WBCHSE Class 12 Physics Notes For Alternating Current

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Direct Current Or DC And Alternating Current Or Ac

Before discussing alternating current thoroughly, it would be helpful to recapitulate on direct current in short.

Direct Current (dc): In an electrochemical cell, the electrical nature of the positive and negative electrodes remains unchanged, i.e., it does not vary with time. The current in a circuit from this cell always remains unidirectional.

  • This kind of current is known as direct current or unidirectional current However, despite being unidirectional, the magnitude of the current may decrease or increase with time.
  • In the graphs, different kinds of direct current concerning time are shown. Of these, the current shown in the is steady current the magnitude of current in this case always remains unchanged.

Alternating Current Different Kinds Of Direct Current With Respect To Time

Alternating Current (ac): An electrochemical cell does not provide large amounts of electrical energy. All electric power plants use a machine called a dynamo or generator for this purpose.

  • The characteristics of this machine are—the electrical nature of its two electrodes does not remain constant, but changes from positive to negative, and from negative to positive periodically.
  • As a result, current through the external circuit connected to this source gets reversed periodically, i.e., the flow of current does not remain unidirectional.
  • The electromotive force obtained from this source is called alternating emf and the current in the circuit is called alternating current.
  • All modern electrical appliances, simple or delicate, are enabled. Hence the importance of the study of ac. Current-time graphs of some alternating currents are shown.

Read and Learn More Class 12 Physics Notes

Alternating Current Current Time Graphs Of Some Alternating Currents

The following characteristics of alternating current are to be noted.

  1. An alternating current is of a wave nature.
  2. The magnitude of the current above the time axis is taken as positive, and that below is taken as negative. This implies that the direction of the current gets reversed periodically, and at the moment of this transition, the instantaneous magnitude of the current becomes zero.
  3. The waveforms of different alternating currents may be different. The waveforms of the three different currents shown are respectively sinusoidal, square, and triangular. Of them, the discussion about the sinusoidal waveform is of particular importance; because, by appropriate mathematical analysis, the square, triangular, or any other waveform can be reduced to a combination of a large number of sinusoidal waves.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Source Of Alternating Current Ac Dynamo

Almost the entire electrical energy requirement of the present-day world is derived from the phenomenon of electromagnetic induction.

The machine employing this phenomenon is called a dynamo or generator. A conducting coil is set to rotate inside a magnetic field, as a result of which a current is induced in the coil. This is the basic mechanism.

Definition: The machine in which the mechanical energy of a rotating conducting coil placed in a magnetic field is converted into electrical energy, is called a dynamo or generator.

Description: The main parts of an alternating current dynamo are shown.

Alternating Current Dynamo Or Generator

N, S: The poles of a strong horseshoe magnet which produces a uniform magnetic field directed from the north to the south pole in the gap between them. Some lines of force are shown in the figure. This magnet is called the field magnet.

ABCD: A rectangular coil called an armature is placed in a uniform magnetic field, which usually contains several turns.

The coil in this case is made to rotate about the axis normal to the magnetic field; the direction of rotation in this case is as shown in the diagram.

R1, R2: Two smooth brass rings called slip rings. They are connected to the coil at its open ends A and D.

T1, T2: Two carbon brushes fitted to the rings with the help of springs, which keep the brushes pressed against the rings R1 and R2.

L: An electric lamp, to indicate the presence of current.

Observation: The lamp glows as long as the coil rotates. From this, we can infer that an electric current is flowing through the external circuit attached to T1 and T2. As soon as the rotation of the coil ceases, the current stops and the lamp goes out.

Working Principle: when the coil ABCD arms AB and CD intersect the magnetic lines of force. As a result, electromagnetic induction takes place.

  • At a certain moment, when the motion of the arm AB is downwards, according to Fleming’s right-hand rule, a current will be induced in the direction BA. At the same time due to the upward motion of the arm CD, the direction of induced current will be along DC.
  • So, a current will be set up in the direction of DCBA in the coil and will flow from T1 to T2 in the external circuit In this situation T1 and T2 behave as the positive and negative poles of a battery, respectively.
  • Here, arms BC and AD do not intersect the magnetic lines of force.
  • When, after half-rotation, the positions of the arms AB and CD are interchanged, by applying Fleming’s right-hand rule, it is observed that current is now induced in the direction ABCD.

As a result, current flows in the second half of rotation from T2 to T1 in the external circuit. Thus, the polarities of T1 and T2 get reversed, reversing the direction of the current.

  • This reversal of the direction of the current goes on periodically. Every time the coil crosses its vertical position, the direction of the current gets reversed. Hence, this dynamo is called an AC dynamo.
  • Because the heating effect of electric current {αI2, i.e., the same for +I and -I) does not depend on the direction of current, the lamp continues to glow as long as the coil rotates.
  • If a DC galvanometer (for example., a moving coil galvanometer) is placed instead of the electric lamp, no deflection of its pointer would be observed.

Factors Affecting The Emf And Current: The electromotive force, and hence the current is directly proportional to

  1. Area of the coil,
  2. Number of turns of the coil,
  3. speed of rotation of the coil and
  4. Strength of the magnetic field.

So, if any one of them is increased, the emf will also increase. The current however will decrease with an increase in the resistance of the circuit.

Electric motor: The principle of action of an electric motor is just opposite to that of a dynamo.

  • A motor is a device in which a current-carrying coil, placed suitably in a magnetic field, rotates on the principle of action of magnets on currents.
  • A motor converts the electrical energy of the coil into its mechanical energy.

Differences Between Dynamo And Motor:

Alternating Current Differences Between Dynamo Amd Motor

However, the discussions on both DC and AC motors are beyond our present syllabus.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Variation Of Alternating Current

Waveform Of Alternating Current: Let a rectangular. coil ABCD is rotating with uniform angular velocity in a uniform magnetic field. The coil is viewed in a way such that only the AD end of the coil becomes visible. The different positions of this end concerning time are shown.

Alternating Current Waveform Of Alternating Current

Let the period of the coil be T. If the coil starts rotating from its vertical position, then at \(\frac{T}{2}\) and T, it again comes to its vertical position.

At these moments both AB and CD are moving parallel to the magnetic field. Considering the equation, e = Blvsind with θ = 0° the induced emf and hence the induced current will be zero.

On the other hand, at the times \(\frac{T}{4}\) and \(\frac{3T}{4}\), the coil lies horizontally. At those positions,r. AB and CD are directed normally to the magnetic field (θ = 90°), and the induced current becomes maximum.

However, at the time \(\frac{T}{4}\), the direction of current is along DCBA, while at the time \(\frac{3T}{4}\), it is along ABCD. So, if the current in the first case is taken as positive, then the current in the second case will be negative.

This figure clearly shows a sine-wave; this is the wave nature of current if the coil starts rotating from its vertical position.

On the other hand, if the coil, starts rotating from its horizontal position, the current Will be a cosine-wave. Note that, both sine and cosine waves are called sinusoidal waves.

The current completes a cycle of change in a time equal to the period of rotation of the coil. A complete wave is thus formed in every cycle.

Expression Of Alternating Current: Let a rectangular coil be rotating with uniform angular velocity ω in a uniform. magnetic field B. At any moment t, the angle between the normal to the coil and the direction of the magnetic field is θ, say.

If the area of the plane of the coil is A, the magnetic flux linked with the coil,

Φ = BAcosθ → (1)

Alternating Current Expression Of Alternating Current

Initially, the angle between the normal to the plane of the coil and the direction of the magnetic field = α (say). After rotation for time t, this angle becomes,

θ = ωt + α

So, from equation (1) we get, magnetic flux,

Φ= BA cos (ωt+α) → (2)

Hence, the magnitude of the induced emf in the coil of a single turn

⇒ \(e=\frac{d \phi}{d t}=\omega B A \sin (\omega t+\alpha)\)

For N turns,

⇒ \(e=\omega B A N \sin (\omega t+\alpha)=e_0 \sin (\omega t+\alpha)\) → (3)

[here, e0 = ωBAN]

If the total resistance of the coil and the external circuit is R, the induced current,

⇒ \(i=\frac{e}{R}=\frac{\omega B A N}{R} \sin (\omega t+\alpha)=\frac{e_0}{R} \sin (\omega t+\alpha)\)

⇒ i0sin(ωt + α) → (4)

Phase: The state of alternating current at any moment is expressed by its phase. In equations (3) and (4), (ωt + α) is the phase of the alternating current.

Phase Difference: If the phases of two alternating currents are δ1 and δ2, then(δ12) is the phase difference of those two currents.

If co is the same for the two currents, the phase difference
becomes (α1– α2)

Peak Value: As -1 ≤ sinθ ≤ 1, it can be concluded that

  • The maximum and minimum values of the electromotive force are e0 and -e0 respectively.
  • Accordingly, the maximum and minimum values of the current are i0 and -i0.

These magnitudes, e0 and i0 are called the peak values of electromotive force and current.

Dependence Of Peak Value On Different Factors: it is obvious from equations (3) and (4) that the peak values of emf as well as current depend directly on each of the quantities involved, viz., co, B, A, N. Moreover, the peak value of current also varies inversely with the resistance (R) of the circuit.

Period And Frequency: The definite time interval in which a complete cycle repeats itself, is called the period (T) of an alternating current.

If the angular velocity, of the coil, is co, then

∴ \(T=\frac{2 \pi}{\omega}\) → (5)

The number of complete waves produced in unit time is called the frequency (n) of an alternating current.

So, \(n=\frac{1}{T}=\frac{\omega}{2 \pi}\) → (6)

Frequency is the most important quantity in the expression of an ac. It is noted that the frequency of domestic electric supply in India is 50 Hz or 50 cycles per second(cps). It indicates that the direction of the current is reversed (50 x 2) or 100 times per second.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Circuit Symbol Of An AC Source

It represents the symbols of an AC source used in a circuit. Any of these symbols may be used for the purpose.

Alternating Current Circuit Symbol Of Ac Source

Average Values Of Alternating Voltage And Current

Alternating voltage (V) or Current (I) always increases, decreases, and changes periodically following the functions since or cos ωt. But the average values of these quantities, do riot change with time.

Calculation: Let an expression for sinusoidal alternating voltage,

V = V0 Sin ωt

Average value of it in a full cycle (i.e., for t = T)

= \(\frac{1}{T} \int_0^T V d t=\frac{1}{T} \int_0^T V_0 \sin \omega t d t=0\)

This is because the voltage in a half cycle is positive and in the next half it is of the same magnitude, but negative.

Alternating Current Average Values Of Alternating Voltage And Current

For convenience, the average value in a half-cycle, instead of in a full-cycle, in ac is considered as the average value \((\bar{V})\) of an alternating voltage.

∴ \(\bar{V}=\frac{1}{T / 2} \int_0^{T / 2} V_0 \sin \omega t d t=\frac{2 V_0}{\omega T}[-\cos \omega t]_0^{T / 2}\)

= \(\frac{2 V_0}{\pi}=0.637 V_0\)

Similarly, the average value of alternating current,

∴ \(\bar{I}=\frac{2 I_0}{\pi}=0.637 I_0\)

RMS Values Of Alternating Voltage And Current

Alternating voltage or current cannot be measured directly with instruments like galvanometers. They can only be measured indirectly following the thermal effect of current.

We know that heat generated in a current-carrying conductor is directly proportional to V2 or l2, i.e., heat thus generated does not depend on the direction of the current.

  • Due to the periodical changes of an alternating current, the heat in the conductor fluctuates from zero to a certain positive value. Hence the average value of heat should be proportional to the average value of V2 or I2.
  • The value most commonly used for an ac is its effective value. The effective value of ac is the amount of ac that produces the same heating effect as an equal amount of dc.
  • This is calculated by squaring all the amplitudes of the sine wave over one period, taking the average of these values, and then taking the square root.

The effective value, being the root of the average (or mean) of the square of the currents, is known as the root mean square (in short, rms) value.

Calculation: Let an expression for sinusoidal alternating voltage,

v = v0 sinωt

So, the mean square of V,

∴ \(\vec{V}^2=\frac{1}{T} \int_0^T V^2 d t=\frac{1}{T} \int_0^T \cdot V_0^2 \sin ^2 \omega t d t\)

= \(\frac{V_0^2}{2 T} \int_0^T(1-\cos 2 \omega t) d t=\frac{V_0^2}{2 T}\left[t-\frac{\sin 2 \omega t}{2 \dot{\omega}}\right]_0^T\)

= \(\frac{V_0^2}{2 T}\left(T-\frac{\sin 4 \pi}{2 \omega}+\frac{\sin 0}{2 \omega}\right)\) [∵ ωt = 2π]

= \(\frac{v_0^2}{2}\)

∴ rms value of V,

⇒ \(V_{\mathrm{rms}}=\sqrt{\bar{V}^2}=\sqrt{\frac{V_0^2}{2}}=\frac{V_0}{\sqrt{2}}=0.707 V_0\)

Similarly, rms value of I,

⇒ \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=0.707 I_0\)

i.e., rms voltage = 70.7% of peak voltage

and rms current = 70.7% of peak current

Relation between peak value, average value, and rms value of alternating voltage and current: From the above discussion we have,

⇒ \(\bar{V}=\frac{2 V_0}{\pi}=0.637 V_0\)

and \(V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=0.707 V_0\)

where V0 is the peak value of alternating voltage.

∴ \(V_{\mathrm{rms}}: \bar{V}=\frac{V_0}{\sqrt{2}}: \frac{2 V_0}{\pi}=\frac{\pi}{2 \sqrt{2}}\)

or, \(V_{\mathrm{rms}}=\frac{\pi}{2 \sqrt{2}} \bar{V}=1.11 \bar{V}\)

∴ \(V_{\text {rms }}>\bar{V}\)

Similarly, \(I_{\mathrm{rms}}>\bar{I}\)

This relation is shown graphically.

Alternating Current Alternating Voltage And Current

We know that the heating effect of electric current is directly proportional to I2. Hence ac ammeters and ac voltmeters, are designed based on the heating effect of current. They directly measure rms values of alternating current and voltage respectively.

Form Factor: The ratio of the rms value and the average value of alternating voltage or current is known as the form factor.

Thus form factor, \(f=\frac{V_{\mathrm{rms}}}{\bar{V}}=\frac{\frac{V_0}{\sqrt{2}}}{\frac{2 V_0}{\pi}}=\frac{\pi}{2 \sqrt{2}}=1.11\).

Note that, the above value of form factor applies only to sinusoidal voltage arid current. For different waveforms, the values of the form factor are different. For example, for a square wave

∴ \(V_{\mathrm{rms}}=V_0 \text { and } \bar{V}=V_0\)

∴ Form factor, \(f=\frac{V_{\mathrm{rms}}}{\bar{V}}=1\)

From the form factor of an alternating voltage or current an effective idea about the waveform can be obtained.

Effects Of Oil An Ac Or DC Currents On The Human Body: the three main factors that determine what kind of shock one experiences are the amplitude, frequency, and duration of the current passing through the body.

Direct current has zero frequency i.e., it has a constant amplitude. On the other hand, the peak value of an ac voltage (V0) is √2. times its rms value (Vrms), for example., a 220V ac supply is going 220√2 or 311 V (approx.) before coming down to zero.

So, one can get a 311V shock from a 220V-50Hz ac supply line for (50 x 2) hor 100 times per second. The calculation indicates the severity of electrocution from ac compared to that from the decision of average for the same duration of time.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Circuit Symbol Of An AC Source Numerical Examples

Example 1. Equation of an ac is \(I=, 10 \sin \left(200 \pi t-\frac{\pi}{15}\right)\) ampere. Determine the frequency and peak value of the current.

Solution:

Comparing with the general equation of ac, I = I0 sin(ωt + α) we get,

the peak value of current, I0 = 10 A

and angular frequency, ω = 200 π Hz

∴ Frequency, \(I=\frac{\omega}{2 \pi}=\frac{200 \pi}{2 \pi}=100 \mathrm{~Hz}\)

Example 2. If an ac is represented by I = 100 sin 200 πt ampere, determine the peak value of the current and period.

Solution:

Comparing with the general equation of ac, I = I0 sin(ωt+α) we get,

the peak value of current, I0 = 100 A

and angular frequency, ω = 200π HZ

∴ Time period, T = \(T=\frac{2 \pi}{\omega}=\frac{2 \pi}{200 \pi}=0.01 \mathrm{~s}\)

Example 3. An alternating current having a peak value of 141 A Is used to heat a metal wire. To produce the same rate of heating effect, another constant current IA is used. What is the value of I?

Solution:

∴ \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{141 \mathrm{~A}}{\sqrt{2}}=100 \mathrm{~A}\)

If a steady DC I produces the same rate of heating, then I = Irms = 100A.

Example 4. The peak value of an alternating magnetic field B is 0.01 T and the frequency is 100 Hz. If a conducting ring of radius 1 m is held normal to the field, what emf will be induced in the ring?

Solution:

n = 100 Hz

∴ Time period, \(T=\frac{1}{n}=\frac{1}{100} \mathrm{~s}\)

The time taken by the field to increase from 0T to, 0.01 T is \(\frac{T}{4}=\frac{1}{400} \mathrm{~s}\).

Now the area, A = π. m2 = π m2

Hence induced emf,

∴ \(|e|=\frac{d \phi}{d t}=\frac{d}{d t}(B A)=A \frac{d B}{d t}\)

or, \(|e|=\pi\left(\frac{0.01-0}{\frac{1}{400}}\right)=4 \pi \mathrm{V}=12.57 \mathrm{~V}\)

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Power Consumed In Ac Circuits Some Important Ac Circuits

Power consumed In ac circuits: In any dc circuit, the potential difference (voltage) and the current are always in phase. But this is not so for ac circuits—in general, a phase difference is evolved between the voltage and the current Mathematically the voltage V and the current I are expressed as,

V = V0 sinωt and I = I0 sin(ωt-θ)

where, θ = phase difference between the voltage and the current. θ always lies between -90° and +90°, i.e., -90° ≤ θ ≤ 90°. Then, power consumed (or power dissipated) in the circuit is,

∴ \(P=V I=V_0 I_0 \sin \omega t \sin (\omega t-\theta)\)

= \(V_0 I_0 \sin \omega t[\sin \omega t \cos \theta-\cos \omega t \sin \theta]\)

=\(V_0 I_0\left[\sin ^2 \omega t \cos \theta-\sin \omega t \cos \omega t \sin \theta\right]\)

= \(V_0 I_0\left[\sin ^2 \omega t \cos \theta-\frac{1}{2} \sin 2 \omega t \sin \theta\right]\)

Over a complete cycle, the average of \(\sin ^2 \omega t=\frac{1}{2}\) and that of sin2ωt = 0. So the average power of the circuit is,

∴ \(\bar{P}=\frac{1}{2} V_0 I_0 \cos \theta=\frac{V_0}{\sqrt{2}} \frac{I_0}{\sqrt{2}} \cos \theta=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta\)

No experiment can measure the instantaneous power P; every measurement leads to the average power \(\bar{P}\). This \(\bar{P}\) is referred to as the effective power, and is usually denoted by the simple symbol P. So the effective power (or true power) of an ac circuit is,

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta\) → (1)

Power Factor: The factor cosθ in equation (1) is vitally important; this factor arises, clearly, due to the voltage-current phase difference θ. This cosθ is called the power factor of an ac circuit.

  • If θ ≠ 0, cos θ< 1; so the power factor, in general, reduces the power consumed in an ac circuit to some extent Also, as -90°≤ θ ≤ 90°, the power factor cosθ is never negative; as a result, the power consumed in the circuit can never be negative.
  • The unit of power P is watt (W); to distinguish between P and the product Vrms Irms, watt is never used as the unit of VrmsIrms — the usual unit of this product is volt. ampere or V.A.
  • Incidentally, from the similarity with the dc expression P = VI, the product Vrms Irms is often called the apparent power of an ac circuit. Clearly,

true power = apparent power x power factor.

If voltage and current are in the same phase, then θ = 0 and cos θ = 1. In this condition, an ac circuit consumes the maximum power.

If the voltage and the current are either -90° or +90° out of phase, then cos θ = 0 and P = 0. An ac circuit of this type consumes no power. The corresponding current is referred to as wattless current.

Time Interval Between The Peak Values Of Voltage And Current:

v = v0 sinωt, I = I0sin(ωt-θ)

So, \(V=V_0, \text { when } \omega t=\frac{\pi}{2}, \text { i.e., } t=\frac{\pi}{2 \omega}\)

Similarly, \(I=I_0 \text {, when }\left(\omega t^{\prime}-\theta\right)=\frac{\pi}{2} \text {, i.e., } t^{\prime}=\frac{\pi}{2 \omega} 4 \frac{\theta}{\omega}\)

Thus, the minimum time interval between the peak values of ac voltage and current is,

∴ \(t_0=t^{\prime}-t=\frac{\theta}{\omega}\)

Purely Resistive Circuit

shows the circuit. Alternating voltage applied in the circuit,

Alternating Current Purely Resistive Circuit

V = V0sinω t → (1)

Here, the peak value of alternating voltage = v0

rms value, vrms = \(\frac{V_0}{\sqrt{2}} ; \text { frequency, } f=\frac{\omega}{2 \pi}\).

According to Ohm’s law,

⇒ \(I=\frac{V}{R}=\frac{V_0}{R} \sin \omega t\)

Or, \(I=I_0 \sin \omega t\)

So, the peak value of ac, \(I_0=\frac{V_0}{R}\)

⇒ rms value, \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{V_0}{\sqrt{2} R}=\frac{V_{\mathrm{rms}}}{R}\);

⇒ frequency, \(f=\frac{\omega}{2 \pi}\)

The factor sinωt in equations (1) and (2) indicates that

  1. Voltage and current are in phase;
  2. There is no change in frequency in this type of circuit.

Alternating Current Purely Resistive Circuit According To Ohms Law

Power in the circuit: Phase difference between voltage and current, θ = 0; so, power factor, cosθ = 1. For this, maximum power is consumed in a purely resistive circuit, which is

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta=\frac{1}{2}\left(I_0 R\right) I_0 \cdot 1=\frac{1}{2} I_0^2 R=\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)

i.e., P = Irms2.R → (3)

Purely Inductive Circuit

shows this circuit. Let the alternating voltage applied in the circuit be,

Alternating Current Purely Inductive Circuit

V = V0sin ωt → (1)

If the instantaneous change of current in the circuit is dl, the emf induced in the two ends of the inductor = \(-L \frac{d I}{d t}\)

This emf reduces the main voltage in the circuit. For this the effective voltage of the circuit = \(V_0 \sin \omega t-L \frac{d I}{d t}\).

A purely inductive circuit has no resistance i.e., R = 0. So according to Ohm’s law,

⇒ \(V_0 \sin \omega t-L \frac{d I}{d t}=0 \quad \text { or, } \frac{d I}{d t}=\frac{V_0}{L} \sin \omega t\)

or, \(\int d I=\int \frac{V_0}{L} \sin \omega t d t\)

or, \(I=-\frac{V_0}{\omega L} \cos \omega t+k\) [k = integration constant] → (2)

Dimensionally k is the same as I. Also, fc is a time-independent quantity. As the source voltage oscillates symmetrically about zero, the current also oscillates symmetrically about zero. For this, no constant or time-independent current can flow through the circuit, i.e., k = 0.

∴ \(I=-\frac{V_0}{\omega L} \cos \omega t=+\frac{V_0}{\omega L} \sin \left(\omega t-90^{\circ}\right)\)

or, I = I0sin(ωt-90°) →(3)

where \(I_0=\frac{V_0}{\omega L}\) = peak value of current.

From the equations (1) and (3) we come to the conclusions

current lags behind the voltage by 90°.

Alternating Current Circuit Lags Behind The Voltage

The quantity coL plays the same role in an inductive circuit as the resistance R in a resistive circuit. This quantity (i.e., ωL) is known as the inductive reactance of an ac circuit. Inductive reactance is the opposition offered by an inductor to the flow of alternating current through it. It is denoted by the symbol XL.

Hence for the above circuit,

XL = ωL = inductive reactance

Similar to that of R its unit is also ohm (Ω).

Power In The Circuit: Phase difference between voltage and current, θ = 90°, so, power factor, cosθ = 0.

Therefore, power consumption, \(P=\frac{1}{2} V_0 I_0 \cos \theta=0\)

The current I in this circuit is wattless.

Purely Capacitive Circuit

shows the circuit. Alternating voltage applied in the circuit,

V = V0sin ωt →(1)

At any moment, if Q is the charge stored in the capacitor C, then its terminal potential difference is \(\frac{Q}{C}\)

This potential difference opposes the applied instantaneous voltage in the circuit.

Alternating Current Purely Capacitive Circuit

So, the effective voltage of the circuit, \(V_e=V_0 \sin \omega t-\frac{Q}{C}\)

As there is no resistance in the circuit, according to Ohm’s law, Ve = IR = 0

and hence, \(V_e=V_0 \sin \omega t-\frac{Q}{C}=0\)

or, Q = CV0sinωt → (2)

Therefore alternating current,

∴ \(I=\frac{d Q}{d t}=\omega C V_0 \cos \omega t=\omega C V_0 \sin \left(\omega t+90^{\circ}\right)\)

= I0sin(ωt+90°) → (3)

Where, \(I_0=\omega C V_0=\frac{V_0}{1 / \omega C}=\frac{V_0}{X_C}\) = peak value of current From the equations (1) and (3) we come to the conclusions

Current leads the applied voltage by 90°;

The quantity \(\frac{1}{\omega C}\)plays the same role in a capacitive circuit as the resistance R in a resistive circuit. This quantity (i.e., \(\frac{1}{\omega C}\) is known as capacitive reactance (XC).

Capacitive reactance is the opposition offered by a capacitor to the flow of alternating current through it. Hence for the above circuit.

∴ \(x_C=\frac{1}{\omega C}\) = capacity reactance

Similar to that of R or XL, its unit is also ohm (Ω).

Alternating Current Current Leads To Applied Voltage

Power In The Circuit: Phase’ difference between voltage and current, θ = -90°; so, power factor, cosθ = 0. Therefore, power consumption \(P=\frac{1}{2} V_0 I_0 \cos \theta=0\).

Here again, this current is wathers.

Wattless or Idle current: Power consumed in an ac circuit, \(P=\frac{1}{2} V_0 I_0 \cos \theta\). Now the phase difference between voltage and current in a pure inductive circuit, θ = 90°.

So power in the circuit is zero. On the other hand phase difference between voltage and current in a pure capacitive circuit, θ = -90°.

So, again power is equal to zero, which means, a purely inductive or capacitive circuit does not dissipate any power; current through the inductor or capacitor is hence called wattless or idle current.

LR Circuit

shows the circuit

Alternating Current LR Circuit

The applied alternating voltage,

V = V0sinωt → (1)

If the instantaneous change in current in of the circuit is dl, then the induced emf in the inductor \(L=-L \frac{d I}{d t}\);

i.e., the effective voltage in the circuit,

⇒ \(V_e=V_0 \sin \omega t-L \frac{d I}{d t}\)

According to Ohm’s law, Ve = IR, where the resistance, if any, of the inductor L is included in R.

