be interesting to note that only the part of the reflector that reflects parallel rays into the observer’s eyes appears brighter to the observer than the other part of the reflector.

2. Diffuse Reflection:

A parallel beam of rays after reflection from a rough surface does the Reflection of light obey the following two laws. not remain parallel. Such type of reflection of light is called diffused reflection, of reflection, reflection.

Inoccurs this type on the surface of every visible object, each ray of the incident parallel beam is reflected in its own way without being parallel to one another. This is because

The normals at the points of incidence are not parallel to each other. Hence no image is obtained.

However, it is because of diffuse reflection the objects in our surroundings are visible. When a beam of rays falls on the rough sur¬ faces of those objects, it is scattered in all directions.

So, whatever may be the position of the observer, quite a few rays will invariably enter the eyes. As a result, the observer sees the object more or less destined. In this case the reflector looks almost equally bright from all directions but no image of a source is seen.

Reflection from any plane, curved or rough surface, follows the two laws of reflection.

3. Comparison between Regular and Diffuse Reflections

Reflection Of Light – Deviation Of A Ray Due To Reflection

When a ray of light changes its original course due to reflection. the angle between the original and the final directions of the ray is a measure of the magnitude of deviation of the ray.

It is obvious from the figure that in the absence of the reflector M₁M₂, the M ray AO would have traveled straight in the direction AOC, but has instead been deviated due to reflection. The magnitude of this deviation,

δ = ∠BOC = 180° – ∠AOB = 180°- 2i

Reflection Of Light – Some Phenomena Of Reflection

When light falls on a black body, practically no reflection takes place. The black body absorbs almost the whole incident light. Hence, optical instruments like cameras, tele¬ scopes, etc.

Are painted black on the insides to avoid unwanted reflection. On the other hand, white objects do not absorb any light but rather reflect it. Hence to stop the absorption of light and to increase the brightness, white surface air is used.

This is the reason why white screens are used for projection In the cinema.

Twilight: 

Twilight Is the time between dawn and surface and the time between sunset and dusk. The sun itself is not actually visible from this ground level because It is below the horizon. However, the suspended dust particles in all upper atmospheres still receive direct fuse reflection, this light spreads In all directions and pat daily illuminates the ground daily.

When light falls on a glass slab only a negligible portion Is reflected: most of the rays pass through It. Hence the glass slab Is coated with aluminium1, to make a mirror. As the coating is opaque the Incident rays are almost completely reflected from the silvered surface, with only a small portion of the incident is reflected from the front surface.

Reflection Of Light Image

 Image Definition:

When rays of light diverging from a point source after reflection or refraction converge to or appear to diverge from a second point, the second point Is called the Image of the first point.

When light rays from an see the object at the place whore it Is actually situated. Hut If the rays come after reflection or refraction we see the object elsewhere. What we see In the new position Is actually Its Image

There are two kinds of images:

1. Real linage and
2. Virtual image.

1. Real image:

When rays of light diverging from a point source after reflection or refraction converge to a second point, the second point is called the real Image of the first point.

A real image can be formed on a screen, as evident from, where a convex lens has formed a real image A’ of a point source A.

Some examples of real image:

1. The image formed on the cinema screen
2. The image formed by a camera, etc

2. Virtual image:

When rays of light diverging from a point source after reflection or refraction appear to diverge from a second point, the second point is called the virtual image of the first point.

A virtual image cannot be formed on a screen as illustrated, where a plane mirror has formed a virtual image A’ of the point source A

The image of a tree standing by the side of a pond, on the surface of water is a virtual image. A mirage is also a virtual image.

Differences between real image and virtual image:

Image in a Plane Mirror:

Imags of a point object: Let A be a point source A ray of light AO is incident at O normally and retraces its path along OA.

Another ray AC follows the path CD after reflection. The reflected rays when produced backwards, meet at A’. It appears that the two reflected rays are coming from A’. So, A’ is the virtual image of A. The line AOA’ joins the object and the image is normal to the mirror.

CN is normal at the point of incidence. Since OA and CN
are parallel,

∠OAC = ∠ACN = i (Alternate angles)

and ∠OA’C = ∠NCD = r (Corresponding angles)

But ∠ACN = ∠NCD (i = r)

∠OAC = ∠OA’C

Hence the two right-angled triangles AOC and A’OC are congruent.

∴ OA = OA’

Thus, object distance from the mirror = image distance from the mirror

Hence, the image formed by a plane mirror lies on the perpendicular from the object to the mirror as far behind the mirror as the object is in front of it.

Image of an extended object:

Let AB be an extended object in front of a plane mirror M₁M₂. Every point of the extended object may be regarded as a source of light. The complete image of the object will be obtained by locating the position of the images of all the point sources.

Drawing of the image of an extended object:

From the topmost point A of the extended object perpendicular A01 is drawn and it is extended up to A’ in such a way that A in the previous case, A’ is the image of A. The topmost point of AB.

Similarly, B’ is the image of the lowermost point of AB, For every intermediate point of AJB, a corresponding image will be formed between A’ and B’. So, A’B’ is the image of AB. Obviously AB and A’B’ will be of the same size.

So, a plane mirror forms a virtual image of the same size of an extended object.

It is also obvious from the figure that the eye can catch the image within the portion PS of the mirror, as long as the relative positions of the eye and the mirror are not changed

Lateral insertion Definition:

An image of an object formed in a plane mirror is inverted sideways. This effect of plane mirrors is called lateral inversion.

The letter P held in front of a plane mirror will be seen as laterally inverted. The lateral turning is due to the fact, that every point image is at the same distance behind the plane mirror B as the point object is in front of it,

Being the size of the image equal to the size of the object. If the mirror is held vertically, it does not. invert the image which means turning an image upside down. Conversely, if a point source is placed at the point / its virtual image is formed at the point O.

This is due to the principle of reversibility of light rays. Only the image is laterally inverted. If we move our’ right hand facing a mirror, we see the image moving its left hand.

The perpendicular distance of every point of the object from the plane mirror = the perpendicular distance of every image point from the mirror.

i.e., AO₁ =  A’O₁; BO₂ = B’O₂.

So, the image is laterally inverted.

The images, due to bodies having symmetrical sides such as a sphere or letters like A, H, M, I, O, T, etc. are not affected by lateral inversion, but the images with non-symmetrical bodies like mug or letters like P, B, C, etc. are affected.

The characteristics of the image formed by a plane mirror:

1. With respect to the mirror, object distance = image distance.
2. The straight line joining the object and its image is perpendicular to the mirror.
3. The image is virtual.
4.  The image is formed behind the mirror and is the same size as that of the object.
5. The image is laterally inverted.

Image in a Plane Mirror due to a Conver- gent Beam:

Suppose a converging beam of light is incident on a plane mirror M₁M₂. In the absence of the mirror, the rays would converge. Due to reflection, the rays PQ and RS meet at point I instead of O. It can be proved that the straight line joining I and O is perpendicular to the mirror and IA = OA. Point O is called the virtual object and point I is its real image. Thus a plane can form a real image of a virtual object

Conversely, if a point source is placed at point I its virtual image is formed at point O. This is due to the principle of reversibility of light rays.

For a given incident ray if the mirror is rotated through an angle, the reflected ray turns through an angle of 2θ.

The minimum length of the plane mirror required to have the full-length image of a person standing in front of it is equal to half the height h of the person i.e. \(\frac{h}{2}\)

If an object moves towards (or away from) a plane mirror with a speed v, the image of the object will move towards (or away from) the mirror with a speed 2v.

4. If two plane mirrors facing each other are inclined at an angle with each other, the number of images formed due to multiple reflections, is given by n = 360°-1. If (36-1) is not an integer, the next integer will indicate the number of images.
360°

5. Due to reflection, the frequency, wavelength and speed of light are not changed.

6. The intensity of light after reflection decreases.

7. For reflection of light from a denser medium a phase change of occurs.

Reflection Of Light – Curved Reflecting Surface

Mirrors used in torches, headlights, or viewfinders of vehicles are curved mirrors. Curved mirrors may be spherical, cylindrical, or parabolic. In this chapter, however, we shall restrict our discussion only to spherical mirrors.

The laws of regular reflection are equally applicable to curved surfaces as well. But in this case, the position of the image and its size differ widely, as we shall see later.

Reflection Of Light Spherical Mirror

A spherical mirror is a part of a hollow sphere or a spherical surface.

There are two types of spherical mirrors:

1. Concave mirror and
2. Convex mirror.

When the inner surface of a spherical mirror acts as a reflector, it is a concave mirror.  When the outer surface of a spherical mirror acts as a reflector, it is a convex mirror.

Some Related Terms:

Pole:

The center of the spherical reflecting surface is called the pole of the mirror. In  O is the pole.

Centre of curvature:

The center of the sphere of which the spherical mirror is a part is called the center of curvature of the mirror. C is the center of curvature of the mirror MOM’. Obviously, the center of curvature of the concave mirror is in front of the reflecting surface while in the case of the convex mirror it is behind the reflecting surface.

The radius of curvature:

It is the radius of that sphere of which the mirror is a part.  OC is the radius of curvature.

Principal axis:

The line passing through the center of curvature and the pole of the mirror is called the principal axis of the mirror.  XX’ is the principal axis.

Aperture:

The line joining the two extreme points on the periphery of a spherical mirror is called the aperture of the mirror. The angle subtended at the center of curvature by the line is called the angular aperture of the mirror. In the line MM’ is the aperture of the spherical mirror and ZMCM’ is its angular aperture.

The discussion in this chapter will be confined to spherical mirrors of small apertures not exceeding 10°, although, for the sake of clarity, illustrative diagrams will indicate larger apertures.

Paraxial and non-paraxial or marginal rays:

1. Paraxial rays:

Rays that are incident very close to the pole and form a very small angle with the principal axis of a spherical mirror are called paraxial rays.

2. Non-paraxial or marginal rays:

Rays that are incident very far away from the pole or near the margin of a spherical mirror and form a comparatively large angle with the principal axis are called non-paraxial or marginal rays.

All rays incident on a spherical mirror whose aperture is negligibly small compared to its radius of curvature, are considered to be paraxial rays. For further discussions, we will assume all spherical mirrors to be of a small aperture and a comparatively large radius of curvature.

Principal focus:

If rays of light parallel to the principal axis are incident on a spherical mirror, the rays after reflection from the mirror converge to a point on the principal axis in case of a concave mirror and appear to diverge from a point on the principal axis behind the mirror in the case of a convex mirror, this point is called the principal focus or simply focus of the mirror.

F is the principal focus of M the concave mirror and convex mirror respectively.  So, the focus of the concave mirror is real and that of the convex mirror is virtual.

It can alternatively be defined as the point on the principal axis of a spherical mirror at which the image of an object placed at infinity is formed.

According to the principle of reversibility of light rays, it can be said:

The rays diverging from the principal focus of a concave mirror proceed parallel to the principal axis after reflection from the mirror

The rays appearing to converge to the principal focus of a convex mirror proceed parallel to the principal axis after reflection from the mirror.

Focal length:

The distance between the principal focus and the pole of the mirror is called the focal length of the mirror. OF is the focal length. The focal length of a spherical mirror does not depend on the color of the incident light.

Focal plane and secondary focus:

The focal plane of a spherical mirror is the imaginary plane passing through the principal focus at right angles to the principal axis of the mirror.

If a parallel beam of rays is incident on a spherical mirror such that it is inclined to the principal axis then, after reflection the reflected rays converge to the point. on the focal plane in the case of a concave mirror and appear to diverge from the point. On the focal plane in case of a convex mirror. The point F₁ is called the secondary focus of the spherical

The principal focus is a fixed point on the principal axis. However, the secondary focus is not a fixed point. If the angle of inclination of the parallel rays with the principal axis is changed, the position of the secondary focus also changes. But a secondary focus always lies on the focal plane.

Relation between Focal Length and Radius of Curvature

1. In the case of the concave mirror:

Let MOM’ be a concave mirror of a small aperture. C, F, and O are the center of curvature, focus, and pole of the mirror respectively. Ray PQ is parallel to the principal axis, hence passing through F after reflection, CQ being the radius of curvature is perpendicular to the mirror at Q.

∴ ∠PQC = ∠FQC

∴  ∠PQC = ∠QCF [alternate angles]

∠FQC = ∠QCF i.e., Δ QCF is an isosceles triangle.

Hence, FQ = FC

Since the aperture of the mirror is very small, Q and O are very close to each other. So, FQ = FO.

∴ FO = FC or, FO = \(\frac{1}{2}\) OC

i.e f= \(\frac{r}{c}\), where f is focal length and r the radius of curvature.

2. In case of convex mirror:

Let MOM₁, be a convex mirror of a small aperture. C, F, and O are the center of curvature, focus, and pole of the mirror respectively. OC is the radius of curvature.

A ray PQ parallel to the principal axis is incident at Q of the mirror. After reflection, the ray QR appears to come from F. The points C and Q are joined and is extended to N. Since the CQ = CO radius of curvature of the mirror, QN is normal at incidence point Q on the mirror.

If a parallel beam of rays is incident on a spherical mirror such that it is inclined to the principal axis then, after reflection, the reflected rays converge to point F, on the focal plane in case of a concave mirror and appear to diverge from the point F,  on the focal plane in case of a convex mir- But r. The point F is called the secondary focus of the spherical

∠PQN = ∠RQN

∠RQN = ∠CQF [vertically opposite]

∴ ∠PQN = ∠CQF

∴  Since PQ and OC are parallel.

∴ ∠PQN = ∠FCQ [corresponding angles]

∴ ∠FCQ = ∠CQF

Hence , FQ = FC

Since the aperture of the mirror is small, Q and O are very 1. Object at infinity: If an object is at infinity, the rays are close to each other.

So, FQ = FO.

∴ FO = FC

Or, FO = \(\frac{1}{2}\) OC

I.e f = \(\frac{r}{2}\)

Hence the focal length of a spherical mirror is small Thus, the image of an object situ- aperture is equal to half of its radius of curvature.

Reflecting power:

Reflecting the power of a spherical mirror, D = \(\frac{1}{f}\) =\(\frac{2}{r}\) = As both focal length and radius of curvature of a plane mirror are infinite, so power of a plane mirror is zero.

Reflection Of Light – Image Formation Of Extended Object By Spherical Mirror

Ray tracing method:

The position, nature, and size of the image of an extended object, formed by a spherical mirror can be determined geometrically. Any extended object can be considered as the sum of point objects and the images of the point objects constitute the image of the extended object.

Any two of the following rays intersecting at a point will indicate the portion of the image:

1. A ray parallel to the principal axis: After reflection passes through the focus, or appears to diverge from the focus.
2. A ray passes through the focus: After reflection emerges In the case of a concave mirror if the object is placed at the center parallel to the principal axis.
3. A ray passing through the center of curvature: After reflection retraces its path in the opposite direction.

Because the ray passing through the center of curvature incidence on the mirror is the normal incidence in this case.

Image Formation by Convex Mirror:

Objectivity at infinity:

If an object is at infinity, the rays coming from it may be assumed to be parallel and hence after reflection will meet at the principal focus F. If the rays are A oblique after reflection they will meet at a secondary focus F’

Thus, the image of an object situated at infinity is formed on the focal plane. The image is real, inverted, and very much diminished in size

2. Object placed between focus and center of curvature:

An object PQ is placed beyond C on the principal axis of the concave mirror MOM’. A ray PA starting Anotherray PB passing through C is reflected back along BP. The two reflected rays AF and BP meet at p.

So, the real image of P is formed at p. Normal pq is drawn on the principal axis. PQ is the image of PQ. The image is situated between F and C. The image is real, inverted, and diminished relative to the object.

3. Object at the center of curvature :

An object PQ stands at C A ray PA parallel to the principal axis is reflected through F along AF. Another ray PB through F is reflected along BD parallel to the principal axis. The two reflected rays AF and BD meet at p. So, the real image. of P is formed at p. pq is D drawn normally on the principal axis. pq is the image of PQ.

The image is real, inverted, and magnified, and of the same size as the object, and formed at the center of curvature itself.

In case of a concave mirror if the object is placed at the center of curvature the image is also formed at the center of curvature. Hence only in this case, the intervening distance between the object and image is minimum and is equal to zero

4. Object placed between focus and centre of curvature

An object PQ stands between F and C. A ray PA parallel to the principal axis is reflected through F along AF. Another ray PB incident normally at B goes back along BC. These two reflected rays meet at p. So, the image of P is formed at p. pq is drawn normally on the principal axis. pq is the image of PQ.

5. Object at focus:

An object PQ is placed at the focus F . A ray PA parallel to the principal axis is reflected through F along AF. Another ray PB incident normally at B goes back along BC. These two reflected rays being parallel to each other meet at infinity producing the image of P.

6. Object placed between focus and pole:

An object PQ is placed between the pole and focus. A ray PA parallel to the principal axis is reflected through F along AF. Another ray PB incident normally at B goes back along BC. These two reflected raj’s which are divergent would not meet anywhere. But when they are produced backwards they meet at p. Thus they appear to diverge from p. pq is drawn normally on the principal axis. Thus pq is the image of PQ

The Image is virtual, erect, magnified, and situated behind the mirror

 Image Formation by Convex Mirror:

Consider an object PQ in front of a convex mirror MOM. A ray. PA parallel to the principal axis goes back along AD. Ray AD appears to come from the focus. Another ray PB incident normally at B is reflected along BP. Reflected rays AD and BP, produced backwards appear to come from p. So. p is the virtual image of P. pq is drawn normally on the principal axis. Thus pq is the image of PQ.

The image is virtual, erect, diminished in size, and situated behind the mirror.

For any portion of the object in front of a convex mirror, the image is real, inverted, and magnified infinitely, and situation the image will always be formed behind the mirror. This image is rated at infinity.

Comparative Study of Real and Virtual Images in the Case of Spherical Mirror

1. Characteristics of real image:

• It is formed on the same side of the mirror as the object. It is always inverted.
• Real image is not formed in a convex mirror

2. Characteristics of virtual image:

• It is always formed on the opposite side of the mirror as the object.
• It is always erect.
• The size of the virtual image becomes larger than the object or equal to it in the case of a concave mirror. Whereas in the case of a convex mirror, it is smaller than the object or equal to it.

Sign Convention for Spherical Mirror

1.  Cartesian sign convention:

• All distances are to be measured from the pole of the spherical mirror.
• All distances measured in a direction opposite to that of the incident rays are to be taken as negative and all distances measured in the same direction as that of the incident rays are to be taken as positive.
• If the principal axis of the mirror is taken as the x-axis, the upward distance along the positive y-axis is taken as positive while the downward distance along the negative y-axis is taken as negative.

See the concave and convex mirrors given in below to understand the above rules.

The nature sign of object distance(u), image distance(v) focal lengh(f), radius of curvature (r), and height of the image in case of the image formation of a real object by a spherical mirror is given in the following table.

Reflection Of Light – Relation Among Object Distance, Image Distance, And Focal Length

1. In case of concave mirror:

Let O, F, C, and OQ be the pole, focus, center of curvature, and principal axis of a concave mirror MOM’ respectively. PQ is an object placed perpendicularly on the principal axis in front of the mirror. A ray PA, parallel to the principal axis, after reflection, does not form a real image that passes through F.

Another ray PA’, passing through C, after reflection goes back following the same path. These two reflected rays cut each other at p. Hence p is the real image of P. pq is the image of PQ. AB is drawn perpendicular to the principal axis.

Triangles PCQ and pCq are similar

∴ \(\frac{P Q}{p q}=\frac{C Q}{C q}\)

Again,  Δ ABF and Δ  pqF are similar

∴ \(\frac{A B}{p q}=\frac{B F}{F q}\)

Or,  \(\frac{P Q}{p q}=\frac{B F}{F q}\)

[ AB = PQ]

From (1) and (2) we get \(\frac{C Q}{C q}=\frac{B F}{F q}\)

Since the mirror is of small aperture It can be assumed BF ≈ OF

∴ \(\frac{C Q}{C q}=\frac{O F}{F q}\)

Object distance, OQ = -u;

Image distance

Oq = -v; focal

Length, OF = -f

Radius of curvature, OC = -r = -2f

∴  CQ = OQ- OC  = -u+ 2f

Cq = OC- Oq = -2f+v

Fq = Oq- OF = -v+f

From (3) we get, \(\frac{-u+2 f}{-2 f+v}=\frac{-f}{-v+f}\)

or, uv- uf- 2fv² + 2f² = fv

or, uv = uf+ vf

Dividing both sides by uvf we get

⇒ \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

Or, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{r}\)

2. In the case of the convex mirror:

Let O, F, C, and CQ be the pole, focus, center of curvature, and principal axis of a convex mirror MOM’ respectively [Fig. 1.28]. PQ is an object placed perpendicularly on the principal axis in front of the mirror. A ray PA parallel to the principal axis and another ray PD proceed- ing to the centre of curvature, form a virtual image p of P after reflection from the mirror. pq is the image of PQ. AB is drawn perpendicular on the principal axis.

Triangles PCQ and pCq are similar

∴ \(\frac{P Q}{p q}=\frac{C Q}{C q}\)

Again ΔABF and ΔpqF are similar

∴ \(\frac{A B}{p q}=\frac{B F}{F q}\)

Or  \(\frac{P Q}{p q}=\frac{B F}{F q}\)

From (5) and (6) we get,

∴ \(\frac{C Q}{C q}=\frac{B F}{F q}\)

Since the mirror is of small aperture It can be assumed BF ≈ OF

∴ \(\frac{C Q}{C q}=\frac{O F}{F q}\)

Object distance, OQ = -u;

Image distance, Oq = +v; focal

Length, OF = +f

Radius of curvature, OC = +r = +2f

∴  CQ = OQ +OC  = -u+ 2f

Cq = OC- Oq = +2f+v

Fq = OF- Oq = +f – v

From (7) we get,

⇒ \(\frac{-u+2 f}{+2 f+v}=\frac{+f}{+f-v}\)

or, 2f²- vf= – uf+2f²+uv-2vf

or, uv = uf+ vf

Dividing both sides by uvf we get

⇒ \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

Or, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{r}\)

It may be noticed that using the same sign convention, the relation between the various distances is the same for both concave mirror and convex mirror. This relation viz., += = 2 is called the mirror equation or spherical mirror equation.

In case of a concave mirror if u and v are the object distance and the image distance of a real object and its real image respectively, then the u-v graph is a rectangular hyperbola and the graph of their reciprocals ( \(\frac{1}{u}\) – \(\frac{1}{v}\) graph) is a straight line.

Equation of plane mirror from the equation of spherical mirror:

The mirror equation is

⇒ \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)= \(\frac{2}{r}\)

For a plane mirror, r→ ∞

∴ \(\frac{1}{v}\) +\(\frac{1}{u}\) = 0

Or, v= -u

This proves that in the case of a plane mirror, the image is as far behind the mirror as the object is in front of it (:: v negative).

Effect of medium on the focal length and image distance for spherical mirror:

The focal length of a spherical mirror is fixed, i.e., independent of the surrounding media. This is because the law of reflection is invariant even if the medium changes.

For a spherical mirror, if object distance u, image distance v, and focal length f, then the relation between them is  \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)= \(\frac{2}{r}\).

For a fixed object distance u, the image distance v is fixed, because f is fixed and independent of medium. So, if the position of the object and mirror are kept fixed and the surrounding medium is changed no change position of the image occur.

Conjugate Foci or Conjugate Points We have the mirror equation:

We have the mirror equation

⇒ \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)= \(\frac{2}{r}\).

If u and v are interchanged, the equation remains the same. This implies that if the object is placed at the position of the image, the image will be formed at the position of the object.

These two points are called conjugate foci and the above equation is alternatively called the conjugate foci relation.

In the case of the virtual image, conjugate foci are situated on two opposite sides of the mirror, and in the case of a real image, conjugate foci are situated on the same side of the mirror.

Newton’s Equation:

Relation among u, v, f with reference to the pole is

⇒ \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)= \(\frac{2}{r}\).

Or, \(\frac{u+v}{u v}=\frac{1}{f}\)

Or, uv-uf-vf+ = 0

Or, uv-uf-vf+f² = 0

Or, u(v-f)-f(v-f) = f²

Or, (u-f)(v-f) = ƒ²

Now, if the object distance and the image distance are measured from the focus, and taken equal to x and y respectively, then we can write,

u-f = x and v-f = y

So, from equation (1) we get,

xy = f²

This equation is known as Newton’s equation. Since ƒ is constant, the graph of x versus y will be a rectangular hyper-bola

Since, f² is a positive quantity, x and y must have the same sign, i.e., the object and the image must be on the same side of the focus.

Magnification of an Image Formed by Spherical Mirrors

Linear or lateral magnification Definition:

The ratio of the height of the image to the height of the object measured in planes that are perpendicular to the principal axis is called the linear or lateral magnification of Linear the image

Linear or lateral magnification is denoted by m

The ratio of the height of the image to the height changed, with no change in the position of the image.

∴ m = \(\frac{\text { height of the image }}{\text { height of the object }}=\frac{I}{O}\) …………………….(1)

1. In the case of a real image formed by a concave mirror:

The ray diagram for the formation of a real image by a concave mirror is shown. Here object distance, RQ-u; image distance, Rq=v, height of the object, PQ = O and height of the image, pq = -I. The Δ PQR and ΔpqR are similar.

∴ \(\frac{p q}{P Q}=\frac{q R}{Q R}\)

Or, \(\frac{-I}{O}=\frac{-v}{-u}\)

Or, \(\frac{I}{O}=-\frac{v}{u}\) ………………(2)

2. In case of a virtual image formed by a concave mirror:

The ray diagram for the formation of a virtual image by a concave mirror is shown.

Here object distance, PQ = -u; image distance, Rq = v, height of the object, PQ =  O and height of the image, pq= I. From ΔPQR and ΔpqR are similar.

∴ \(\frac{p q}{P Q}=\frac{q R}{Q R}\)

Or, \(\frac{I}{O}=\frac{v}{-u}\)

Or, \(\frac{I}{O}=-\frac{v}{u}\) ………………(3)

3. In the case of a virtual image formed by a convex mirror:

In Again the equation can be written as, the ray diagram for the formation of a virtual image by a convex mirror is shown. Here object distance, RQ-u; image distance Rq= y; height of the object, since, PQ = O and height of the image pq = I.

According to the PQR and pqR are similar

∴ \(\frac{p q}{P Q}=\frac{q R}{Q R}\)

Or, \(\frac{I}{O}=\frac{v}{-u}\)

Or, \(\frac{I}{O}=-\frac{v}{u}\) ………………(4)

From equations (2), (3), and (4) it follows that the magnification produced by both kinds of mirror is given by

m= \(\frac{I}{O}=-\frac{v}{u}\) ………………(5)

Some useful hints:

1. The relation m.= – \(\frac{v}{u}\)– is applicable both for concave and convex mirrors.
2. In solving numerical problems, values of u, v, and f should be put with appropriate signs in the mirror equation. No sign for the unknown quantity should be used.
3. Using the appropriate sign of u and v, if
   • m becomes negative, the image will be inverted
   • m becomes positive, the image will be erect
4.  If |m|>1; the size of the image> size of the object
   • If |m|<1; the size of the image< size of the object
   • If |m| = 1; the size of the image size of the object

Magnification in terms of focal length, object distance, and image distance:

From the mirror equation, we get,

⇒ \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)

Or, \(\frac{u}{v}\) + 1 = \(\frac{u}{v}\)

Or, \(\frac{u}{v}\) = \(\frac{u-f}{f}\)

Or, \(\frac{u}{v}\) = \(\frac{f}{u-f}\)

Since, m= – \(\frac{u}{v}\)

∴ m = ( \(\frac{f}{u-f}\) )

Or, ( \(\frac{f}{f-u}\) )

Again the equation can be written as,

1+ \(\frac{v}{u}\)  = \(\frac{v}{f}\)

Or, \(\frac{v}{u}\) = \(\frac{v-f}{f}\)

m= – \(\frac{v-f}{f}\) = \(\frac{f-v}{f}\)

Areal Magnification Definition

Areal magnification for spherical mirrors is the ratio between the image of an area of a plane, and the area of that plane placed perpendicular to the principal axis of the mirror.

Let us consider, that the length and breadth of a plane of a rectangular object are 1 and b respectively.

∴ Area of the plane, A = lb

If the linear magnification of the image of the object by spherical mirror be m, then

length of the image, l’ = m × 1

and breadth, b’ = m × b

∴ Area of the image, A’ = l’b’ = m²lb = m²A Therefore, areal magnification,

m’ = \(\frac{A^{\prime}}{A}\) = m²

Longitudinal or axial magnification of the image of an object kept along the principal axis Definition:

If any object is placed along the principal axis of a spherical mirror then the ratio of the image length and object length is the longitudinal or axial magnification of that image.

Let an object ADEB is placed in front of a spherical mirror MM’.

From the figure, the distance of farther point A of an object, OA = u₁ and that of the nearer point B of the object, OB = u₂

Now, the distance of the farther point of the image, OB’ =v₁ that of the nearer point, OA’=v₂

Longitudinal Magnification , m= \(\frac{v_1-v_2}{u_1-u_2}=\frac{\Delta v}{\Delta u}\)

Here, Δu and Δv are the lengths of the object and its image respectively, along the principal axis of the mirror. For very small magnitudes of Au and Av, these can be considered as du and du respectively.

So,  m” = \(\frac{d v}{d u}\)

Differentiating the equation \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

\(-\frac{1}{v^2} \frac{d v}{d u}-\frac{1}{u^2}\) = 0

Where, [f is constant]

Or, \(\frac{d v}{d u}=-\frac{v^2}{u^2}\)

∴ m” = \(\frac{d v}{d u}\) = – m²

∴  Longitudinal magnification = -(linear magnification)²

Note that, if object length and image length are very small dv then,

∴ m” = \(\frac{d v}{d u}\)

Im may be positive or negative but m” is always negative. This implies that irrespective of whether the object is virtual or real, the image is formed along the principal axis, with opposite alignment. This pianomeonon is called axial inversion.

Formation of Image of a Virtual Object: 

 

In a beam of converging rays is incident on the mirror. In the absence of the mirror, in either case, the

A converging beam of rays would meet at P behind the mirror. But the beam of rays meets at P’ after reflection. Here point P is the virtual object and P’ is the real image of point P. Obviously, object distance OP is positive.

Thus, a convex mirror can also form a real image, but only if the object is virtual. Now consider the case, where the virtual object distance OP is greater than the focal length OF of a convex mirror.

In this case, image P’ will be virtual. with respect to u, we get, In the case of the concave mirror, a real image always be formed for a virtual object and this image is situated between the pole and the focus of the mirror.

Reflection Of Light – Method Of Identifying Mirrors

If an object is placed in front of a plane mirror, a virtual, erect image of the same size as the object is formed. If an object is placed very close to a concave mirror a virtual, erect, and magnified image is formed. A convex mirror forms a virtual, erect image smaller than the object.

For identification, one can hold a pen or a finger very close to the mirror. If an erect image of the same size as the object is formed, then the mirror is a plane one. If an erect image larger than the object is formed, the mirror is concave. If it is smaller than the object the mirror is convex.

Reflection Of Light – Spherical Aberration And Its Remedy

Spherical mirror:

Spherical aberration: The mirror equation is applicable only for spherical mirrors of small aperture to increase the intensity of illumination of the reflected rays to get a brighter image, we often use spherical mirrors of large aperture.

If a beam of rays is incident parallel to the principal axis of a concave mirror of large aperture, not all the rays meet after reflection at a single point. Rather, the reflected rays meet at various points of the principal axis between F to F₁.

So the image becomes indistinct. The larger the aperture of the mirror, the more indistinct the image. This defect of the image is called spherical aberration.

Remedy for spherical aberration: 

If the shape of the mirror is changed from spherical to parab- poloidal, it is possible to get an image free from spherical aberration. Because, according to the geometrical properties of the parabola, all the rays parallel to the principal axis of a paraboloidal mirror meet at its focus after reflection from it.

Reflection Of Light  – Uses Of Spherical Mirrors

Concave mirror:

• Concave mirrors are often used as shaving glasses (mirrors) to see the magnified image of the face, its distance being less than the focal length of the mirrors.
• Small concave mirrors are used by doctors to focus a parallel beam of light on the affected parts like the eye, ear, throat etc. to examine them.
• A small electric lamp placed at the focus of a concave mirror produces a parallel beam of light. So concave mirrors are used as reflectors in torches and car headlights. For better results, paraboloidal mirrors can be used as car headlights. Concave mirrors are used in solar cookers.
• The Palomar Observatory in California has the best reflecting telescope which uses a concave mirror for studying distant stars.

Convex mirror:

Convex mirrors are used as rear-view mirrors in automobiles and other vehicles, designed to allow the driver to see through the rear windshield. This is because a convex mirror forms erect and diminished images of objects and give a wider field of view compared to that of a plane mirror.

Paraboloidal mirror:

According to the geometrical properties of the parabola, the rays parallel to the principal axis of a paraboloidal mirror meet at its focus after reflection from the mirror.

So, if a source of light is placed at the focus F of a paraboloidal mirror, the reflected rays proceed ahead parallel to the principal axis.  For this reason, paraboloidal mirrors are used in car headlights and searchlights.