∴ \(V_0 \sin \omega t-L \frac{d I}{d t}=I R\)

or, \(L \frac{d I}{d t}+R I=V_0 \sin \omega t\) → (2)

Let I = I0sin(ωt+ α)

or, \(\frac{d I}{d t}=\omega I_0 \cos (\omega t+\alpha)\)

So, putting the values of I and \(\frac{d I}{d t}\) in identity (2) we get

ωLI0 cos(ωt+α) + RI0 sin(ωt+α) = V0 sinωt

or, I0{R sin((ωt+ α) + L cos(ωt+α)} = V0 sinωt

or, \(I_0 \sqrt{R^2+(\omega L)^2}\{\frac{R}{\sqrt{R^2+(\omega L)^2}} \sin (\omega t+\alpha)+\frac{\omega L}{\sqrt{R^2+(\omega L)^2}} \cos (\omega t+\alpha)\}=V_0 \sin \omega t\)

or, \(I_0 Z\left\{\frac{R}{Z} \sin (\omega t+\alpha)+\frac{\omega L}{Z} \cos (\omega t+\alpha)\right\}=V_0 \sin \omega t \text { [where } Z=\sqrt{R^2+(\omega L)^2} \text { ] }\)

or, \(I_0 Z\{\sin (\omega t+\alpha) \cos \theta+\cos (\omega t+\alpha) \sin \theta\}=V_0 \sin \omega t \text { [where } \cos \theta=\frac{R}{Z} \text { and } \sin \theta=\frac{\omega L}{Z} \text { ] } \)

or, I0Zsin(ωt+ α + θ) = V0sin ωt

From this identity, comparing both sides we get,

⇒ \(I_0 Z=V_0 \quad \text { or, } I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+(\omega L)^2}}\)

and α + θ = 0 or, α = -θ

So, the ac in the circuit,

∴ \(I=I_0 \sin (\omega t-\theta)=\frac{V_0}{Z} \sin (\omega t-\theta)\) → (3)

From equations (1) and (3) we conclude,

The current lags behind the applied voltage by a phase angle θ given by \(\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\omega L}{Z} \times \frac{Z}{R}=\frac{\omega L}{R}\)

i.e., \(\theta=\tan ^{-1}\left(\frac{\omega L}{R}\right)\)

This phase relation is shown. We know that in a purely resistive circuit V and I are in the same phase i.e., phase difference, θ = 0. On the other hand, in a pure inductive circuit, I always lag behind V by θ = 90°. So in an LR circuit, I should lag behind V by 0 < θ < 90°.

In this circuit, Z plays the same role as R in a pure resistive circuit. Z is known as the impedance of an ac circuit.

Impedance, \(Z=\sqrt{R^2+(\omega L)^2}=\sqrt{R^2+\left(X_L\right)^2}\) → (4)

where, XL = ωL = inductive reactance

Alternating Current Phase Relation

Impedance in an LR circuit is the effective resistance of the circuit arising from the combined effects of ohmic resistance and inductive reactance.

We can express the relation among R, XL, and Z with a suitable right-angled triangle. This triangle is called the impedance triangle.

Note that, R, ωL, and Z have, the same unit ohm(Ω).

Alternating Current Impedance Triangle

Power In The Circuit: 

Here, the power factor of the circuit, \(\cos \theta=\frac{R}{Z}\)

∴ \(P=\frac{1}{2} V_0 I_o \cos \theta=\frac{1}{2}\left(I_0 Z\right) I_0 \cdot \frac{R}{Z}=\frac{1}{2} I_0^2 R=\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)

i.e., \(P=I_{\mathrm{rms}}^2 R\)

This means that the power is dissipated only in the resistance R. Current through the inductor is wattless.

CR Circuit

It shows the circuit. Alternating voltage applied to the circuit,

Alternating Current CR Circuit

V = V0sin ωt → (1)

At any moment if Q is the charge stored in the capacitor C, then, the effective voltage in the circuit,

∴ \(V_e=V_0 \sin \omega t-\frac{Q}{C}\)

According to Ohm’s law

∴ \(V_0 \sin \omega t-\frac{Q}{C}=I R\)

or, \(R I+\frac{1}{C} Q=V_0 \sin \omega t\)

Now let us assume that Q = Q0 sin(ωt+ α)

∴ \(I=\frac{d Q}{d t}=\omega Q_0 \cos (\omega t+\alpha)\)

So, from the equation (2) we get,

∴ \(\omega R Q_0 \cos (\omega t+\alpha)+\frac{1}{C} Q_0 \sin (\omega t+\alpha)=V_0 \sin \omega t\)

or, \(\omega Q_0\left[R \cos (\omega t+\alpha)+\frac{1}{\omega C} \sin (\omega t+\alpha)\right]=V_0 \sin \omega t\)

or, \(\omega Q_0 Z\left[\frac{R}{Z} \cos (\omega t+\alpha)+\frac{1 /(\omega C)}{Z} \sin (\omega t+\alpha)\right]=V_0 \sin \omega t\)

[where \(Z=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}\)(say)]

or, \(\omega Q_0 Z\{\cos (\omega t+\alpha) \cdot \cos \theta+\sin (\omega t+\alpha) \cdot \sin \theta\}=V_0 \sin \omega t\)

[where \(\frac{R}{Z}=\cos \theta \text { and } \frac{1 / \omega C}{Z}=\sin \theta\)(say)]

or, \(\omega Q_0 Z \cos (\omega t+\alpha-\theta)=V_0 \sin \omega t\)

or, \(\omega Q_0 Z \sin \left(\omega t+\alpha-\theta+90^{\circ}\right)=V_0 \sin \omega t\)

Comparing both sides we get,

∴ \(\omega Q_0 Z=V_0 \quad \text { or, } Q_0=\frac{V_0}{\omega Z}\)

and α – θ + 90° = 0 or, α = θ – 90°

So, alternating current in the circuit,

I = ωQ0cos(ωt+ α)

= \(\omega \frac{V_0}{\omega Z} \cos \left(\omega t+\theta-90^{\circ}\right)=\frac{V_0}{Z} \sin (\omega t+\theta)\)

i.e., \(I=I_0 \sin (\omega t+\theta)=\frac{V_0}{Z} \sin (\omega t+\theta)\) → (3)

Here, \(I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}\)

From equations (1) and (3) we conclude the following.

1. The current I leads the voltage V by a phase angle θ, where \(\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{1 / \omega C}{Z} \times \frac{Z}{R}=\frac{1}{\omega C R}\),

i.e., \(\theta=\tan ^{-1}\left(\frac{1}{\omega C R}\right)\)

This phase relation is shown. We know that in a purely resistive circuit, V and I are in the same phase, i.e., phase difference, θ = 0. On the other hand, in a pure capacitive circuit, I always lead V by θ = 90°. So, in a CR circuit, I should lead V by 0 < θ < 90°.

Alternating Current I n A Pure Capacitive Circuit

2. Z plays the same role as R in a pure resistive circuit. This Z is known as the impedance of the circuit.

Impendance, \(Z=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}=\sqrt{R^2+X_C^2}\) → (4)

where \(X_C=\frac{1}{\omega C}\) = capacitive reactance.

Impedance in a CR circuit is the effective resistance of the circuit arising from the combined effects of ohmic resistance and capacitive reactance.

shows the impedance triangle for the circuit.

Alternating Current Impedance In A CR Circuit

Power In The Circuit: Here, the power factor of the circuit, \(\cos \theta=\frac{R}{Z}\)

so, \(P=\frac{1}{2} V_0 I_0 \cos \theta=\frac{1}{2}\left(I_0 Z\right) I_0 \cdot \frac{R}{Z}=\frac{1}{2} I_0^2 R=\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)

∴ P = I2rms.R2

Hence, power is dissipated only in the resistance R . Current through the capacitor C is wattless.

Series LCR Circuit

shows the circuit. Alternating voltage applied in the circuit,

V = V0sin ωt → (1)

If the instantaneous change in current in the circuit is dI, then the emf induced in the inductor

= \(-L \frac{d I}{d t}\).

Alternating Current Series LCR Circuit

On the other hand, if Q is the instantaneous charge stored in the capacitor, the opposite emf thus developed in the circuit = \(-\frac{Q}{C}\).

So, the effective voltage in the circuit,

∴ \(V_e=V_0 \sin \omega t-L \frac{d I}{d t}-\frac{Q}{C}\)

According to Ohm’s law,

∴ \(V_0 \sin \omega t-L \frac{d I}{d t}-\frac{Q}{C}=I R\)

or, \(L \frac{d I}{d t}+R I+\frac{Q}{C}=V_0 \sin \omega t\)

Now, let us assume, Q = Q0sin(ωt+ α)

So, current, \(I=\frac{d Q}{d t}=\omega Q_0 \cos (\omega t+\alpha)\)

or, \(\frac{d I}{d t}=-\omega^2 Q_0 \sin (\omega t+\alpha)\)

Putting these values in equation (2) we get,

= \(-\omega^2 L Q_0 \sin (\omega t+\alpha)+\omega R Q_0 \cos (\omega t+\alpha)+\frac{Q_0}{C} \sin (\omega t+\alpha)=V_0 \sin \omega t\)

or, \(\omega Q_0\left\{R \cos (\omega t+\alpha)-\left(\omega L-\frac{1}{\omega C}\right) \sin (\omega t+\alpha)\right\}=V_0 \sin \omega t\)

or, \(\omega Q_0 Z\left\{\frac{R}{Z} \cos (\omega t+\alpha)-\frac{\omega L-1 / \omega C}{Z} \sin (\omega t+\alpha)\right\}=V_0 \sin \omega t\)

[Where \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)]

or, \(\omega Q_0 Z\{\cos (\omega t+\alpha) \cos \theta-\sin (\omega t+\alpha) \sin \theta\}=V_0 \sin \omega t=\)

[Where \(\frac{R}{Z}=\cos \theta \text { and } \frac{\omega L-1 / \omega C}{Z}=\sin \theta\)]

or, \(\omega Q_0 Z \cos (\omega t+\alpha+\theta)=V_0 \sin \omega t\)

or, \(\omega Q_0 Z \sin \left(\omega t+\alpha+\theta+90^{\circ}\right)=V_0 \sin \omega t\)

Comparing both sides we get,

⇒ \(\omega Q_0 Z=V_0 \quad \text { or, } Q_0=\frac{V_0}{\omega Z}\)

and α + θ + 90° = 0 or, α = -θ-90°

Then, cos(ωt+α) = cos(ωt- θ – 90°) = sin(ωt – θ)

So, alternating current in the circuit,

⇒ \(I=\omega Q_0 \cos (\omega t+\theta)\)

= \(\frac{V_0}{Z} \sin (\omega t-\theta)=I_0 \sin (\omega t-\theta)\) → (3)

Here \(I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}\) → (4)

From equations (1) and (3) we conclude,

Current lags behind voltage by a phase angle θ where \(\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{(\omega L-1 / \omega C)}{Z}}{\frac{R}{\bar{Z}}}=\frac{\omega L-\frac{1}{\omega C}}{R}\)

i.e., \(\theta=\tan ^{-1}\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)\) → (5)

This phase relation is shown. Now if VL < VC i.e., \(\omega L<\frac{1}{\omega C}\) will be negative. In such case, current I leads voltage V by angle θ. The voltage across the resistance R i.e., VR and current I are in the same phase. On the other hand, the current I lags behind VL by 90° but leads VC by a 90° phase angle.

Alternating Current Phase Angle

It is clear that M VL< VC i.e., if \(\omega L<\frac{1}{\omega C}\), I leads V by an angle θ.

2. In an LCR circuit, Z plays the same role as R in a pure resistive circuit. So, Z is the impedance of the circuit.

Impedance, \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}=\sqrt{R^2+X^2}\)

where \(X=\omega L-\frac{1}{\omega C}=X_L-X_C\) = reactance of the circuit.

Impedance in an LCR circuit is the effective resistance of the circuit arising from the combined effect of ohmic resistance and reactance of the circuit.

shows the impedance triangle for the circuit.

Alternating Current Series LCR Circuit Impedance Triangle

Power In Series LCR Circuit:

Here, the power factor of the circuit, \(\cos \theta=\frac{R}{Z}\)

Therefore power consumed in the circuit

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta=\frac{1}{2}\left(I_0 Z\right) I_0 \cdot \frac{R}{Z}=\frac{1}{2} I_0^2 R=\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)

i.e., P = I2rms R

This indicates that in this circuit power is dissipated neither in the inductor nor in the capacitor, but in the resistor only.

Hence currents through L or C are wattless.

The effective resistance of an LCR alternating current circuit is essentially the impedance of that circuit.

The inverse of impedance is known as admittance, i.e., admittance = \(\frac{1}{Z}\). Its unit is ohm” 1 or Siemens.

A pure resistance R opposes the current in any circuit and electrical energy is dissipated through it.

An inductive reactance XL and a capacitive reactance XC also oppose this current in a circuit, but no energy is dissipated through a pure inductor or a pure capacitor.

Both XL = coL and XC = depend on the frequency \(X_C=\frac{1}{\omega C}\) a) of the applied voltage. Clearly, for a dc voltage, (o = 0 . so that XL = 0, but XC = ∞. This means that dc passes freely through a pure inductor, but is blocked by a pure capacitor, which acts like an open switch.

Units of impedance and reactance: Both quantities have the same unit ohm (Ω).

Advantages Of The Capacitor Dependent Regulator Over A Resistor Dependent Regulator: In a resistor-dependent regulator the resistance also dissipates some energy which heats the regulator.

  • On the other hand, only a capacitor but no resistor is used in an electronic regulator. Current in the circuit can be changed by regulating the capacitance of the capacitor and hence the speed of an electric fan can be controlled at will.
  • As the current flowing through a pure capacitive ac circuit is wattless, such a regulator almost does not dissipate any power.
  • So power is saved more in a capacitor-dependent regulator than in a resistor-dependent regulator. An electric regulator is useful in many devices—running an electric fan is just one of them.

Alternating Current Advantages Of The Capacitor Dependent Regulator Over A Resistor Dependent Regulator

Series resonance: It is observed from equation (4) that for \(\omega L=\frac{1}{\omega C}\), the denominator becomes minimum, and hence the current I will be maximum. This phenomenon is called the series resonance of the LCR circuit. If f0 is the frequency for which the circuit reaches the above state, the condition for resonance, then

\(\omega_0 L=\frac{1}{\omega_0 C} \quad\left[\omega_0=2 \pi f_0\right]\) → (6)

i.e., \(\omega_0=\frac{1}{\sqrt{L C}}\)

or, \(f_0=\frac{\omega_0}{2 \pi}=\frac{1}{2 \pi \sqrt{L C}}\) → (7)

This frequency f0 is called resonant frequency. The particular frequency of current In an LCR series circuit for which inductive reactance (XL) and capacitive reactance (XC) become equal to each other is called resonant frequency.

Under this condition, the circuit is termed a resonant circuit. Thus, whatever the value of frequency other than f0, current I i.e., Irms is always less than its maximum value I.

The change of ac with angular frequency in an LCR circuit is shown graphically. It is known as the resonance curve. As \(I_m=\frac{V_0}{R}, I_m\) increases with the decrease of R.

Alternating Current Series Resonance

Properties:

  1. In the case of resonance,

∴ \(\omega_0 L=\frac{1}{\omega_0 C} \text { or, } X_L=X_C\)

which indicates that the inductive and capacitive reactance balance each other. Hence the circuit becomes equivalent to a pure resistive circuit.

2. According to equation (5), for resonance,

∴ \(\tan \theta=\frac{0}{R}=0 \text { or, } \theta=0\)

i.e., in a resonance circuit, alternating voltage V and alternating current I are in equal phases. In this condition, according to equation (4) the maximum value of I0 i.e., Im = V0/R, which is equivalent to a pure resistive circuit.

3. LCR series circuit finds use in the receiver of a radio set. The resonant frequency of this LCR circuit is tuned with the frequency of the signal transmitted from a radio station.

Hence resonance occurs. As a result, the magnitude of the current increases a lot and the transmitted signal can easily be received. LCR series circuit is also known as acceptor circuit.

Sharpness Of Resonance And Q-Factor: Power consumed at LCR series circuit,

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta\)

where cosθ = power factor = \(\frac{R}{Z}\)

∴ \(P=\frac{1}{2} V_0 I_0 \cdot \frac{R}{Z}=\frac{1}{2} V_0 \cdot \frac{V_0}{Z} \cdot \frac{R}{Z}\)

= \(\frac{1}{2} V_0^2 \frac{R}{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\) → (8)

At resonance, \(\omega=\omega_0 \text { and } \omega_0 L=\frac{1}{\omega_0 C}\) and as a result power dissipation becomes maximum. From equation (8) we come to the value of maximum power dissipation, i.e.,

∴ \(P_m=\frac{V_0^2 R}{2 R^2}=\frac{V_0^2}{2 R}\) → (9)

From (8) and (9) we get,

∴ \(P=P_m \cdot \frac{R^2}{R^2+\left(\omega L-\frac{L}{\omega C}\right)^2}\) → (10)

Alternating Current Sharpness Of Resonance And Q-Factor

This shows the average power (P) versus frequency (ω) curve using the same circuit parameters.

The points A and B have some special importance. Each of these points denotes half maximum power \(\left(\frac{P_m}{2}\right)\).

The rapidity with which resonance phenomena arise and then disappear is a measurement of the sharpness of resonance. Resonance will be sharp if the value of bandwidth (Δω) is small.

This is of course possible only when the resonance curve falls steeply around ω = ω0.

Putting \(P=\frac{P_m}{2}\) in equation (10) we get,

= \(\frac{1}{2}=\frac{R^2}{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \quad \text { or, }\left(\omega L-\frac{1}{\omega C}\right)= \pm R\)

or, \(\omega^2-\frac{1}{L C}= \pm \frac{R}{L} \omega\) → (11)

But we have \(\omega_0^2=\frac{1}{L C}\) (from series resonance condition)

Therefore from equation (11),

= \(\omega^2-\omega_0^2= \pm \frac{R}{L} \omega\)

i.e., \(\omega^2-\frac{R}{L} \omega-\omega_0^2=0 \quad \text { and } \omega^2+\frac{R}{L} \omega-\omega_0^2=0\)

Solving these quadratic equations neglecting the negative values of ω we get,

∴ \(\omega_1=\frac{R}{2 L}+\left(\omega_0^2+\frac{R^2}{4 L^2}\right)^{1 / 2}\)

and \(\omega_2=-\frac{R}{2 L}+\left(\omega_0^2+\frac{R^2}{4 L^2}\right)^{1 / 2}\)

Hence, bandwidth \((\Delta \omega)=\omega_1-\omega_2=\frac{R}{L}\) → (12)

and Q-factor = \(\frac{\omega_0}{\Delta \omega}=\frac{\omega_0 L}{R}=\frac{1}{\sqrt{L C}} \cdot \frac{L}{R}=\frac{1}{R} \sqrt{\frac{L}{C}}\) → (13)

Q-factor measures the sharpness of resonance in an LCR circuit. Inevitably, as Acy goes lesser, the Q-factor becomes greater and so resonance becomes sharper.

From equation (13), \(Q=\frac{\omega_0 L}{R}=\frac{X_L}{R}=\frac{I X_L}{I R}=\frac{V_L}{V_R}\)

where VL = voltage difference across inductor

and VR = voltage difference across the resistor

At resonance, VR = applied voltage ( V)

and VL = VC [VC = voltage difference across capacitor]

∴ \(Q=\frac{V_L}{V}=\frac{V_C}{V}\)

i.e., voltage difference (also called voltage drop) across inductor or capacitor concerning applied voltage in LCR series circuit is termed as Q-factor of the circuit.

The Q-factor is often much greater than 1. This means that in a series resonant circuit, VL and VC are much greater than the applied voltage V. This signifies a prominent voltage amplification across the inductor L and capacitor C.

Q-factor is a dimensionless parameter, which denotes the degree of damping of a resonator or oscillator. The more the value of Q, the less the rate of dissipation of energy concerning die energy stored, and the less the damping of the oscillator.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Power Consumed In Ac Circuits Numerical Examples

Example 1. In an LCR series ac circuit R = 10Ω, L = 50mH, and C = 5μV. Find out the resonant frequency and Q-factor. Find also the bandwidth and half-power frequencies.

Solution:

Here, L = 50mH = 5 x 10-2H;

C = 5μF = 5 X 10-6F

∴ Resonant frequency,

⇒ \(f_0=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \times 3.14 \times \sqrt{\left(5 \times 10^{-2}\right) \times\left(5 \times 10^{-6}\right)}}\)

⇒ \(\frac{10^4}{2 \times 3.14 \times 5}=318.5 \mathrm{~Hz}\)

and Q-factor = \(\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{10} \sqrt{\frac{5 \times 10^{-2}}{5 \times 10^{-6}}}=\frac{1}{10} \times 100=10\)

Now, \(Q=\frac{\omega_e}{\Delta \omega}=\frac{f_0}{\Delta f}\)

∴ Bandwidth, \(\Delta f=\frac{f_0}{Q}=\frac{318.5}{10}=31.85 \mathrm{~Hz}\)

Half-power frequencies are,

⇒ \(f_1=f_0-\frac{\Delta f}{2}=318.5-\frac{31.85}{2}=302.6 \mathrm{~Hz}\)

and \(f_2=f_1+\frac{\Delta f}{2}=302.6+15.92=318.52 \mathrm{~Hz}\)

Example 2. In an LCR series combination, R = 400Ω, L = JOOmH, and C = 1 μF. This combination is connected to a 25 sin 2000t volt source. Find

  1. The impedance,
  2. Peak value of current,
  3. The phase difference of voltage and current,
  4. Power factor and
  5. Dissipated power in the circuit.

Solution:

Applied ac voltage V = 25 sin 2000 t volt

Peak value of voltage, V0 = 25V;

angular frequency, ω = 2000 Hz

and L = 100 mH = 0.1H

So, ωL = 2000 x 0.1 =200Ω

Here, \(C=1 \mu \mathrm{F}=10^{-6} \mathrm{~F} \quad \text { So, } \frac{1}{\omega C}=\frac{10^6}{2000}=500 \Omega\)

1. Impedance of the circuit,

∴ \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

= \(\sqrt{(400)^2+(200-500)^2}=500 \Omega\)

2. Peak value of current, \(I_0=\frac{V_0}{Z}=\frac{25}{500}=0.05 \mathrm{~A}\)

3. If the phase difference between voltage and current is θ then,

∴ \(\tan \theta=\frac{\omega L-\frac{1}{\omega C}}{R}=\frac{200-500}{400}=-\frac{3}{4}=\tan \left(-36.9^{\circ}\right)\)

i.e., current leads voltage by a phase angle of 36.9°.

4. Power factor of the circuit,

∴ \(\cos \theta=\frac{R}{Z}=\frac{400}{500}=0.8\)

5. Power dissipated,

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta=\frac{1}{2} \times 25 \times 0.05 \times 0.8=0.5 \mathrm{~W}\)

Example 3. The power factor of an LR circuit is \(\frac{1}{\sqrt{2}}\). If the frequency of ac is doubled, what will be the power factor?

Solution:

Power factor, \(\cos \theta=\frac{R}{Z}=\frac{1}{\sqrt{2}}\)

∴ Z2 = 2R2 or, R2 + (ωL)2 = 2R2 or, (ωL)2 = R2

Now if the angular frequency co is doubled,

(ω’L)2 = (2ωL)2 = 4(ωL)2 = 4R2

So, impendance, \(Z^{\prime}=\sqrt{R^2+\left(\omega^{\prime} L\right)^2}=\sqrt{R^2+4 R^2}=R \sqrt{5}\)

Hence, power factro, \(\frac{R}{Z^{\prime}}=\frac{R}{R \sqrt{5}}=\frac{1}{\sqrt{5}}\)

Example 4. f the value of inductor L is 1 mH and the applied ac source frequency is 50 Hz, find the inductive reactance in the above case.

Solution:

Here,1 = 1 mH = 10-3 H

and ω = 2πf= 2 x 3.14 x 50 = 314 Hz

∴ XL = ωL = 314 x 10-3 = 0.314 Ω

Example 5. A series LC circuit has L = 0.405-H and C = 25 μF. The resistance R is zero. Find the frequency of resonance.

Solution:

Here, L = 0.405 H, C = 25 μF = 25 x 10-6 F

∴ \(f_r=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \times 3.14 \times \sqrt{0.405 \times 25 \times 10^{-6}}}=50 \mathrm{~Hz}\)

Example 6. An inductor and a capacitor of reactances 25Ω and 75Ω, respectively, are connected across a 250 V ac source in series. Find the potential difference between the inductor and the capacitor. Establish their relationship with the main voltage.

Alternating Current An Inductor And A Capacitor Of Reactances

Solution:

The impedance of the series circuit having capacitor and inductor,

Z = XC-XL = (75-25) Ω = 50Ω

So, current, \(I=\frac{250}{50}=5 \mathrm{~A}\)

∴ Potential differences across the inductor,

VL = 5 x 25V = 125V

and potential differences across the capacitor,

VC = 5 x 75 V = 375 V

Now, main voltage,

V = 250 V

Hence, the relationship between V, VL and VC is

V = VC – VL

Example 7. A capacitor, a resistor of 5Ω, and an inductor of 50 mH are in series with an ac source marked 100 V, 50Hz. It is found that the voltage is in phase with the current. Calculate the capacitance of the capacitor and the impedance of the circuit.

Solution:

Since both the voltage and current of the circuit are in the same phase, the circuit is purely resistive. So impedance of the circuit, Z = R = 5 Ω

Frequency of the resonant circuit,

∴ \(f=\frac{1}{2 \pi \sqrt{L C}}\)

∴ \(C=\frac{1}{4 \pi^2 L f^2}=\frac{49}{4 \times 484 \times 50 \times 10^{-3} \times 50 \times 50}\)

= 2.03 x 10-4 F

Hence, the capacitance is 2.03 x 10-4 F and the impedance is 5Ω.

Example 8. A capacitor and a resistor are connected in series with an ac source. If the potential differences across C, R are 120 V, 90 V respectively and If the rms current of the circuit; is 3 A, calculate

  1. The impedance and
  2. The power factor of the circuit.

Solution:

Alternating voltage in the circuit,

∴ \(V=\sqrt{V_R^2+V_C^2}=\sqrt{90^2+120^2}=150 \mathrm{~V}\)

Now, current through the circuit, I = 3 A

  1. Impedance of the circuit, \(Z=\frac{\dot{V}}{I}=\frac{150}{3} \Omega=50 \Omega\)
  2. Power factor of the circuit, \(\cos \theta=\frac{V_R}{V}=\frac{90}{150}=0.6\)

Example 9. A 200μF capacitor in series with a 50 Ω resistor is connected to a 220 V, 50 Hz ac source.

  1. What is the maximum current in the circuit?
  2. What is the difference in time when the current and the voltage attain maximum values?

Solution:

Angular frequency of the source,

ω = 2πf=2 x 3.14 x 50Hz

C = 200μF = 2 x l0-4 F

Maximum current passing through the circuit

∴ \(I_0=\frac{V_0}{\sqrt{R^2+\frac{1}{C^2 \omega^2}}}\)

= \(\frac{\sqrt{2} \times 220}{\sqrt{(50)^2+\frac{1}{\left(2 \times 10^{-4}\right)^2 \times 4 \times(3.14)^2 \times(50)^2}}}\)

= 5.93 A

Now, if θ is the phase angle, then

= \(\tan \theta=\frac{1}{\omega C R}=\frac{1}{2 \pi f C R}=\frac{1}{2 \times 3.14 \times 50 \times 2 \times 10^{-4} \times 50}\)

= 0. 3185

or, \(\theta=\tan ^{-1}(0.3185)=17.67^{\circ}=\frac{17.67 \times \pi}{180} \mathrm{rad}\)

If the voltage and the current attain maximum value at a time different from t, then

θ = ωt

or, \(t=\frac{\theta}{\omega}=\frac{17.67 \times \pi}{180 \times 2 \pi \times 50}=9.82 \times 10^{-4} \mathrm{~s}\)

Example 10. A resistor, R = 300 Ω, and a capacitor, C = 25 μF are connected in series with an ac source. The peak value of voltage ( V0) and the frequency (f) of the source is 150 V and \(\frac{50}{\pi}\) Hz respectively. Find the peak value of the current and the power dissipated in the circuit.