Sign rules that have been followed here:

To solve the numerical problems, a few sign conventions have been followed here.

• The value of u, v, f, or r  is used with their proper sign-in in the mirror equation, viz \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{r}\)
• The focal length for the concave mirror is considered negative and that for the convex mirror is positive.
•  If the image distance is negative, it indicates that the image is formed on the same side as the object, and the image is real and inverted.
• If the image distance is positive, it indicates that the image is formed behind the mirror, and the image is virtual and erect.
• Though mis the standard formula for any type of spherical mirror, to find u or v from this equation, we only take the mod value of m.
• For the determination of the nature of the image, to find the value of m, the associated formula is used following the proper sign convention of u and v. If m becomes positive, the image will be erect, and m becomes negative, the image will be inverted.

Reflection Of Light Numerical Examples

Example 1.  An object is placed 60 cm away from a convex mirror. The size of the image is rd the size of the object. Determine the radius of curvature of the mirror.
Solution:

We have from the mirror equation,

f = \(\frac{u v}{u+v}\)

According to the question

m = \(\left|-\frac{v}{u}\right|=+\frac{1}{3}\)

v = \(\left|-\frac{u}{3}\right|=\left|\frac{60}{3}\right|\)

U = 60 cm

v= 20 cm

Now, substituting the values of u and v with their proper sign in equation (1) we get,

f = \(\frac{-60 \times(+20)}{-60+20}\)

= + 30 cm

The radius of curvature of the spherical mirror,

r = 2f = 2 × (+30) = +60 cm cm

Example 2. An object of size 5 cm is placed on the principal axis of a convex mirror at a distance of 10 cm from it. The focal length of the convex mirror is 20 cm. Determine the nature, position, and size of the image formed.
Solution:

u=-10 cm; f = +20 cm

The focal length of the convex mirror is positive

Mirror equation:

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

= \(\frac{1}{20}+\frac{1}{10}\)

= \(\frac{1+2}{20}\)

= \(\frac{3}{20}\)

Or, v=\(\frac{20}{3}\)

= 6.67 cm

From the positive sign, it can be inferred that the image is formed 6.67 cm behind the mirror. So the image is virtual.

Magnification, m = – \(\frac{v}{u}=-\frac{20 / 3}{-10}\)

= \(\frac{2}{3}\)

As magnification is positive, the image is erect.

The size of the image = \(\frac{2}{3}\)  × 5

= \(\frac{10}{3}\)

= 3.33 cm.

Example 3. The image of an object placed 50 cm in front of a concave mirror is formed 2 m behind the mirror. Deter- mine its principal focus and radius of curvature.
Solution:

Here, u=-50 cm; v = +2 m = +200 cm [since the image is formed behind the mirror, v is positive]

We have,  \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(f=\frac{u v}{u+v}\)

= \(\frac{-50 \times(+200)}{-50+(+200)}\)

= \(\frac{-200}{3}\)

= – 66. 67 cm

∴ r = 2f

= 2 × \(\frac{-200}{3}\)

= – \(\frac{-400}{3}\)

= – 133.3 cm

So, the focal length of the concave mirror = 66.67cm and its radius of curvature = 135.3 cm

Example 4.  An object of length 5 cm is placed perpendicularly on the principal axis at a distance of 75 cm from a concave mirror. If the radius of curvature of the mirror is 60 cm, calculate the image distance and its height.
Solution:

Here, u=-75 cm; r = -60 cm;

∴ f =  \(\frac{r}{2}\)

= \(\frac{-60}{2}\)

=-30 cm

We know,    \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-75}=\frac{1}{30}\)

Or, \(\frac{1}{v}\)

= – \(\frac{1}{30}+\frac{1}{75}\)

= – \(\frac{1}{50}\)

Or, v= -50 cm

So, the image is formed at a distance of 50 cm in front of the
mirror. The image is real.

Again , m= \(\frac{\text { height of the image }(I)}{\text { height of the object }(O)}=\frac{v}{u}\)

here we take only the mod value of m as we no need to know the natire of image thus formed

Or, \(\frac{I}{5}=\frac{50}{75}\)

Or, I = 5 × \(\frac{2}{3}\)

= 3.33 cm

So, the height of the images = 3.33cm.

Example 5. A beam of converging rays is incident on a convex mirror of focal length 30 cm. In the absence of the mirror, the converging rays would meet at a distance of 20 cm from the pole of the mirror. If the mirror is situated at the said position where will the converging rays meet? Solution: In the absence of the mirror the converging rays would meet at P. So, P is the virtual object in the case of the mirror.
Solution:

In the absence of the mirror, the converging rays would meet at P. So, P is the virtual object in case of the mirror

In the case of the virtual object, OP = u = +20 cm , f = +30 cm, image distance, v =?

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{20}=\frac{1}{30}\),

Or, \(\frac{1}{v}\) = \(-\frac{1}{20}+\frac{1}{30}\),

= \(\frac{-1}{60}\)

Or, v= -60 cm

As v is negative, the converging rays will meet at Q at a distance
of 60 cm in front of the mirror

Example 6. An image of size \(\frac{1}{n}\) times that of the object is formed in 1 P a convex mirror. If r is the radius of curvature of the ‘HIJL 1 mirror, calculate the object distance.
Solution:

Considering the mod value only, magnification,

m = \(\frac{1}{n}\)

= \(\frac{v}{u}\)

Or, v = \(\frac{u}{n}\)

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\)

[ Here object distance = -u]

Or, \(\frac{n}{u}-\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{(n-1)}{u}=\frac{1}{f}\)

= (n-1)f

Again, f= \(\frac{r}{2}\) , Then u= (n-1) \(\frac{r}{2}\)

∴ Object distance = (n-1)\(\frac{r}{2}\)

Example 7. An object of height 2.5 cm is placed perpendicularly on the principal axis of a concave mirror of focal length f at a distance off. What will be the nature of the image of the object and its height?
Solution:

Here, u = \(\frac{3}{4}\) f= Focal Length = -f

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}-\frac{4}{3 f}=-\frac{1}{f}\)

Or, \(\frac{1}{v}=-\frac{1}{f}+\frac{4}{3 f}\)

= \(+\frac{1}{3 f}\)

Or, v= +3f

The positive sign of v indicates that the image formed by the concave mirror is virtual in nature and is situated behind the mirror at a distance of 3f.

Magnification, m = – \(\frac{v}{u}\)

= \(-\frac{3 f}{-\frac{3}{4} f}\)

= 4

Again, m= \(=\frac{\text { height of the image }}{\text { height of the object }}\)

Or, 4 = \(\frac{\text { height of the image }}{2.5}\)

∴ Height of the image = 10 cm

Example 8.  The image of the flame of a candle due to a mirror is formed on a screen at a distance of 9 cm from the candle. The image is magnified 4 times. Determine the nature, position, and focal length of the mirror.
Solution: 

As a magnified image image is formed on a screen, the image is real.

Let u = – x cm and hence v = -(x+9) cm

Here, m = 4 or \(\frac{v}{u}\) = 4 [taking only the mod value of m]

Or, \(\frac{x+9}{x}\)  = 4

Or, 4x = x+9 or, x = 3

∴ u = -3 cm

So, the mirror is situated at a distance of 3 cm from the flame

Now, v= -(3+9) = -12 cm

We, know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\) .

Or, \(\frac{1}{-12}+\frac{1}{-3}=\frac{1}{f}\) .

Or, \(-\left(\frac{1+4}{12}\right)=\frac{1}{f}\) .

or, f = \(-\frac{12}{5}\)

= -2.4 cm

So, the focal length of the mirror is 2.4 cm. Its negative sign indicates the nature of the mirror as a concave one [Fig. 1.40].

Example 9. The focal length of a concave mirror is f. A point object is placed beyond the focal length at a distance from the focus. Prove that the image will be formed beyond the focal length at a distance  \(\frac{f}{x}\)from the focus and the magnification of the image will be (\(\frac{1}{x}\)
Solution:

Here, object distance, u = -(f+xf)

We, know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-(f+x f)}\)

Or, \(\frac{1}{v}=\frac{1}{f+x f}-\frac{1}{f}\)

= – \(\frac{1}{v}=\frac{1}{f}\left[\frac{1}{1+x}-1\right]\)

= \(\frac{-x}{f(1+x)}\)

∴ v =  \(\frac{f(1+x)}{-x}=-\left(f+\frac{f}{x}\right)\)

So, the image is formed beyond the focal length at a distance of be u. \(\frac{f}{x}\) from the focus

∴ Magnification , m= \(-\frac{v}{u}\)

=- \(\frac{f(1+x)}{x}{-f(1+x)}\)

=  \(\frac{1}{x}\)

⇒ \(-\frac{v}{u}=-\frac{-\frac{f(1+x)}{x}}{-f(1+x)}\)

Example 10. A thin glass plate is placed in between a convex mirror of a length 20 cm and a point source. The distance between the glass plate and the mirror is 5 cm.  The image formed by the reflected rays from the front face of the glass plate and that due to the reflected rays by the convex mirror coincides at the same point. What is the distance of the glass plate from the source? Draw the ray diagram

The ray diagram has been shown. The distance between the glass plate M and the point source

P = x cm (say). Light rays starting from P form the image at Q by reflection at the convex mirror. The glass plate also forms the image of P at Q.

∴ Distance of the image from the glass plate = x cm

∴ Distance of the image from the convex mirror = (x-5) cm glass plate also forms the image of P at Q.

∴ Distance of the object from the convex mirror=-(x+5) cm 

We know , \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{x-5}+\frac{1}{-(x+5)}=+\frac{1}{20}\)

Or, \(\frac{+x+5-x+5}{(x-5)(x+5)}=+\frac{1}{20}\)

Or, \(\frac{10}{x^2-25}=+\frac{1}{20}\)

∴ The distance of the glass plate from the source = 15 cm

Example 11. The sun subtends an angle 0.5° at the center of a concave mirror having a radius of curvature 1 m. What will be the diameter of the image of the sun formed by the mirror?
Solution:

MM’ is a concave mirror Let the diameter of the sun be D and the distance of the sun from the mirror be u,

∴ \(\frac{D}{u}=\frac{\pi}{360}\)

0.5 ° = \(\frac{1}{2} \times \frac{\pi}{180}=\frac{\pi}{360}\) radian

Or, \(\frac{360 D}{\pi}\)

The sun is at a large distance from the mirror. So, the rays coming from the sun are assumed to be parallel and hence its image will be formed at the focal plane.

∴ v= \(f\frac{r}{2}\)

= \(\frac{1}{2}\)

= 0.5 cm

Now magnification, m = \(-\frac{v}{u}=\frac{-0.5 \pi}{-360 D}\)

⇒ \(\frac{0.5 \pi}{360 D}\)  [since in this case both u and v are negative]

Again \(\frac{I}{O}=\frac{\text { diameter of the image of the sun }}{\text { diameter of the sun }(D)}\)

∴  Diameter of the image of the sun

= m × D = \(\frac{0.5 \pi}{360 D}\) × D= 0.004363

m = 0.4363 cm. [ taking only the numerical of m]

Example 12. An object is placed at a distance of 25 cm from a con- cave mirror and a real image is formed by the mirror at a distance of 37.5 cm. What is the focal length of the mirror? Now if the object is moved 15 cm towards the mirror, what will be the image distance, its nature and magnification?
Solution:

Here, u = -25 cm, v = -37.5 cm

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{-37.5}+\frac{1}{-25}=\frac{1}{f}\)

= \(-\frac{10}{375}-\frac{1}{25}\)

= \(-\frac{25}{375}=-\frac{1}{15}\)

Or, f= – 15cm

Now if the object is moved 15 cm towards the mirror, the object distance will be (25-15) = -10 cm

=  \(\frac{1}{v}+\frac{1}{-10}=-\frac{1}{15}\)

Or, \(-\frac{1}{15}+\frac{1}{10}\)

= \(\frac{-2+3}{30}=+\frac{1}{30}\)

= + \(\frac{1}{30}\)

Or, v= + 30 cm

The positive sign of v indicates that the image formed by the mirror is virtual and it is formed at a distance of 30 cm behind the mirror.

Magnification, m = \(-\frac{v}{u}\)

= \(-\frac{30}{-10}\)

= 3

∴ The image will be magnified 3 times. The positive sign of m indicates the image as erect.

Example 13. An object is placed at a distance of 50 cm in front of a convex mirror. Now a plane mirror is placed in between the object and the convex mirror, covering the lower half of the convex mirror. If the distance of the plane mirror from the object is 30 cm, it is seen that there is no parallax of the images formed by the two mirrors. What is the radius of curvature of the convex mirror?
Solution:

MM’ is a convex mirror and A is a plane mirror. P’Q’ is the object and P’Q’ is the image formed at the same place by the two mirrors.

The radius of curvature,

Here OR = 50 cm , AR = 30 cm

Image distance from the plane mirror = O’A = 30 cm

According to the figure,

OA = OR-AR

= 50-30

= 20 cm

∴ Image distance from the convex mirror,

O’O = O’A-OA = 30-20 = 10 cm

We have \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Here, v= +10 cm , u= -50 cm

∴ \(\frac{1}{10}+\frac{1}{-50}=\frac{1}{f}\)

Or,  \(\frac{4}{50}=\frac{1}{f}\)

Or,  \(f=+\frac{25}{2}\)

Radius of curvature,

r = 2f

\(\frac{25}{2}\)

=  25 cm

The radius of curvature of the convex mirror = 25 cm

Example 14. A concave mirror of focal length 10 cm and a convex mirror of focal length 15 cm are held co-axially face to face at a distance 40 cm apart. An object of height 2 cm is placed perpendicularly on the common axis in between the two mirrors. The distance of the object from the concave mirror is 15 cm. Considering the first reflection occurs in the concave mirror and the second reflection in the convex mirror, calculate the position, nature, and height of the final image.
Solution:

In the case of the first reflection in the concave mirror: u = -15 cm; f = -10 cm

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{y_1}+\frac{1}{-15}=\frac{1}{-10}\)

Or, \(\frac{1}{v}=-\frac{1}{10}+\frac{1}{15}\)

Or, \(\frac{3_1+2}{30}\)

Or, \(-\frac{1}{30}\)

v= -30 cm

So, the image formed by the concave mirror is real and formed at a distance of 30 cm from the mirror. This image acts as the object of the convex mirror.

In case of the second reflection in the convex mirror:

u = -(40- 30) = -10 cm , f = +15 cm

We know \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-10}=+\frac{1}{15}\)

Or, \(\frac{1}{v}=+\frac{1}{15}+\frac{1}{10}\)

= + \(\frac{1}{6}\)

Or, v= + 6cm

So the final image is virtual and is formed at a distance of 6 cm
behind the convex mirror.

Magnification by the concave mirror,

m₁ = \(\frac{v}{u}\)

= \(\frac{-30}{-15}\)

= -2

Magnification by the convex mirror,

m₂ = – \(\frac{v}{u}\)

= – \(\frac{6}{-10}\)

= \(\frac{3}{5}\)

∴ Total magnification.

m = m₁ × m₂

= -2 ×\(\frac{3}{5}\)

=  – \(\frac{6}{5}\)

As m is negative, so the final image is inverted with respect to the object.

∴ Height- of the final image (taking only the magnitude of m)

= \(\frac{6}{5}\) × height of the object

= \(\frac{6}{5}\) × 2

= \(\frac{12}{5}\)

= 2. 4 cm

Example 15. The focal length of a concave mirror is 30 cm. An object is placed at a distance of 45 cm in front of the mirror. A plane mirror is placed perpendicularly on the principal axis of the concave mirror in such a way that the object is situated in between the two mirrors. Light rays from the object at first are reflected from the concave mirror and then from the plane mirror. As a result the final image coincides with the object. What is the distance between the two mirrors?
Solution:

In case of the concave mirror M₁ M₂, P is the object and P’ is its image.

Here, u = -45 cm,  f= – 30 cm

We know \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}-\frac{1}{45}\)

= \(-\frac{1}{30}\)

Or, \(\frac{1}{v}=-\frac{1}{30}+\frac{1}{45}\)

= \(\frac{-3+2}{90}\)

= \(\frac{1}{90}\)

Or, v= -90 cm

Op’ = 90 cm

According to the question, the reflected rays from the concave mirror form the final image at P after reflection from the plane
mirror M.

∴ PM=  P’M = x (say)

Here, OP’ = OP+PM+P’M

Or, 90 = 45+x+x

Or, x = \(\frac{45}{2}\)

= 22.5 cm

∴ Distance between the mirrors = 45+22.5 67.5 cm.