Solution:

If ω is the angular frequency of the ac source; then

⇒ \(\frac{1}{\omega C}=\frac{1}{2 \pi \times \frac{50}{\pi} \times 25 \times 10^{-6}}=400 \Omega\)

Thus peak value of the current,

⇒ \(I_0=\frac{V_0}{\sqrt{R^2+\frac{1}{\omega^2 C^2}}}=\frac{150}{\sqrt{300^2+400^2}}=0.3 \mathrm{~A}\)

Hence, the power dissipated in the circuit

⇒ \(\frac{1}{2} I_0^2 R=\frac{1}{2} \times(0.3)^2 \times 300=13.5 \mathrm{~W}\)

Example 11. A series LCR circuit containing a resistance of 120 Ω has an angular resonance frequency of 4 x 105 rad/s. At resonance, the voltages across resistance and inductance are 60 V and 40 V, respectively. Find the values of L and C. At what frequency, does the current in the circuit lag behind the voltage by 45°?

Solution:

At resonance, XL = XC

∴ \(I=\frac{V_R}{R}\)

= \(\frac{60}{120}\) [voltage across the resistance, VR = 60 v]

= 0.5 A

Now, voltage across the inductor,

VL = IXL = IωL

∴ or, \(L=\frac{V_L}{I \omega}=\frac{40}{0.5 \times 4 \times 10^5}\) [∵ angular frequency, w = 4 x 105 rad/s] [angular frequency, = 4 x 105 rad/s]

= 2 x 10-4 H

We know that at resonance,

⇒ \(X_L=X_C \quad \text { or, } \omega L=\frac{1}{\omega C}\)

or, \(C=\frac{1}{\omega^2 L}=\frac{1}{\left(4 \times 10^5\right)^2 \times 0.2 \times 10^{-3}}=3.125 \times 10^{-8} \mathrm{~F}\)

Let the angular frequencyÿ ω1 when the current lags behind the voltage by 45°.

∴ \(\tan 45^{\circ}=\frac{\omega_1 L-\frac{1}{\omega_1 C}}{R}\)  \(C=3.125 \times 10^{-8} \mathrm{~F}=\frac{1}{32} \times 10^{-6} \mathrm{~F}\)

or, \(1 \times 120=\omega_1 \times 2 \times 10^{-4}-\frac{1}{\omega_1\left(\frac{1}{32}\right) \times 10^{-6}}\)

or, \(\omega_1^2-6 \times 10^5 \omega_1-16 \times 10^{10}=0\)

The physically meaningful solution of the above equation is,

∴ \(\omega_1=\frac{6 \times 10^5+10 \times 10^5}{2}=8 \times 10^5 \mathrm{rad} / \mathrm{s}\)

Example 12. In an LR series circuit, a sinusoidal voltage V = V0Sin ωt is applied. It is given that L = 35 mH, R = 11Ω, Vrms = 220V, \(\frac{\omega}{2 \pi}=50 \mathrm{~Hz} \text { and } \pi=\frac{22}{7}\). Find the amplitude of the current in the steady state and obtain the phase difference between the current and the voltage. Also, plot the variation of current for one cycle on the given graph.

Alternating Current LR Series Circuit A Sinusoidal Voltage

Solution:

Inductive reactance,

∴ \(X_L=\omega L=(2 \pi)(50)\left(35 \times 10^{-3}\right) \approx 11 \Omega\)

Impedance, Z = \(\sqrt{R^2+X_L^2}=\sqrt{11^2+11^2}=11 \sqrt{2} \Omega\)

Given Vrms = 220V

Hence, amplitude of voltage, V0 = 2 Vrms = 22072 V

∴ Amplitude of current, \(I_0=\frac{V_0}{Z}=\frac{220 \sqrt{2}}{11 \sqrt{2}}=20 \mathrm{~A}\)

The phase difference between the current and the voltage in the circuit,

∴ \(\theta=\tan ^{-1}\left(\frac{X_L}{R}\right)=\tan ^{-1}\left(\frac{11}{11}\right)=\frac{\pi}{4}\)

In the LR circuit, voltage leads the current by phase angle θ. Thus current in the current circuit,

∴ \(I=I_0 \sin (\omega t-\theta)=20 \sin \left(\omega t-\frac{\pi}{4}\right)\)

Alternating Current LR Circuit Voltage Leads The Current

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current LG Oscillations

In the circuit shown, S1S2S3 is a two-way switch. To charge the capacitor C first, one has to connect the switch S1S2. If S1S3 is connected, the charged capacitor and the inductance L constitute an LC circuit.

  • As soon as the switch is disconnected from S2, no current flows through the circuit anymore. In this condition if the amount of charge in the capacitor C is Q, the energy stored in the electric field of the capacitor is \(E_C=\frac{1}{2} \frac{Q^2}{C}\), but the inductor possesses no energy, i.e., EL = 0.
  • Due to the accumulation of charge in the capacitor, it acts as a battery. Hence immediately after connecting the switch S1S3, the capacitor starts to send current in the LC circuit thus forming.
  • As this discharging current gradually rises, a magnetic field starts developing in the inductor. At this stage, it may be said that the capacitor is being discharged gradually through the inductor L.
  • After some time, the capacitor C becomes completely discharged and the current of the LC circuit reaches a peak value I0 i.e., Ec becomes zero and \(E_L=\frac{1}{2} L I_0^2\).

It means that the electrical energy in the capacitor has been transferred fully into the magnetic energy in the inductor.

Alternating Current LG Oscillations

Again, due to the flow of current in the LC circuit, the capacitor is being charged in the reverse direction i.e., the capacitor is now charged with a polarity opposite to its initial state.

  • Obviously with the increase of charge in the capacitor, the energy stored in its electric field begins to increase and the energy stored in the magnetic field of the inductance gradually decreases.
  • Hence the magnitude of the current decreases and finally comes to zero. Thus after some time, EL becomes zero again, and \(E_C=\frac{1}{2} \frac{Q^2}{C}\). This process of discharging and charging of the capacitor occurs alternately.
  • We know that a capacitor can store electric energy whereas an inductor can store magnetic energy in it. Now a charged capacitor with an inductor is connected to an ac circuit, and a periodical energy transformation starts.

The energy of the capacitor is converted to the energy of the inductor and back again. This phenomenon is called LC oscillations.

In the above discussion, one half-cycle of this oscillation has been discussed. At the end of the next half-cycle, the circuit comes back to its initial condition.

  • The polarity of the plates becomes equal to its initial state and the current circuit becomes zero again. The direction of the current in the second half cycle is just opposite to that in the first half cycle.
  • So the current in an LC circuit fluctuates periodically between the peal values I0 and -I0.
  • The resistance of a pure inductor is zero so there is no energy loss due to the Joule effect. So as time elapses, no loss of total energy takes place, i.e., there is no damping of LC oscillations.
  • So the peak value of the alternating current in the LC circuit remains unchanged. But practically, the resistance of any coil cannot be ignored. Thus, there is always some resistance in the circuit due to which some energy is lost in the form of heat.
  • So the current remains oscillatory, but is damped, To maintain the alternating current, the circuit must be supplied with the same amount of energy as is being lost during each cycle, from some external source.
  • If damping is absent, the frequency of LC oscillations, \(f=\frac{1}{2 \pi \sqrt{L C}}\). When this frequency becomes equal to the frequency of the applied alternating emf then resonance occurs in that circuit.

Oscillators convert direct current (dc) from a power supply to an alternating current signal. LC circuits are used in many cases as an important component of an oscillator.

Alternating Current Damping Of LC Oscillations

Class 12 Physics Electromagnetic Induction And Alternating Current Chapter 2 Alternating Current LG Oscillations Numerical Examples

Example 1. A 220 V, 50 Hz ac source is connected to an inductance of 0.2H and a resistance of 20 Ω in series. What is the current in the circuit?

Solution:

rms value of current, \(I_{\mathrm{rms}}=\frac{E_{\mathrm{rms}}}{\sqrt{R^2+(\omega L)^2}}\)

Here, = 220 V,

ω = 2πf = 2 x 3.14 x 50 = 314 Hz

∴ \(\sqrt{R^2+(\omega L)^2}=\sqrt{(20)^2+(314 \times 0.2)^2}\)

∴ \(I_{\mathrm{rms}}=\frac{220}{\sqrt{(20)^2+(314 \times 0.2)^2}}=3.34 \mathrm{~A}\).

Example 2. An ac source of frequency 50 Hz is connected with a resistance (R = 36Ω) and L of 0.12 H in series. What is the phase difference between current and voltage?

Solution:

If θ is the phase difference, then

∴ \(\tan \theta=\frac{\omega L}{R}\)

or, \(\theta=\tan ^{-1} \frac{\omega L}{R}=\tan ^{-1} \frac{314 \times 0.12}{36}=\tan ^{-1}(1.047)=46.3^{\circ}\)

[here ω = 2π x 50 = 2 x 3.14 x 50 = 314 Hz]

So, phase difference = 46.3°.

Example 3. A current of 1 A flows in a coil when connected to a 100 V dc source. If the same coil is connected to a 100 V, 50 Hz ac source, a current of 0.5 A flows in the coil. Calculate the inductance of the coil.

Solution:

If R is the resistance of the coil, in dc circuit \(\frac{E}{I}=R\)

or, \(R=\frac{100}{1}=100 \Omega\)

In ac circuit, \(I_{\mathrm{rms}}=\frac{E_{\mathrm{rms}}}{\sqrt{R^2+(\omega L)^2}}\)

or, \(\sqrt{R^2+(\omega L)^2}=\frac{E_{\mathrm{rms}}}{I_{\mathrm{rms}}}=\frac{100}{0.5}=200 \Omega\)

∴ R2 + (ωL)2 = (200)2

or, \(\omega L=\sqrt{(200)^2-(100)^2}=100 \sqrt{3} \Omega\)

or, \(L=\frac{100 \sqrt{3}}{\omega}=\frac{100 \times 1.732}{2 \times 3.14 \times 50}=0.55 \mathrm{H}\)

Example 4. A lamp in which 10 A current can flow at 15 V is connected with an alternating source of potential 220 V and frequency 50 Hz. What should be the inductance of the choke coil required to light the bulb?

Solution:

Resistance of the lamp, \(R=\frac{15}{10}=1.5 \Omega\)

To send 10 A current through the lamp, the required impedance of the ac circuit, \(Z=\frac{220}{10}=22 \Omega\)

Now if L is the inductance of the choke coil and its resistance is negligible, then

∴ \(Z=\sqrt{R^2+(\omega L)^2} \text { or, } \omega L=\sqrt{Z^2-R^2}\)

∴ \(L=\frac{1}{\omega} \sqrt{Z^2-R^2} ;[\omega=2 \pi f=2 \times 3.14 \times 50=314 \mathrm{~Hz}]\)

= \(\frac{1}{314} \sqrt{(22)^2-(1.5)^2}=0.07 \mathrm{H}\)

Example 5. What will be the peak value of alternating current when a condenser of 1 μF is connected to an alternating voltage of 200 V, 60 Hz?

Solution:

c = 1μF = 10-6F; ω = 2πf = 2 x 3.14 x 60 Hz

Peak value of current,

\(I_0=I_{\mathrm{rms}} \times \sqrt{2}=\frac{E_{\mathrm{rms}}}{\frac{1}{\omega C}} \cdot \sqrt{2}=E_{\mathrm{rms}} \cdot \omega C \sqrt{2}\)

= 200 x (2 x 3.14 x 60) x 10-6 x 1.414

= 0.106 A (approx.)

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Transformer

The electrical appliance used to increase or decrease alternating voltage is called a transformer.

The transformer, which increases the voltage is called a step-up transformer, and the transformer used to decrease the voltage is called a step-down transformer.

Transformer works on the principle of mutual induction between a pair of colls.

Description: The core of a transformer is constructed by several thin laminated sheets of soft iron placed one Over the other. It is known as a laminated core.

A core of a special shape is so chosen that no part of the magnetic flux is wasted and hence the density of lines of induction inside the core becomes maximum. Two insulated wires are wound in many turns on the middle arm of the core very close to each other.

Open coil acts as the primary (P) and the other as the secondary (S).

Alternating Current Transformer

Working Principle: An alternating voltage (Vp) is applied to the primary coil from an alternating current source.

The alternating current in coil P generates induced emf in the secondary coil S, i.e., an alternating voltage Vs is generated between the ends of S.

If the dissipation of magnetic flux and loss of energy due to heating is neglected in this transformer (called an ideal transformer), it can be proved that,

∴ \(\frac{V_s}{V_p}=\frac{N_s}{N_p}=k\)

where Np and Ns are the total numbers of turps of the primary and secondary coils, respectively, and k is called the turns ratio or transformer ratio.

  • Because of the special nature of the winding, it can be assumed fairly correctly that the magnetic flux ΦB associated with each turn of primary and secondary is the same. If e is the induced emf in each turn, \(e=-\frac{d \phi_B}{d t}\).
  • Hence the emf induced in the primary oil \(V_p=N_p e=-N_p \frac{d \phi_B}{d t}\). Similarly, \(V_s=-N_s \frac{d \phi_B}{d t}\). Now, the emf induced in the primary must necessarily be equal to the voltage applied. Then, by dividing the two \(\frac{V_s}{V_p}=\frac{N_s}{N_{\dot{p}}}=k\).

If Ns > Np, i.e., k > 1, Vs > Vp we get a step-up transformer.

If Ns < Np, i.e., k < 1, Vf < Vp we get a step-down transformer.

Uses: Transformers are widely used in our daily lives. With its help, a high voltage can be converted into a low voltage and vice versa, whenever necessary. For example,

  1. An electrical power station uses a step-up transformer to produce and transmit large amounts of ac electrical energy over long distances.
  2. The energy is transmitted at high voltage (such as 66000V- 132000 V) to reduce the loss of energy due to heating.
  3. But supply to domestic area needs low voltages (such as 110 V-440 V). Such conversions of voltage at different levels are facilitated by step-down, transformers.
  4. Radio, television, electric bell,s and other electrical appliances require small-sized transformers.

Energy Loss In Transformer: In an ideal transformer, the power dissipated in the primary coil (IpVp) = power dissipated in the secondary coil (IsVs).

But no transformer, in practice, is ideal. Generally, Some input energy is wasted in any transformer and hence VsIs < VpIp .The term VsIs / VpIp is called the efficiency of a transformer.

The main causes of energy loss and their remedies are given below.

  1. Copper Loss: Generally copper wire is used to make primary and secondary coils. Due to Joule’s heating, some energy is wasted as heat energy.
    • Remedies: Thick wire should be used to reduce this loss of energy.
  2. Iron Loss: Iron core should be used in primary and secondary cords.
    1. Due to the eddy current in this core, energy loss is unavoidable.
      • Remedies: To reduce such loss, a laminated core is used.
    2. The change of magnetization cycle in the core fails to synchronize with ac and some energy is necessarily wasted in the core—known as hysteresis loss.
      • Remedies: Due to high coercivity, the core should not be made of steel. The iron core is more effective, in reducing energy loss.
  3. Loss due to magnetic flux leakage: The flux generated by the primary coil may not be wholly linked to the secondary coil due to possible defective design of the core.
    • Remedies: Obviously, special care should be taken in the construction of the core.

Classification Of Transformer: The two most common designs of the transformer are given below:

Core-Type Transformer: In this type, primary and secondary coils are wound around the core ring. Here every limb is occupied with both primary and secondary winding placed successively around them.

Alternating Current Classification of Transformer

Shell-Type Transformer: Here the primary and secondary windings pass inside the steel magnetic circuit (core) which forms a shell around the windings, The main frame is constructed with three limbs. Both the primary and secondary windings are wound around the central limb.

Besides these, transformers can be classified based on their and these are audio frequency transformers, radio frequency transformers, etc.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Transformer Numerical Example

Example 1. The number of turns in the primary aid secondary coils of an ideal transformer is 140 and 280, respectively. If the current through the primary coil is 4 A, what will be the current in the secondary coil?

Solution:

In an ideal transformer, the secondary and primary coils are equal,

i.e. VsIs ± VpIp

∴ \(I_s=I_p \cdot \frac{V_p}{V_s}=I_p \cdot \frac{N_p}{N_s}=4 \times \frac{140}{280}=2 \mathrm{~A}\)

Example 2. The initial voltage and Input power of a transformer of efficiency 80% are 100 V and 4 kW, respectively. If the voltage of the secondary coil Is 200 V, determine the currents flowing through the primary and the secondary coil.

Solution:

Power of the primary coil i.e., input power

Pp = VpIp

or, \(I_p=\frac{P_p}{V_p}=\frac{4 \times 1000}{100}=40 \mathrm{~A}\)

Power of the secondary coil, \(P_s=P_p \times \frac{80}{100}\)

Again, Ps = VsIs

So, \(I_s=\frac{P_s}{V_s}=\frac{80}{100} \times \frac{P_p}{V_s}=\frac{80}{100} \times \frac{4 \times 1000}{200}=16 \mathrm{~A}\)

 

Class 12 Physics Electromagnetic Induction And Alternating Current  Alternating Current Very Short Questions And Answers

Properties Of Alternating Voltage And Current

Question 1. If the frequency of an alternating emf is 50 Hz, how many times the direction of emf will be reversed per second?

Answer: 100

Question 2. What percentage of its peak value is the rms value of an ac?

Answer: 70.7

Question 3. What is the peak value of the voltage of a 220 V ac line?

Answer: 311 V

Question 4. If an alternating current is represented by I = sin l00 π mA, what is its peak value?

Answer: 1 mA

Question 5. If an alternating current is represented by, I = sin 100 πt mA, then what is the frequency of that current?

Answer: 50 Hz

Question 6. After what time will the direction of current in an electric supply of frequency 50 Hz be reversed?

Answer: 0.01 s

Question 7. An alternating source of emf E = E0 sinwt and negligible resistance is connected directly to an ac voltmeter. What reading will it show?

Answer: \(\left[\frac{E_0}{\sqrt{2}}\right]\)

Question 8. What changes are observed in the rms value of an ac with changes in the frequency

Answer: No change

Question 9. What is the rms value of an alternating current, I = I0 sin ωt?

Answer: \(\left[\frac{I_0}{\sqrt{2}}\right]\)

Question 10. What is the ratio between the peak value and the average value of a sinusoidal emf?

Answer: \(\left[\frac{\pi}{2}\right]\)

Question 11. The instantaneous i current in an ac circuit is I = 6 sin 314t A. What is the rms value of current?

Answer: 4.24 A

Question 12. An alternating current is I = cos 100 πt A. Find out its frequency, peak value, and rms value.

Answer: 50 Hz, 1 A, 0.707 A

Question 13. Why a dc voltmeter and dc ammeter cannot read ac?

Answer: The average is zero in a cycle

Question 14. What will be the phase difference between current and emf when 220 V, 50 Hz ac source is connected to a circuit containing pure resistor?

Answer: Zero

Series AC Circuits With R, L, C

Question 15. What is the unit of impedance?

Answer: ohm

Question 16. What is the reactance of pressure resistances in an ac circuit?

Answer: Zero

Question 17. If an LCR circuit is connected to a dc source, what will be the current through the circuit?

Answer: zero

Question 18. What will be the reactance if a current of frequency f flows through an inductor of self-inductance L?

Answer: 2 π fL.

Question 19. What will be the reactance if a current of frequency f flows through a capacitor of capacitance C?

Answer: \(\left[\frac{1}{2 \pi f C}\right]\)

Question 20. If the frequency of an ac circuit is increased, how would the reactance of an inductor change?

Answer: Increase

Question 21. If the frequency of an ac circuit is increased, how would the reactance of a capacitor change?

Answer: Decrease

Question 22. In an LR circuit, the alternating current ________ the alternating emf by a phase current the alter angle.

Answer: Lags behind

Question 23. In a CR circuit, the alternating current ________ the alternating emf by a certain phase angle.

Answer: Lead

Question 24. In an alternating series LCR circuit, what is the phase difference between the voltage drops across L and C?

Answer: 180

Question 25. When does the LCR series circuit have minimum impedance?

Answer: At resonance

Question 26. What is the reactance of a capacitor of capacitance C at f Hz?

Answer: \(\left[-\frac{1}{2 \pi f C}\right]\)

Power In Ac Circuits

Question 27. What is the power factor of a circuit having pure resistance only?

Answer: Zero

Question 28. What is the power dissipated in an ac circuitin which voltage and current are given by \(V=230 \sin \left(\omega t+\frac{\pi}{2}\right)\) and I = I0 sinωt?

Answer: Zero

LC Oscillations

Question 29. What is the natural frequency of an LC oscillator?

Answer: \(\left[\frac{1}{2 \pi \sqrt{L C}}\right]\)

Ac Generator And Transformer

Question 30. Indicate the change in emf produced by an ac dynamo in the following cases:

  1. The magnetic field is doubled,
  2. The angular velocity of the coil is decreased.

Answer: Will be doubled, will decrease

Question 31. If the area of the coil of an ac dynamo is halved, how would the emf generated change?

Answer: Halved

Question 32. If the angular velocity of the coil of an ac dynamo is doubled, how would the emfproduced change?

Answer: doubled

Question 33. By what factor would the output voltage of an ac generator change, if the number of turns in its coil is doubled?

Answer: 2

Question 34. The turns ratio of an ideal transformer is 4: 1. What will be the current in the secondary if that in the primary is 1.2A?

Answer: 4.8 A

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Synopsis Conclusion

In a dynamo, mechanical energy is converted into electrical energy.

In an electric motor, electric energy is converted into mechanical energy.

  • The current whose direction in an electrical circuit reverses periodically in a definite time interval is called alternating current.
  • The emf or potential difference whose direction reverses periodically in a definite time interval is called alternating emf or alternating potential difference.
  • The electrical machine used to increase or decrease an alternating voltage is called a transformer. The transformer which increases the voltage is called a step-up transformer and which decreases the voltage is called a step-down transformer.
  • The amount of power dissipation in any part of an AC circuit depends not only on ac voltage and current but also on their phase difference.
  • The inductor or capacitor in an ac circuit resists the current just like a resistor. This resistance is called reactance. The pure resistance, inductive reactance, and capacitive reactance do not remain in the same phase.

The effective resistance against current in an ac circuit due to a combination of pure resistance, inductive reactance, and capacitive reactance is known as the impedance of the circuit. The magnitude of this impedance depends on the ac voltage and current.

  • With the change of ac voltage frequency current also changes simultaneously.
  • The frequency for which current becomes maximum is known as resonant frequency.
  • In an LC circuit periodic interchange occurs between the stored energy in the electric field of the capacitor and that in the magnetic field of the inductor. This is LC oscillation.
  • The power factor ofpure resistor is 1 i.e., it is the resistance that dissipates maximum power.
  • The power factor of a pure inductor or capacitor is zero, i.e., they do not dissipate any power. Current through them is called wattless.

The emf induced in a coil rotating with uniform angular velocity ω in a uniform magnetic field intensity B, about an axis perpendicular to the field is,

e = NABω sin(ωt+ α) = e0sin (ωt+α)

[The coil is of cross-sectional area A having N turns]

If the total resistance of the coil and the external circuit is R, the induced current

∴ \(i=\frac{e}{R}=\frac{\omega B A N}{R} \sin (\omega t+\alpha)=\frac{e_0}{R} \sin (\omega t+\alpha)\)

= i0sin(ωt+ α)

Equation of an alternating emf,

V = V0sin(ωt+ α) where V0 = ωABN

Equation of an alternating current,

∴ \(I=\frac{V}{R}=\frac{V_0}{R} \sin (\omega t+\alpha)=I_0 \sin (\omega t+\alpha)\)

The average values of alternating voltage and current are respectively,

∴ \(\bar{V}=\frac{2 V_0}{\pi} \text { and } \bar{I}=\frac{2 I_0}{\pi}\)

rms values of the alternating voltage and current are respectively,

∴ \(V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}} \text { and } I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}\)

Form factor for a sinusoidal wave, \(f=\frac{V_{\mathrm{rms}}}{\bar{V}}=\frac{\pi}{2 \sqrt{2}}=1.11\)

If the number of turns in the primary coil of a transformer = Np, the number of turns in its secondary coil = Ns, and the ratio of the number of turns = k, then in the case of an ideal transformer,

∴ \(\frac{I_p}{I_s}=\frac{V_s}{V_p}=\frac{N_s}{N_p}=k\)

In an ideal transformer, input power (VpIp) = Output power ( VsIs ).

The efficiency of a transformer = \(\frac{V_s I_s}{V_p I_p}\); the efficiency of an ideal transformer = 1 or 100%.

Inductive reactance, XL = ωL.

Capacitive reactance, \(X_C=\frac{1}{\omega C}\)

The impedance of an LCR series circuit,

∴ \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

In an LCR series circuit, if ac voltage V = V0 sin ωt then alternating current, I = I0 sin(ωt- θ),

Where \(\dot{I}_0=\frac{V_0}{Z} \text { and } \tan \theta=\frac{\omega L-\frac{1}{\omega C}}{R}\)

Condition for resonance in an LCR series circuit,

∴ \(\omega L=\frac{1}{\omega C}\)

Resonant frequency, \(f_0=\frac{1}{2 \pi \sqrt{L C}}\)

Effective voltage magnification for series resonance,

∴ \(Q=\frac{V_L}{V_R}=\frac{V_C}{V_R}=\frac{\omega_0 L}{R}=\frac{1}{\omega_0 C R}=\frac{1}{R} \sqrt{\frac{L}{C}}\)

The natural frequency of an LC oscillator,

∴ \(f=\frac{1}{2 \pi \sqrt{L C}}\)

Power dissipation in an ac circuit

= \(V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta\)

where cos θ = power factor.

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Assertion Reason Type Question And Answers

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 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, Statement 2 is false.
  4. Statement 1 is false Statement 2 is true.

Question 1. Statement 1: The peak values by alternating voltage and alternating current in a circuit are V0 andd0 respectively.

The phase difference between voltage and current is θ. Then the power consumed is V0 I0 cosθ.

Statement 2: The consumed power in an alternating circuit depends on the phase difference between the emf and current.

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

Question 2. Statement I: Q-factor of a series LCR circuit is \(\frac{1}{R} \sqrt{\frac{L}{C}}\).

Statement 2: The resonant frequency of an LCR circuit does not depend on the resistance of the circuit.

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

Question 3. Statement 1: If the total energy in an LC oscillator is equally distributed between the magnetic and electric fields then the charge stored in the capacitor is \(\frac{1}{\sqrt{2}}\) fraction of the maximum charge stored in the capacitor during oscillation.

Statement 2: The charge stored in the capacitor becomes maximum at a time when the total energy of the LC circuit is stored in the electric field.

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

Question 4. Statement 1: Form factor becomes different for different waveforms of alternating voltage and current

Statement 2: The mean value of alternating voltage or current = \(\frac{2}{\pi}\) x peak value and rms value = \(\frac{1}{\sqrt{2}}\) x peak value for any waveform.

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

Question 5. Statement 1: A series LCR circuit when connected to an ac source gives the terminal potential difference 50 V across each of resistor R, inductor L and capacitor C. Then the terminal potential difference across LC is zero.

Statement 2: The terminal alternating voltages across the inductor and capacitor in a series LCR Circuit in an opposite phase.

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

Question 6. Statement 1: If the value of the output voltage of an ideal transformer is half the value of the input voltage, then the output current will become twice.

Statement 2: No energy is dissipated in an ideal transformer

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

Question 7. Statement 1: The alternating current lags behind the voltage by a phase angle \(\frac{\pi}{2}\) when ac flows through an inductor.

Statement 2: The inductive reactance increases as the frequency of ac source decreases.

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

Question 8. Statement 1: An inductor acts as a perfect conductor for dc.

Statement 2: dc remains constant in magnitude and direction.

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

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Match the following

Question 1. Match the columns for a series LCR circuit.

Alternating Current A Series LCR Circuit

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

Question 2. An LCR circuit (R = 40 Ω, L = 100mH, C = 0.242 μF) is connected with an ac voltage source of peak voltage 200 V and frequency 1000 Hz.

Alternating Current LCR Circuit Is Connected With An Ac Voltage Source Of Peak Voltage And Frequency

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

Question 3. Column I describes some action and column 2 the required device.

 

Alternating Current Some Actions And The Required Device

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

Question 4. In an LR circuit instantaneous voltage and instantaneous current are V = 100 sin100t and i = I0 sin(100t – \(\frac{\pi}{4}\) respectively.

 

Alternating Current LCR Circuit Instantaneous Voltage And Instantaneous Current

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

Question 5. Referring to the given circuit, match the following.

image

Alternating Current Referring The Given Circuit

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

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Comprehension Type Questions And Answers

Read the following passage carefully and answer the questions at the end of It.

Question 1. A series combination of an inductor of self-inductance L, capacitor of capacitance C, and resistor of resistance R is connected to an alternating voltage source of V=V0 sin ωt. The current through the circuit is I = I0 sin(t-0), where \(I_0=\frac{V_0}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}\) and \(\theta=\tan ^{-1} \frac{1}{R}\left(\omega L-\frac{1}{\omega C}\right)\).

Note that, the frequency of both voltage and, current is \(f=\frac{\omega}{2 \pi}\). The rms value of these parameters during one complete cycle are \(V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}} \text { and } I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}\) respectively. These values are shown in alternating voltmeter and ammeter.

The power consumed by the circuit P = VI. The mean value i.e., the effective power of the circuit in a complete cycle is \(\bar{P}=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta\). This cos θ is termed the power factor.

1. V = V0 sin ωt electromotive force is applied to an alternating circuit consisting of resistance R’ and an inductor of self-inductance L. The phase difference between the voltage and current is

  1. 90°
  2. \(\tan ^{-1} \frac{\omega L}{R^{\prime}}\)
  3. \(\tan ^{-1} \frac{R^{\prime}}{\sqrt{\left(R^{\prime}\right)^2+\omega^2 L^2}}\)
  4. \(\tan ^{-1} \frac{\sqrt{R^{\prime 2}+\dot{\omega}^2 L^2}}{R^{\prime}}\)

Answer: 2. \(\tan ^{-1} \frac{\omega L}{R^{\prime}}\)

2. The power factor of the circuit in question (1) is

  1. Zero
  2. \(\frac{\omega L}{R^{\prime}}\)
  3. \(\frac{R^{\prime}}{\sqrt{R^{\prime 2}+\omega^2 L^2}}\)
  4. \(\frac{\sqrt{R^{\prime 2}+\omega^2 L^2}}{R^{\prime}}\)

Answer: 3. \(\frac{R^{\prime}}{\sqrt{R^{\prime 2}+\omega^2 L^2}}\)

3. In the circuit in question (1) the inductor is replaced by a pure capacitor; the phase difference between the current and terminal voltage of the capacitor is

  1. -90°
  2. zero
  3. Between -90° and zero
  4. +90°

Answer: 1. -90°

4. The power factor of the circuit in question (3) is

  1. -1
  2. Zero
  3. Between zero and 1
  4. 1

Answer: 3. Between zero and 1

5. The voltage applied in an LCR circuit having R = 10Ω, L = 10 mH and C = 1 μF is V = 20 sin ωt volt. For what frequency of the applied voltage will the current reach Its peak value?

  1. 159 Hz
  2. 1592 Hz
  3. 1.59 x 104 Hz
  4. 1.59 x 105 Hz

Answer: 2. 1592 Hz

6. The phase difference between the voltage and peak current in question (v) is

  1. Zero
  2. -90°
  3. +90°
  4. 180°

Answer: 1. Zero

7. Which element is responsible for the power consumption in an alternating current circuit?

  1. Only resistor
  2. Only Inductor
  3. Only capacitor
  4. Resistor, inductor, and capacitor

Answer: 2. Only Inductor

8. The frequency of the applied alternating voltage In an. ac circuit is 50 Hz. The resistance and self-inductance of the circuit are 37.6 fL and 120 mH. The phase difference between the voltage and current is

  1. Zero
  2. 45°
  3. 60°
  4. 90°

Answer: 2. 45°

Question 2. A transformer is a device used to increase or decrease the voltage in the transmission line according to requirements. Generally, the input line voltage is fed into a primary coil and the output line voltage is obtained from the terminals of another coil. In an ideal transformer, the primary and secondary coils are linked in such a way that there is no loss of magnetic flux and electrical energy.

In an ideal transformer, if the number of turns and input voltage across the terminals of the primary coil are N1 and V1, then the output voltage at the two terminals of the secondary coil \(V_2=V_1 \cdot \frac{N_2}{N_1}\), where N2 is. the number of turns in the secondary coil.

1. The ratio of the number of turns of the primary and secondary coils of an ideal transformer is 2: 1. If the input voltage is 440 V, then the output voltage is

  1. 220 V
  2. 440 V
  3. 88 W
  4. None of these

Answer: 1. 220 V

2. In question (1) if the input power of the transformer is 44 W, then the output power is

  1. 22 W
  2. 44 W
  3. 88 W
  4. None of these

Answer: 2. 44 W

3. In the above-mentioned transformer the input and output currents are respectively,

  1. 100 mA, 100 mA
  2. 200 mA, 200 mA
  3. 100 mA, 200 mA
  4. 200 mA, 100 mA

Answer: 3. 100 mA, 100 mA

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Integer Answer Type Questions And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. A resistance and a capacitor are connected in series with an alternating voltage of rms value 13 V. The terminal voltage of the resistor is 12 V and that across the capacitor is (n + 0.38) V. What is the value of n?

Answer: 6

Question 2. The current and voltage in an ac circuit are \(I=\sin \left(100 t+\frac{\pi}{3}\right) \mathrm{A}\) and V = 20 sin 100 tV. Calculate the power of the circuit in W.

Answer: 5

Question 3. In a series LCR circuit, the capacitance C is replaced by 2C. To keep the resonance frequency unchanged, the inductance has to be replaced by an inductance of L’. Find the ratio of L and L’.

Answer: 2

Question 4. An alternating voltage of 5 V of frequency 50 Hz is connected to a series LCR circuit. The potential difference across the inductor and resistor is 6 V and 4 V respectively. What is the voltage across the capacitor (in V)?

Answer: 3

Question 5. In a series LCR circuit R = 1 kΩ, C = 2μF, and the potential difference across R is 2 V. At resonance ω = 200 rad.s-1. What is the potential difference (in V) across L at resonance? across

Answer: 5

Question 6. In a series LCR circuit R = 25Ω, L = 10 mH, and C = 1μF. The circuit is connected to an AC source of variable frequency. What is the Q-factor of the circuit?

Answer: 4

Question 7. A current of 50 mA flows through a 4 μF capacitor connected to a 500 Hz ac source. The terminal potential difference (in V) across the capacitor is (η + 0.98). What is the value of η? (π = 3.14)

Answer: 3

Question 8. In the figure, an LCR series circuit is shown. What would be the ammeter reading ampere?

Alternating Current Ammeter Reading In Ampere

Answer: 8

WBCHSE Class 12 Physics  Alternating Current Short Answer Questions

Class 12 Physics Electromagnetic Induction And Alternating Current

Question 1. The 60pF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. What is the net power absorbed by the circuit over a complete cycle?

Answer:

∴ \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}\)

Where, \(X_c=\frac{1}{2 \pi f C}=\frac{1}{2 \pi \times 60 \times 60 \times 10^{-6}} \Omega\)

and Vrms = 110V

Irms = 110(2π x 60 x 60 x 10-6)A = 2.49 A

The net power absorbed is zero as in the case of an ideal capacitor there is no power loss.

Question 2. Determine the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF, and R = 10 Ω. What is the Q-value of this circuit?

Answer:

= \(\omega_r=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{2 \times 32 \times 10^{-6}}}=125 \mathrm{~s}^{-1}\)

∴ \(Q=\frac{X_L}{R}=\frac{\omega_r L}{R}=\frac{125 \times 2}{10}=25\)

Question 3. A charged 30 μF capacitor is connected to a 27 mH inductor.

  1. What is the angular frequency of free oscillations of the circuit?
  2. If the initial charge on the capacitor is 6 mC then what is the total energy stored in the circuit initially? What is the total energy at a later time?

Answer:

  1. \(\omega_r=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{27 \times 10^{-3} \times 32 \times 10^{-6}}}=1.1 \times 10^3 \mathrm{~s}^{-1}\)
  2. \(E=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} \frac{\left(6 \times 10^{-3}\right)^2}{30 \times 10^{-6}}=0.6 \mathrm{~J}\)

There will be no change in total energy.

Question 4. A series LCR circuit with R = 20 Ω, L = 1.5 H, and C = 35 μF F is connected to a variable frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Answer:

When natural frequency and supply frequency are equal, resonance occurs.

∴XL = XC    ∴Z = R

P = \(\frac{V^2}{Z}=\frac{V^2}{R}=\frac{200 \times 200}{20}=2000 \mathrm{~W}\)

Question 5. A radio can tune over the frequency range of a portion of the MW broadcast band: 800 kHz to 1200 kHz. If its LC circuit has an effective inductance of 200μH, what must be the range of the variable capacitor?

[Hint: for tuning, the natural frequency of the LC circuit should be equal to the frequency of the radio wave.]

Answer:

∴ \(f=\frac{1}{2 \pi \sqrt{L C}} \quad \text { or, } C=\frac{1}{4 \pi^2 f^2 L}\)

when L = 200 μH and f = 800 kHz,

∴ \(C_1=\frac{1}{4 \pi^2 \times 800 \times 800 \times 10^6 \times 200 \times 10^{-6}}\)

= 87.9 pF

when f = 1200 kHz,

∴ \(C_2=\frac{1}{4 \pi^2 \times 1200 \times 1200 \times 10^6 \times 200 \times 10^{-6}}\)

= 87.9 pF

The range of the capacitance should be between 87.9 pF and 197.8 pF, i.e., between 88 pF and 198 pF.

Question 6. A series LCR circuit is connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40Ω.

  1. Determine the source frequency for which resonance occurs in the circuit.
  2. Obtain the impedance of the circuit and amplitude of current at resonance.
  3. Determine the rms potential drop across L and C.
  4. Show that at resonance the potential drop across LC combination is zero.

Answer:

1. \(\omega_r=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{5 \times 180 \times 10^{-6}}}=50 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

2. At resonance impedance, Z = R = 40 H

∴ \(I_{\max }=\frac{V_{\max }}{R}=\frac{\sqrt{2} \times 230}{40}=8.1 \mathrm{~A}\)

3. Across L, VL = I x XL = IωL

∴ VL = 8.1 x 50 x 5 = 2025 V

Across \(C, V_C=\frac{I}{\omega C}=\frac{8.1}{50 \times 80 \times 10^{-6}}=2025 \mathrm{~V}\)

The potential drop across LC combination = VL – VC = 0.

Question 7. An LC circuit has L = 20 mH, C = 50μF and initial. charge 10 mC the resistance being negligible.

  1. What is the total energy stored initially? Is it conserved during the LC oscillator?
  2. What is the natural frequency of the circuit?
  3. After what time interval from the moment the circuit is switched on the energy stored is
    1. Completely electrical i.e., stored only in the capacitor and
    2. Completely magnetic, i.e., stored only in the inductor?
  4. At what time is the total energy shared equally between the inductor and the capacitor?
  5. If a resistor is inserted in the circuit, how much energy is dissipated as heat?

Answer:

1. Total Initial energy = \(\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} \frac{10^{-4}}{50 \times 10^{-6}}=1 \mathrm{~J}\)

If R = 0, then total energy is conserved.

2. \(\omega=\frac{1}{\sqrt{L C}}\)

∴ Resonance frequency,

⇒ \(\omega=\frac{1}{\sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}}=10^3 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

⇒ \(f=\frac{\omega}{2 \pi}=\frac{10^3}{2 \times 3.14}=159 \mathrm{~Hz}\)

3. q = q0 cos ωt

1. When \(t=0, \frac{T}{2}, T, \frac{3 T}{2}, \ldots \ldots\)

then q = ±q0 i.e., the energy is completely electrical.

2. When \(t=\frac{T}{4}, \frac{3 T}{4}, \frac{5 T}{4}, \ldots \ldots\)

then q = 0 i.e., the energy is purely magnetic.

4. \(E=\frac{q^2}{2 C}=\frac{1}{2 C} q_0^2 \cos ^2 \omega t\)

when ωt = 45°, \(\cos \omega t=\frac{1}{\sqrt{2}}\)

∴ \(E=\frac{1}{2 C} \cdot q_0^2 \cdot \frac{1}{2}=\frac{1}{2}\left(\frac{q_0^2}{2 C}\right)=\frac{1}{2}\) x total energy

so, when \(\omega t=45^{\circ} \text { or } t=\frac{T}{8}, \frac{3 T}{8}, \frac{5 T}{8}, \cdots\) the energy is shared equally between the capacitor and inductor.

5. When a resistor is connected in the circuit, all the energy stored in the circuit (i.e., 1 J ) will be dissipated as heat energy since the LC oscillation will be damped and stop ultimately.

Question 8. A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz AC supply.

  1. What is the maximum current in the coil?
  2. What is the time lag between the voltage maximum and current maximum?
  3. If the circuit is connected to a high-frequency supply (240 V, 10 kHz ), what will be the answer to (1) and (2). From the answer explain the statement that at a very high frequency the presence of a inductor in the circuit nearly amounts to an open circuit.
  4. How does an inductor behave in a dc circuit after the steady state?

Answer:

1. \(I_{\max }=\frac{V_{\max }}{\sqrt{R^2+\omega^2 L^2}}=\frac{\sqrt{2} \times 240}{\sqrt{100^2+(2 \pi \times 50 \times 0.5)^2}}\)

= 1.82A

2. \(\tan \phi=\frac{X_L}{R}=\frac{2 \pi f L}{R}=\frac{2 \pi \times 50 \times 0.5}{100}=1.571\)

∴ Φ = tan-11.571 = 57.5°

∴ Time interval = \(\frac{T}{360} \times \phi=\frac{\phi}{360 f}\)

= \(\frac{57.5}{360 \times 50}=3.19 \times 10^{-3} \mathrm{~s}\)

3. ω = 2πf = 2π x 104 rad s-1.

∴ \(I_{\max }=\frac{V_{\max }}{\sqrt{R^2+\omega^2 L^2}}=\frac{240 \sqrt{2}}{\sqrt{100^2+4 \pi^2 \cdot 10^8 \cdot 5^2}}=0.011 \mathrm{~A}\)

∴ \(\phi=\tan ^{-1}\left\{\frac{X_L}{R}\right\}=\tan ^{-1}\left(\frac{2 \pi f L}{R}\right)\)

= \(\tan ^{-1} \frac{2 \pi \times 10^4 \times 0.5}{100}\)

= \(\tan ^{-1}(100 \pi) \approx \frac{\pi}{2}\)

∴ Time interval = \(\phi \cdot \frac{T}{360}=\frac{\phi}{360 f}\)

= \(\frac{90}{360 \times 10 \times 10^3}=0.25 \times 10^{-4} \mathrm{~s}\)

Imax is very small. So it can be concluded that at high frequencies an inductance behaves as an open circuit.

4. In a steady dc circuit, f = 0.

The inductance acts as a simple conductor.

Question 9. A circuit containing an 80 mH inductor and a 60μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.

  1. Obtain the current amplitude and rms values.
  2. Obtain rms values of potential drops across the inductor and capacitor.
  3. What is the average power transferred to the inductor?
  4. What is the average power transferred to the capacitor?
  5. What is the total average power absorbed by the circuit?

Answer:

1. \(I_{\max }=\frac{V_{\max }}{\sqrt{R^2+\left(X_L-X_C\right)^2}}\)

∴ XL = ωL = 100π x 80 x 10-3 = 25.12 Ω

∴ \(X_C=\frac{1}{\omega C}=\frac{1}{100 \pi \times 60 \times 10^{-6}}=53.03 \Omega\)

∴ \(I_{\max }=\frac{230 \sqrt{2}}{\sqrt{(25.12-53.03)^2}}=11.6 \mathrm{~A}\)

= \(I_{\mathrm{rms}}=\frac{I_{\max }}{\sqrt{2}}=\frac{11.6}{\sqrt{2}}=8.20 \mathrm{~A}\)

2. \(V_L=I_{\mathrm{rms}} \cdot \omega L=8.20 \times 25.12 \approx 206 \mathrm{~V}\)

∴ \(V_C=I_{\mathrm{rms}} \cdot \frac{1}{\omega C}=8.20 \times 53.03 \approx 435 \mathrm{~V}\)

3. Average power transferred to the inductor,

∴ \(P_L=V I \cos \phi=V I \cos \frac{\pi}{2}=0\)

4. Average power transferred to the capacitor,

∴ \(P_C=V I \cos \phi=V I \cos \frac{\pi}{2}=0\)

5. Average power absorbed by the circuit = 0

Question 10. A series LCR circuit with L = 0.12H, C = 480nF, and R = 23 Ω is connected to a 230 V variable frequency supply.

  1. What is the source frequency for which the current amplitude is maximum? Obtain this maximum value.
  2. What is the source frequency for which the average power absorbed by the circuit is maximum? Obtain the value of this maximum power.
  3. What is the Q-factor of the circuit?

Answer:

1. The current amplitude and average absorbed power both are maximum at the resonant frequency.

∴ \(f_0=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \times 3.14 \sqrt{0.12 \times 480 \times 10^{-9}}}\)

= 663 Hz

The current amplitude is maximum when the source frequency is 663 Hz.

∴ \(I_{\max }=\frac{V_{\max }}{R}=\frac{230 \sqrt{2}}{23}=14.14 \mathrm{~A}\)

2. The average absorbed power is maximum when the source frequency is 663 Hz.

∴ \(P_{\mathrm{av}}=\frac{1}{2} I_{\max }^2 R=\frac{1}{2} \times(14.14)^2 \times 23=2300 \mathrm{~W}\)

3. \(Q=\frac{X_L}{R}=\frac{2 \pi f_0 \cdot L}{R}=\frac{2 \pi \times 663 \times 0.12}{23}=21.7\)

Question 11.

  1. In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit?
  2. Is the same true for rms voltage?

Answer:

  1. Yes.
  2. No, it is not true for rms voltage because the potential differences across various parts of the circuit may not be in the same phase.

Question 12. Why a capacitor is used in the primary circuit of an induction coil?

Answer:

Whenever there is a break in the current, a large emf is induced in the circuit which is utilized in charging the capacitor. This prevents any spark in the circuit.

Question 13. When a choke is connected In series with a lamp in the DC line, the lamp shines brightly. The insertion of an Iron core in the choke does not affect the brightness. What happens In the case of ac line?

Answer:

An inductance acts as a simple conductor in a dc line, and reducing its self-inductance by introducing an iron core, does not affect the brightness of the lamp. In ac line the presence of inductance results in a drop of voltage. So the brightness of the lamp decreases. The introduction of the iron core further reduces the brightness of the lamp.

Question 14. Why a choke is needed with a fluorescent lamp with ac mains? Why a normal resistor cannot be used in place of the choke?

Answer:

An inductor can introduce a voltage drop in a circuit without any loss of power but a resistor gets heated during the process and some power is lost. A choke acts as an inductor in a circuit, so a choke is used in place of a resistor.

Question 15. A small town with a demand of 800 kW of power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two-wire line carrying power is 0.5H.km-1. The line gets power from the line through a 4000-220 V step-down transformer at a substation in the town.

  1. Estimate the line power loss in the form of heat.
  2. How much power must the plant supply, assuming there is negligible power loss due to leakage?
  3. Characterize the step-up transformer at the plant.

Answer:

1. Total resistance of the line = 0.5 x 15 x 2 = 15 Ω rms current through the line,

∴ \(I_{\mathrm{rms}}=\frac{\text { power }}{\text { voltage }}=\frac{800 \times 1000}{4000}=200 \mathrm{~A}\)

∴ Power loss due to the generation of heat

= I2rms.R = (200)2 x 15 = 600 kW

2. Power supplied by the plant = 800 + 600 = 1400 kW

3. Voltage dropin the line

= Irms .R = 200 x 15 = 3000 V

∴ The plant transformer should supply (4000 + 3000) = 7000 V

∴ A 440 V/7000V step-up transformer should be used.

Question 16. An ac having a peak value of 1.41 ampere is used to heat a wire. A dc producing the same heating rate will be approximately.

  1. 1.41 A
  2. 2.0 A
  3. 0.705 A
  4. 1.0 A

Answer: 4. 1.0 A

The rms value of the alternating current,

∴ \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{1.41}{1.41}=1 \mathrm{~A}\)

Hence, the dc required to produce the same heating rate =1 A.

The option 4 is correct.

Question 17. An ac voltage e = E0 sin cot is applied across an ideal inductor of self-inductance [L]. Write down the peak current.

Answer:

Peak value of the current = \(\frac{E_0}{\omega L}\)[Inductive reactarpe = ωL]

Question 18. The instantaneous voltage from an ac source is given by e = 200 sin 314t volt. Find the rms voltage. What is the frequency of the source?

Answer:

Peak voltage = 200 V

Then, rms voltage = \(\frac{200}{\sqrt{2}}=141.4 \mathrm{~V}\)

Angular frequency, \(\omega=314 \mathrm{~Hz}\)

The frequency of the source = \(\frac{\omega}{2 \pi}=\frac{314}{2 \times 3.14}=50 \mathrm{~Hz}\)

Question 19. State the condition trader in which the phenomenon of resonance occurs in a series LCR circuit when ac voltage is applied. In a series LCR circuit, the current is in the same phase as the voltage. Calculate the value of self-inductance if the capacitor is 20 μF and the resistance used is 10 ohm with the ac source of frequency 50 Hz.

Answer:

Condition of resonance in a series LCR circuit:

∴ \(\omega L=\frac{1}{\omega C}\)

where a) = angular frequency of the source, L = self-inductance of the coil, and C = capacitance.

According to the question, electric current and emf are in the same phase in the LCR circuit, i.e., the circuit is in the resonance condition.

Here, ω = 2π x 50 = 2 x 3.14 x 50 = 314 Hz

C = 20 μF = 20 x 10-6 F = 2 x 10-5 F

Now, \(\omega L=\frac{1}{\omega C}\)

or, \(L=\frac{1}{\omega^2 C}=\frac{1}{(314)^2 \times\left(2 \times 10^{-5}\right)}=0.507 \mathrm{H}\)

Question 20. Whatis Q-factor?

Answer:

Q-factor is a dimensionless parameter that describes how underdamped an oscillator or a resonator is.

Question 21.

  1. Compare between inductive reactance and capacitive reactance.
  2. In an LCR series combination, R = 400Ω, L = 100 mH and C= 1μF. This combination is connected to a 25 sin 2000t volt voltage source. Find the impedance of the circuit and the peak value of the circuit current.

Answer:

1. Inductive reactance, XL = coL increases with the increase of either frequency or inductance or both. Capacitive reactance, \(X_C=\frac{1}{\omega C}\) decreases with the increase of either frequency or capacitance or both.

2. 1st Part: Frequency of the source, ω = 200 rad/s

∴ Impedance,

∴ \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

= \(\sqrt{400^2+\left(2000 \times 100 \times 10^{-3}-\frac{1}{200 \times 10^{-6}}\right)^2}\)

= \(\sqrt{16 \times 10^4+9 \times 10^4}=500 \Omega\)

2nd Part: Peak value of current,

∴ \(I_0=\frac{V_0}{R}=\frac{25}{400}=62.5 \mathrm{~mA}\)

Question 22.

  1. Why is the use of ac voltage preferred over dc voltage?
  2. The power factor of the LR circuit is \(\frac{1}{\sqrt{3}}\). If the frequency of ac is doubled, what will be the power factor?

Answer:

1. Ac voltage can be stepped up or stepped down using a transformer, which is essential for power transmission. and power consumption in daily life. In addition, any capacitor or inductor can be used as an active component in an ac circuit,.These advantages are not provided by dc voltages.

2. If the phase difference between the voltage and the current in the LR circuit is θ,

∴ \(\tan \theta=\frac{\omega L}{R}\)

From the figure, power factor,

∴ \(\cos \theta=\frac{R}{\sqrt{R^2+\omega^2 L^2}}\)

In the first case,

∴ \(\frac{R}{\sqrt{R^2+\omega^2 L^2}}=\frac{1}{\sqrt{3}} \quad \text { or, } \frac{R^2}{R^2+\omega^2 L^2}=\frac{1}{3}\)

or, ω2L2 = 3R2 – R2 = 2R2

In the second case, the frequency Is doubled. The new frequency Is 2ω.

∴ Power factor = \(\frac{R}{\sqrt{R^2+4 \omega^2 L^2}}=\frac{R}{\sqrt{R^2+4 \times 2 R^2}}\)

= \(\frac{R}{\sqrt{9 R^2}}=\frac{1}{3}\)

Alternating Current Phase Difference

Question 23. If the rotating speed of a dynamo is doubled, the induced electromotive force will be

  1. Doubled
  2. Halved
  3. Four times as much
  4. Unchanged

Answer: 1. Doubled

e = ωBANsin(ωt + α); if w is doubled then e is also doubled.

The option 1 is correct.

Question 24. The number of turns of the primary and secondary of a transformer are 500 and 5000 respectively. The primary is connected to a 20 V, 50 Hz ac supply. The output of the secondary will be

  1. 2 V, 50 Hz
  2. 200 V, 50 Hz
  3. 200 V, 5 Hz
  4. 200 V, 500 Hz

Answer: 2. 200 V, 50 Hz

Output voltage = \(20 \times \frac{5000}{500}=200 \mathrm{~V}\); the frequency will not change.

The option 2 is correct.

Question 25. What is the rms value of the current i = 5√2 sin 100 t A?

Answer:

rms value = \(\frac{\text { peak value }}{\sqrt{2}}=\frac{5 \sqrt{2}}{\sqrt{2}}=5 \mathrm{~A}\)

Question 26. In the circuit shown below, the switch Is kept In position a for n long time and is then thrown to position b. The amplitude of the resulting oscillating current Is given by

Alternating Current Resulting Oscillating Current

  1. \(B \sqrt{L / C}\)
  2. E/R
  3. Infinity
  4. \(E \sqrt{C / L}\)

Answer: 4. \(E \sqrt{C / L}\)

If the switch Is kept In position for a long time, the charge accumulated on the capacitor plates = EC.

When the switch Is thrown to position b, It behaves as a source of emf due to the accumulated charges.

Electromotive force = \(\frac{E C}{C}=E\)

As a result, the LC circuit changes to an oscillating circuit, whose angular frequency, \(\omega=\frac{1}{\sqrt{L C}}\)

∴ Maximum current through the circuit

= \(\frac{E}{\omega L}=\frac{E}{\frac{1}{\sqrt{L C}} \cdot L}=E \cdot \sqrt{\frac{C}{L}}\)

or, \(\frac{E}{1 / \omega C}=E \omega C=E \frac{1}{\sqrt{L C}} C=E \sqrt{\frac{C}{L}}\)

The option 4 is correct.