Example 16. A concave mirror forms a real image magnified two times. If both the object and the screen are moved a real Image magnified three times that of the object is formed. If the screen is moved through a distance of 25cm, then determine the displacement of the object and focal length of the mirror.
Solution:

In the first case,

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-u}=\frac{1}{-f}\)

Or, \(1-\frac{v}{u}=-\frac{v}{f}\)

∴ 1+m = –  \(-\frac{v}{u}\)…………………….(1)

Since m= – \(-\frac{v}{f}\)

In the first case, magnification = 2

∴ 1+2 = \(-\frac{v}{f}\)

Or,  – \(\frac{v}{f}\) = 3…………………….(2)

In the second case, magnification = 3

∴ 1+3 = – \(\frac{(v+25)}{f}\)

Or,  4 = – \(\frac{v}{f}-\frac{25}{f}\)

∴ 4= 3 – \(3\frac{25}{f}\)

Or, f= – 25 cm

The focal length of the concave mirror = 25cm

From equation (2) we have,

v =3f = -3 × (-25) = 75 cm

According to the equation of a spherical mirror,

⇒ \(\frac{1}{u}=\frac{1}{f}-\frac{1}{v}=\frac{1}{75}-\frac{1}{25}\)

[since the image is real, hence v is taken negative]

Or,

u = -37.5 cm

In the first case object distance 37.5 cm

In the second case image distance

v₁= v+25= 75+25= 100 cm

Suppose, object distance = u₁

Since in this case magnification =3

∴ \(u_1=-\frac{1}{3} v_1\)

= \(\frac{100}{3}\)

= – 33.33 cm

∴ Object distance in the second case =33.33cm

∴ Displacement of the object = 37.5-33.33

= 4.17cm

So, the displacement of the object = 4.17 cm, and the focal length of the mirror = 25 cm

Example 17.  A cube of side 2 m is placed in front of a large concave mirror of focal length 1 m in such a way that the face A of the cube is at a distance of 3 m and the face B at a distance of 5 m from the mirror.

1. Calculate the distance between the images of the faces A and B.
2. Determine the height of the images of faces A and B. 
3. Will the image of the cube be a cube

Solution:

1. Distance of the face A from the mirror, u₁ = -3 m, focal length of the mirror, f= -1 m

Let the image distance of the face A from the mirror be v₁ m

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or,   \(\frac{1}{v_1}-\frac{1}{3}=-\frac{1}{1}\)

Or,= v₁ = -1.5 m

Distance of the face B from the mirror, u=-5 m; the image distance of this face = v₂ m

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v_2}-\frac{1}{5}=-\frac{1}{1}\)

Or,= v₂ = -1.25 m

So, the distance between the images of the faces A and B

= v₁– v₂ = 1.5-1.25 = 0.25 m [Taking magnitude of v₁ and v₂ ]

2. Magnification in the first case

m₁ = \(\frac{1}{v_1}-\frac{1}{u_1}\)

= \(\frac{-1.5}{-3}\)

= – 0.5

∴ \(\frac{\text { height of the image of the face } A\left(I_A\right)}{\text { length of the face } A\left(O_A\right)}\)

Or, I_A= -5.5 × 2 = -1m

Again, magnification in the second case

⇒ \(\frac{1}{v_2}-\frac{1}{u_2}\)

m₂ = \(\frac{1}{v_2}-\frac{1}{u_2}\)

= \(\frac{-1.25}{-5}\)

= – 0.25

∴ Height of the image of the face B,

Or, I_B= -5.5 × 2 = -1m

I_B =  m₂ ×  height of the face B

= -0.25 × 2

= -0.5 m

3. So, it is seen that the height of the image of face A and that of the face B are not equal. So, the image of the cube will no longer be a cube.

Example 18. A is a point object in a circular track. A light ray starting from object A is reflected twice by the circular track and returns again to A. The angle of incidance is a. The distance of A from the center of the circular track is x and the diameter of the circular track from A intersects the path of the ray at a point D whose distance from the center of the circular track is y.
Show that, tan α = \(\sqrt{\frac{x-y}{x+y}}\)

Solution:

In Δ OBC, OB = OC; so, ∠OBC= ∠OCB = α

Also ∠ABC = ∠ACB= 2α; so, AB = AC

Since Δ ABC is isosceles, hence median AD ⊥ BC

∴ tan α =  \(\frac{y}{B D}\) and tan 2α = \(\frac{x+y}{B D}\)

Or, \(\frac{\tan 2 \alpha}{\tan \alpha}=\frac{x+y}{y}\)

Or, \(\frac{2 \tan \alpha}{\tan \alpha\left(1-\tan ^2 \alpha\right)}=\frac{x+y}{y}\)

Or, \(1-\tan ^2 \alpha=\frac{2 y}{x+y}\)

Or, \(\tan ^2 \alpha=1-\frac{2 y}{x+y}\)

= \(\frac{x-y}{x+y}\)

Or, tanα = \(\sqrt{\frac{x-y}{x+y}}\)

So, the distance between the images of the faces A and B

Example 19. A concave mirror and a convex mirror are placed co-axially face to face. The focal length of each of them is f and the distance between them is 4f. A point source is so placed on their common axis in between the two mirrors that if the first reflection is considered to take place on the convex mirror, the final image coincides with the point source. Determine the position of the source.
Solution:

Let the point source O be situated at a distance x from the convex mirror

If the image distance is v₁ then

⇒ \(\frac{1}{v_1}+\frac{1}{-x}=\frac{1}{f}\)

Or, \(\frac{1}{v_1}=+\frac{1}{f}+\frac{1}{x}\)

= \(+\frac{1}{f}+\frac{1}{x}=\frac{(x+f)}{f x}\)

Or, \(v_1=\frac{f x}{x+f}\)

This image will be formed at Oj behind the convex mirror. Now u X this image will act as the object of the concave mirror.

∴ Object distance for the concave mirror,

u₂= 4f+ v₁

= \(4 f+\frac{f x}{x+f}\)

= \(\frac{5 f x+4 f^2}{x+f}\)

∴ Image distance v₂ = 4f- x

Since the final image coincides with the object

∴ \(\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f}\)

Or, \(\frac{1}{4 f-x}+\frac{x+f}{5 f x+4 f^2}=\frac{1}{f}\)

Or, \(\frac{1}{4 f-x}=\frac{1}{f}-\frac{x+f}{5 f x+4 f^2}\)

= \(\frac{4 x+3 f}{f(5 x+4 f)}\)

Or = x² – 2fx – 2f² = 0

Or, \(\frac{2 f \pm \sqrt{4 f^2+8 f^2}}{2}=f \pm f \sqrt{3}\)

∴ x = \(f(1+\sqrt{3})\)  [neglecting the negative value]

The distance of the source from convex mirror = \(f(1+\sqrt{3})\)

Example 20.  The diameter of the moon is 3450 km and its distance from the earth is 3.8 x 105 km, What will be the diameter of the image of the moon formed by a convex mirror of focal length 7.6 m? As the moon is very far away from the earth the image of the moon will be formed in the focal plane of the mirror
Solution:

Hence, v = f = 7.6 m

Now, u= 3.8 × 105 km and magnification, m = \(\frac{v}{u}\)

[Taking the mod value of m]

∴ The diameter of the image

= m × diameter of the moon

= \(=\frac{7.6 \mathrm{~m}}{3.8 \times 10^5 \mathrm{~km}} \times 3450\)km

= 00.69m

= 6.9 cm

Example 21. Three times magnified image of an object is formed on a screen placed at a distance of 8 cm from the object with the help of the spherical mirror. Determine the nature of the mirror, focal length, and the distance of the mirror from the object.]
Solution:

Since the image is formed on a screen, the image is real and the mirror is concave.

Suppose, u= x cm and so v = (x+8)cm

Here,m = 3 or, \(\frac{v}{u}\) = 3 [taking the mod value of m]

Or, \(\frac{v}{u}\) = 3

Or, \(\frac{(x+8)}{x}\) = 3

Or,   x= 4cm

So the mirror is situated at a distance of 4 cm from the object

∴ v = (4+8) = 12 cm

We know , \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or,  \(\frac{1}{-12}+\frac{1}{-4}=\frac{1}{f}\)

Or, f= -3 cm

∴  Focal length of the mirror

= 3 cm. A negative sign in the value of/ supports the mirror as concave in nature

Example 22. An object is placed just at the middle point between a concave mirror of radius of curvature 40 cm and a convex mirror of radius of curvature 30 cm facing each other. The mirrors are situated 50 cm apart from each other. Considering the first reflection occurs in the concave mirror, determine the position and nature of the image formed by this mirror. Next, find the position and nature of the image formed by the convex mirror considering the first image of concave mirror as its object.
Solution:

Focal length of the concave mirror = \(\frac{40}{2}\) = 20 CM

Focal length of the convex mirror = \(\frac{30}{2}\)  = 15cm

In case of the concave mirror, u= – 25 cm: f= -20 cm.

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-25}=\frac{1}{-20}\)

Or, \(\frac{1}{v}=\frac{1}{25}-\frac{1}{20}=\frac{4-5}{100}\)

Or, \(\frac{1}{v}=-\frac{1}{100}\)

Or, v= – 100 cm

So, the image formed by the concave mirror in the

absence of the convex mirror would be real and inverted and situated at Q₁ at a distance of 100 cm from the concave mirror. This image will act as a virtual object for the convex mirror.

Her O₁Q₁ = 100 cm

Now for the presence of the convex mirror M₃M₄ object distance, M₃M₄ = , object distance , u₁= (100-50) cm = 50 cm;

Focal length, f₁= 15 cm.

According to the equation of mirror,

⇒ \(\frac{1}{v_1}+\frac{1}{50}=\frac{1}{15}\)

Or, \(\frac{1}{v_1}=\frac{1}{15}-\frac{1}{50}\)

Or, \(\frac{10-3}{150}=\frac{7}{150}\)

Or,  v₁= \(\frac{150}{7}\)

= 21.43cm

So, the final image will be formed at Q₂ at a distance of 21.43 cm from the convex mirror. A positive value of v₁ indicates that this image is virtual.

Example 23. An arrow of height 2.5 cm is situated vertically at a distance of 10 cm from a convex mirror of a focal length 20 cm. Where will the image be formed? Determine its height. If the arrow is moved away from the mirror what will happen to its image?
Solution:

Here, u = -10 cm and f= 20 cm

We know,  \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-10}=\frac{1}{20}\)

Or, \(\frac{1}{v}=\frac{1}{10}+\frac{1}{20}\)

Or, \(\frac{2+1}{20}=\frac{3}{20}\)

or, v= \(\frac{20}{3}\)

= 6. 66cm

So, the image will be formed at a distance of 6.66 cm behind the convex mirror.

Magnification, m =\(\frac{I}{O}=-\frac{v}{u}\)

I = \(-O \times \frac{v}{u}\)

= \(-2.5 \times \frac{20}{3 \times(-10)}\)

= \(\frac{5}{3}\)

= 1.66cm

So, the height of the image = 1.66 cm. If the arrow is moved away from the convex mirror the image will move towards the focus of the mirror and will be diminished in size

Example 24. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. What is the length of the image?
Solution:

Here for the side A, f = -10 cm, u = -20 cm.

From the equation, \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) we get v = -20 cm

For the side B, u – -30 cm

Similarly, v = -15 cm

Length of the image = 20- 15 = 5 cm

Reflection Of Light Synopsis

Reflection of light:

1. Coming through a medium when light incidents on the upper surface of another medium, then a part of that incident light returns to the first medium by changing its direction.
2. This phenomenon is known as the reflection of light.

Laws of reflection:

1. The incident ray, the reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane.
2. The angle of incidence is equal to the angle of reflection.

Image:

When rays of light diverging from a point source after reflection or refraction converge to or appear to diverge from a second point, the second point is called the image of the first point.

Focus in case of reflection from a spherical surface:

When a beam of rays parallels to the principal axis is incident on a spherical mirror, it is seen that the reflected rays either converge to or appear to diverge from a fixed point on the principal axis. This point is called the principal focus or focal point or briefly the focus of the mirror.

Conjugate foci:

1. If a pair of points is such that when an object is placed at one of them, its image is formed at the other by a fixed mirror, then that pair of points are called conjugate foci of that mirror.

2. The focal length of a concave mirror is taken as positive and the focal length of a convex mirror is taken as negative.

3. The distance measured from the pole of the mirror and in ‘front of it is taken as positive but the distance measured in the back side of the mirror is taken as negative.

4. If the aperture is small then the focal length of a spherical mirror becomes half of its radius of curvature

5. Real image is formed in front of the mirror but virtual is formed at the back of the mirror.

6. The ratio of the length of the image to that of the object is called linear magnification of the image.

7. The ratio of the area of an image to that of the two-dimensional object is called areal magnification.

8. The ratio of the length of the image to that of the object along the principal axis is called longitudinal or axial mag- notification.. da

9. In the case of a spherical mirror longitudinal magnification is equal to the square of the linear magnification.

10. Angle of reflection (r) = angle of incidence (1)

11. Angle of deviation for a ray of light after reflection from a plane mirror,

δ = 180°-2i [i= angle of incidence]

12. In case of the image formed by a plane mirror, the distance of the image from the mirror

= distance of the object from the mirror.

13. In the case of a spherical mirror of a small aperture the relation between the focal length of the mirror () and its radius of curvature (r) is

f = \(\frac{r}{2}\)

14. In the case of a spherical mirror of a small aperture, the relation among the object distance (u), image distance (v), and focal length (f) is

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

15. Newton’s equation:

xy = f² [where, x = u-f and y = v-f]

16. Linear magnification:

m = \(\frac{\text { length or height of the image }(I)}{\text { length or height of the object }(O)}\)

= \(\frac{\text { image distance }(v)}{\text { object distance }(u)}\)

= \(\frac{f}{f-u}=\frac{f-\nu}{f}\)

17. Areal magnification for a two-dimensional object:

m’ = \(\frac{\text { area of image }}{\text { area of object }}\)

18. Longitudinal or axial magnification:

m” = \(\frac{\text { length of the image along the principal axis }}{\text { length of the object along the principal axis }}\)

Rules for solving numerical problems relating to spherical mirrors

To solve the numerical problems in connection with spherical mirrors the following rules are to be followed:

• In the general equation of mirrors, put the numerical values of u, v, f etc. with their proper signs.
• Then after solving the equation, draw inference about the distance from the sign.
• A positive sign of the focal length will suggest that the mirror is convex while a negative sign will suggest that the mirror is concave.
•  If the image distance is negative, we have to understand that the image has been formed in front of the mirror and it is real and inverted.
• Again, if the image distance is positive, we have to understand that the image has been formed behind the mirror and it is virtual and erect.