Question 27. When the frequency of the ac voltage applied to a series LCR circuit is gradually increased from a low value, the impedance of the circuit

  1. Monotonically increases
  2. First increases and then decreases
  3. First decreases and then increases
  4. Monotonically decreases

Answer: 3. First decreases and then increases

We know Z = \(\sqrt{\left(\omega L-\frac{1}{\omega C}\right)^2+R^2}\)

If \(\frac{d Z}{d \omega}=0 \text { then, } \omega=\frac{1}{\sqrt{L C}}\)

When \(\omega<\frac{1}{\sqrt{L C}}\) i.e., Z is a decreasing function. Again, when \(\omega>\frac{1}{\sqrt{L C}} \text { then } \frac{d Z}{d \omega}>0\) i.e., Z is a increasing function.

The option 3 is correct.

Question 28. An alternating current is flowing through a series LCR circuit. It is found that the current reaches a value of 1 mA at both 200 Hz and 800 Hz frequencies. What is the resonance frequency of the circuit?

  1. 600 Hz
  2. 300 Hz
  3. 500 Hz
  4. 400 Hz

Answer: 4. 400 Hz

For LCR circuit, \(I=\frac{V_0}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}\)

Since currents for frequencies 200 Hz and 800 Hz are the same,

∴ \(\frac{V_0}{\sqrt{R^2+\left(\omega_1 L-\frac{1}{\omega_1 C}\right)^2}}=\frac{V_0}{\sqrt{R^2+\left(\omega_2 L-\frac{1}{\omega_2 C}\right)^2}}\)

or, \(\left(\omega_1 L-\frac{1}{\omega_1 C}\right)= \pm\left(\omega_2 L-\frac{1}{\omega_2 C}\right)\) (1)

By interchanging the value of ω1 and ω2, the sign of the value of \(\left(\omega L-\frac{1}{\omega C}\right)\) may change. So, by considering’+’ sign in the left-hand side and the sign on the right-hand side of equation (1), we have

∴ \(\left(\omega_1 L-\frac{1}{\omega_1 C}\right)=-\left(\omega_2 L-\frac{1}{\omega_2 C}\right)\)

or, \(L C=\frac{1}{\omega_1 \omega_2} \quad \text { or, } \omega_1 \omega_2=\frac{1}{L C} \quad \text { or, } \omega_1 \omega_2=\omega_0^2\)

or, f1f2 = f20 [f0 – resonance frequency]

or, f0 = 400 Hz

The option 4 is correct

Question 29. An inductor (L = 0.03 H) and a resistor (R = 0.15kΩ) are connected in series to a battery of 15 V in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1 ms, the current in the circuit will be (e5 ≈ 150)

Alternating Current An Inductor And A Resistor Connected In Series To A Battery

  1. 100 mA
  2. 67 mA
  3. 6.7 mA
  4. 0.67 mA

Answer: 4. 0.67 mA

When the key K1 is closed, current through the inductor,

∴ \(I_0=\frac{e}{R}=\frac{15}{0.15 \times 10^3}=0.1 \mathrm{~A}\)

Here, t = 1 ms = 1-3 s

The time constant of the LR circuit,

∴ \(t_0=\frac{L}{R}=\frac{0.03}{0.15 \times 10^3}=2 \times 10^{-4} \mathrm{~s}\)

∴ \(\frac{t}{t_0}=\frac{10^{-3}}{2 \times 10^{-4}}=5\)

∴ \(I=I_0 e^{-t / t_0}=0.1 e^{-5}=\frac{0.1}{150}=0.67 \times 10^{-3} \mathrm{~A}\)

= 0.67 mA

The option 4 is correct

Question 30. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit, the capacitor is charged to Q0 and then connected to the L and R as shown below:

Alternating Current LCR Circuit Is Equivalent To A Damped Pendulum

If a student plots graphs of the square of maximum charge (Q2max) on a capacitor with time (t) for two different values L1 and L2(L1 > L2) of L then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale)

Alternating Current Plots Graphs

Answer: 1.

Damping of the charge of the capacitance occurs at an exponential rate.

Also, the greater the value of L, the greater the energy loss due to R.

The option 1. is correct.

Question 31. An arc lamp requires a direct current of 10 A at 80V to function. It is connected to a 220 V (rms), 50Hz ac supply, and the series inductor needed for it to work close to

  1. 80 H
  2. 0.08 H
  3. 0.044 H
  4. 0.065 H

Answer: 4. 0.065 H

When the clamp functions with dc, V = 80V, I = 10 A

Therefore, resistance of the lamp, \(R=\frac{V}{I}=\frac{80}{10}=8 \Omega\)

When the lamp and the inductor are connected to an ac source, the impedance of the circuit,

∴ \(Z=\sqrt{R^2+\omega^2 L^2}\)

The electric current in the circuit,

∴ \(I=\frac{V^{\prime}}{Z}=\frac{V^{\prime}}{\sqrt{R^2+\omega^2 L^2}} \text { or, } R^2+\omega^2 L^2=\left(\frac{V^{\prime}}{I}\right)^2\)

or, \(8^2+\omega^2 L^2=\left(\frac{220}{10}\right)^2 \text { or, } \omega^2 L^2=(22)^2-8^2\)

or, \(L=\frac{\sqrt{22^2-8^2}}{\omega}=\frac{\sqrt{30 \times 14}}{2 \pi \times 50}=0.065 \mathrm{H}\)

The option 4 is correct.

Question 32. For an RLC circuit driven with a voltage of amplitude vm and frequency \(\omega_0=\frac{1}{\sqrt{L C}}\) the current exhibits resonance. The quality factor, Q is given by

  1. \(\frac{R}{\left(\omega_0 C\right)}\)
  2. \(\frac{C R}{\omega_0}\)
  3. \(\frac{\omega_0 L}{R}\)
  4. \(\frac{\omega_0 R}{L}\)

Answer: 3. \(\frac{\omega_0 L}{R}\)

Q factor of RLC circuit = \(\frac{\omega_0}{\Delta \omega}=\frac{\omega_0}{\frac{R}{L}}=\frac{\omega_0 L}{R}\)

The option 3 is correct.

Question 33. In an ac circuit, the instantaneous emf and current is given by e = 100 sin 30t; \(i=20 \sin \left(30 t-\frac{\pi}{4}\right)\).

In one cycle of ac, the average power consumed by the circuit and the wattless current are, respectively

  1. \(\frac{50}{\sqrt{2}}, 0\)
  2. 50,0
  3. 50,10
  4. \(\frac{1000}{\sqrt{2}}, 10 \)

Answer: 4. \(\frac{1000}{\sqrt{2}}, 10 \)

⇒ \(P_{\mathrm{avg}}=e_{\mathrm{rms}} i_{\mathrm{rms}} \cos \phi_{3 \mathrm{~b}}=\frac{100}{\sqrt{2}} \times \frac{20}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\left [∵ \phi_{9 \mathrm{rg}}, \frac{\pi}{4}\right]\)

= \(=\frac{1000}{\sqrt{2}}\)

The wattless current, \(i=i_{\mathrm{rms}} \cos \phi=\frac{20}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=10\)

The option 4 is correct.

Question 34. A transformer having an efficiency of 90% is working on 200V and 3kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and the current in the primary coil respectively are

  1. 300V, 15A
  2. 450V, 15A
  3. 450V, 13.5A
  4. 600V, 15A

Answer: 2. 450V, 15A

Power of the secondary coil, Ps = VpIp

∴ \(I_p=\frac{3 \times 1000}{200}=15 \mathrm{~A}\)

Power of the secondary coil, Ps = V$IS

∴ \(V_s=\frac{P_s}{I_s}=\frac{P_p \times \frac{90}{100}}{6}=3 \times 1000 \times \frac{90}{100} \times \frac{1}{6}=450 \mathrm{~V}\)

The option 2 is correct.

Question 35. A resistance R draws power P when connected to an ac source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes Z, the power drawn will be:

  1. \(P\left(\frac{R}{Z}\right)^2\)
  2. \(P \sqrt{\frac{R}{Z}}\)
  3. \(P\left(\frac{R}{Z}\right)\)
  4. P

Answer: 1. \(P\left(\frac{R}{Z}\right)^2\)

The rms currents in the first and second cases are \(I=\frac{E}{R} and I^{\prime}=\frac{E}{Z}\), respectively.

∴ The pure inductor does not draw any power, then the respective powers drawn are \(P=I^2 R=\left(\frac{E}{R}\right)^2 R=\frac{E^2}{R}\)

and \(P^{\prime}=I^{\prime 2} R=\left(\frac{E}{Z}\right)^2 R=\frac{E^2}{R}\left(\frac{R}{Z}\right)^2=P\left(\frac{R}{Z}\right)^2\)

Option 1 is correct.

Question 36. A small signal voltage V(t) = V0 sinty t is applied across an ideal capacitor C.

  1. Over a full cycle, the capacitor C does not consume any energy from the voltage source
  2. Current I(t) is in phase with voltage V(t)
  3. Current I(t) leads voltage V(t) by 180°
  4. Current I(t) lags voltage V(t) by 90°

Answer: 1. Over a full cycle, the capacitor C does not consume any energy from the voltage source

The option 1 is correct

Question 37. An inductor 20mH, a capacitor 50 μF, and a resistor 40 Ω connected in series across a source of emf V = 10 sin 340 t. The power loss in ac circuit is

  1. 0.67W
  2. 0.76W
  3. 0.89W
  4. 0.51 W

Answer: 4. 0.51 W

Here, ω = 340 Hz

Inductive reactance, XL = ωL = 340 x 20 x 10-3 = 6.8H

Capacitive reactance, \(X_C=\frac{1}{\omega C}=\frac{1}{340 \times 50 \times 10^{-6}}=58.8 \Omega\)

Resistance, R = 40 Ω

Therefore, the impedance of the circuit,

∴ \(Z=\sqrt{R^2+\left(X_C-X_L\right)^2}\)

= \(\sqrt{40^2+(58.8-6.8)^2}=65.6 \Omega\)

Now, rms value of current, \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}=\frac{10}{\sqrt{2} \times 65.6}\)

Power dissipated in the circuit,

P = \(V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta=V_{\mathrm{rms}} I_{\mathrm{rms}} \frac{R}{Z}\)

= \(\frac{10}{\sqrt{2}} \times \frac{10}{\sqrt{2} \times 65.6} \times \frac{40}{65.6}=0.46 \mathrm{~W} \approx 0.51 \mathrm{~W}\)

The option 4 is correct.

Question 38. An inductor 20 mH, a capacitor 100 μF, and a resistor 50 Ω are connected in series across a source of emf, V= 10 sin 314t. The power of the circuit is

  1. 2.74 W
  2. 0.43 W
  3. 0.79 W
  4. 1.13 W

Answer: 3. 0.79 W

∴ \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

= \(\sqrt{(50)^2+\left[314 \times 20 \times 10^{-3}-\frac{1}{314 \times 100 \times 10^{-6}}\right]^2}\)

= 56.16 Ω

The power loss in the circuit,

∴ \(P=\left(\frac{V_{\mathrm{rms}}}{Z}\right)^2 \cdot R=\left(\frac{\frac{10}{\sqrt{2}}}{56.16}\right)^2 \times 50=0.79 \mathrm{~W}\)

The option 3 is correct.

Question 39. An alternating voltage given by V = 140 sin 314t is connected across a pure resistor of 50 Ω. Find

  1. The frequency of the source
  2. The rms current through the resistor.

Answer:

V = 140 sin 314t = V0 sin ωt

1. Frequency of the source,

∴ \(n=\frac{\omega}{2 \pi}=\frac{314}{2 \times 3.14}=50 \mathrm{~Hz}\)

2. \(I=\frac{V}{R}=\frac{140}{50} \sin 314 t=2.8 \sin 314 t=I_0 \sin \omega t\)

∴ \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}} \approx \frac{2.8}{1.4}=2 \mathrm{~A}\)

Question 40.

  1. For a given ac i = im sin ωt, show that the average power dissipated in a resistor R over a complete cycle is \(\frac{1}{2} i_m^2 R\)
  2. A light bulb is rated at 100 W for a 220 V ac supply. Calculate the resistance of the bulb.

Answer:

1. i = im sin cot, so v = im R sin cot

Power dissipated, P = vi = i2mRsin2cot

The average of sin2tot over a complete cycle = \(\frac{1}{2}\)

∴ Average power dissipated,

= \(\bar{P}=i_m^2 R \cdot \frac{1}{2}=\frac{1}{2} i_m^2 R\)

2. \(P=\frac{V^2}{R} \quad \text { or, } R=\frac{V^2}{P}=\frac{(220)^2}{100}=484 \Omega\)

Question 41. Why is the use of ac voltage preferred over dc voltage? Give two reasons.

Answer:

  1. Alternating voltages can be stepped up or down quite easily.
  2. Alternating power can be transmitted over a long distance with very small thermal loss using a high voltage-low current ac supply.

Question 42. A voltage V = V0sincut is applied to a series LCR circuit Derive the expression for the average power dissipated over a cycle. Under what condition is

  1. No power dissipated even though the current flows through the circuit,
  2. Maximum power dissipated in the circuit?

Answer:

  1. No power is dissipated if the pure resistance R in the circuit is zero.
  2. Maximum power is dissipated if the inductive and capacitative reactances cancel each other i.e., when the circuit impedance becomes equal to the pure resistance of the circuit(Z = R).

Question 43. In a series LR circuit, XL = R, and the power factor of the circuit is P1. When a capacitor with capacitance C such that XC = XL is put in series, the power factor becomes P2. Find out P1/P2.

Answer:

In a series LR circuit, power factor \(\left(P_1\right)=\frac{R}{Z}\)

Here, Z = impedance = \(\sqrt{R^2+X_L^2}\)

Here, \(Z=\sqrt{R^2+R^2}=\sqrt{2} R\)

∴ \(P_1=\frac{1}{\sqrt{2}}\)

In a series LCR Circuit, power factor \(\left(P_2\right)=\frac{R}{Z}\)

Where \(Z=\sqrt{R^2+\left(X_L-X_C\right)^2}=R\)

∴ P2 = 1

Hence, \(\frac{P_1}{P_2}=\frac{1}{\sqrt{2}}\)

Question 44.

  1. When an ac source is connected to an ideal capacitor, show that the average power supplied by the source over a complete cycle is zero.
  2. A bulb is connected in series with a variable capacitor and an ac source as shown. What happens to the brightness of the bulb when the key is plugged in and the capacitance of the capacitor is gradually reduced?

Alternating Current A Bulb Is Connected In Series With A Variable Capacitor And An Ac Source

Answer:

1. The instantaneous power supplied to the capacitor,

P = I0 cos(ωt) V0 sin(ωT)

P = I0 V0 cos(ωt) sin(ωt)

∴ \(P=\frac{I_0 V_0}{2} \sin (2 \omega t)\)

Therefore, the average power,

∴ \(\bar{P}=\frac{I_0 V_0}{2} \sin (2 \omega t)=\frac{I_0 V_0}{2} \sin (2 \omega t)\)

Now, the average of sin( 2<wt) over the cycle is zero.

∴ \(\bar{P}=0\)

2. The capacitance of the capacitor is gradually reduced. Therefore, the capacitive reactance \(X_C=\frac{1}{2 \pi f C}\) increases.

Therefore, essentially, the overall resistance of the circuit increases. This causes a reduction in the amount of current flowing through the circuit. Therefore, the brightness of the bulb reduces.

Question 45. A device X is connected to an ac source V = V0 sin ωt. The variation of voltage, current, and power in one cycle is shown in the following graph:

Alternating Current The Variation Of Voltage, Current And Power In One Cycle

  1. Identify the device X
  2. Which of the curves A, B, and C represent the voltage, current, and power consumed in the circuit? Justify your answer.
  3. How does its impedance vary with the frequency of the ac source? Show graphically.
  4. Obtain an an expression for the current in the circuit and its phase relation with ac voltage

Answer:

1. Device X is a capacitor.

2. B represents voltage because it is a sine wave. C represents current because current leads voltage by \(\frac{\pi}{2}\). A represents power because the average power cycle is zero.

3. Impendance, \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)

Alternating Current Impedance Vary With Frequency Of The Ac Source

4. \(C=\frac{q}{V}\)

∴ q = CV = CV0 sinωt [∵ V = V0sin t]

∴ Current, \(I=\frac{d t}{d q}=\frac{d}{d t}\left(C V_0 \sin \omega t\right)=\omega C V_0 \cos \omega t\)

= \(\frac{V_0}{\frac{1}{\omega C}} \cos \omega t\)

∴ \(I=\frac{V_0}{\dot{X}_C} \sin \left(\omega t+\frac{\pi}{2}\right) \quad \text { or, } I=I_0 \sin \left(\omega t+\frac{\pi}{2}\right)\)

In a pure capacitive circuit current leads voltage by \(\frac{\pi}{2}\).

Question 46. Find the value of the phase difference between the current and the voltage hi the series LCR circuit shown below. Which one leads in phase: current or voltage?

 

Alternating Current The Value Of Phase Difference Between The Current And The Voltage In The Series LCR Circuit

Answer:

Given, V = V0 sin(1000t +Φ), R = 4000, C = 2μF,

L = 100 mH.

The standard equation is given by,

V = V0 sin (ωt + Φ)

∴ ω = 1000 Hz

Now, XL = ωL= 1000 x 100 x 10-3 = 102 = 100Ω

and \(X_C=\frac{1}{\omega C}=\frac{1}{1000 \times 2 \times 10^{-6}}=500 \Omega\)

The phase difference between the current and the voltage in the series LCR circuit is given by,

∴ \(\phi=\tan ^{-1}\left(\frac{X_C-X_L}{R}\right)\)

or, \(\phi=\tan ^{-1}\left(\frac{500-100}{400}\right)=\tan ^{-1} 1\)

∴ Φ = 45°

Since, XC > XL, therefore current leads in phase.

Question 47. Without making any other change, find the value of the additional capacitor, to be connected in parallel with the capacitor C, to make the power factor of the circuit unity.

 

Alternating Current Power Factor Of The Circuit Unity

Answer:

To make the power factor of the circuit unity,

Alternating Current Make Power Factor Of The Circuit Unity

XC = XL

∴ \(\frac{1}{\omega\left(C+C_1\right)}=100 \text { or, } \frac{1}{1000\left(C+C_1\right)}=100\)

or, \(C+C_1=\frac{1}{10^5}\)

or, \(C_1=\frac{1}{10^5}-C=10^{-5}-0.2 \times 10^{-5}=0.8 \times 10^{-5}\)

or, C1 = 8μF

Question 48.

  1. Draw a labeled diagram of a step-up Transformer. Obtain the ratio of secondary to primary voltage in terms of several turns and currents in the two coils.
  2. A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V.

Answer:

Alternating Current Step-up Transformer

1. When an alternating potential Vp is applied to the primary coil, an emf (ep) is induced in it.

∴ \(e_p=-N_p \frac{d \phi}{d t}\)

It resistance of the primary coil is low, Vp = ep

∴ \(V_p=-N_p \frac{d \phi}{d t}\)

As the same flux is linked with the secondary coil with the help of a soft iron core due to mutual induction, emf (es) is induced in it.

∴ \(e_s=-N_s \frac{d \phi}{d t}\)

If the output circuit is open then, Vs = es

∴ \(V_s=-N_s \frac{d \phi}{d t}\)

Thus, \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\)

For an ideal transformer, Pout = Pin

or, IsVs = IpVp

∴ \(\frac{V_s}{V_p}=\frac{I_p}{I_x}=\frac{N_s}{N_p}\)

For setup transformer \(\frac{N_s}{N_p}>1\).

In the case of dc voltage, flux does not change. Thus no emf is induced in the circuit.

2. Given, Vp = 2200 V, Np = 3000, Vs = 220 V

We know, \(\frac{V_s}{V_p}=\frac{N_s}{N_p} \quad \text { or, } N_s=\frac{N_p \cdot V_s}{V_p}=\frac{3000 \times 220}{2200}\)

∴ Ns = 300

Question 49. A device X is connected across an ac source of voltage V= V0sincot. The current through X is given as \(I=I_0 \sin \left(\omega t+\frac{\pi}{2}\right)\).

  1. Identify the device X and write the expression for its reactance.
  2. Draw graphs showing variation of voltage and current with time over one cycle of ac, for X.
  3. How does the reactance of device X vary with the frequency of the ac? Show this variation graphically.
  4. Draw the phasor diagram for the device X.

Answer:

1. Device X is a capacitor

∴ Capacitive reactance, \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)

Alternating Current Variation Of Voltage And Current With Time Over One Cycle Of Ac

Alternating Current Reactance Of The Device With Frequency

Alternating Current Phasor

Question 50. The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is conveyed to a higher or lower voltage.

  1. Name the device used to change the alternating voltage to a higher or lower value. State one cause for power dissipation in this device.
  2. Explain with an example, how power loss is reduced if the energy is transmitted over long distances as an alternating current rather than a direct current.
  3. Write two values each shown by the teachers and Geeta.

Answer:

1. 1st part: A Transformer is a device used to change the alternating voltage to a higher or lower value. 2nd part: One cause for power dissipation in this device is copper loss. Generally, copper wire is used to make primary and secondary coils. Due to Joule’s heating, some energy is dissipated as heat energy.

2. At the electric power producing station, a set-up transformer is used to increase the alternating voltage up to several kilovolts. So the electric current flowing through transmission wires decreases. As Joule’s heating is proportional to the square of current, the loss of electrical energy across transmission wires decreases due to a decrease in current.

3. Two values shown by teachers are

  1. sense of responsibility,
  2. good practical knowledge. Two values shown by Geeta are—
    • curiosity to learn,
    • critical thinking.

 

WBCHSE Class 12 Physics Alternating Current Questions And Answers

Physics Electromagnetic Induction And Alternating Current

Alternating Current Long Questions And Answers

Question 1. In an oscillating LC circuit the maximum charge on the capacitor is Q. When the charge Is stored equally between the electric and magnetic fields, what Is the charge on the capacitor?

Answer:

Maximum charge of the capacitor = Q

So total energy of the circuit = \(\frac{1}{2} \frac{Q^2}{C}\)

When the’ chat ge of the capacitor is q, then the energy stored in the electric field is \(\frac{1}{2} \frac{q^2}{C}\). At this stage energy stored in the magnetic field is \(\frac{1}{2} L i^2\).

If it is \(\frac{1}{2} \frac{q^2}{C}\), equal to the total energy

= \(\frac{1}{2} \frac{q^2}{C}+\frac{1}{2} L i^2=\frac{1}{2} \frac{q^2}{C}+\frac{1}{2} \frac{q^2}{C}=\frac{q^2}{C}\)

so, \(\frac{q^2}{C}=\frac{1}{2} \frac{Q^2}{C} \quad \text { or, } q^2=\frac{Q^2}{2} \quad \text { or, } q=\frac{Q}{\sqrt{2}}\)

Question 2. In an LCR circuit, capacitance is changed from C to 2C. For the resonant frequency to remain unchanged what should be the change in the value of inductance L?

Answer:

Resonant frequency \(f_0=\frac{1}{2 \pi \sqrt{L C}}\)

For f0 to be constant, LC = constant.

If C is made 2C and the changed value of L is L’

then, \(L C=L^{\prime} \cdot 2 C \quad \text { or, } L^{\prime}=\frac{L}{2}\)

Question 3. If the emf of an ac circuit is E = E0 sin ωt and current I = I0 cos ωt, what is the power dissipated in the circuit?

Answer:

∴ \(E=E_0 \sin \omega t ; I=I_0 \cos \omega t=I_0 \sin \left(\omega t+90^{\circ}\right)\)

So, phase difference, θ = 90°

Therefore, power factor = cos θ – cos90° = 0

i.e., power dissipated = 0

Question 4. What should be the nature of the graph of the impedance Z concerning frequency in an alternating LCR circuit?

Answer:

If f is the frequency, ω = 2πf

Impedance, \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

If frequency, \(f_0=\frac{1}{2 \pi \sqrt{L C}}\)

then \(\omega_0=2 \pi f_0=\frac{1}{\sqrt{L C}} \quad \text { or, } \omega_0 L=\frac{1}{\omega_0 C}\).

Under this condition Z is minimum, i.e., Z = R.

If the frequency increases or decreases concerning this resonant frequency f0, Z will vary accordingly, as shown.

Alternating Current Nature Of Graph Of The Impedance

Question 5. How does the wattless current conform to the principle of energy conservation?

Answer:

An alternating current through a pure inductor or pure capacitor is witnessed, which means that no part of the input energy is lost due to heating.

Instead, the electrical energy supplied from the alternating source is stored in the magnetic field of the inductor or the electrostatic field of the capacitor.

Because of this stored energy, the current continues to flow in the circuit, even if the alternating spruce is removed from the circuit.

Question 6. The current through a circuit is given by \(I=I_0 \sin (\omega t+\pi / 6)\) when the applied emf is V = V0sinωt. Find the power dissipated in the circuit in one complete cycle. Draw the. phasor diagram for the given current and voltage. What are the possible two elements in the circuit?

Answer:

Here, the current leads the voltage by a phase angle, \(\theta=\frac{\pi}{6}\)

So, the power dissipated in one complete cycle

∴ \(P=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta=\frac{V_0}{\sqrt{2}} \frac{I_0}{\sqrt{2}}\cos \frac{\pi}{6}\)

= \(\frac{1}{2} V_0 I_0 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} V_0 I_0\)

This shows the phase relation for the given current and; voltage.

If the circuit contains two elements only, then these are a pure resistor R and a capacitor C, because the current leads the voltage in a CR circuit.

Alternating Current The Phase Relation For The Current And Voltage

Question 7. How does the resistance R change with the change in frequency of AC? Show graphically.

Answer:

As R is independent of frequency (f), the nature of the graph is a straight line, parallel to the frequency axis.

Alternating Current Resistance Change In Frequency

Question 8. Sketch a graph to show how the reactance of an inductor varies as a function of frequency.

Answer:

We know, \(X_L=\omega L=2 \pi f L\) [∵\(f=\frac{\omega}{2 \pi}\)]

where f is the frequency of ac supply. For a given coil, L is constant.

∴ \(X_L \propto f\)

So, Inductive reactance is directly proportional to the frequency of the current. Thus f-XL graph is a straight line passing through the origin.

Alternating Current Reactance Of An Inductor Varies As A Function Of Frequency

Question 9. Sketch a graph to show how the reactance of a capacitor varies as a function of frequency.

Answer: We know, \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\) [∵ \(f=\frac{\omega}{2 \pi}\)]

For constant C, \(X_C \propto \frac{1}{f}\)

So, the capacitive reactance is inversely proportional to the frequency of the current. Thus, the f-XC graph is a rectangular hyperbola.

Alternating Current Reactance Of An Capacitor Varies As A Function Of Frequency

Question 10. How does an inductor behave in a dc circuit?

Answer:

An inductor, having a pure resistance R and a self-inductance L, has an impedance, \(Z=\sqrt{R^2+\omega^2 L^2}\) Here, ω = 2πf, where f = frequency of the electric source. Naturally, f = 0 for dc circuits; so, ω = 0, and Z = R.

This means that the inductor behaves as a pure resistance in dc circuits, whereas its inductance L, plays no role.

Question 11. The LC oscillations can be compared to mechanical oscillations of a block of mass m attached to a spring of force constant k. Explain.

Answer:

The LC oscillations are comparable to mechanical oscillations of a spring-mass system since both of them are simple harmonic motions.

Alternating Current LC Oscillations Of A Block Of Mass

The general equation for the oscillation of a pure LC circuit,

∴ \(\frac{d^2 q}{d t^2}+\frac{q}{L C}=0\) → (1)

Again, the equation for the mechanical oscillation of a spring-mass system,

∴ \(\frac{d^2 x}{d t^2}+\omega_0^2 x=0\) → (2)

Where, \(\omega_0=\sqrt{\frac{k}{m}}\)

Comparing equations (1) and (2), we may write that the LC oscillations and the mechanical oscillations of the spring-mass system are analogous.

The analogy is given below in tabular form.

Alternating Current LC Oscillations And The Mechanical Oscillations Of Spring Mass System

Question 12. In the given circuit, the switch K2 iS opened, and the switch If, is closed at time t = 0. At time t = t0, the switch K1 is opened, and the switch K2 is simultaneously closed. Sketch the variation of the inductor current I with time.

Alternating Current The Variation Of The Inductor Current With Time

Answer:

Between time t = 0 and t = t0, the equation for the inductor current, \(L \frac{d I}{d t}=E \quad \text { or, } I=\frac{E}{L} t\)

For time t > t0, the equation for the inductor current,

∴ \(L\left(\frac{d I}{d t}\right)=0\)

or, \(I=I\left(t_0\right)=\frac{E t_0}{L} \text { (constant) }\)

Thus I increase linearly with a slope for t = 0 to t = t0.

For time t ≥ t0, I becomes a constant \(\left(=\frac{E t_0}{L}\right)\)

The variation of current (I) with time (t) is shown.

Alternating Current Inductor Current

Question 13. Show that In the free oscillation of an LC circuit, the sum of energies stored In the capacitor and the Inductor is constant in time.

Answer:

Energy stored in the capacitor at time t,

⇒ \(U_E=\frac{q^2}{2 C}=\frac{q_m^2}{2 C} \cos ^2 \omega t\)

Energy stored in the inductor at time t,

⇒ \(U_B=\frac{1}{2} L I^2=\frac{L}{2}\left[q_m \omega \sin \omega t\right]^2\)

= \(\frac{q_m^2}{2} L \omega^2 \sin ^2 \omega t=\frac{q_m^2}{2 L C} L \sin ^2 \omega t=\frac{q_m^2}{2 C} \sin ^2 \omega t\)

Therefore total energy,

∴ \(U=U_E+U_B=\frac{q_m^2}{2 C}\left(\cos ^2 \omega t+\sin ^2 \omega t\right)=\frac{q_m^2}{2 C}\).

Therefore, for free oscillations of an LC circuit, the net energy remains constant at all times.

Alternating Current Energy Is Stored In The Capacitor And The Inductor Is Constant In Time

WBCHSE Class 12 Physics Alternating Current Multiple Questions And Answers

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Multiple Questions And Answers

Question 1. The internal resistance and internal reactance of an alternating current generator are Rg and Xg respectively. Power from this source is supplied to a load consisting of resisting Rg and reactance XL. For maximum power to be delivered from the generator to the load. The value of XL is equal to

  1. Zero
  2. Xg
  3. -Xg
  4. Rg

Answer: 3. -Xg

For maximum power, total reactance is zero.

Question 2. To reduce the resonant frequency in an LCR series circuit with a generator,

  1. The frequency of the generator should be reduced
  2. Another capacitor should be connected in parallel with the first capacitor
  3. The iron core of the inductor should be removed
  4. The dielectric in the capacitor should be removed

Answer: 2. Another capacitor should be connected in parallel with the first capacitor

∴ \(f=\frac{1}{2 \pi \sqrt{L C}}\)

To reduce f either L or C or both has to be increased.

Question 3. Which of the following combinations should be selected for fine-tuning an LCR circuit used for communication?

  1. R = 20 Ω, L = 1.5 H, C= 35 μF
  2. R = 25 Ω, L = 2.5 H, C = 45 μF
  3. R = 15 Ω, L = 3.5 H, C= 30 μR
  4. R = 25 Ω, L = 1.5 H, C = 45 μF

Answer: 3. R = 15 Ω, L = 3.5 H, C= 30 μR

LCR circuit used for communication should have a high Q-factor, \(Q=\frac{1}{R} \sqrt{\frac{L}{C}}\).

Question 4. The current through an ac circuit first increases and then decreases as its frequency is increased. Which among the following are most likely combination of the circuit?

  1. Inductorand capacitor
  2. Resistorandinductor
  3. Resistor and capacitor
  4. Resistor, inductor, and capacitor

Answer:

2. Resistorandinductor

4. Resistor, inductor and capacitor

Question 5. The current through an ac (series) circuit increases as the source frequency is increased. Which of the following is the most suitable combination of the circuit?

  1. Only resistor
  2. A resistor and an inductor
  3. A resistor and a capacitor
  4. Only capacitor

Answer:

3. Resistor and a capacitor

4. Only capacitor

Question 6. When an ac voltage of 220 V is applied to a capacitor C

  1. The maximum voltage between plates is 220 V
  2. The current is in phase with the applied voltage
  3. The charge on the plates is in phase with the applied voltage
  4. The power delivered to the capacitor is zero

Answer:

3. The charge on the plates is in phase with the applied voltage

4. Power delivered to the capacitor is zero

P = vrms Irms cos Φ

∴ P = 0 [∴ Φ = 900 ]

Question 7. The line that draws the power supply to your house has

  1. Zero average current
  2. 220 V average voltage
  3. Voltage and current out of phase by 90°
  4. Voltage and current possibly differ in phase Φ such that \(|\phi|<\frac{\pi}{2}\)

Answer:

1. Zero average current

4. Voltage and current possibly differing in phase <p such that \(|\phi|<\frac{\pi}{2}\)

Since the line draws ac, the average current is zero. Again, since the line has some resistance (R ≠ 0), there is some phase difference between the voltage and current.

Question 8. An alternating current is given by the equation \(I=i_1 \sin \omega t+i_2 \cos \omega t\). The rms current is given by

  1. \(\left(i_2+i_1\right) / \sqrt{2}\)
  2. \(\left(i_2-i_1\right) / \sqrt{2}\)
  3. \(\sqrt{\left\{\left(i_1^2+i_2^2\right) / 2\right\}}\)
  4. \(\sqrt{\left\{\left(i_1^2+i_2^2\right) /(\sqrt{2})\right\}}\)

Answer: 3. \(\sqrt{\left\{\left(i_1^2+i_2^2\right) / 2\right\}}\)

Question 9. An ac having a peak value of 1.41 A is used to heat a wire. A dc producing the same heating rate will be of

  1. 1.41 A
  2. 2.0 A
  3. 0.705 A
  4. 1.0 A

Answer: 4. 1.0 A

Question 10. The general equation for the instantaneous voltage of a 50 Hz generator with a peak voltage 220 V is

  1. 220 sin 50 πt
  2. 220 sin l00 πf
  3. ± 220 sin l00 πt
  4. 220 sin 25 π t

Answer: 2. 220 sin l00 πf

Question 11. The relation between angular velocity (ω) and driving frequency (f) of an alternating current is

  1. ω = 27 πf
  2. \(\omega=\frac{2 \pi}{f}\)
  3. \(f=\frac{2 \pi}{\omega}\)
  4. f =27 πω

Answer: 1. ω = 27 πf

Question 12. Form factor of an alternating voltage is the ratio of

  1. Peak value and rms value
  2. Peak value and average value
  3. rms value and average value
  4. rms value and peak value

Answer: 3. rms value and average value

Question 13. The value of an ac voltage at time 0 < t < \(\frac{\pi}{\omega}\) is given by V= V0 Sinot and at time \(\frac{\pi}{\omega} < t < \frac{2 \pi}{\omega}\) is given by V = -V0 Sin cyf. The average value of V for a complete cycle is

  1. \(\frac{V_0}{\sqrt{2}}\)
  2. \(\left(\frac{2}{\pi}\right) \mathrm{v}_0\)
  3. \(\frac{V_0}{2}\)
  4. Zero

Answer: 2. \(\left(\frac{2}{\pi}\right) \mathrm{v}_0\)

Question 14. The rms value of the potential difference V is shown.

Alternating Current Potential Difference

  1. \(\frac{V_0}{\sqrt{3}}\)
  2. \(V_0\)
  3. \(\frac{V_0}{\sqrt{2}}\)
  4. \(\frac{V_0}{2}\)

Answer: 3. \(\frac{V_0}{\sqrt{2}}\)

Hint: In this case, V = V0, when 0 ≤ t ≤ \(\frac{T}{2}\)

= 0, when \(\frac{T}{2}\) ≤ t ≤ T

∴ \(V_{\mathrm{rms}}^2=\frac{\int_0^T V^2 d t}{\int_0^T d t}=\frac{1}{T} V_0^2\left[\int_0^{T / 2} d t\right]=\frac{V_0^2}{2}\)

or, \(V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}\)

Question 15. The rms value and frequency of an AC are 5A and 50 Hz respectively. The value of the current after \(\frac{1}{300}\)s from the time when its value becomes zero is

  1. \(5 \sqrt{2} \mathrm{~A}\)
  2. \(5 \sqrt{\frac{3}{2}} \mathrm{~A}\)
  3. \(\frac{5}{6} \mathrm{~A}\)
  4. \(\frac{5}{\sqrt{2}} \mathrm{~A}\)

Answer: 2. \(5 \sqrt{\frac{3}{2}} \mathrm{~A}\)

Hint: \(I=I_0 \sin \omega t=5 \sqrt{2} \sin \left(100 \pi \times \frac{1}{300}\right)\)

= \(5 \sqrt{2} \times \frac{\sqrt{3}}{2}=5 \sqrt{\frac{3}{2}} \mathrm{~A}\)

Series AC Circuits with R, L, C

Question 16. In an ac circuit containing capacitance, only the current

  1. Leads the voltage by 180°
  2. Is in phase with the voltage
  3. Leads the voltage by 90°
  4. Lags behind the voltage by 90°

Answer: 3. Leads the voltage by 90°

Question 17. In an LR circuit, the phase angle between alternating voltage and alternating current is 45°. The value of inductive reactance will be

  1. \(\frac{R}{4}\)
  2. \(\frac{R}{2}\)
  3. R
  4. Data insufficient

Answer: 3. R

Question 18. In an LCR series circuit, the capacitance is reduced to one-fourth, when in resonance. What change should be made in the inductance, so that the circuit remains in resonance?

  1. 4 times
  2. \(\frac{1}{4} \text { times }\)
  3. 8 times
  4. 2 times

Answer: 1. 4 times

Question 19. The phase difference between V and 1 of an LCR circuit in series resonance is

  1. π
  2. \(\frac{\pi}{2}\)
  3. \(\frac{\pi}{4}\)
  4. 0

Answer: 4. 0

Question 20. The reactance of an inductor of inductance \(\frac{1}{\pi}\) at frequency 50 Hz is

  1. \(\frac{50}{\pi} \Omega\)
  2. \(\frac{\pi}{50} \Omega\)
  3. 100
  4. 50

Answer: 3. 100

Question 21. which quantity in an ac circuit is not dependent on frequency?

  1. Resistance
  2. Impedance
  3. Inductive reactance
  4. Capacitative reactance

Answer: 1. Resistance

Question 22. The condition of getting maximum current in an LCR series circuit is

  1. \(X_L=0\)
  2. \(X_C=0\)
  3. \(X_L=X_C\)
  4. \(R=X_L-X_C\)

Answer: 3. \(X_L=X_C\)

Question 23. The series resonant frequency of an LCR circuit is f. If the capacitance is made,4, times the initial value, then the resonant frequency will become

  1. f/2
  2. 2f
  3. f
  4. f/4

Answer: 1. f/2

Question 24. A coil has a resistance 30 Ω and inductive reactance 20 Ω at 50 Hz frequency. If an ac source of 200V, 100 Hz is connected across the coil, the current in the coil will be

  1. 2.0 A
  2. 4.0 A
  3. 8.0 A
  4. \(\frac{20}{\sqrt{13}} \mathrm{~A}\)

Answer: 2. 4.0 A

Hint: In this case, R = 30Ω and XL = 20 Ω

∴ XL = ωL = 2nfL \f- frequency]

∴ \(\frac{X_L}{X_L^{\prime}}=\frac{f}{f^{\prime}} \quad \text { or, } \quad X_L^{\prime}=X_L \times\left(\frac{f^{\prime}}{f}\right)=20 \times\left(\frac{100}{50}\right)=40 \Omega\)

∴ \(Z=\sqrt{R^2+X_L^{\prime 2}}=\sqrt{(30)^2+(40)^2}=50\)

∴ [laex]I=\frac{V}{Z}=\frac{200}{50}=4 \mathrm{~A}[/latex]

Question 25. A fully charged capacitor C with initial charge q0 is connected to a coil of self-inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic field is

  1. \(\frac{\pi}{4} \sqrt{L C}\)
  2. \(2 \pi \sqrt{L C}\)
  3. \(\sqrt{L C}\)
  4. \(\pi \sqrt{L C}\)

Answer: 1. \(\frac{\pi}{4} \sqrt{L C}\)

Hint: During discharging of capacitor C through inductance L, let at any instant, charge in capacitor be Q.

∴ Q = Q0 sin ωt

Maximum energy storedin capacitor \(=\frac{1}{2} \frac{Q_0^2}{C}\)

Let at an instant t, the energy be stored equally between electric and magnetic field. The energy stored in the electric field instant t is

⇒ \(\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2}\left[\frac{1}{2} \frac{Q_0^2}{C}\right] \quad \text { or, } Q^2=\frac{Q_0^2}{2}\)

or, \(Q=\frac{Q_0}{\sqrt{2}} \quad \text { or, } Q_0 \sin \omega t=\frac{Q_0}{\sqrt{2}}\)

or, \(\omega t=\frac{\pi}{4} \quad \text { or, } t=\frac{\pi}{4 \omega}=\frac{\pi \sqrt{L C}}{4}\)  [∵\(\omega = \frac{1}{\sqrt{L C}}\)

Question 26. A voltage V0 sincot is applied across a series combination of resistance R and inductor L. The peak value of the current in the circuit is

  1. \(\frac{V_0}{\sqrt{R^2+\omega^2 L^2}}\)
  2. \(\frac{V_0}{\sqrt{R^2-\omega^2 L^2}}\)
  3. \(\frac{V_0}{\sqrt{R^2+\omega^2 L^2}} \sin \omega t\)
  4. \(\frac{V_0}{R}\)

Answer: 1. \(\frac{V_0}{\sqrt{R^2+\omega^2 L^2}}\)

Question 27. When an ideal choke is connected to an ac source of 100 V and 50 Hz, a current of 8 A flows through the circuit A current of 10A flows through the circuit when a pure resistoris connected instead of the choke coil. If the two are connected in series with an ac supply of 100V and 40 Hz, then the current in the circuit is

  1. 10 A
  2. 8 A
  3. \(5 \sqrt{2} \mathrm{~A}\)
  4. \(10 \sqrt{2} \mathrm{~A}\)

Answer: 3. \(5 \sqrt{2} \mathrm{~A}\)

⇒ \(X_L=\omega L=\frac{100}{8}\)

∴ \(L=\frac{100}{8 \omega}=\frac{1}{8 \pi} \text { and } R=\frac{100}{10}=10 \Omega\)

When R and L are connected in series,

∴ \(Z=\sqrt{\left(\frac{1}{8 \pi} \times 2 \pi \times 40\right)^2+10^2}=10 \sqrt{2}\)

∴ \(I=\frac{V}{Z}=\frac{100}{10 \sqrt{2}}=5 \sqrt{2} \mathrm{~A}\)

Question 28. In an LCR circuit voltages across R, L, and C are 10V, 10V and 20V respectively. The voltage between the two endpoints of the whole combination is

  1. 30 V
  2. 10√3 V
  3. 20 V
  4. 10√2 V

Answer: 4. 10√2 V

Question 29. In an ac circuit alternating voltage E = 200√2 sin100t volt is connected to a capacitor of capacity 1μF. The rms value of the current in the circuit is

  1. 10 mA
  2. 100 mA
  3. 200 mA
  4. 20 mA

Answer: 4. 20 mA

Hint: From the given equation, we may write

E0 = 200√2 V and ω = 100 rad/s

∴ \(X_C=\frac{1}{\omega C}=\frac{1}{100\left(1 \times 10^{-6}\right)}=10^4 \Omega\)

∴ \(I_{\mathrm{rms}}=\frac{E_{\mathrm{rms}}}{X_C}=\frac{\frac{E_0}{\sqrt{2}}}{X_C}=\frac{\frac{200 \sqrt{2}}{\sqrt{2}}}{10^4}=20 \mathrm{~mA}\)

Question 30. In the given network, readings of the ammeter (A) and the voltmeter (V) are respectively

Alternating Current Readings Of Ammeter And The Voltmeter

  1. 800 V, 2 A
  2. 220 V, 2.2 A
  3. 300 V, 2 A
  4. 100 V, 2 A

Answer: 2. 800 V, 2 A

Power in AC Circuits

Question 31. The power factor of an LR circuit carrying an ac of angular frequency ω is

  1. \(\frac{R}{\omega L}\)
  2. \(\frac{\omega L}{R}\)
  3. \(\frac{R}{\sqrt{R^2+\omega^2 L^2}}\)
  4. \(\frac{R}{\sqrt{R^2-\omega^2 L^2}}\)

Answer: 3. \(\frac{R}{\sqrt{R^2+\omega^2 L^2}}\)

Question 32. One of the conditions for getting a wattless current in an ac circuit is

  1. I = 0
  2. C = 0
  3. R = 0
  4. L = C

Answer: 3. R = 0

Question 33. If an emf E = E0 cos ωt is applied to a circuit, the current becomes I = I0 cos ωt What is the power factor of the circuit?

  1. Zero
  2. \(\frac{1}{\sqrt{2}}\)
  3. 1

Answer: 3. 1

Question 34. In an ac circuit, V and I are given by V = 100sin(100 t) V, and I = \(100 \sin \left(100 t+\frac{\pi}{3}\right)\) A respectively. The power dissipated in the circuit is

  1. 104 W
  2. 10 W
  3. 2500 W
  4. 5 W

Answer: 3. 2500 W

Hint: From the given equations for V and we get,

v0  = 100 v, I0 and θ = \(\frac{\pi}{3}\)

Power, P = \(\frac{V_0 I_0}{2} \cos \theta=\frac{100 \times 100}{2} \cos \frac{\pi}{3}=2500 \mathrm{~W}\)

LC Oscillations

Question 35. The inductance and capacitance in a closed circuit are 20 mH and 2μF respectively. The natural frequency will be

  1. 796 Hz
  2. 5000 Hz
  3. 40 Hz
  4. 31400 Hz

Answer: 1. 796 Hz

Question 36. For an LC oscillator which one of the following is not true?

  1. It converts DC to AC current
  2. It can be used as a filter
  3. It can sustain stable oscillations only for frequencies less than the resonance frequency
  4. The resonance frequency is radians per second

Answer: 3. It can sustain stable oscillations only for frequencies less than the resonance frequency

AC Generator and Transformer

Question 37. An ideal transformer is used to decrease an alternating voltage from 880 V to 220 V. If the number of turns of its primary coil is 4000, then what is that in the secondary coil?

  1. 16000
  2. 4000
  3. 2000
  4. 1000

Answer: 4. 1000

Question 38. The core of any transformer is laminated to

  1. Increase the secondary voltage
  2. Reduce the energy loss due to eddy currents
  3. Reduce the energy loss due to hysteresis
  4. Make it robust

Answer: 2. Reduce the energy loss due to eddy currents

Question 39. In a non-ideal transformer, the primary and secondary voltages and currents are V1, I1, and V2, I2 respectively. The efficiency of the transformer is

  1. \(\frac{V_2}{V_1}\)
  2. \(\frac{I_2}{I_1}\)
  3. \(\frac{V_2 I_2}{V_1 I_1}\)
  4. \(\frac{V_1 I_1}{V_2 I_2}\)

Answer: 3. \(\frac{V_2 I_2}{V_1 I_1}\)

Question 40. The turns ratio of an ideal transformer is 1: n. The input-to-output power transfer ratio is

  1. 1:1
  2. l:n
  3. n:1
  4. 1: n2

Answer: 1. 1:1

Question 41. For the circuit,

Alternating Current The Circuit

  1. Mean value = I0
  2. rms value = \(\frac{I_0}{\sqrt{2}}\)
  3. Form factor = 1
  4. Form factor = \(\frac{1}{\sqrt{2}}\)

Answer:

1. Mean value = I0

3. Form factor = 1

Question 42. An emf of V = V0sin ωt is applied on a series LCR circuit. If there is no phase difference between the voltage and current, then

  1. \(I=\frac{V_0}{R} \sin \omega t\)
  2. \(\omega L=\frac{1}{\omega C}\)
  3. Effective power = \(\frac{V_0^2}{R}\)
  4. Ratio of terminal potential difference across L and R = \(\frac{1}{\omega C R}\)

Answer:

1. \(I=\frac{V_0}{R} \sin \omega t\)

2. \(\omega L=\frac{1}{\omega C}\)

4. ratio of terminal potential difference across L and R = \(\frac{1}{\omega C R}\)

Question 43. A coil of resistance 8 Ω and self-inductance 19.1 mH is connected with an ac source of peak voltage 200 V and frequency 50 Hz

  1. Reactance due to induction = 0.955 Ω
  2. The impedance of the circuit = 10 Ω
  3. rms value of current = 10√2 Ω
  4. Power dissipated = 2000 W

Answer:

2. The impedance of the circuit = 10 Ω

3. rms value of current = 10√2 Ω

4. Power dissipated = 2000 W

Question 44. If only a capacitor is connected to an ac circuit

  1. Wattless current is obtained
  2. The current is 90° ahead of the voltage
  3. The current lags the voltage by 90°
  4. Effective power is inversely proportional to cuC

Answer:

1. Wattless current is obtained

2. The current is 90° ahead of voltage

Question 45. The alternating current in an alternating circuit is given by I – I0 sin ωt. In this case

  1. The time taken by the current to reach maximum value I0 from zero is \(\frac{\pi}{2 \omega}\)
  2. The time taken by the current to reach maximum value I0 from zero is \(\frac{\pi}{4 \omega}\)
  3. The time taken by the current to reach rms value from zero is \(\frac{\pi}{4 \omega}\)
  4. The time taken by the current to reach -I0 from zero is \(\frac{\pi}{\omega}\)

Answer:

1. The time taken by the current to reach maximum value I0 from zero is \(\frac{\pi}{2 \omega}\)

3. The time taken by the current to reach rms value from zero is \(\frac{\pi}{4 \omega}\)

Question 46. In a series LCR circuit the resonant frequency f0, alternating voltage V = V0 sin ωt and current I = I0 sin(ωt+ θ). So if frequency

  1. f < f0 then θ > 0
  2. f < f0 then θ < 0
  3. f > f0 then θ > 0
  4. f > f0 then θ <0

Answer:

1. f < f0 then θ > 0

4. f > f0 then θ <0

Question 47. In an ideal transformer, the number of turns in the primary and secondary is N1 and N2, and current and power in the input and output are I1, I2, and P1, P2 respectively. Then

  1. \(I_2=I_1 \frac{N_1}{N_2}\)
  2. \(I_2=I_1 \cdot \frac{N_2}{N_1}\)
  3. P2 = P1
  4. \(P_2=P_1 \cdot \frac{N_1}{N_2}\)

Answer:

1. \(I_2=I_1 \frac{N_1}{N_2}\)

3. P2 = P1

Question 48. L, C, and R represent the inductance, capacitance, and reactance respectively. Which of the following combinations have the same dimensions as that of frequency?

  1. \(\frac{1}{R C}\)
  2. \(\frac{R}{L}\)
  3. \(\frac{1}{\sqrt{L C}}\)
  4. \(\frac{C}{L}\)

Answer:

1. \(\frac{1}{R C}\)

2. \(\frac{R}{L}\)

3. \(\frac{1}{\sqrt{L C}}\)

Question 49. In a resonant LCR circuit,

  1. The power factor is zero
  2. The power factor is one
  3. Dissipated in the resistor is zero
  4. The power dissipated in the capacitor is zero

Answer:

2. The Power factor is one

4. The Power dissipated in the capacitor is zero

Question 50. Two LR circuits are shown. The change in current in this circuit is shown. Choose the correct options.

Alternating Current Two LR Circuits

  1. R1 > R2
  2. R1 = R2
  3. L1 > L2
  4. L1 < L2

Answer:

2. R1 = R2

4. L1 < L2

WBCHSE Class 12 Physics Notes For Dual Nature Of Matter And Radiation

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

Dual Nature Of Matter And Radiation Lenards Experiment

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.

Dual Nature Of Matter And Radiation Demonstrative Experiment

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.

Dual Nature Of Matter And Radiation Keeoing Frequency Of Light Constant

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

Dual Nature Of Matter And Radiation Keeping Intensity Of Light Constant

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 (Emax) = 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,

Emax = eV0 …………………….. (1)

Also, if the maximum initial velocity of the electron is vmax, 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, Emax = 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 Emax 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.

Dual Nature Of Matter And Radiation Stopping Potential Or Maximum Kinetic Energy

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.

Dual Nature Of Matter And Radiation Photo Electric Cell

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 (RL) 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-

Dual Nature Of Matter And Radiation Solar Cell

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+ Emax

Emax= hf – W0 ………………………..(1)

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

½mv²max =  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, Emax = 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 Emax

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 Emax. = 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, Emax  = hf – W0 ,we get,

Emax  = hf – hf0 = h( f- f Emax  = 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

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

Dual Nature Of Matter And Radiation Graphical Representation Of Einsteins Equation

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

From equation (1)

E0max = hf – W0

The graph obtained by plotting Emax 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 Frequency Versus Maximum Kinetic Energy

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 = Emax + 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,

Emax  = 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 × 1034 J s
Solution:

Maximum kinetic energy of photoelectron,

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

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

= 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  × 1027 erg.s )

Solution:

Threshold wavelength

λ0 = 3800 A° =  3800 × 108  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, Emax = 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

Emax = 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  × 1027 erg.s , e = 1.6 × 1019 C)
Solution:

For the first radiation, stopping potential V0 = 0.6V

∴ Maximum kinetic energy of photoelectron,

Emax  = eV0 = 0. 6eV

Wavelength, A = 4940 A° = 4940 × 108 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

Emax= hf – W0

Or,  W0 = hf – Emax

= 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.m2 is incident on a platinum surface. The area of the surface is 1.0 × 104 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 × 1019 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 m2 , A = 1.0 × 104 m2

hf = 10.6 eV = 10.6 × 1.6 × 1019 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, Emax = 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 × 1023J K1, Plank’s constant , h = 6.625 × 1034 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,

Emax = hf-W0

Or hf = Emax  + 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 × 1012 erg

We know, h = 6.60 × 1027 ergs. And

And c = 3 × 1010 cm .s1

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

λ0 = 3.09375 × 105 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 = \(\frac{h c}{\lambda}\) – W

In the second case,

K2 = \(\) – W0 = \(\) – W

It is given that, K2 = 3K1

Or, 2hc/λ = W

= 3(hc/λ – W)

2W = hc/λ

∴ W= 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.

Dual Nature Of Matter And Radiation Double Slit Experiments

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 .s1), v = 107 m. s1. 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 .s1

= 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 .s1m

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

Dual Nature Of Matter And Radiation Davission And Germer Experiment

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, θ2, and the corresponding intensities  I1, I2…….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.

Dual Nature Of Matter And Radiation Scattering Angles

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.

Dual Nature Of Matter And Radiation Incident Beam And Nickel 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

Dual Nature Of Matter And Radiation The Nature Of Matter Wave

  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.

Dual Nature Of Matter And Radiation 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. mol1.K1. 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 me = 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. m²

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 vmax 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 mv²max = 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

Dual Nature Of Matter And Radiation Photoelectric Effect

Answer: 1-B, 2 – C, 3-D, 4-A

Question 2.