Reflection Of Light Very Short Question And Answers

Question 1. Can an image be formed due to diffused reflection? 
Answer: No

Question 2. Can a plane mirror form a real image?
Answer: Yes

Question 3. If a light ray is incident on a horizontal plane mirror making an angle 30° with the mirror, what angle will the reflected ray make with the mirror?
Answer: 30

Question 4. What type of reflection takes place on the screen of a cinema hall?
Answer: Diffuse

Question 5. What is the angle of deviation of a ray incident perpendicularly on a reflecting surface?
Answer:  180°

Question 6. Focus of which spherical mirror is virtual.
Answer:  Convex

Question 7. Two concave mirrors have same focal length but different apertures. Both the mirrors form an image of the sun on a screen. For which image formation will the temperature of the screen become higher?
Answer:  Concave

Question 8. The secondary focus of a concave mirror is a fixed point. Is this statement correct?
Answer: No

Question 9. The principal focus of a concave mirror is a fixed point. Is this statement correct?
Answer:  Yes

Question 10. Two concave mirrors have the same aperture but different focal lengths. Both form images of the sun on a screen. For which image formation will the temperature of the
Answer:  In both cases, hotness will be the same

Question 11. The radius of curvature of a concave mirror is 30 cm. What is its focal length?
Answer: 15 cm

Question 12. Which spherical mirror is called a divergent mirror-concave or convex?
Answer:  Convex

Question 13. The focal length of a spherical mirror is 40 cm. What is its radius of curvature?
Answer: 80 cm

Question 14. What is the value of the focal length of a plane mirror?
Answer: Infinite

15. What is the power of a plane mirror?
Answer: Zero

Question 16. What is the relation between the radius of curvature (r) and focal length (f) of a spherical mirror?
Answer: r = 2f

Question 17. Does the size of mirror affect the nature of the image?
Answer: No

Question 18. The focal length of a concave mirror is equal for all colors of light. Is the statement true or false? [true] [yes] 21. What type of mirror is to be used for getting parallel rays from a small source of light?
Answer: True

Question 19. What will happen to the focal length of the concave mirror when it is immersed in water? Will remain unchanged
Answer: Will remain unchanged

Question 20. Where should an object be placed in front of a concave mirror to get a magnified image?
Answer:  Between f and 2f 

Question 21. Why do we sometimes use a concave mirror instead of a plane mirror as a common mirror? [to get a magnified erect image]
Answer: To get a magnified erect image

Question 22. What is the magnification of the image of an object placed at the center of curvature of a concave mirror?
Answer: 1

Question 23. What is the magnification of an object placed at the focus Answer: Infinity

Question 24. Can a convex mirror ever form a real image of a real object? of a concave mirror?
Answer:  No

Question 25. What will the image of an object placed before a convex screen be higher? mirror-erect or inverted?
Answer: Erect

Question 26. Two concave mirrors are placed face to face and they have the same centre of curvature. A point source of light is placed at their common center of curvature. Where will the image be formed?
Answer: At their common center of curvature

Question 27. If an object is placed between the pôle and the focus of a concave mirror, will the size of the image be magnified with respect to the size of the objector?
Answer: Yes

Question 28. What is the minimum distance between an object and its image formed by a concave mirror?
Answer:  Zero

Question 29. If a concave mirror is immersed in water will its focal length change?
Answer: No

Question 30. In the case of a concave mirror, what is the shape of the uv graph?
Answer:  Rectangular hyperbola]

Question 31. In the case of a concave mirror, what is the shape of the graph?
Answer: Straight line

Question 32. The focal length of a concave mirror in vacuum is 2 m. What will be the focal length of the concave mirror in a medium of refractive index 2.76?
Answer: 2 m

Question 33. At what distance in front of a concave mirror (f = 10 m) an object is to be placed so that the size of the image will be halved of the size of the object?
Answer: 30 cm

Question 34. If the conjugate foci of a spherical mirror lie on the same side of the mirror then, what is the nature of the mirror?
Answer: Concave

Question 35. What is the magnification produced in a plane mirror? 40. A cube is placed in front of a large concave mirror. Will the image of the cube be a cube?
Answer: 1

Question 36. Can a convex mirror form a real image?
Answer: Can form real image of a virtual object

Question 37. For a spherical mirror if the linear magnification of the image be m what will be its lateral magnification?
Answer: m²

Question 38. The image of a candle formed by a concave mirror is cast on a screen. What will happen if the mirror is covered partly?
Answer: The brightness of the image will be reduced]

Question 39. Will the focal length of a spherical mirror be affected if the wavelength of the light used is increased?
Answer: No

Question 40. What type of mirror do car drivers use to view the traffic at the back of the car?
Answer: Convex

Question 41. What type of mirror do dentists use?
Answer: Concave

Question 42. What type of mirror is used in searchlights and headlights of vehicles?
Answer: Paraboloidal

Reflection Of Light Fill In The Blanks

Question 1. The reflecting surface from which regular reflection of light takes place is called__________________  object 
Answer:  Smooth plane reflector

Question 2. A plane mirror can form a real image of a____________ object
Answer: Virtual

Question 3. A smaller virtual image is formed by the blank). mirror. [Fill in Answer: Convex

Question 4. An object is moving towards a convex mirror from a large distance. The image will move with _________________ velocity than  the object  ___________  the mirror 
Answer:  Less, towards

Reflection Of Light Assertion Reason Type

Direction: These questions have statement I and statement II. the four choices given below, choose the one that best scribes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 2 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 3 is true, statement 2 is false.
4. Statement 4 is false, and statement 2 is true.

Question 1.

Statement 1: The formula connecting u, v, and ƒ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.

Statement 2: Laws of reflection are strictly valid for plane surfaces but not for large spherical surfaces.

Answer:  3. Statement 3 is true, statement 2 is false.

Question 2.

Statement 1: A concave mirror is preferred to a plane mirror for shaving.

Statement 2: When a man keeps his face between the pole and the focus of the mirror, an erect and highly magnified virtual image is formed.

Answer:  1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3. 

Statement 1: A virtual image can not be directly photographed.

Statement 2: A virtual image can be produced by using a convex mirror.

Answer: 2. Statement 2 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 4.

Statement 1: In the absence of diffuse reflection an object would appear either dazzlingly bright or quite dark.

Statement 2: The angle of incidence is not equal to the angle of reflection in this case.

Answer:   3. Statement 3 is true, statement 2 is false.

Question 5.

Statement 1: Convex mirror is used as driver’s mirror.

Statement 2: Convex mirror gives an index wider field of view of the traffic.

Answer:  1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Reflection Of Light Match The Columns

Question 1. 

Answer: 1-B, 2- D, 3- A, 4. C

Question 2. The nature of the image and the type of mirror are given in column A and column B respectively. Match the column.

Answer:  1-D, 2-C, 3 A, 4 – B

Question 3. Some relations related to spherical mirror and their field of application are given in column A and column B respectively (The symbols have their usual meanings). Match the column.

 

Answer:  1- D, 2- C, 3- B, 4- A

Question 4. The position of an object with respect to a concave mirror and the position of the image are given in column 1 and column 2 respectively.

Answer: 1- B, 2-D, 3- E, 4- C, 5-A

Question 5. Match the corresponding entries of column I with column [where m is the magnification produced by the mirror].

1. 1. (A, C ) 2. (A, D)  3. ( A, B)  4. (C, D)
2. 1. (A, D)  2. (B, C) 3. (B, D) 4. (C, D)
3. 1. (C, D) 2. (B, D) 3. (B, C) 4. (A, D)
4. 1. (B, C) 2. (B, C )3. (B, D) 4. (A, D)

Answer: 1. (B, C)  2. (B, C) 3. (B, D) 4. (A, D)

Magnification of image produced by a spherical mirror,

For real images, m<0, and m> 0 for virtual images.

The convex mirror always forms a virtual image, so m> 0 always and the size of the image is less or equal to the size of object 1.e., m≤1.

In the case of the concave mirror, if the real image is formed, then there can be |m|≥ 1 and |m | ≤1, and for a virtual image m>1. The option is correct.

WBCHSE Class 12 Physics Constituents Of An Atom

Early nineteenth century, John Dalton proposed that all matter is made up of tiny, indivisible particles called ‘atoms! This statement along with other similar statements was later proved to be wrong and discarded

Towards the end of the nineteenth century, William Crookes, Joseph John Thomson, Philipp Lenard, and others, while studying the silent electric discharge in gases at low pressure paved the way for the discovery ofelectrons. Among them, J J Thomson is credited with the discovery of the electron, a subatomic particle. After the discovery of electrons, the myth of the indivisibility of ‘atom’ was dispelled

Electron:

It has been possible to show the emission of electrons from almost all matters through suitable experimental arrangements. Electrons are negatively charged particles carrying a charge which is denoted by e or e^-1

Millikan determined the amount of charge of an electron, e = 1.6 × 10^-19 coulomb = 4.8 × 10^-10 esu J. J Thomson determined the specific charge of an electron, i.e., charge to mass ratio for an electron 1.76 × 10¹¹ C . kg^-1.

Therefore, the mass of the electron

Read and Learn More Class 12 Physics Notes

m_e = \(\frac{\text { amount of charge of an electron }(e)}{\text { specific charge of an electron }\left(\frac{e}{m_e}\right)}\)

9.1 × 10^-31 kg^-1

= 9.1×10^-28 g

Positive charge:

Most materials are electrically neutral.  From various electron emission experiments, it was proved that (since matter is neutral) there must be some positive charge also in the atom. Later experiments confirmed the presence of a positive charge on an atom.

Rutherford’s Atomic Model

Alpha parties scattering experiment:

Rutherford and his co-workers performed the famous alpha par¬ ticle scattering experiment. Alpha particles are emitted from a radioactive source with considerable energy. These a -particles were collimated into a narrow parallel beam which was made incident on a thin foil of a heavy metal like gold (Z = 79), silver (Z = 47), etc.

These metals being ductile, can be easily drawn into a thin foil of width about 10⁷ m. The thinness ensured that eadi a -particle could interact with a single atom in each collision.

Due to collision with the foil, die a -particles were scattered in different directions which were detected by a fluorescent detection screen. Very few a -particles were scattered in a backward direction without penetrating the foil, but being deflected through angles greater than 90°.

Observation and inference:

• Most of the a -particles passed straight through the metal foil without suffering any deflection. This observa¬ tion leads to the conclusion that most of the space inside the atom is empty.
• Low angle scattering: Some of the alpha particles were scattered through small angles i.e., the scattering angle Here, it is assumed that this scattering
• This takes place due to coulomb attraction between an alpha particle of charge +2e and an electron of charge -e. The deflections of α -particles are so small that an α -particle is about 7000 times heavier than an electron. From this, it is concluded that electrons are embedded discretely inside the atom
• Large angle scattering: Some a -particles, though very few suffered deflection by 90° or larger angles. Some of these a -particles were even deflected through 180° Rutherford quantitatively analyzed the

A number of these large angle deflections. He argued that, to deflect the a -particle backward, it must experience a large repulsive force which was possible if the entire positive charge and almost total mass of the atom were concentrated in a small space. This confirmed the presence of the nucleus.

Thus the strong electrostatic repulsion between an a -particle of charge +2e and a nucleus of charge +Ze was the cause of these large deflections.

By quantitative analysis, it was also concluded that the diameter ofthe nucleus is about 10^-14 m which is about \(\frac{1}{1000}\) times the atomic diameter of 10^-10 m. Hence the volume of the nucleus is only about 1 in 10¹² parts of the atom

Rutherford’s atomic model is often compared with the solar system, but there are more dissimilarities than resemblances. Similarities are only marginal.

Similarities:

Dissimilarities:

From the observations, Rutherford proposed his atomic model.

The salient features of the model are:

• The nucleus exists at the center of an atom and electrons orbit around the nucleus in circular paths.
• The necessary centripetal force for the orbital motion is provided by the electrostatic attraction between the opposite charges of the nucleus and the electron

Drawbacks of the Rutherford model:

InstabilHy of the atom: An orbiting electron in an atom is subjected to centripetal acceleration. According to Max¬ Well’s classical electromagnetic theory, any accelerated charged particle emits electromagnetic radiation.

Since orbiting electrons are accelerated, they should also emit radiation. If this were to happen, the energy of the orbiting electron would keep on decreasing. It would follow a spiral path and ultimately collide with the nucleus.

Theoretical calculations show that under this condition no atom would survive for more than 10-8 s. However, matter is stable implying that atoms too cannot be unstable.

Continuous atomic spectrum: If electron energy in an atom had converted into radiant energy, we would get a con¬ continuous spectrum. Interestingly, atoms of hydrogen, helium, etc. produce a line spectrum instead of a continuous spectrum

WBCHSE Class 12 Physics Atom Bohr’s Atomic Model

Rutherford’s atomic model was modified by Niels Bohr. The postulates Rutherford model which has no technical difficulties remain intact in the Bohr model of the atom. They are  ISM Positively charged nucleus occupies a negligible space at the center of the atom.

Negatively charged electrons revolve around the nucleus in circular orbits. If the mass, speed, and radius of the orbit of an electron are m, v, and r, respectively, the Centripetal force necessary to revolve electrons in a circular orbit,

F = \(\frac{m v^2}{r}\) ……………..(1)

The electrostatic force of attraction acts between the nucleus and electrons. Now, if the atomic number of an atom is Z and the charge of an electron is, the total charge of the nucleus is Ze. So, the electrostatic force of attraction between the nucleus and an electron

F₂ = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e \cdot e}{r^2}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r^2}\)

On the other hand, resolving the drawbacks of Rutherford’s model related to the

• Stability of atoms and
• Atomic line spectrum, Bohr introduced some revolutionary ideas that are not consistent with classical physics.

From this, Bohr’s theory of the atom was  established and the consequent structure of the atom “is called Bohr’s atomic model

Postulates of Bohr’s Theory

Bohr’s model is based on three postulates. These are a[so known as Bohr’s quantum postulates.

• Electrons inside an atom can only revolve in some allowed orbits. When an electron revolves in an allowed orbit, it does not radiate energy.
• Note that, according to classical electromagnetic theory, a rotating (i.e., accelerated) charge radiates energy.
• According to this postulate, the energy of an electron in any allowed orbit remains constant; hence these orbits are called stationary orbits or stable orbits.
• Transition of electrons from one stable orbit to another is possible.
• During the transition, an emission or an absorp¬ tion of radiation occurs.

Its frequency f is determined from the relation, hf = E₁~ E₂ where h is Planck’s constant and (E₁ ~E₂) is the energy difference of the electron in the two stable orbits.

When the energy (E₁) of an electron in its initial orbit is more than its energy (E₂) in its final orbit, i.e., when E₁> E₂, the difference in energy is converted into the energy of an emitted photon. As a result, the atom radiates energy. In this case,

E₁– E₂ = hf ……………………………..(1)

Again, if E₁ < E₂, an external photon should supply the difference in energy. So, in this case, the atom absorbs energy.

E₂– E₂ = hf ……………………………..(2)

Equations (1) and (2) are called Bohr’s frequency conditions

The orbit. where the angular momentum of the electron Is an integral multiple of \(\frac{k}{2 \pi}\) , is known as a stationary or stable orbit

Now, if the radius of any stable orbit is r_n, the speed of
electron in drat orbit be v_n, angular momentum,

L_n = momentum x radius of the circular path

So, according to this postulate,

⇒ \(m v_n r_n=\frac{n h}{2 \pi}\)

where, n = 1.2.3,…… This equation is called Bohr’s quantum condition, n is called the principal quantum number of an electron or its orbit. These stable orbits are called Bohr orbits.

It is dear that, none of Bohr’s postulates is consistent with classical physics.

• Yet, the success of Bohr’s theory is borne out by die analysis of the atomic spectrum of hydrogen and some other elements.
• Moreover, a dear qualitative picture related to the atomic structure of any element can be obtained from Bohr’s theory.
• Hence, in spite, of some inconsistency relating to some classical experiments, the Bohr model is considered the basis of the atomic structure of all elements.

Bohr’s Quantum Condition from de Broglie’s Hypothesis

• According to de Broglie’s hypothesis, a stream of any parties can be considered to be matter waves
• If the stream of particles advances freely, the corresponding matter wave behaves as a progressive wave
• On the other hand, if the stream of particles is confined within a definite region, then naturally the corresponding matter wave behaves as a stationary wave
• From these considerations, de Broglie assumed that,
• A matter wave of electrons confined within an atom is a stationary wave.
• The electronic orbits in an atom should be such that, an integral number of electron waves is present in a complete orbit.
• Otherwise, after a complete revolution, the
• Wave will reach n different point mul It once a Stationary wave will not be formed.

If the radius of the die circular path is r, circumference = 2πr. So, if the wavelength of the electron wave is λ, then

2πr = nλ [n = 1,2,3, ……… ]

Here,  four complete waves arc shown In n complete
orbit i.e., n = 4.

Again, according to de Broglie’s hypothesis, if the mass of electron = m and its speed = v_n, then

= \(\lambda=\frac{h}{m v}\)

h = planks constant

So, 2πr = \(n \frac{h}{m v}\)

Where n = 1,2,3…….

Or mvr = \(n \frac{h}{2 \pi}\)

For different values of n, the values of r and v will be different; if these values are taken as r_n and v_n for the definite value of n, then

⇒ \(m v_n r_n=n \frac{h}{2 \pi}\)

Where n = 1,2,3 ……………

This is Bohr’s quantum condition. So the quantum condition is consistent with the idea of de Broglie’s matter wave. But it should be kept in mind that, as an explanation of unstable orbits of an atom, the above-mentioned discussion is a kind of oversimplification. A true explanation can only be made with quantum mechanics.

Application of Bohr’s Theory

According to this condition, the radii of different Bohr orbits and orbital speeds of electrons are different. In the case of n -th orbit

⇒ \(m v_n r_n=\frac{n h}{2 \pi} ; \text { where } n=1,2,3 \ldots\)

Accroding \(v_n^2=\frac{n^2 h^2}{4 \pi^2 m^2 r_n^2}\) …………………….(1)

The radius of the n-th Bohr orbit (r„): Electrostatic force of attraction between electron and nucleus of nth orbit

⇒ \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n^2}\)

This force of attraction supplies the necessary centripetal force \(\frac{m v_n^2}{r_n}\)

So, \(\frac{m v_n^2}{r_n}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n^2}\)

Or, \(v_n^2=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{m r_n}\) …………………….(2)

Comparing equation (1) and (2), we get

⇒ \(\frac{n^2 h^2}{4 \pi^2 m^2 r_n^2}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{m r_n}\)

Or, \(r_n=\frac{\epsilon_0 n^2 h^2}{\pi m Z e^2}\) ……………………………… (3)

This is the expression of the radius of n -th Bohr orbit

Total energy of the electron in n-th orbit (E_n):

The kinetic energy of the electron in n -th orbit,

⇒ \(E_k=\frac{1}{2} m \nu_n^2=\frac{1}{2} \cdot \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{2 r_n}\)

Now, we know that the potential energy of an electron in n -th orbit due to the electrostatic force of attraction

⇒ \(E_p=-\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n}\)

So, the total energy of the electron in n -th orbit

⇒ \(E_n=E_k+E_p=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n}\left(\frac{1}{2}-1\right)\)

⇒ \(-\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{2 r_n}\) ………………………. (4)

Hence, E_n= -E_n and E_p = 2E_n

Substituting the value of rn from equation (3) into equation (4), we get

⇒ \(-\frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2 \cdot \pi m Z e^2}{2 \cdot \epsilon_0 n^2 h^2}=-\frac{m e^4 Z^2}{8 \epsilon_0^2 n^2 h^2}\) …………………….(5)

Hydrogen atom:

The structure of the hydrogen atom is the simplest of all. For hydrogen Z = 1, i.e., the electric charge of its nucleus is +e, and only one electron having charge -e revolves around it.