Dual Nature Of Matter And Radiation Photoelectric Effect.

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

Dual Nature Of Matter And Radiation A Number Of Stationary Charged Particles

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

Dual Nature Of Matter And Radiation Function And Threshold Frequency

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

Dual Nature Of Matter And Radiation Short Answer Questions

Dual Nature Of Matter And Radiation Short Answer Questions

Question 1. Photoelectric current continues when the anode is even at a slight negative potential concerning the photocathode. Explain.
Answer:

Emitted photoelectrons possess some initial kinetic energy. Hence, a few electrons can reach the anode overcoming the repulsive force of the negative potential. If the value of the negative potential is not too high, the photoelectric current continues to flow

Question 2. When radiation of wavelength 2000A° is incident on a nickel plate, the plate gets positively charged. But when the wavelength of Incident radiation is raised to 3440 A°, the plate remains neutral. Explain
Answer:

The photoelectric effect takes place for incident radiation of wavelength 2000A and due to the emission of photoelectrons, the plate becomes positively charged. No photoelectric effect takes place when radiation of wavelength 3440A is incident on the plate, as the threshold wavelength for photoelectric effect is more than 2000 A but less than 3440A

Question 3. Give examples of the production of

  1. An electron by a photon and
  2. Photon by an electron.

Answer:

  1. When a photon is incident on the metal surface, the electron is emitted in a photoelectric effect.
  2. In the process of production of X-rays, when high-speed electrons in cathode rays are incident on a metal surface, an X-ray photon is emitted.

(Hence photoelectric effect and X-ray production are two opposite site effects)

Question 4. A metal plate that emits photoelectrons under the influence of blue light, may not emit photoelectrons under the influence of red light. Explain.
Answer:

  • For the emission of electrons from a metal surface, the frequency of the incident light should be higher than the threshold frequency ofthe metal.
  • The frequency of blue light is greater than the frequency of red light.
  • If the frequency of blue light is higher than the threshold frequency for the metal, electrons will be emitted.
  • But if the frequency of red light is less than the threshold frequency, red light will not be able to emit electrons from that metal surface

Question 5. Radiation of frequency 1015 Hz is Incident separately on two photosensitive surfaces P and Q. The following observations were made:

  1. Surface P: Photoemission occurs but the photoelectrons have zero kinetic energy
  2. Surface Q: Photoemission occurs and electrons have non-zero kinetic energy.

Question 6. Which of these two has a higher work function? If the frequency of incident light Is reduced, what will happen to photoelectron emission In the two cases?
Answer:

The energy of the incident photon, hf = Ek + W0; here W0 – work function of the emitting surface, and Ek = kinetic energy of the emitted photoelectron

. Given, for the surface P, Ek = 0; but for the surface Q, Ek> 0. For the same incident light, hf= constant. So the work function W0 is higher for the surface P.

If incident light of a lower frequency is taken, he would reduce. Then it would be less than the work function of the surface P; so, this surface would emit no photoelectron. The surface Q may emit photoelectrons, provided hf > W0 for that surface

Question 7. State two important properties of photons which
Answer:

The properties of photons used to write Einstein’s photoelectric equation are

  1. The rest mass of the photon is zero,
  2. The energy ofthe photon is E = hv

Question 8. An electron (charge = e, mass = m ) is accelerated from rest through a potential difference of V volt. What will be the de Broglie wavelength ofthe electron?
Answer:

The kinetic energy gained by the electron

= eV = ½mv²

Or, v = \(\sqrt{\frac{2 e V}{m}}\)

Hence, de Broglie wavelength,

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

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

Question 9. When light incident on a metal ha* energy le** than the work function of the metal, then no electron h emitted from the surface of the metal. Mathematically justify this statement
Answer:

Einstein’s photoelectric equation,

½mv²max = hf – W0 = work function

If hf < W0, the maximum kinetic energy of electrons (½mv²max )is negative and so v²max   is negative. However, the magnitude of the square of a physical quantity cannot be negative. Hence photoelectrons are not emitted from the surface ofthe metal

Question 10. Is matter wave an electromagnetic wave?
Answer:

No, matter wave is not an electromagnetic wave, because the matter waves are not associated with periodic vibrations of the electric and magnetic fields

Question 11. A beam of photons of energy 5.0 eV falls on a free metal surface of work function 3.0 eV. As soon as the photo-electrons are emitted, they are removed. But the emission comes to a stop after a while explain the reasons, i
Answer:

When the photons of energy 5.0 eV fall on a free metal surface of work function 3.0 eV, electrons are emitted from the surface instantaneously and the surface becomes positively charged. After a while, the number of free electrons in the metal surface decreases and more energy is required for the emission of electrons from the metal surface, i.e., the work function of the metal increases.

When the work function becomes greater than 5.0 eV, the photons of energy 5.0 eV are not able to eject electrons from the metal surface. So, the emission of electrons ceases after some time. To overcome this difficulty/the metal surface is connected to a negative pole of an electric source to make it negatively charged.

Question 12. Does the photoelectric emission take place due to the incidence of visible light and ultraviolet rays on the faces of different types of metals in our everyday experience?
Answer:

Yes, the photoelectric emission does take place. But in the absence of any positively charged collector, the emitted photoelectrons accumulate on the metal surface and make a layer of negative charges. Within a very short period, the emission of photoelectrons comes to a stop due to the repulsion of the negatively charged layer. Hence the photoelectric emission stops after a while

Question 13. Why the photoelectric emission can not be performed by using X-rays and gamma rays?
Answer:

When a photon of sufficient energy falls on a metal surface, the photon vanishes by imparting all its energy to an electron and the electron comes out of the metal surface. This phenomenon is known as the photoelectric effect. But in the case of X-rays and gamma rays, the photon energy is too high for the electron of the metal to absorb so that the photon, is not annihilated by imparting its whole energy to an electron. Hence the photoelectric emission does not take place with X-ray or gamma-ray photons.

Question 14. What is the relation of the de Broglie wavelength of a moving particle with temperature?
Answer:

If the thermal energy is not a source of kinetic of a proving,partic)e, then the de Broglie wavelength of that particle is independent of the temperature. Hence the de Broglie wavelength emitted particles in photoelectric effect radioactive radiation etc. have no relation with temperature. T is given by v ∝ \(\nu \propto \sqrt{T}\).

Therefore, the de Broglie wavelength

λ = \(\lambda=\frac{h}{m v} \quad \text { or, } \lambda \propto \frac{1}{\sqrt{T}}\)

Question 15. The threshold frequency for a certain, metal is 3.3 × 10 14 Hz. If the light of frequency 8.2 × 10 -14 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:

We know, eV0 = hf-hf0

Or, V0 = \(\frac{h}{e}\left(f-f_0\right)\)

= \(\frac{6.626 \times 10^{-34}\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}}\)

= 2.03V

Question 16. What is the de Broglie wavelength of a bullet of mass 0.040 kg traveling at the speed of 1.0 km s-1?
Answer:

Here, m = 0.04 J kg, v = 1.0 km. s-1  = 103 m. s-1

∴ λ = \(\frac{h}{m v}=\frac{6.626 \times 10^{-34}}{0.040 \times 10^3}\)

= 1.66 ×10-35m

Question 17. Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic?
Answer:

The work function of a metal is the minimum energy required to knock out an electron from the highest Tilled level of the conduction band. The conduction band comprises different energy levels forming a continuous band of levels. Electrons in different energy levels need different energy for emission. Thus, the electron, once emitted has different kinetic energies depending on the energy supplied to the emitter.

Question 18. Write the expression for the de Broglie wavelength associated with a charged particle having charge q and mass m, when it is accelerated by a potential V.
Answer:

Kinetic energy acquired, E = qV

Momentum, p = \(\sqrt{2 m E}=\sqrt{2 m q V}\)

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

Question 19. If light of wavelength 412.5 nm is incident on each of the metals given in the table, which ones will show photoelectric emission and why?
Answer:

Dual Nature Of Matter And Radiation Metal And Work Function

The energy of incident light,

E = 12400/λ(in Å)

= \(\frac{12400}{4125}\)

= 3 eV

Na and K will show photoelectric emission because their work functions are less than the energy of incident light

Question 20. Does the stopping potential depend on

  1. The intensity and
  2. The frequency of the incident light? Explain

Answer:

  1. Stopping potential does not depend on the intensity of the incident light.
  2. Stopping potential Is directly proportional to the frequency of the incident light

WBCHSE Class 12 Physics Dual Nature Of Matter And Radiation Question And Answers

Dual Nature Of Matter And Radiation Long Questions And Answers

Question 1. The kinetic energy of a proton is the same as that of an α  -particle. What is the ratio of their respective de Broglie wavelengths?
Answer:

Kinetic energy

E = ½mv²=  \(\frac{1}{2} m v^2=\frac{(m v)^2}{2 m}\)

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

Hence, de Broglie wavelength,

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

Since the kinetic energy of a proton equals that of a-parade

⇒ \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{m_2}{m_1}}=\sqrt{\frac{4}{1}}\)

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

Question 2. What should be the percentage increase or decrease of kinetic energy of an electron so that its de Broglie wavelength is halved?
Answer:

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

[p = momentum]

∴ p = \(\sqrt{2 m E}\)

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

∴ \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{E_2}{E_1}}\)

Here \(\lambda_2=\frac{\lambda_1}{2}\) Or, \(\frac{\lambda_1}{\lambda_2}\) = 2

Hence \(\sqrt{\frac{E_2}{E_1}}\) = 2 Or, E2 = 4E1

∴ Increase in kinetic energy

= \(\frac{E_2-E_1}{E_1} \times 100 \%\) × 100%

= \(\frac{4 E_1-E_1}{E_1} \times 100 \%\) × 100%

= 300%

Question 3. The kinetic energy of a free electron is doubled. By how many times, would its de Broglie wavelength increase
Answer:

Kinetic energy, E = \(\frac{p^2}{2 m}\) [p= momentum of electron]

∴ p = \(\sqrt{2 m E}\)

de Broglie wavelength

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

i.e \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{E_2}{E_1}} \text { or, } \lambda_2=\lambda_1 \sqrt{\frac{E_1}{E_2}}\)

=  \(\lambda_1 \sqrt{\frac{1}{2}}=\frac{\lambda_1}{\sqrt{2}}\)

Hence, de Broglie wavelength will be \(\) times Original One.

Question 4. Energy of a photon = E; kinetic energy of a proton is also the same as energy of the photon. The corresponding wavelength of the photon and de Broglie wavelength of the proton are λ1 and λ2, respectively. What is the relation between E and the ratio of λ1 and λ2? 
Answer:

In the case of photon, E = hf = hc/λ1

Or, λ1 = hc/E

In case of proton, E = \(\frac{p^2}{2 m}\) [p = momentum of proton]

Hence, p = \(\sqrt{2 m E}\) and, \(\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

i.e \(\frac{\lambda_2}{\lambda_1}=\frac{h}{\sqrt{2 m E}} \cdot \frac{E}{h c}\)

= \(\frac{1}{c} \sqrt{\frac{E}{2 m}} \quad \text { or, } \frac{\lambda_2}{\lambda_1} \propto \sqrt{E}\)

Question 5. Distinguish between matter wave and light wave
Answer:

Dual Nature Of Matter And Radiation Electromagnetic Waves And Matter Waves

Question 6. The maximum kinetic energies of photoelectrons emitted from a metal surface when illuminated by radiations of wavelengths λ1 and λ2 are K1 and K2, respectively. If λ1 = 3λ2, show that,\(K_1<\frac{K_2}{3}\)
Answer:

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

K1 = \(K_1=\frac{h c}{\lambda_1}-W \text { and } K_2=\frac{h c}{\lambda_2}-W\) – W

Hence,K1  – \(K_1-\frac{K_2}{3}=h c\left(\frac{1}{\lambda_1}-\frac{1}{3 \lambda_2}\right)-W+\frac{W}{3}\)

= hc. 0 –  \(\frac{2 W}{3}\)

Since λ1=  3λ2

= – \(\frac{2 W}{3}\)<0

K1 < \(\frac{K_2}{3}\)

Question 7. A proton and an electron have the same kinetic energy. Which one has a larger de Broglie wavelength?
Answer:

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

Since h and K are constants, λ∝ \(\frac{1}{\sqrt{m}}\)

If I and mp are the masses of an electron and a proton respectively.

Then,  \(\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_p}{m_e}}>1\)

∴ λe > λp

So, the de Broglie wavelength of electrons is greater

Question 8. A proton and an electron have the same de Broglie wave¬ length. Which one has higher kinetic energy?
Answer:

de Broglie wavelength,

λ = \(\frac{h}{\sqrt{2 m K}}\) [ K= kinetic energy]

Or, mK = \(\frac{h^2}{2 \lambda^2}\)

For the same λ, mK = constant

i.e \(K \propto \frac{1}{m} \quad \text { or, } \frac{K_e}{K_p}=\frac{m_p}{m_e}>1\)

∴ Ke>Kp

∴  Electron has higher kinetic energy.

Question 9. A photon and an electron have the same de Broglie wavelength. Which one has greater total energy?
Answer:

The total energy of a photon of wavelength λ is given by

Ep = hf = hc/λ ……………………………………………… (1)

de Broglie wavelength J of an electron (same as bf photon) of mass m moving with velocity v is given by

λ = \(\frac{h}{m v} \text { or, } m=\frac{h}{\lambda v}\)

The total energy of electron of mass m

Ee = mc²  (According to the theory of relativity)

= \(\frac{h c^2}{\lambda v}\) ………………………………… (2)

Dividing equation (2) by equation (1) we hive

⇒ \(\frac{E_e}{E_p}=\frac{\frac{h c^2}{\lambda v}}{\frac{h c}{\lambda}}=\frac{c}{v}\)

Since c>c.

∴ Ee >Ep

So the total energy of the electron is greater than that of a photon

Question 10. Find the exp-session of de Broglie wavelength of a particle moving with a velocity close to the velocity of light.
Answer:

When a particle moves with a velocity v close to the velocity of light c, then the effective mass of the particle

m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

Where mQ = rest mass of the particle

Hence, the de Broglie wavelength associated with the particle is

λ = \(\frac{h}{m v}=\frac{h \sqrt{1-\frac{v^2}{c^2}}}{m_0 v}\)

= \(\frac{h}{m_0 v} \sqrt{1-\frac{v^2}{c^2}}\)

Question 11. If another the speed particle of anis 3v, electron ratio iso f the v and de Brogliethe speed wavelength of the electron to that of the particle Is 104/8.13. Is the particle a proton or an α  particle? Justify your answer
Answer:

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

[ h = Planck’s constant]

∴ For the electron and the other particle

⇒ \(\frac{\lambda_1}{\lambda_2}=\frac{m_2}{m_1} \cdot \frac{v_2}{v_1}\)

Or, = \(\frac{m_2}{m_1}=\frac{\lambda_1}{\lambda_2} \cdot \frac{v_1}{v_2}=\frac{10^4}{8.13} \times \frac{v}{3 v}\)

= \(\frac{410}{1}\)

∴ The other particle is 410 times heavier than an electron. It is neither a proton nor an α-particle

Question 12. Prove that the product of the slope of the V0-f graph and electronic charge gives the value of Planck’s constant.
Answer:

From Einstein’s photoelectric equation, we know that,

½mv² = hf – W0

But, ½mv²max  = eV0

∴ eV0 = hf – W0

Or,  \(\frac{h}{e} f-\frac{W_0}{e}\) ………………………………………….. (1)

Dual Nature Of Matter And Radiation Slope Of V And f Graph

So the plot of stopping potential ( V0) versus frequency (f) will be a straight line

Comparing equation (1) with (the equation of a straight line y = mx + c, we get the slope of V0– f graph m = h/e

Planks constant h = me

Question 13.  The given graphs show the variation of photoelectric current (I) with the applied voltage ( V) when light of different Intensi¬ ties Is Incident on surfaces of different materials Identify the pairs of curves that correspond to different materials but the same intensity of incident radiations.

Dual Nature Of Matter And Radiation Variation Of Photoelectric Current

Answer:

Stopping potential depends on the material of the photocathode, but Is independent of the Intensity of the Incident radiation. So the pairs (1,3) and (2, 4) correspond to different materials but the same Intensity of incident radiation.

Question 14. The potential energy U of a moving particle of mass m varies with x as shown In the figure. The de Broglie wavelengths of the particle In the regions 0 <x< 1 and x > 1 are λ1 and λ2 respectively. If £ the total energy of the particle Is nE, find λ12

Dual Nature Of Matter And Radiation Potential Energy Of U

Answer:

Total energy =nE = potential energy + kinetic energy

In the region 0 < x < 1, potential energy, U1= E

∴ Kinetic energy Is K1 = nE- E = (n- 1 )E

∴ de Broglie wavelength,

λ1=  \(\frac{h}{\sqrt{2 m K_1}}=\frac{h}{\sqrt{2 m(n-1) E}}\) ………………………………………….(1)

In the region x > 1 , potential energy, U2 = 0

Kinetic energy, K2 = nE- 0 = nE

de Broglie wavelength

λ2 = \(\frac{h}{\sqrt{2 m K_2}}=\frac{h}{\sqrt{2 m n E}}\)………………………………..(2)

From equations (1) and (2) we get

⇒ \(\frac{\lambda_1}{\lambda_2}=\frac{h}{\sqrt{2 m(n-1) E}} \times \frac{\sqrt{2 m n E}}{h}\)

= \(\sqrt{\frac{n}{n-1}}\)

Question 15. Monochromatic light of wavelength 632.8 nm Is produced by a helium-neon laser. The power emitted is 9.42 mW.

  1. Find the energy and momentum of each photon in the light beam.
  2. How many photons per second, on average, arrive at a target irradiated by this beam? (Assume the entire beam to be incident on a small area of the target)
  3. What should be the velocity of a hydrogen atom to have the same momentum ofthe photon?

Answer:

1. Wavelength, λ  = 632.8 nm = 632.8 × 10 -9m

∴ The energy of each photon

E = \(\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{632.8 \times 10^{-9}}\)

= 3.14 × 10-19 J and momentum

p = \(\frac{h}{\lambda}=\frac{6.626 \times 10^{-34}}{632.8 \times 10^{-9}}\)

= 1.05 × 10-27 kg. m. s-1

2. Emission power

P = 9.42 mW = 9.42 × 10 -19 W

Number of photons emitted per second

n = \(\frac{P}{E}=\frac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}}\)

3. Mass of hydrogen atom

= Mass of proton = 1.67 × 10 -27 kg

∴ Velocity of the hydrogen atom

= \(\frac{\text { momentum of photon }}{\text { mass of hydrogen }}=\frac{1.05 \times 10^{-27}}{1.67 \times 10^{-27}}\)

= 0.63 m. s-1

Question 16. The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W .m-2. How many photons (nearly) per square meter are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Wavelength, μ  = 550 nm = 550 × 10 -9  m

Energy flux, Φ= 1.388 × 10 3 W .m-2

The energy of each photon,

E = \(\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}}\)

= 3.61× 10 -19J

∴ Number of photons incident on earth per square meter per second,

n = \(\frac{\phi}{E}=\frac{1.388 \times 10^3}{3.61 \times 10^{-19}}\)

= 3.85 × 10 21

Question 17. The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer:

Work function, W0 = 4.2 eV = 6.72 × 10-19 J

Threshold frequency, f0  = \(\frac{W_0}{h}\)

⇒ \(\frac{6.72 \times 10^{-19}}{6.606 \times 10^{-34}}\)

= 1.101 × 1015 Hz

∴ Threshold wavelength,

λ0 = \(=\frac{6.72 \times 10^{-19}}{6.626 \times 10^{-34}}\)

= \(\frac{c}{f_0}=\frac{3 \times 10^8}{1.01 \times 10^{15}}\)

= 297 nm

∴ λ0 < λ

Where λ = incident wavelength

i.e f0 >f

∴ Photoelectric emission is not possible in this case.

Question 18. Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut-off) potential of photoelectrons Is 0.38 y and the work functional the material from which staff cathode Ht made

Frequency of Incidental light,

f = \(\frac{c}{\lambda}=\frac{3 \times 10^8}{488 \times 10^{-9}}\)

= 6.15 × 1014 Hz

∴ Work function,

W0 = hf-eV0

= 6.626 × 10-34 × 6.15 × 104 – 1.6 × 10-19 × 0.38

= 3.467 × 10-19 J = 2.167.eV

Question 19. The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which

  1. An electron and
  2. A neutron would have the same de Broglie wavelength

Answer:

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

Or, E = \(\text { or, } E=\frac{1}{2 m} \cdot \frac{h^2}{\lambda^2}\)

1. The kinetic energy of the electron

Ee= \(\frac{1}{2 \times 9.1 \times 10^{-31}} \times\left(\frac{6.626 \times 10^{-34}}{589 \times 10^{-9}}\right)^2\)

= 6.95 ×10-25 J

2. Kinetic energy of neutron

En=  \(\frac{1}{2 \times 1.67 \times 10^{-27}} \times\left(\frac{6.626 \times 10^{-34}}{589 \times 10^{-9}}\right)^2\)

= 3.79 ×10-28J

Question 20. The wavelength of each photon and electron is 1.00 nm, Find their

  1. Momentum and
  2. Energy.

Answer:

1. Momentum, p = \(\frac{h}{\lambda}\)

Or, p = \(\frac{6.626 \times 10^{-34}}{1 \times 10^{-9}}\)

= 6.626 ×10-25J kg. m. s-1

2. Energy of photon

E = \(\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{10^{-9}}\)

= 1.24 keV

The kinetic energy of the electron

½mv² = \(\frac{p^2}{2 m}\)

= \(\frac{\left(6.626 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31}} \mathrm{~J}\)

= 1.51

Question 21. The average kinetic energy of a neutron at 300 K is \(\frac{3}{2}\) kT.  Find Its do Broglie wavelength.
Answer:

de Broglie wavelength

λ = \(\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{3 m k T}}\)

Since E = \(\frac{3}{2}\) KT

= \(=\frac{6.626 \times 10^{-34}}{\left(3 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300\right)^{1 / 2}}\)

K = 1.38 × 10-23

= 1.456A°

Question 22. What is the dc Broglie wavelength of a nitrogen molecule at 300 K ? Assume the molecule is moving with its rms speed. (Atomic mass of nitrogen = 14.0076 u J)
Answer:

vrms = \(\sqrt{\frac{3 k T}{m}}\)

de Broglie wavelength of nitrogen molecule

λ = \(\frac{h}{m v_{\mathrm{rms}}}=\frac{h}{\sqrt{3 m k T}}\)

= \(\frac{6.626 \times 10^{-34}}{\left(3 \times 28.0152 \times 1.38 \times 10^{-23} \times 300\right)^{1 / 2}}\)

= 0.275 ×10-10 m

Question 23. In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event Is interpreted as the annihilation of an electron-positron pair of total energy 10.2 GeV into two ϒ – rays of equal energy. What is the wavelength associated with each γ-  ray ( lGeV = 109 eV )
Answer:

The energy of two γ-  ray = 10.2 GeV

∴ A Energy of each γ-  ray = 5.1 GeV = 5.1 ×109eV

λ = \(\frac{h c}{E}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5.1 \times 10^9 \times 1.6 \times 10^{-19}}\)

= 2.44 ×10-16 m

Question 24. Estimating the following two numbers should be interesting. The first number will tell you why radio engineers need not worry much about photons! The second number tells you why our eye can never ‘count photons! even In barely detectable light.

  1. The number of photons emitted per second by a medium wave transmitter of 10 kW power, emitting radiowaves of wave-length 500 m
  2. The number of photons entering the pupil of our eye per second corresponds to the minimum intensity of white light that we humans can perceive (-10-10W.m-2). Take the area ofthe pupil to be about 0.4 cm2 and the average frequency of white light to be about 6 ×1014 Hz.

Answer:

1. Number of photons emitted per second,

n = \(n=\frac{\text { power of transmitter }}{\text { energy of each photon }}\)

= \(\frac{P}{h c / \lambda}=\frac{10^4 \times 500}{6.626 \times 10^{-34} \times 3 \times 10^8}\)

= 2.52 ×1031

This number is so very large that radio engineers need not worry much about photons.

2. Minimum intensity, I = 10-10 W.m-2

Area of pupil = 0.4 cm2 = 0.4 ×10-4m2

Average frequency, f  = 6 × 10-4 Hz

E = hf = 6.626 ×10-34 × 6 × 1014

= 3.98 ×10-19J

∴ Number of photons incident on the eye,

n = \(\frac{I A}{E}=\frac{10^{-10} \times 0.4 \times 10^{-4}}{3.98 \times 10^{-19}}\)

= 1.01 × 104

This is quite a small number, but still large enough to be counted

Question 25. Monochromatic radiation of wavelength 640.2 nm? from a neon lamp irradiates photosensitive material made of calcium or tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photocell. Predict the new stopping voltage
Answer:

The work function of the photoelectric cell

W0 = \(\frac{h c}{\lambda}\)eV0

= \(\left(\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{640.2 \times 10^{-9}}-1.6 \times 10^{-19} \times 0.54\right)\)

= 1.40 eV

In the second case, A = 427.2 nm

V0 = \(\frac{1}{e}\left(\frac{h c}{\lambda}-W_0\right)\)

= \(\frac{1}{1.6 \times 10^{-19}}\) \(\times\left(\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{427.2 \times 10^{-9}}-2.24 \times 10^{-19}\right)\)

= 1.51 V

Question 26. An electron microscope uses electrons accelerated by a voltage of 50 kV. determine the die deBroglie wavelength associated with the electrons. If other factors (such as numerical aperture etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope with yellow light?
Answer:

Wavelength of electron

λe = \(\frac{h}{\sqrt{2 m e V}}\)

= \(\frac{6.626 \times 10^{-34}}{\left(2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 50 \times 10^3\right)^{\frac{1}{2}}} \mathrm{~m}\)

= 0.0549 A°

∴ Resolving power of a microscope ∝  1/ λ

The wavelength of yellow light, λy = 5900 A°

∴ \(\quad \frac{\text { resolving power of electron microscope }}{\text { resolving power of optical microscope }}\)

= \(\frac{\lambda_y}{\lambda_e}=\frac{5900}{0.0549}\)

= 1.07 × 105

Question 27. Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure.
Answer:

Here, T = 27°C = 300 K,

P = 1 atm = 1.01 × 105 N . m-2

Mass of He atom,

mHe = \(\frac{4}{\text { Avogadro’s number }}=\frac{4}{6.023 \times 10^{23}}\)

= 6.64 × 105  N.m-2

∴ Typical de Broglie wavelength

λ = \(\frac{h}{\sqrt{3 m_{\mathrm{He}} k T}}\)

= \(\frac{6.626 \times 10^{-34}}{\left(3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300\right)^{1 / 2}} \mathrm{~m}\)

= 0.73A°

Question 28. Einstein’s photoelectric equation shows how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable photograph.
Answer:

Einstein’s equation: eV0 = h(f-f0)

Where, V0 = cut-off voltage due to frequency/ of incident light,

f0 = threshold frequency

Now, V0=  \(\frac{h f}{e}-\frac{h f_0}{e}\)

This is of the form y = mx + c. So the graph of V0 vs f will be a straight line AB of slope – v0,

The point of intersection of AB with the x-axis gives the value of f. For any incident frequency, corresponding to any point C on AB, the ordinate OD gives the value of the cut-off voltage V0.