All the information obtained from Bohr’s theory for this atom is given below

The radius of n – th Bhor orbit:

Substituting Z = 1 in the equation (3) we get,

r_n = \(\frac{\epsilon_0 n^2 h^2}{\pi m e^2}\) ……………………………… (6)

Here, ∈₀ = permittivity of vacuum

= 8.854 × 10^-12C² . N^-1 . m^-2

h = Planck’s constant = 6.63 × 10^-34 J.s

m = mass of electron = 9.1 × 10^-31 kg

e = charge of an electron = 1.6 × 10^-19 C

Hence, r_n α n²

The information obtained about Bohr orbits from equation (6) is:

1. For n = 1, r becomes minimum, i.e., the orbit lies closest to the nucleus. This is known as the first Bohr orbit or K – shell of the atom. The radius of this orbit is called the first

Bohr radius is denoted by the sign a₀, i.e., r₁ = a₀.

Putting n = 1 in equation (6), we get,

a₀ = \(\frac{\epsilon_0 h^2}{\pi m e^2}\)………………..(7)

Substituting the values of the constants in this expression we get

a₀ = 0.53 × 10^-10 m

= 0.53Å

= 0.053 nm

Angstrom and nanometre units:

1 angstrom (Å) = 10^-10m; 1 nanometre (nm) = 10^-9m

So, 1nm = 10Å  or, 1Å = 0.1 nm. Nowadays nm unit Is more commonly used instead of Å in SI.

2. Using the value of aQ in equation (6), we get, r_n = n²a₀

Substituting n = 2, 3, 4, in this equation, the radii of the orbits, L, M, N,….. are obtained, respectively.

Since rn the ratio of the radii of the orbits K, L, M, N…. is 1: 4:9:16………………………….

For example,

The radius of the L -orbit, r_n = 2². a₀ = 4 × 0.53 = 2.12Å;

The radius of the M -orbit, r₃ = 3² . a₀ = 9 × 0.53 = 4.77Å ; etc.

3. The orbits having radii a₀, 4a₀, 9a₀, ……………………….are permissible and no orbit exists in between them.

For example, the radii of the first and the second Bohr orbits are 0.53 Å and 2.12 Å, respectively. So, no electron can revolve along any circular path having a radius >0.53 A but <2.12 Å. So, the Bohr orbits are discrete, due to integral values of n. Thus n is called the principal quantum number

The energy of the electron to the n-th orbit (E_n):

Substituting the value of rn from equation (8) in equation (4), we get for hydrogen Z = 1

(E_n) = \(E_n=-\frac{1}{4 \pi \epsilon_0} \frac{e^2}{2 n^2 a_0}\) ………………. (9)

This is the expression for the total energy of the electron revolving in the n-th orbit of a hydrogen atom.

The inferences made from equation (9) are:

The value of En is negative; hence higher the value of n, i.e., the greater the distance greater the energy of the electron. So in the first Bohr orbit or the K-orbit, the energy of the electron is the least This is called the ground state of the hydrogen atom. Putting n = 1 in equation (9), we get

⇒ \(E_1=-\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{2 a_0}\) ………………………(10)

Using the values of the constants we get, E₁ = -13.6eV; hence, the ground state energy of the hydrogen atom = -13.6eV.

Again, putting the value of a₀ from equation (7) into equation (10), we get,

E₁ = \(-\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2 \pi m e^2}{2 \epsilon_0 h^2}\)

= \(-\frac{m e^4}{8 \epsilon_0^2 h^2}\)

Where c is the speed of light in a vacuum

Where, r = \(\frac{m e^4}{8 \epsilon_0^2 c h^3}\) = constant

Here, R is a constant called the Rydberg constant Substituting the accepted values of the physical constants, the value of

R comes out to be 1.09737 × 10⁷m^–

From equations (9) (10) and (11), we get

⇒ \(E_n=-\frac{R c h}{n^2}=-\frac{13.6 \mathrm{eV}}{n^2}\)

Or, \(E_n \propto \frac{1}{n^2}\)

Thus the energies in the orbits K, L, M, N.. are as \(1: \frac{1}{4}: \frac{1}{9}: \frac{1}{16}\)

So, in the orbits, of hydrogen atoms, the energies are E₁ = -13.6eV, E₂ = -3.4eV, E₃ = -1.51eV, and E₄ = -0.85eV, respectively.

Hence, these energy levels are discrete, i.e., the electron cannot exist in any intermediate energy state.

Clearly, as n increases, E_n becomes less negative i.e., the energy increases. The energy levels of the hydrogen atoms are represented in the energy level diagram the highest energy state corresponds to n = and has energy

⇒ \(E_{\infty}=\frac{13.6}{\infty^2}=0 \mathrm{eV}\)

Note that an electron can have any total energy above QeV. In such a situation, the electron is free and there is a continuum of energy state above

E_∞ = \(\frac{13.6}{\infty^2}\)

= 0eV

Speed of the electron in n th orbit (v):

In the case of the hydrogen atom, putting Z = 1, we get from equation (2)

⇒ \(v_n^2 \propto \frac{1}{r_n} \text { or, } v_n \propto \frac{1}{\sqrt{r_n}}\)

Again from equation (3), we get

⇒  \(r_n \propto n^2 \text { so, } v_n \propto \frac{1}{n}\)

Hence, the ratio of the speeds of electrons in K, L, M, N … orbit is

r_n  ∝ n²

So, r_n  ∝ 1/ n

Hence speed of the electron is highest in the first Bohr orbit

In the case of the hydrogen atom, putting Z = 1 in equation (2), we get

⇒ \(v_n^2=\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{m r_n}\)

Or, \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{m v_n r_n}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{n \cdot\left(\frac{h}{2 \pi}\right)}\)

= \(=\frac{c}{n}\left[\frac{e^2}{4 \pi \epsilon_0 c\left(\frac{h}{2 \pi}\right)}\right]=\frac{c}{n} \alpha\)

c is the speed of light in a vacuum

Where = \(\alpha\left(=\frac{e^2}{4 \pi \epsilon_0 c\left(\frac{h}{2 \pi}\right)}\right)\) is a dimensionless constant

a is called Sommerfeld’s fine structure constant. Substituting the accepted values of the physical constants, the value of a comes out to be a

⇒ \(\frac{1}{137}\) approximately.

For the first Bohr orbit n = 1, the speed of an electron,

⇒  \(v_1=\frac{c}{137}\) = 2.18 × 10⁶ m.s^-1

This speed is approximately part of the speed of light (c) in a vacuum.

So, the speed of an electron in the next orbit is:

⇒  \(v_2=\frac{1}{2} v_1=\frac{c}{274}\)

= \(v_3=\frac{1}{3} v_1=\frac{c}{411}\)

Orbital angular momentum of the electron in n-th orbit (L_n):

From Bohr’s quantum condition, we directly obtain

L_n= nh/2π

Naturally, Ln oc n, i.e., L₁: L₂: L₃: …………….. = 1: 2: 3: ………………….

So, the greater the distance of the orbits, the greater will be the value of

For example,

L₁= \(\frac{h}{2 \pi}=\frac{6.63 \times 10^{-34}}{2 \pi}\)

= 1.06 × 10^-34m. s^-1

L₁= 2 L₁= 2.12 ×10^-34 J.s :  etc

Hydrogen Spectrum

1. Atomic spectrum:

Line spectrum:

If electric discharge occurs inside any elementary gas or vapor kept in a discharge tube at a few mm of mercury pressure, the tube acts as a bright source of light. It is generally called a discharge lamp. Discharge lamps of neon, sodium, mercury, and halogen gases are used in our daily lives.

With the help of a prism or other instruments, different fundamental colors of different wavelengths are obtained in the dispersion of light emitted from a discharge lamp. This is known as the atomic spectrum. In this spectrum, there are ultraviolet rays and infrared rays along with visible light.

With the help of a special experimental arrangement, the wavelength of each fundamental ray can be determined. Atomic spectra for different elements are different

The characteristics of the atomic spectra are that these are usually a combination of some thin, bright and discrete lines; i.e., in between two bright lines there is a dark space. This spectrum is known as line spectrum

On the other hand, the spectrum obtained from a heated solid (e.g., an incandescent tungsten lamp) is a continuous spectrum; the different colored lights present in it form continuous illumination on the screen and there is no dark space in between them

Again, the molecular spectrum is usually a band spectrum. Instead of discrete lines, closely spaced groups of lines are so formed that each group appears to be a band. Between two consecutive bands, there is a dark space

2. Balmer series of hydrogen spectrum:

In the visible region of the atomic spectrum of hydrogen, there are four bright lines. The experimental values of their wavelengths are 6563A, 4861 Å, 4341 Å, and 4102 Å. These four lines constitute the Balmer series of the hydrogen spectrum. A Swiss mathematics teacher, Balmer tried to express these wavelengths by a simple relation in 1884, many years before the proposal of the Bohr model, which is

⇒ \(\frac{1}{\lambda}=A\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) …………………… (1)

Here, λ = wavelength of spectral line,

A = constant and n = 3, 4, 5, …………………………..

The number of complete waves present in unit length is denoted by the reciprocal of wavelength, I; hence j is called the number and is sometimes expressed by the symbol.

Substituting A = 1.09678 × 10⁷ m^-1 in equation (1), the experimental values of the wavelength of the spectral lines can be found.

For example,

For n = 3 , λ = 6563 Å; for n = 4,  λ = 4861 Å

For n = 5, λ = 4341 Å ; for n = 6, λ = 4102 Å

Moreover, substituting n = 7, 8, 9, different values of /I are obtained and these also belong to the Balmer series. But these wavelengths lie in the ultraviolet region, not in the visible region

Balmer could arrange the spectral lines of hydrogen in a definite pattern, but could not determine relation (1) theoretically

3. Other series of hydrogen spectrum:

Lyman series:

The relation denoting this series is

⇒ \(\frac{1}{\lambda}=A\left(\frac{1}{1^2}-\frac{1}{n^2}\right) ; n=2,3,4, \cdot \cdot\)

Using the same value of A, the values of A obtained from this relation, resemble that of the lines obtained in the ultraviolet region of the hydrogen spectrum. For example, if n

If n = 2, then λ = 1216 Å; if n = 3, then  λ = 1026Å

Paschen series:

The relation denoting this series is

⇒ \(\frac{1}{\lambda}=A\left(\frac{1}{3^2}-\frac{1}{n^2}\right) ; n=4,5,6, \cdots\)

From this relation, the wavelength of some spectral lines of the infrared region of the hydrogen spectrum is obtained. For example,

If n = 4 then λ = then 18751 Å

Brackett series and Pfund series:

In the infrared region of the hydrogen spectrum, some more series are present except the Paschen series; these are the Brackett series and the Pfund series. But in this case, the brightness (or intensity) ofthe corresponding spectral lines is negligibly small.

The relation denoting the Brackett series is,

⇒ \(\frac{1}{\lambda}=A\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\) : n = 5,6,7 ………………..

The relation denoting the Pfund series is

⇒ \(\frac{1}{\lambda}=A\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\) : n = 6,7,8 ………………..

4. Rydberg formula:

Just after the discovery of the Balmer series, Rydberg expressed the following general equation related to the series of spectrums. This is known as the Rydberg formula

⇒ \(\frac{1}{\lambda}=\frac{R}{(m+a)^2}-\frac{R}{(n+b)^2}\) …………………… (2)

Where, R is constant (Rydberg constant) for a particular element, a and b are the characteristic constants for a particular series, m is a whole number that is constant for a given series and n is a varying whole number whose different values indicate the different lines of the series. With the help of equation (2), the series of spectra of most of the elements can be expressed accurately.

Explanation of Hydrogen Spectrum from Bohr’s Theory

Rydberg constant:

The discrete energy levels of the hydrogen atom

E_n = \(-\frac{R c h}{n^2}\) ……………………. (1)

Here, R = \(\frac{m e^4}{8 \epsilon_0^2 c h^3}\) ……………………. (2)

Where, c = speed of light in vacuum = 3 × 10⁸ m .s^-1 and
n = 1,2,3………………..

∴ Ground state energy, E₁ = -Rch

Substituting the values of different constants in equation (2), we get, R ≈ 1.09737 × 10⁷m^-1

Note that, in the analysis of the Balmer series of the hydrogen spectrum, the value of the constant A obtained (1.09678 × 10⁷ m ^-1 ) is slightly less than the above value of R.

This difference becomes negligible if the mass of the hydrogen nucleus is corrected. So, we can say that, the constant A is the Rydberg constant, and equation (2) denotes its expression

Calculation of Rydberg constant in the CGS system:

The expression for the Rydberg constant in the CGS system can be obtained by substituting in place of e0 in equation (2). Using the CGS values of the constants, we get

R = \(\frac{2 \pi^2 m e^4}{c h^3}\)

= 109737 cm ^-1

The wavelength of the emitted radiation:

Let the electron in a hydrogen atom make a transition from a higher energy state E_ni to a lower energy state E_ni According to Bohr’s postulate, a photon will be emitted from the hydrogen atom due to this transition.

If the frequency of this photon is f (wavelength \(\) ) then

⇒ \(E_{n_i}-E_{n_f}=h f=\frac{h c}{\lambda}\) …………….(3)

Substituting n = n_i and n = n_f respectively in equation (1), we get

⇒ \(E_{n_i}=-\frac{R c h}{n_i^2} \text { and } E_{n_f}=-\frac{R c h}{n_f^2}\)

So, from equation (3), we get,

⇒ \(\frac{h c}{\lambda}=-\frac{R c h}{n_i^2}+\frac{R c h}{n_f^2}\)

= \({Rch}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\)

Or \(\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\) ……………………….. (4)

1. Balmer series:

If the electron in a hydrogen atom jumps from any one of the energy states E₃, E₄, E₅,……………. etc. to the energy state ,E₂ then putting n_i = 3,4, 5, …………… and n_f= 2 in equation (4), we get

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right) ; n=3,4,5, \cdots\) …………. (5)

This relation indicates the Balmer series of the atomic spectrum of hydrogen

2. Other series:

Similarly, substituting n_i = 2, 3, 4, and n_f = 1 in equation (4), we get the Lyman series :

⇒\(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\): n =2,3,4 ………………. (6)

Again, substituting nt = 4, 5, 6— and nÿ= 3 , we get Paschen series

⇒\(\frac{1}{\lambda}=R\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\): n = 4,5,6 ………………. (7)

Substituting n_i = 5, 6, 7, ………..  and n_f= 4, we get Brackett series

⇒\(\frac{1}{\lambda}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\): n = 5,6,7  ………………. (8)

Substituting n_i = 6,7,8,……..  and n_f= 5, Pfund series is obtained:

⇒\(\frac{1}{\lambda}=R\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\): n = 6,7,8 ……………… (9)

So the relations, shown by Balmer and other scientists for different wavelengths of the atomic spectrum of hydrogen, can be established accurately from Bohr’s theory. The basis of the success of Bohr’s theory lies in the accurate explanation of the hydrogen spectrum, although it has deviations from classical physics Different series of the atomic spectrum of hydrogen are

Absorption and emission spectrum of hydrogen:

We have seen that if an electron jumps from a higher energy state to a lower one, energy equal to the difference between the states is emitted where AE = hc/ λ. So the reverse process, if the atom absorbs a photon of wavelength A, the electron will jump from the lower energy state to the higher energy state. Since the difference in energy states is fixed, the wavelength of the absorbed photon and that of the emitted photon are exactly equal.

Thus, if the light coming from a source (e.g., an incandes¬ cent tungsten lamp) passes through hydrogen gas, due to absorption, some dark lines are formed in its spectrum. These dark or black lines form an absorption spectrum The bright lines present in the emission spectrum obtained from the hydrogen gas discharge tube, become dark in the continuous spectrum of other sources while absorbed by hydrogen

WBCHSE Class 12 Physics Atom Bohr’s Atomic Model Numerical Examples

Example 1.  If the value of the Rydberg constant of hydrogen is 109737 cm^-1, determine the longest and the shortest Paschen wavelength of the Balmer series.
Solution:

If the wavelength of the Balmer series is λ, then

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) n = 3,4,5 ………….

R = Rydberg constant

Substituting the minimum value of n, i.e., n = 3, we get

⇒  \(\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R \times \frac{5}{36}\)

Or, \(\lambda=\frac{36}{5 R}=\frac{36}{5 \times 109737}\)

= \(\frac{36 \times 10^8}{5 \times 109737}\)

= 6561 Å

This is the longest wavelength.

Again, by substituting the maximum value of n, i.e n = we get

⇒  \(\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{\infty}\right)=R \times \frac{1}{4}\)

Or, \(\) cm-1

= \(\lambda=\frac{4}{R}=\frac{4}{109737} \mathrm{~cm}\)

= \(\frac{4 \times 10^8}{109737}\)

= 3645 Å

This is the shortest wavelength

Example 2. As a result of a collision with a 20 eV energy, a hydrogen atom is excited to the higher energy state, and the electron is scattered with a reduced velocity. Subsequently, a photon with a length of 1216 Å is emitted from the hydrogen wave atom Determine the velocity of the electron after collision
Solution:

The wavelength of the emitted photon,

λ = 1216 Å = 121 6 × 10^-8 cm

∴ The amount of energy gained by the hydrogen atom from the electron

E₁ – E₂ = hf

= \(\frac{h c}{\lambda}=\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{1216 \times 10^{-8}}\)

= 1.634 × 10^-11  erg

According to the questions, the initial kinetic energy of the electron

= 20 eV = 20 × 1.6  × 10^-12  erg

= 3.2 × 10^-11erg.

∴ The remaining kinetic energy of the electron after its collision with a hydrogen atom

⇒ \(\frac{1}{2} m v^2=(3.2-1.634) \times 10^{-11}\) = (3.2 – 1.634)

= 1.566  × 10^-11  erg

Or, v = \(\sqrt{\frac{2 \times 1.566 \times 10^{-11}}{9.1 \times 10^{-28}}}\)

= 1.855 × 10 cm .s^-1

Example 3. Light from a discharge tube containing hydrogen atoms is incident on the surface of a piece of sodium. The maximum kinetic energy of the photoelectrons emitted from sodium is 0.73 eV. The work function of sodium is 1.82 eV. Find

1. The energy of photons responsible for the photoelectric emission 
2. The quantum numbers of the two orbits in the hydrogen atom involved in the emission of photons and
3. The change in angular momentum ofthe electron of a hydrogen atom in the two orbits

Solution:

1. According to the photoelectric equation, energy of the photon, hf = E_max + W₀ = 0.73 + 1.82 = 2.55 eV

2. The energy difference between the two orbits is 2. 55 eV. Now in case of the hydrogen atom

Energy in the ground state (n= 1) , E₁ = -13.6 eV

Energy in n = 2 state E₂ = \(\frac{E_1}{2^2}=-\frac{13.6}{4}\)

= -3.4 eV

Energy in n = 3 state  E₃ = \(\frac{E_1}{3^2}=-\frac{13.6}{9}\)

= 1.51 eV

Energy in n = 4 state E₃ = \(\frac{E_1}{2^2}=-\frac{13.6}{16}\)

E₄– E₂ = 2.55 eV

So, the quantum numbers of the two orbits are 4 and 2.

3. According to Bohr’s postulate, change in angular momentum

= \(4 \cdot \frac{h}{2 \pi}-2 \cdot \frac{h}{2 \pi}=\frac{h}{\pi}\)

= \(\frac{6.625 \times 10^{-27}}{\pi}\)

= 2.11 × 10^-27 erg. s

Example 4. When ultraviolet light of wavelengths 800Å and 700Å are incident on the hydrogen atom at ground state, electrons are emitted with energies 1.8 eV and 4 eV, respectively. Determine the value of Planck’s constant
Solution:

Let the ground state energy of the hydrogen atom = -E₀. Hence, the minimum amount of energy E₀ is required to liberate its electron, i.e., the work function of the hydrogen atom, W₀ = E₀.

So, if the incident photon can provide E amount of kinetic energy to the electron, then

hf = E + E₀  or, hc/λ = E + E₀

In the first case,

λ₁ = 800 Å = 800 × 10^-8  cm ,

E₁ = 1.8 eV = 1.8 × 1.6 × 10^-12 erg

In the second case,

λ₂  = 700 A = 700 × 10^-8   cm ,

E₂ = 4.0 eV = 4.0 × 1.6 × 10^-12   erg

From equation (1), we get,

⇒ \(\frac{h c}{\lambda_1}-\frac{h c}{\lambda_2}=E_1-E_2\)

Or, \(h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)=E_1-E_2\)

= E₁ – E₂

Or, \(h=\frac{E_2-E_1}{c} \cdot \frac{\lambda_1 \lambda_2}{\lambda_1-\lambda_2}\)

∴ h = \(\frac{(4.0-1.8) \times 1.6 \times 10^{-12}}{3 \times 10^{10}} \times \frac{800 \times 700 \times 10^{-16}}{(800-700) \times 10^{-8}}\)

= 6.57 × 10^-27  erg.s

Example 5. In absorbing 10.2 eV energy, the electron of a hydrogen atom Jumps from Its Initial orbit to the next higher energy orbit. As the electron returns to the former orbit, a photon Is emitted. What Is the wavelength of this photon?
Solution:

According to Bohr’s postulate,

hν = E_i – E_f

Or, \(\frac{h c}{\lambda}=E_l-B_f\)

Or,\(\lambda=\frac{h c}{E_l-E_f}\)

The difference between the two energy levels,

E_i – E_f = 10.2 eV = 10.2 ×  1.6 × 10^-12 erg

λ = \(\frac{6.55 \times 10^{-27} \times 3 \times 10^{10}}{102 \times 16 \times 10^{-12}}\)

= 1204 × 10⁸cm

= 1204 Å

WBCHSE Class 12 Physics Atom Ionisation Energy And Ionisation Potential

Ionization energy:

The minimum amount of energy required to ionize an atom of an element at its ground state is called the ionization energy of that element.

Example:

The ground state energy of hydrogen atoms = -13.6 eV. In a H⁺ ion, the only electron of the corresponding hydrogen atom has been removed. In this state, the electron is no longer attracted by the atom, i.e., its potential energy becomes zero.

Again, the condition for minimum energy to be possessed by the electron is that its kinetic energy is zero; as a result its total energy = kinetic energy + potential energy = 0. Naturally, if the electron is brought from a -13.6 eV energy state to a zero energy state, the atom can be converted into an ion. So, the minimum amount of energy supplied = 0- (-13.6) = 13.6 eV

Hence, the ionization energy of hydrogen = 13.6 eV.

Ionization potential:

The minimum potential to be applied to an atom of an element in its ground state to convert it into a positive ion is called the ionization potential of that element.

Example:

According to the definition, if 1 V potential is applied to an electron having charge -e, the gain in energy of the elec¬ tron = 1 eV.

Ionization energy of hydrogen = 13.6 eV; so, to supply this 13.6 eV energy to the electron of the hydrogen atom, a minimum 13.6V potential should be applied to it.

Hence, the ionization potential of hydrogen = 13.6 V

Atom Ionisation Energy And Ionisation Potential Numerical Examples

Example 1. A hydrogen atom in its ground state, is excited using monochromatic radiation of wavelength 975 Å. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. Given, ionization energy of the hydrogen atom is 13.6 eV
Solution:

Wavelength of incident radiation,

A = 975 Å = 975 × 10^-8cm

∴ Energy of this photon

hf = \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{975 \times 10^{-8}}\)

= \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{975 \times 10^{-8}} \mathrm{erg}\)

= \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{975 \times 10^{-8} \times 1.6 \times 10^{-12}} \mathrm{eV}\)

= 12.47 eV

The ionization energy of the hydrogen atom = 13.6 eV, i.e., the ground state energy of this atom = -13.6 eV. So, the energy ofthe excited state in which the atom is raised under the influence of incident radiation = – 13.6 + 12.74 = -0.86 eV.

The quantum number in the ground state = 1. So, if the quantum number in the excited state be n, then

– 0.86 = \(-\frac{13.6}{n^2}\)

Or = \(\frac{13.6}{0.86}\)

= 16 (approx)

So, n = 4, i.e., the excited state is the fourth Bohr orbit. During the transition from the fourth Bohr orbit to the ground state, the decrease in energy ofthe atom may occur in 6 different ways. As a result, 6 lines will be obtained in the spectrum

Of them, during the transition from n = 4 to n = 3, the energy difference is the least and hence in this case, the wavelength of the emitted spectral line will be the maximum

Energy of the third Bohr orbit = – \(\frac{13.6}{3^2}\)

= -1.51 eV

∴ Decrease in energy due to transition from n = 4 to n = 3

=-0.86 -(-1.51) = 0.65 eV

Hence, the relation hf = hc/ λ

= E₄ – E₃ gives

= \(\frac{h c}{E_4-E_3}\)

= \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{0.65 \times 1.6 \times 10^{-12}}\)

= 19110 × 10^-8  cm

= 19110 Å

The relation between the energy of photon E and wavelength A after substituting the values of different constants is E

eV =12400/λ (in Å)

In any calculation, this relation can be used directly.

Success and Failure of the Bohr’s Theory

1. With the help of Bohr’s theory, the spectrum of hydrogen or a hydrogen-like atom can be explained almost accurately. QQ With the help of Bohr’s theory, the main characteristics of the atomic spectrum of alkali metals can be explained.

2. Bohr’s theory cannot analyze the energy of atoms having more than one electron, nor can it explain the wavelengths of spectral lines of such atoms quantitatively. Still Bohr model is the most suitable model supporting the electronic configuration of any atom.

3. Actually, every spectral line of the spectrum of hydrogen or other atoms consists of several spectral lines; they remain so close to each other that with the help of an ordinary prism disperser, they cannot be identified separately.

4. This fine structure of every primary spectral line cannot be explained with the help of Bohr’s theory. This can be called the limitation of Bohr’s theory instead of its failure.

5. If a charged particle possesses acceleration, it radiates electromagnetic radiation. This theory of classical physics has been proved by different experiments. However, Bohr’s theory fails to explain why no electromagnetic radiation is emitted from the electron having centripetal acceleration while revolving in a Bohr orbit.

WBCHSE Class 12 Physics Atom X rays

Roentgen’s discovery:

In 1895, German physicist Wil helm Roentgen observed that, when high-velocity cathode rays were obstructed by any solid target, an invisible ray came out from that target.

The characteristics of this ray that he observed are. The rays:

• Can produce fluorescence,
• Can affect photographic plates,
• Can penetrate thin sheets of light materials like paper, glass, wood, aluminum, etc.
• Cannot penetrate heavy metals like iron, lead, etc. Rather. it casts a shadow behind them

At that time, the nature of this ray was unknown, and hence eV = hfx Amin Roentgen called this ray X-ray

Nature of X-rays

X-rays are not deflected by electric or magnetic fields; from this, we can conclude that X-rays are not streams of charged particles.

So, an X-ray is either a stream of high-velocity uncharged particles or a kind of wave

If an X-ray is considered to be a kind of wave, it should show properties like interference, diffraction, etc. But in any traditional experimental setup, these properties were not observed, until finally Max von Laue and others after him observed the diffraction of X-ray by passing it through three-dimensional crystals: Thus it was proved that X-ray is a kind of wave. The wavelength of this ray is so small that its diffraction is possible only by crystals of regular intermolecular space (10^-8 cm approximately).

Later on, it was possible to observe the wave properties of X-rays like reflection, refraction, interference, and polarisation.

So, it has been observed, that an X-ray is not a stream of any high-velocity uncharged particles, but an electromagnetic wave like visible light. Though compared to visible light, the wavelength of X-ray is very small, almost \(\frac{1}{1000}\) parts or even less than that of the former. It was not until 1923 when A. H. Compton, using his X-ray scattering experiment, established the particle nature of photons, and hence of X-rays

Frequency and Wavelength of X-ray

According to quantum theory, electromagnetic radiation is a stream of massless and chargeless photons which travels

With the speed of light (c = 3 × 10⁸ m . s^-1). The energy of each photon,

E = hf = hc/λ ………………(1)

Here, h = Planck’s constant =6.63 × 10^-34J . s;

f = frequency and λ = wavelength of the electromagnetic wave

When an electron (charge = e ) overcomes a potential difference V, energy gained by it = eV. If this energy is spent entirely to eject an X-ray photon, the energy and frequency of the photon become maximum. If the corresponding wavelength is taken as λ_min, then

eV = \(h f_{\max }\) = \(h \frac{c^{\circ}}{\lambda_{\min }}\)

Or, \(\lambda_{\min }=\frac{h c}{e V}\) ………………(2)

In Sl, e = 1.6 × 10^-19 C, c = 3 × 10^-8 m.s^-1 and

h = 6.63 × 10^-34J . s

If the potential difference between the anode and the cathode is V, putting these values in equation (2) we get

= \(\frac{1.243 \times 10^{-6}}{V} \mathrm{~m}\)

= \(\frac{12430}{V}\) Å

= \(\frac{1243}{V}\) ……………(3)

For example,

If V = 50 kV = 50000V, then λ_min = 0.2486 Å

If V = 10 kV = 10000 V, then λ_min = 1.243 Å

Discussions:

1. When the whole energy of an electron (eV) belonging to cathode rays is utilized to bombard the target to produce a photon, only then the wavelength of the X-ray becomes equal to λ_min as given in equation (2). Generally, the whole energy is not converted into the energy of an X-ray photon; hence the wavelength becomes more than λ_minin For this, is called the minimum wavelength of X-ray.

2. The higher the value of the potential difference V_i smaller the wavelength of X-rays and hence greater its penetrating power. From equation (3), we see that V should be more than 10⁶ volts to produce hard X-rays equivalent to the wavelength 0.01 Å

3. The wavelength of X-ray \(\frac{1}{1000}\) Is part of the wavelength of visible light. Hence, from the equation, E = \(\frac{h c}{\lambda}\), we can say that, the energy of an X-ray photon is about 1000 times greater.

4. Soft and hard X-rays:

The wavelength range of X-rays is from 0.01 Å to 10Å. If λ ≈ 10 A, according to the relation E = hc/λ, the energy of the X-ray photon becomes 1240 eV or 1.24 keV (approx.); this kind of X-ray has less penetrating power and is called soft X-ray. On the other hand, if λ  =; 0.01 Å, the energy of the X-ray photon becomes 1.24 MeV (approx.). Due to this high energy, the penetrating power of X-ray becomes very high. It is known as a hard X-ray

Uses of X-rays:

1. Important uses of X-rays in medical science:
   • Radiography: X-rays are used for the detection of fractured bones and kidney stones.
   • Radiotherapy: Hard X-rays are used in radiotherapy to destroy the cancer-affected cells.
   • Fluoroscopy: Fluoroscopy is an imaging technique in which X-rays are used to take real-time moving images of the internal structures of a living body.

Important uses of x rays in other fields:

• X-rays are used to analyze the structures of different crystals.
• In metallurgy, X-rays are used to determine the defects in metallic castings

WBCHSE Class 12 Physics X-rays And Atomic Numbers

When a solid target (made of solid copper or tungsten) is bombarded with a stream of high-velocity electrons having a kinetic energy of some keV or higher X-rays are emitted from the target. Sometimes, this kind of material absorbs X-rays. Analyzing the emitted or absorbed X-rays, we come to know many aspects of the atomic structure of these elements

Consider a molybdenum (Mo) target being bombarded by a stream of electrons having a kinetic energy of 35 keV. The spectrum of the wavelengths of X-rays is thus produced.

This spectrum is formed due to the superposition of two kinds of spectra: O continuous spectrum, characteristic or line spectrum. A continuous X-ray spectrum superimposed with some brighter lines is formed on the photographic plate. The origins of the two spectra are different and explained in the next article.

Continuous X-ray Spectrum

During the discussion of continuous spectrum, the characteristic spectrum consisting of two sharp peaks will be overlooked.

Suppose an electron traveling with kinetic energy K₀ undergoes collision with an atom of molybdenum  Assume that the electron loses Δk amount of energy by this collision. This energy is converted into the energy of an X-ray photon. It should be mentioned here that, the atom being much heavier than an electron, the amount of energy transferred to the atom is neglected.

The electron may collide again with another atom after its colli¬ sion with the first atom. In this case, the electron collides with (K₀-ΔK) amount of energy. The X-ray photon thus emitted, generally possesses less energy than that of the previous photon.

In this way, the electron may suffer successive collisions with different atoms until it comes to rest. The photons thus emitted during these collisions having different energies, form a part of the continuous spectrum of X-rays. But in actual practice, a tar

Get is hit with innumerable electrons. Hence, the entire X-ray spectrum looks like that. This kind of spectrum has an important characteristic. It has a definite cut-off wavelength (say, λ_min). Below this cut-off

Wavelength, there is no existence ofthe spectrum. If the electron loses its whole kinetic energy (XQ) in the first impact, an X-ray of wavelength is emitted. If/ be the frequency ofthe emitted X-ray photon, then

⇒ \(K_0=h f=\frac{h c}{\lambda_{\min }}\)

⇒  \(\lambda_{\min }=\frac{h c}{K_0}\)………………(1)

So, the die value does not depend on the nature ofthe solid used as the target. So, if copper is used instead of molybdenum as the die target material, the continuous X-ray spectrum changes but the die value of Am;n remains the same

WBCHSE Class 12 Physics Atom X-rays And Atomic Numbers Numerical Examples

Example 1. If a stream of electrons of kinetic energy 36 keV is bombarded on a molybdenum target, what will be the cut-off wavelength of the emitted X-ray? nucleus
Solution:

We know that,  \(\lambda_{\min }=\frac{h c}{K_0}\)

Where h = 4.14 × 10^-15 eV. s, c = 3 ×  10⁸ m .s¹

Given. K₀ = 36 keV= 3.6 × 10^-4 eV

∴  λ_min  = \(\frac{4.14 \times 10^{-15} \times 3 \times 10^8}{3.6 \times 10^4}\)

= 3.45  × 10^-11  m

= 0.0345 nm

Example 2. What minimi terminal potential difference of a Coolidge tube should be maintained to produce an X-ray of wavelength 0.8 Å? [h = 6.62 × 10^-34 J.s, e = 1.60 × 10^-19  r= C = 3 × 10⁸ m. s ^-1]
Solution:

The energy of X-ray photons,

E = hc/λ

[here,  λ = 0.8 Å = 0.8 × 10^-10 m]

If electrons are accelerated in an X-ray tube by a potential difference V, the energy of an electron = eV; if this energy is completely converted into the energy of an X-ray photon, the value of V will be the minimum

So, eV_min = \(\frac{h c}{\lambda}\)

V_min = \(\frac{h c}{e \lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{1.60 \times 10^{-19} \times 0.8 \times 10^{-10}}\)

= 15516 V

Characteristic X-ray Spectrum

The two sharp peaks in the spectrum of characteristic X-rays are named as K_α and K_β  These two peaks mainly form the spectrum of the characteristic X-rays of molybdenum.

In an X-ray tube (Coolidge tube), the target is bombarded by high-energy cathode rays. The electrons present in the rays are of very high energy, their effect is not limited to the outer electron levels; the levels adjacent to the nucleus are also affected.

If an electron from any of these levels comes out of the atom, a deficit of electrons occurs in that orbit. If this deficit occurs in the X-orbit (the orbit closest to the nucleus), an elec¬ tron from a higher energy state will transit to this X-orbit.

Now, during the transition of an electron from L -orbit to Korbit, it radiates energy as a photon. This radiation forms the K_α -peak of the characteristic X-ray spectrum of molybdenum. The energy level diagram of molybdenum is  How different peaks of the spectrum are formed has been shown in this diagram. Again, during the transition of the electron from M orbit to K-orbit, the radiation thus emitted forms the Kβ-peak of the spectrum.

This spectrum contains several smaller peaks and the brightness of these lines in the spectrum is very low. Formation of L_α and L_β peaks among the small peaks is als

Moseley’s Law

In 1913, British physicist H. G. I. Moseley analyzed the spectrum of characteristic X-rays of all the elements he knew as targets. He observed that the spectra of different elements, particularly the K_α -peaks follow a particular rule. According to the position of different elements in the periodic table, he drew a graph of the square root of the frequency of K_α and obtained a straight line.

A part of the graph based on which, Moseley concluded that “there was a fundamental quantity which increases by regular steps as we pass from one element to the next”. In 1920 Rutherford identified this quantity as the atomic number Z of the element which denotes the number of positive charges present in the nucleus. So, only from the atomic number of an element can its identity be known, not from its atomic weight.

Statement of Moseley’s law:

The square root of the frequency of a peak of the characteristic X-ray spectrum of

Any element is directly proportional to the atomic number of that element.

If the frequency of K_α -line of any element having atomic num¬ ber Z be f. Then according to Moseley’s law,

√f ∝ Z

Explanation of Moseley’s plot from Bohr’s theory:

With the help of the equation of the n-th energy state of an atom obtained from Bohr’s theory, this plot can be explained.

Equation (13)

⇒ \(E_n=-\frac{R c h}{n^2}=-\frac{13.6}{n^2} \mathrm{eV}\)………………(1)

Where, n = 1,2,3,

We know that in an atom containing two or more electrons, only two are in the K-orbit Let any of these come out of the atom. Then the electrons present in other orbits like L, and M, would experience not only the effect of the positive charge of the nucleus but also the influence of the negative charges of the remaining electron in the kT-orbit.

This is because the radius of the kT-orbit in an atom is the least compared to that of other orbits. So, we can assume that, relative to the surface of the sphere on which the electron of the K-orbit lies, the other electrons lie outside.

So, the effective amount of positive charge that attracts an electron of L, M, orbits is (Z- 1)e, where Z is the atomic number. The above equation (1) is applicable for hydrogen atoms. In the case of an atom having atomic number Z, its changed form is applicable. The equation for the n-th energy of the atom is

En = \(E_n=-\frac{13.6(Z-1)^2}{n^2}(\text { in } \mathrm{eV})\) ……………. (2)

The earlier discussion shows that Ka of the spectrum is produced due to the transition of an electron from the L(n =2) orbit to the K(n = 1) orbit. Due to this transition, if the frequency of the emitted X-ray is f, then

⇒ \(h f=\left(E_n\right)_{n=2}-\left(E_n\right)_{n=1}=E_2-E_1\)

= \(-\frac{13.6(Z-1)^2}{2^2}+\frac{13.6(Z-1)^2}{1^2}\) (in eV)

∴ hf = 10.2(Z- 1)²

√f ∝  (Z – 1)…………………………………. (3)

Equation (3) is the equation of a straight line. Hence, the graph of the square root of the frequency of the peak K_α concerning the atomic number of the atom is a straight line. In this way, the theoretical basis of Moseley’s plot is established in the light of Bohr’s theory.

Importance of Moseley’s work:

1. According to Moseley’s law, the characteristic X-ray spectrum is regarded as the identifying character of an element.

2. Before 1913, the elements were arranged in the periodic table according to the increasing order of their atomic weights.

3. Despite that, according to the basis of chemical tests, some elements were placed before the elements hav¬ ing comparatively lower atomic weights in the periodic table. The reason for this exception was not understood before Moseley’s analysis.

4. Moseley showed that there would not be any exception to the arrangements of the periodic table if we arranged the elements according to their increasing atomic numbers.

5. There were many vacant places in the periodic table in 1913. After the discovery of Mosley’s law, it has been possible to fill those gaps accurately.

6. Lanthanide elements are more or less identical in respect of their chemical properties. Thus, it was difficult to identify

WBCHSE Class 12 Physics Atom Conclusion

Bhor model of an atom is based on three postulates

1. For the revolutions of electrons inside an atom, there are some allowed orbits. When an electron revolves in an allowed orbit, it does not radiate energy. Since the energy of an electron in any allowed orbit is constant, die orbits are called stable orbits.

2. An electron can transit from one stable orbit to another. During this transition, homogeneous absorp¬ tion or emission of radiation occurs whose frequency is determined from the relation, hf = E₁~ E₂. [h = Planck’s constant, E₁~ E₂ energy difference of the electron in the two stable orbits]

3. The orbits, where the angular momentum of the electron is an integral multiple of h/2λ are the only allowed orbits and are called Bohr orbits.

4. In the visible region of the atomic spectrum of hydrogen, four bright lines can be observed. Their wavelengths, as obtained from the experiment are 6563 Å, 4861 Å, 4341 Å, and 4102 Å. These four lines are known to form the Balmer series of hydrogen spectra.

5. The number of complete waves of radiation present in unit length is 1/λ, and hence 1/λ (=f) is called wave number.

6. In the first Bohr orbit, Le., in K-orbit, the energy electron becomes minimum. This is known as the ground state energy or the lowest energy level of the atom.

7. The minimum amount of energy required to ionize an atom of an element in its ground state, is called the ionization energy of that element.

The ionization energy of hydrogen = 13.6 eV

8. The minimum potential applied to an atom of an element at its ground state to convert it into a positive ion, is called the ionization potential of that element Ionisation potential of hydrogen = 13.6 V

9. X-ray is not a stream of fast-moving charged or uncharged particles, rather it is an electromagnetic wave like light

10. The energy of X-ray is more than that oflight Spectrum of X-rays is formed due to die superposition of two spectra

1. Continuous spectrum and
2. Characteristic tic spectrum.

A continuous X-ray spectrum has a definite cut-off wavelength below which no radiation exists.

11. Moseley’s law: The square root of the frequency of a peak of the characteristic X-ray spectrum of an element is directly proportional to the atomic number of that element

12. Energy difference of an electron in two stable orbits,

E₁~ E₂ = hf

13. According to Bohr’s quantum condition, if rn = radius of n -th orbit and vn = velocity of the electron in n -th orbit, then

Here, the principal quantum number, n = 1, 2, 3,…………

14. First Bohr radius of ground state (rt = 1)

⇒ \(a_0=\frac{\epsilon_0 h^2}{\pi m e^2}\)

= 0.53 Å

= 0.053 nm

And \(E_1=\frac{m e^4}{8 \epsilon_0^2 c h^3}\) . ch = -Rch

Where , R = \(\frac{m e^4}{8 \epsilon_0^2 c h^3}\)

Rydberg constant = 1.09737 × 10⁷ m^-1

15. Ground state energy of hydrogen atom =-13.6 eV

16. For the hydrogen spectrum, if A is the wavelength of the spectral line then

⇒ \(\frac{1}{\lambda}=R\left[\frac{1}{\left(n^{\prime}\right)^2}-\frac{1}{n^2}\right]\)

17. According to Bohr’s postulate, when the electron in a hydrogen atom transits from a higher energy state lower energy state Enÿ, a photon is emitted from the hydrogen atom. If the frequency of this photon is (wavelength, λ = then

⇒ \(E_{n_i}-E_{n_f}=h f=\frac{h c}{\lambda}\)

The minimum wavelength of X-ray

⇒ \(\lambda_{\min }=\frac{h c}{e V} .\)

In SI, e = 1.6C, c = 3× 10⁸m. s^-1 , and

h = 6.63 × 10^-34  J.s^-1

⇒ \(\frac{1.243 \times 10^{-6}}{V} \mathrm{~m}\)

= 12340/v Å

18. If h, f, and c are Planck’s constant, the frequency of the emitted X-ray photon, and the speed of the X-ray respectively, then the kinetic energy of the incident electron

⇒ \(K_0=h f=\frac{h c}{\lambda_{\min }}\)

Or,\(\lambda_{\min }=\frac{h c}{K_0}\)

λ_min = cut-off wavelength

19. If the electron in a hydrogen atom is excited to the n -th state, then the number of possible spectral lines it can emit in transition to the ground state is

⇒  \({ }^n C_2=\frac{1}{2} n(n-1)\)

20. Speed of electron in the n -th orbit of the hydrogen atom

⇒  \(v_n=\frac{c}{137 n}\)

WBCHSE Class 12 Physics Atom Very Short Question And Answers

Question 1. In the Bohr model of the hydrogen atom, what Is the ratio of the kinetic energy to the total energy of the electron in any quantum state
Answer:

According to the Bohr model, if the kinetic energy of the electron in any quantum state is E, total energy becomes -II. So, the ratio of these two energies is

Question 2. How are X-rays produced?
Answer:

When a stream ofelectrons, having a kinetic energy of a few keV or more, hits a solid target (like copper, tungsten, molybdenum, etc.), X-rays are emitted from that target

Question 3. In which part of the electromagnetic do the spectral lines of a hydrogen atom given by the Balmer series occur?
Answer: Visible range.

Question 4. What is Bohr’s quantum condition for the angular momentum of an electron in a hydrogen atom?
Answer:

Angular momentum = nh/ 2: n = 1,2,3,……………..

Question 5. Name the different types of X-ray spectrum
Answer:  Continuous spectrum and characteristics spectrum

Question 6. The nucleus contains the entire ____________ charge and nearly the entire
_______ of an atom
Answer: Positive And Mass

Question 7. In Rutherford’s experiment, which particle is responsible for the low-angle scattering of a -particles?
Answer: Electron

Question 8. The total energy of the electron in the first Bohr orbit of a hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron?
Answer: 13.6 eV, -27.2 eV

Question 9. The wavelength of a spectral line found in the atomic spectrum of hydrogen is 4861 Å. Between which two quantum states does the transition of electrons occur to generate this line? Rydberg constant = 1.097 × 10⁸. m^-1
Answer: From the fourth to the second

Question 10. What is the order of magnitude of the ratio between the volume of? an atom and that of its nucleus?
Answer:  10¹²: 1

Question 11. The energy of an electron in the first excited state of a hydrogen atom is -3.4 eV what is the kinetic energy of this electron
Answer: + 3.4 eV

Question 12. What is the ratio of the areas of the first and the second orbits of a hydrogen atom?
Answer: 1:16

Question 13. What is the approximate diameter of a hydrogen atom
Answer: 1.06A

Question 14. If the radius of the first Bohr orbit is r, what would be the radius of the second?
Answer: 4r

Question 15. The radius of the first electron orbit of a hydrogen atom is 5.3 × 10^-11m. What is the radius of its second orbit?
Answer: 21.2 m

Question 16. An electron beam hits a target to produce a continuous X-ray spectrum. If E is the kinetic energy of each electron in the beam, what would be the lowest wavelength of the emitted X-rays?
Answer: hc/E

WBCHSE Class 12 Physics Atom Assertion Type

Direction: These questions have statement 1 and statement 2 Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true, and statement II is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, and statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The total positive charge and almost all the mass of an atom are confined in the nucleus.

Statement 2: In Rutherford’s a -particle scattering experiment, the majority of the a -particles penetrate the target without any deflection.

Answer:   2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 2.

Statement 1: The circular orbit of the electron as stated in Rutherford’s atomic model can never be a stable orbit

Statement 2: Any accelerated charged particle radiates energy

Answer: 1. Statement 1 is true, statement 2 is true, and statement II is a correct explanation for statement 1.

Question 3. 

Statement 1: The distance of the electron from the nucleus is minimal when a hydrogen atom is in the ground state.

Statement 2: According to Bohr’s theory the radius of circular motion of an electron in n -th energy state, rn oc n.

Answer: 3. Statement 1 is true, and statement 2 is false.

Question 4.

Statement 1: The kinetic energy of an electron in the first excited state of a hydrogen atom is 6.8 eV.

Statement 2: The total energy of the first excited state of a hydrogen atom is -3.4 eV.

Answer:  4. Statement 1 is false, and statement 2 is true.

Question 5.

Statement 1: All lines in the Balmer series of the hydrogen spectrum are in the visible region.

Statement 2: Balmer series is formed due to the transition of electrons from 2, 3, 4- – permitted energy levels to the ground level.

Answer: 3. Statement 1 is true, and statement 2 is false.

Question 6.

Statement 1: The ionization potential of a hydrogen atom is 13.6 eV.

Statement 2: The Ground state energy of the hydrogen atom is 13.6 eV.

Answer: 3. Statement 1 is true, and statement 2 is false.

Question 7.

Statement 1: Magnetic moment of election In the n-th orbit of the hydrogen atom

Statement 2: The magnetic moment of a particle of charge  rotating in an orbit of radius r with velocity v is given by \(\mu=\frac{1}{2} e v r\)

Answer: 1. Statement 1 is true, statement 2 is true, and statement II is a correct explanation for statement 1.

WBCHSE Class 12 Physics Atom Match The Columns

Question 1. Match the columns for the hydrogen atom.

Answer: 1- C, 2-D, 3- C, 4- A

Question 2. Match the columns for an electron rotating in the n-th orbit of an atom

Answer: 1- A, 2 – C, 3 – D, 4 – B

Question 3. 

Answer: 1- D, 2 – C, 3 – A, 4- B

Question 4. Match Column A (fundamental experiment) with Column B (its 1 conclusion) and select the correct option from the choices given below the list:

1. 1- A, 2- D, 3 – C
2. 1- B, 2- D, 3 – C
3. 1- B, 2- A, 3 – C
4. 1- D, 2- C, 3 – B

Answer: 3. 1- B, 2- A, 3 – C