Dual Nature Of Matter And Radiation Straight Line AB Of Slope

Question 29. An electron microscope uses electrons, accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture, etc. to be the same, how does the resolving power of an electron microscope compare with that of an optical microscope that uses yellow light? C Kinetic energy of the electron

E = 50 keV = 50 × 1.6  ×10-16 J

∴ de Broglie wavelength

λ = \(\frac{h}{\sqrt{2 m E}}\)

= \(\frac{6.626 \times 10^{-34}}{\sqrt{2 \times\left(9.1 \times 10^{-31}\right) \times\left(50 \times 1.6 \times 10^{-16}\right)}} \times 10^{10} \mathrm{~A}\)

= 0.055 A°

λ<< λ’

The wavelength of yellow light,

λ’≈ 5900 A°

λ<<λ’

So, λ<<λ’

The least distance (d) between two points that a microscope can observe separately is proportional to the wavelength of light used, i.e, d ∝ λ

As λ<<λ’ we have d<<d’. So an electron microscope can resolve two points very much closer. Then its resolving power is much higher than that of an optical microscope

Question 30. The graph shows the variation of stopping potential with the frequency of incident radiation for two photosensitive metals A and B. Which one ofthe two has a higher value of work function? Justify your answer.

Dual Nature Of Matter And Radiation Stopping Potential With Frequency

Answer:

For metal A, W’0 = hf’0; For metal B , W0 = hf0

As f’ >f0

∴ W’0 >W0

So, the function of metal A is greater than that of metal B

Question 31. Determine the value of the de Broglie wavelength associated with the electron orbiting in the ground state of the hydrogen atom (given E0 = -(13.6/n²) eV and Bohr radius r0 = 0.53 A ). How will the de Broglie wavelength change when it is in the first excited state?
Answer:

We know, the de Broglie wavelength

λ = \(\frac{h}{m v}=\frac{h r}{m v r}\)

= \(\frac{h r}{\frac{n h}{2 \pi}}=\frac{2 \pi r}{n}\)

For ground State, n= 1

∴  \(\lambda_1=\frac{2 \pi r_0}{1}\)

For ground state, n = 1.

∴ λ1 = 2π × 0.53 ×  10-10

= 3.3 m ×  10-10= 3.3 Å

Again , λn=  \(\frac{2 \pi r_n}{n}=\frac{2 \pi n^2 r_0}{n}\)

= n2πr0

= n λ1

When n = 2 (for first excited state) = λ1 = 2λ1

Hence, de Broglie wavelength doubles in the first excited state

Question 32. Define the term ‘intensity of radiation’ in a photon picture of light Ultraviolet light of wavelength 2270 A° from 100 W mercury source irradiates a photocell made of a given metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photocell respond to a high intensity ( ~ 105 W. m-2 J red light of wavelength 6300 A° produced by a laser?
Answer:

The intensity of radiation can be defined as the amount of light energy incident per square meter per second.

V0 = 1.3 V; λ = 2270 × 10-10 m

We know that hf= hf0 + ±½ mv²max

Also, hf = W0+eV0

W0 = hf- ev0

W0 = hc/λ – eV0

W0 = \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{2270 \times 10^{-10}}-1.3 \times 1.6 \times 10^{-19}\)

= \(6.669 \times 10^{-19} \mathrm{~J}=\frac{6.669 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\)

= 4.168 eV

Again W0 = \(h f_0=\frac{h c}{\lambda_0}\)

Or, λ0 = \(\frac{h c}{W_0}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6.669 \times 10^{-19}}\) m

= 2.98 m = 2980 Å

The photocell would not respond to a high intensity ( ~ 105 W. m-2 ) red light of wavelength 6300 A° produced by a laser because this wavelength is greater than 2980 A°

Question 33. A proton and a particle are accelerated through the same potential difference. Which one of the two has (a) greater de Broglie wavelength, and (b) less kinetic energy? Justify your answer
Answer:

The de Broglie wavelength is related to the accelerating potential as

λ = \(\frac{h}{\sqrt{2 m e V}}\)

λp = \(\frac{h}{\sqrt{2 m_p e_p V}} \text { and, } \lambda_\alpha=\frac{h}{\sqrt{2 m_\alpha e_\alpha V}}\)

λα =  And \(\)

∴ \(\frac{\lambda_p}{\lambda_\alpha}=\frac{h}{\sqrt{2 m_p e_p V}} \times \frac{\sqrt{2 m_\alpha e_\alpha V}}{h}=\frac{\sqrt{m_\alpha e_\alpha}}{\sqrt{m_p e_p}}\)

Now, the mass and charge of a -particle is greater than that of a proton

λp > λα for the same potential difference.

The kinetic energy is related to accelerating potential as

K = eV

∴ Kp = epV and  Kα = eα V

∴ \(\frac{K_p}{K_\alpha}=\frac{e_p}{e_\alpha}\)

Since the charge on the proton is less than that on the alpha particle, Kp < Ka for the same potential difference

Question 34. In the study of a photoelectric effect the graph between the. stopping potential V and frequency u of the incident radiation on two different metals P and Q is shown in:

  1. Which one of the two metals has a higher threshold frequency?
  2. Determine the work function of the metal which has greater value.
  3. Find the maximum kinetic energy of electron emitted by light of frequency 8 × 10-33 Hz Hz for this metal.

Dual Nature Of Matter And Radiation Stopping Potential V And Frequency

Answer:

1. Threshold frequency of P, νp = 3 × 1014 Hz

The threshold frequency of Q, νQ = 6  × 1014 Hz

So, metal Q has a higher threshold frequency.

2. Workfunction of metal Q,

W = hν0 .= 6.6  × 10-34  × 6 × 1014

= 39.6 × 10-20 J

= 2.47 eV

3. The maximum kinetic energy of electron emitted from metal Q by light of frequency 8 × 10-14  Hz is,

= 6.6 × 10-34  ×  (8 × 1014– 6 ×1014 )

=13.2 × 10-20 J

= 0.825 eV

Question 35. The following graph shows the variation of photocurrent for a photosensitive metal:

Dual Nature Of Matter And Radiation Photocurrent For A Photosensitive Metal

  1. Identify the variable X on the horizontal axis.
  2. What does point A on the horizontal axis represent?
  3. Draw this graph for three different values of frequencies of incident radiation, ν1, ν2, and ν3( ν123 )same intensity.
  4. Draw this graph for three different values of intensities of incident radiation I1 , I2 and I3(I1> I2> I3) having

Answer:

1. The variable X on the horizontal axis is the potential difference between the anode and the cathode.

2. The point A on the horizontal axis represents stopping potential

Dual Nature Of Matter And Radiation Different Values Of Frequencies And Intensities Of Incident Radiation

Question 36. The work functions of the following metals are given; Na = 2.75 eV, K = 2.3 eV, Mo = 4.17 eV, and Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 A from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away?
Answer:

Condition for photoelectric emission, hf> W0

Or, \(\frac{h c}{\lambda}>W_0\)

For λ = 3300 Å

⇒ \(\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} \mathrm{~J}\)

= \(\frac{6.03 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\)

Mo and Ni will not cause photoelectric emission. If the laser source is brought nearer placed 50 cm away, then also there will be no photoelectric emission from Mo and Ni, since it depends upon the frequency of the source

Question 37. Radiation of frequency 1015 Hz Is Incident on two photosensitive surfaces P and Q. There is no photoemission from surface PhotoemUslon occurs from surface Q but photoelectrons have zero kinetic energy. Explain these observations and find the value of work function for surface
Answer:

There is no photoemission from surface I* because the frequency of incident radiation is less than its threshold frequency. Since photoemission occurs from surface Q and the photoelectrons have , zero kinetic energy, then the frequency of incident radiation is equal to the threshold frequency for surface Q .

Work function for surface Q, WQ = hf

= 6.6 × 10-34 × 1015

= 6.6 × 10-195

= 4.125 eV

Question 38. Find the energy required by an electron to have its de Broglie wavelength reduced from 10-10 m to 0.5 × 10-10 m.
Answer:

de Broglie wavelength

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

Or,  \(2 m E=\frac{h^2}{\lambda^2} \quad \text { or, } E=\frac{h^2}{2 m} \cdot \frac{1}{\lambda^2}\)

∴ The energy required by the electron

E1 – E2 = \(\frac{h^2}{2 m}\left(\frac{1}{\lambda_2^2}-\frac{1}{\lambda_1^2}\right)\)

= \(\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31}} \cdot\left[\frac{1}{\left(0.5 \times 10^{-10}\right)^2}-\frac{1}{\left(10^{-10}\right)^2}\right]\)

= 7.2 × 10-17J

E1 – E2 = \(\frac{7.2 \times 10^{-17}}{1.6 \times 10^{-19}} \mathrm{eV}\)

= 450 eV

Question 39.

  1. Draw the curve showing the variation of de Broglie wavelength of a particle with its momentum. Find the momentum of a photon of wavelength 0.01A.
  2. Mention the inference of Davisson Garmer’s experiment.

Answer:

1. If momentum =p, then the de Broglie wavelength of a particle

λ = h/p

Or, λp = h = Planck’s constant Hence the λ-p graph will be a rectangular hyperbola,

Given, λ = 0.01 A° = 10-12 m

Dual Nature Of Matter And Radiation de Broglie Wavelength Of A Particle

Momentum ofthe photon

p = \(\frac{E}{c}=\frac{h f}{c}=\frac{h}{\lambda}=\frac{6.63 \times 10^{-34}}{10^{-12}}\)

= 6.63 ×10-22  kg.m.s-1

2. The inference of this experiment is that an electron beam behaves as a matter wave. Hence the experiment proves de Broglie’s hypothesis.

Question 40. When light of wavelengths A and 2\ are incident on a metal surface, the stopping potentials are V0 and VQ/4 respectively. If c be the velocity of light in air, find the threshold frequency of photoelectric emission
Answer:

From Einstein’s photoelectric equation

eV0 =  hf-W0 or , W0 = hf – eV0 = \(\frac{h c}{\lambda}\) – eV0

In first case W0= \(\frac{h c}{\lambda}-e V_0\) …………………….(1)

In the Second case

W0 = \(\frac{h c}{2 \lambda}-\frac{e V_0}{4}\)

Or, W0 = \(\frac{2 h c}{\lambda}-e V_0\) …………………………(2)

From equations (1) and (2) we get,

4 W0 – W0 = 3W0 = \(\frac{h c}{\lambda} \text { or, } w_0=\frac{h c}{3 \lambda}\)

Threshold frequency \(\frac{W_0}{h}=\frac{c}{3 \lambda}\) =

∴ Threshold wavelength = 3λ

WBCHSE Class 12 Physics Notes For Light Interference

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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Cylindrical Wavefront

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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light 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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Huygens Principle

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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Laws Of Reflection

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.

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Reflected Wavefront On The Arc

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 (AM1′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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Laws Of Refraction

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.

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Refractive Indices Of Media

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 FM1′) 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.

Light Wave 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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Plane Wave Front Incident On A Plane Reflector

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 μg = 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) nλg = cg (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, \({ }_a \mu_g=\frac{c_a}{c_g}\)

∴ \(1.51=\frac{3 \times 10^8}{c_g}\)

cg = 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, cw = 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,

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Youngs Double Slit Experiment

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.

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Analytical Treatment Of Interference

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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Interferences

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

Io= 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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Variation Of Intensity

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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light 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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Formation Of Interference Fringes

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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Width Of Interference Fringes

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)

xn 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 Qn 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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Velocity Of Light In Air 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°

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Youngs Double Slit Experiment On Interference

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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Youngs Double Slit Experiment On Width Of Fringe

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

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

= 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 Imax and Imin 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, μgxg= μwxw

Here, xg = 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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Wavefronts

Answer:  1- C, 2-A, 3-B

Question 2.

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Distance Between Slits

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

Question 3.

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Distance Between Slits Decreased

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.

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Glass Slab

Now from the above information match the columns.

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light 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

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Youngs Double Slit Experiment Arrangement With Screen

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

WBCHSE Class 12 Physics Light Interference Short Answer Questions

Light Wave And Interference Of Light Short Question And Answers

Question 1. What change will he observe in the interference pattern produced in Young’s double silt experiment, If a blue colour of the same intensity is used Instead of a yellow colour?

Answer: We know, the fringe width wavelength of incident light. The wavelength of blue light is less than that of yellow light. Hence on using blue light, fringe width will decrease

Question 2. In a double slit experiment using white light a white fringe is noticed on the screen. What will be the change in the position of the white fringe if the screen is shifted 0.05 m from the slits?
Answer: The white fringe is the central bright fringe. Its position does not change even when shifting the screen

Question 3. Why did Huygens Introduce the concept of secondary wavelets?
Answer:

Huygens introduced the concept of secondary wavelets to define the new position of a wavefront

Question 4. What is the distance between the first bright fringe and the first dark fringe in an interference pattern? (use con vocational symbols)
Answer:

The distance between the first bright fringe and the first dark fringe is half of the fringe width. Fringe width = A, where 2d = distance between the sources, A = wavelength of inci¬ dent light and D = distance of screen from the pair of sources.

Requred distance = \(\frac{D}{2 d} \cdot \lambda \times \frac{1}{2}=\frac{D}{4 d} \cdot \lambda\)

Question 5. In all interference patterns, the width of a dark fringe Is y1 and that of a bright fringe is y2. What will be the relation between y1 and y2?
Answer:

In an interference pattern width of a dark fringe is equal to that of a bright fringe. Hence, in this case y1 = y2

Question 6. Monochromatic light was used in Young’s double slit experiment, for producing Interference fringes. If a thin mica sheet Is held on the path of any of the super-posing light beams what change will be noticed in the fringe pattern? ,
Answer:

On placing a thin mica sheet the path of any of the superposing light beams fringe width will remain the same but the entire fringe pattern will shift.

Question 7. What is the justification in applying the principle of linear superposition of wave displacement in explaining the distributions in interference and diffraction patterns?
Answer:

The linear combination of the wave equation is also an equation. This is the very basis of the superposition of waves. a wave

Question 8. When a low-flying aircraft passes overhead, sometimes a slight shaking of the pictures on our TV screen is observed. Why?
Answer:

The metallic body of the low-flying aircraft reflects the TV signal. Interference takes place between these reflected rays and the direct rays transmitted by the TV. Hence slight shaking of the pictures is observed.

Question 9. The width of an interference fringe is 1.5mm. What would be the width of the fringe if the separation between the slits is made twice the original value?

In the first case, y = \(\frac{D}{2 d} \lambda\)

= 1.5 mm

Where 2d = distance between the slits

In the second case, the distance between the slits = 2 (2d)

∴ y’ = \(=\frac{D}{2 \times 2 d}\)λ

= \(\frac{1}{2}\left(\frac{D}{2 d} \lambda\right)\)

= \(\frac{1}{2}\) × 1.5 = 0.75mm

= 0.75 mm

Question 10.

  1. State one defect of Huygens’ wave theory.
  2. Prove the laws of reflection by using Huygens’ principle

1. According to Huygens’ principle each point on the second are wavefront emanating from the source acts as a source and from this source wavelets are supposed to spread out uniformly in all directions including in the direction of the source But in reality, no such back wavefront exists and Huygens principle fails to explain ythe reason behind the absence of back wavefront.

Question 11. In a certain medium, the path difference of 5 × 10-5 cm corresponds to a phase difference n. Estimate the speed of the light waves of frequency 3 × 1014 Hz in the medium
Answer:

Wavelength (λ) of the light waves used = path difference due to a phase difference of 2π = 2x (5 × 10-5)

= 10-4cm = 10-6 m

∴ Speed of the light waves  = νλ = (3 × 1014) × 10-6

= 3 × 108 m.s-1

Question 12.  In Young’s double slit experiment, the fringe width is 2.0 mm. Determine the separation between the 9th bright fringe and the 2nd dark fringe

Fringe width =2.0 mm. So the distance between the second dark fringe and the third bright fringe

⇒ \(\frac{\text { fringe width }}{2}\)

= 1.0 mm

Again, the distance between third bright fringe and 9th bright fringe =(9 – 3) × 2.0 = 12.0 mm

So the distance between the second dark fringe and 9th bright fringe

= 1.0+12.0 = 13.0 mm = 1.3 cm

Question 13. How does fringe the width, distance in Young’s double slit experiment change the when of separation double between slit expert
Answer:

Fringe width, y = \(\frac{\lambda D}{2 d}\) where D is the distance between the slit and the screen.

So, if D is doubled then y is also doubled.

WBCHSE Class 12 Physics Light Interference Questions and Answers

Light Wave And Interference Of Light Questions and Answers

Question 1. A point object t is placed on the axis of a convex lens at a distance greater than the focal length of the lens. What is the shape of the refracted wavefront?
Answer:

The wavefront of light rays, emitted born a point source it spherical convex. After being refracted through the Iens, the wavefront becomes spherical concave. This implies that after refraction through the lens, light rays converge

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Concave And Convex Spherical Wavefront

Question 2. What will be the nature of the wavefront of the direct sunlight and why will it be so?
Answer:

Plane wavefronts. Actually, the spherical wavefronts with the sun at the centre be have effective as plane wavefronts on the earth, at a very large distance from the sun.

Question 3. Interference fringe does not contradict the law of conservation energy – justify
Answer:

In an interference pattern, there is no loss or destruction of light energy in the dark fingers area. The energy just gets shifted from the region of the dark band to the region of the bright band. Total energy remains the same.It can be shown that the average intensity of a set of simultaneous consecutive dark and bright fringes is the same as the intensity of usual illumination in the same region.

Hence, interference fringe does not contradict the law of conservation of energy

Question 4. Two media of refractive indices μ1 and μ21 and μ2 ) are separated by a plane surface. If some part of a plane wavefront is in the first medium and other port of the wavefront Is the second medium, show the shape of the wavefront In this position diagrammatically. stance colours reach here in the same phase and bri
Answer:

AB and BC are two plane wavefronts inclined to each other

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Two Plane Wavefronts

Question 5. Why don’t two light sources of the same type produce an Interference pattern?
Answer:

The primary condition for the formation of interference fringes is that the two sources must be coherent and the phase difference between the intended waves for interference pattern on superposition, must remain constant

Extended light sources are aggregations of point sources where temperature varies from point to point which affects the radiation. Due to the random vibration of particles emitting light, there is a continuous change in phase. Thus no two-point sources from extended sources, maintain a constant phase difference, hence are not coherent. Interference can be produced by waves from coherent sources only. Therefore a general illumination is found all over the screen rather than any interference fringes

Question 6. What are non-localised fringes?
Answer:

Interference fringes can be found on a screen placed anywhere in front of the coherent sources. Thus interference pattern is not localised in a region and hence these are called non-localised fringes

Question 7. Explain the change that occurs In the interference pattern in Young’s double slit experiment,if white light is used instead of monochromatic light
Answer: 

If white light is used instead of monochromatic light In Youngs’ double slit experiment, there will be bright-coloured fringes on either side of the white central fringe. White light of different coloured lights has different wavelengths Each of these waves produces a different characteristic interference pattern.

So the interference pattern looks coloured. But the line at the centre of this coloured. But the line at the centre of this interference coloured pattern remains at equal disfrom each of the coherent sources; hence light of all colour eeach here in the same phase and bright white light is produced.

Question 8. If all experiments related to Young’s double-slit Interference is performed underwater, what change in
Answer:

The wavelength oflight is smaller in water than in air. Since fringe width is proportional to the wavelength of light, fringe width will decrease and lines of interference pattern will be thinner

Question 9. Two waves whose Intensities are in the ratio 9: 1 interfere. Find the ratio of the intensities of bright and dark fringes.
Answer:

Let A1 and A2 be the amplitudes of two waves of intensities fj and J2 respectively

Given that \(\frac{I_1}{I_2}=\frac{9}{1}\)

Again \(\frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}=\frac{9}{1}\)

Or, \(\frac{A_1}{A_2}=\frac{3}{1}\)

If Amax  and Amin are the amplitudes of bright and dark bands, then

⇒ \(\frac{A_{\max }}{A_{\min }}=\frac{A_1+A_2}{A_1-A_2} \text { or, } \frac{A_{\max }}{A_{\min }}=\frac{3+1}{3-1}\)

= \(\frac{4}{2}=\frac{2}{1}\)

⇒ \(\frac{I_{\max }}{I_{\min }}=\frac{A_{\max }^2}{A_{\min }^2}=\frac{4}{1}\)

Question 10. In Young’s double-slit experiment, one of the covered

  1. With translucent e silts is paper and an opaque plate.
  2. What changes will be observed In the Interference pattern in each case

Answer: 

1. An interference pattern will be formed and the fringe width will be unchanged. the difference in intensities of dark and bright fringes trill decreases i.e., bright fringes trill become less bright and dark fringers will become less dark.

2. Interference pattern trill vanishes and a continuous illumination trill be seen on the screen

Question 11. What Is the effect on the interference pattern In Young’s double silt experiment if,

  1. The screen is moved away from the silts
  2. Separation between the slits is increased

Answer:

The width of interference fringes in Young’s double slit experiment, y = \(\frac{D}{2 d} \lambda\)

Here, D = distance of the screen from the slits,

2d = separation between the two slits,

λ = wavelength of light.

If the screen is moved away, D would increase, and so this fringe width would also increase.

If the separation 2d is increased, this fringe width will decrease

Question 12. In Young’s double-slit experiment, if the distance between the two slits is halved and the distance between the screen a and plane of slits is doubled, how will the interference pattern be affected?
Answer: 

Fringe width, y = \(\frac{\lambda D}{2 d}\), 2d = separation between the slits, D = distance between screen and slits, A = wavelength of incident light wave.

Now if the distance between slits is d and the distance between the screen and the slit is changed to 2D then the fringe width will become,

⇒ \(\frac{\lambda D}{2 d}\)

Hence, the fringe width will become 4 times the earlier width.

Question 13. In Young’s double slit experiment, what is the path difference between the two light waves forming the 5th bright band on the screen?
Answer:

Path difference for the n-th bright band,

δ = 2n

\(\frac{\lambda}{2}\) = nλ

Given n = 5

So, δ = 5

Question 14. What is the significance of the optical path?
Answer:

The optical path is the distance travelled by light rays in a vacuum in the same time that it takes to traverse a certain distance (x) in a medium of refractive index. Its value is given by x.

Further, the change in phase for two rays of the same frequency will remain the same if they cover equal optical path lengths.

Question 15. Two light beams of intensities I and 4I, respectively form interference fringes on a screen. For the two beams, the phase difference at point A is \(\) and point B is. Find the difference in result intensities as A And B
Answer:

Resultant intensity at A,

\(I_A=I+4 I+2 \sqrt{I \cdot 4 I} \cos \frac{\pi}{2}\) = 5I

Resultant intensity at B,

⇒ \(I_B=I+4 I+2 \sqrt{I \cdot 4 I} \cos \pi=I\)

Hence, difference in resultant intensity = 5I – I = 4I

Question 16. The ratio of the amplitudes of two waves emitted from a pair of coherent sources is 2: 1. If the two waves superpose, what will be the ratio of the maximum and minimum intensities? What would have been the intensity at different points on the screen if the sources were not coherent?
Answer:

Let the amplitudes of the waves be A and A2 respectively.

By hypothesis

⇒ \(\frac{A_1}{A_2}=\frac{2}{1} \quad \text { or, } \frac{A_1+A_2}{A_1-A_2}=\frac{2+1}{2-1}=\frac{3}{1}\)

Or, \(\frac{A_{\max }}{A_{\min }}=\frac{3}{1}\)

⇒ \(\frac{I_{\max }}{I_{\min }}=\frac{A_{\max }^2}{A_{\min }^2}=\frac{9}{1}\)

Ratio of intensities =9:1.

For incoherent sources, at any point on the screen, the intensity

would be the sum of the intensities of two waves.

If amplitudes are 2A and A respectively then, intensities

7 = 4A² and 2 = A²

Resultant intensity at any point on screen = 4A²+ A² – 5

Question 17. If the ratio of maximum to minimum intensities of the fringes, produced in Young’s double slit experiment is 4:1, what is the ratio of the amplitudes of light In a wave of coherent sources?
Answer: In this case

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

Or, \(\frac{4}{1}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2} \text { or, } \frac{A_1+A_2}{A_1-A_2}=\frac{2}{1}\)

Or, \(\frac{A_1+A_2+A_1-A_2}{A_1+A_2-A_1+A_2}=\frac{2+1}{2-1}\)

Or, \(\frac{A_1}{A_2}=\frac{3}{1}\)

Question 18. Light waves of different intensities from two coherent sources superpose to interfere. If the ratio of the maximum intensity to minimum intensity is 25, find the ratio of the intensities of the sources.
Answer:

Here

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

Or, 25 = \(=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2} \quad \text { or, } \frac{A_1+A_2}{A_1-A_2}\) = 5

Or, \(\frac{A_1+A_2+A_1-A_2}{A_1+A_2-A_1+A_2}=\frac{5+1}{5-1}\)

Or, \(\frac{A_1}{A_2}=\frac{6}{4}=\frac{3}{2}\)

Or, \(\frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}=\frac{9}{4}\)

I1= I2= 9:4

Question 19.  In a laboratory, interference fringes are observed In air medium. Tlie laboratory is now evacuated by removing air. If other conditions remain unaltered, what changes will be observed in the fringe pattern?
Answer:

The refractive index of air is slightly more than 1 . labora¬ tory being evacuated, the refractive index of the medium decreases. Hence wavelength increases. As fringe width is directly propor¬ tional to the wavelength of light used, fringes of marginally increased width will be observed

Question  20. In an Interference pattern by two identical slits, the intensity of the central maximum is I. What will be the intensity at the same spot if one of the slits is close

Let the amplitudes of the waves be a and a

amax = a+a = 2a

So, Imax  = a²max = 4a²

= 4I0

[I0 = Intensity due to each slit]

When one of the slits is closed, the intensity at the same spot is \(I_0=\frac{I_{\max }}{4}=\frac{I}{4}\)

Question 21. In a double-slit experiment, Instead of taking slits of equal width, one slit is made twice as wide as the other. Then how will the maximum and minimum intensities change?
Answer:

In case 0f interference of two waves with the same amplitude (a)

amax = a+a = 2a;

Amin = a-a = 0

Maximum intensity ∝ a² : minimum intensity = 0

In case of interference of two waves with amplitudes a and A (A>a)

amax = a+A: amin = A – a ≠ 0

Minimum intensity  0 and maximum intensity (a+A)> 4 about

Hence, the maximum and minimum intenpattern will increase

Question 22. Monochromatic light of wavelength 589 nm is incident on a water surface from air What are the wavelength, frequency and speed of

  1. Reflected and
  2. Refracted light?

1. For reflected waves, the length and speed remain the same.

∴  \(\frac{c}{\lambda}=\frac{3 \times 10^8}{589 \times 10^{-9}}\) and c= 3 × 108m .s-1

∴  Frequency f = \(\frac{c}{\lambda}=\frac{3 \times 10^8}{589 \times 10^{-9}}\) = \(\)

= 5.09 × 1014 HZ

2. In the case of refraction only the frequency if fixed

= \(\frac{3 \times 10^8}{1.33}\)

= \(=\frac{v}{f}=\frac{2.20}{5.09 \times 10^{14}}\)

= 44 nm

Question 23. What is the shape of the wavefront in each of the following cases?

  1. Light diverges from a point source.
  2. Light emerges out of a convex lens when a point source is placed at its focus.
  3. The portion of the wavefront oflight from a distant star intercepted by the earth. Pt

Answer:

  1. Spherical
  2. Plane
  3. Plane

Question 24. In Young’s double slit experiment using monochromatic light of wavelength A, the intensity oflight at a point on the screen where the path difference is A is k units. What is the intensity oflight at a point where the path difference is
Answer: 

Resultant intensity at a point in Youngs double slit

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

When path difference = λ, phase difference

β = \(\frac{2 \pi}{\lambda} \cdot \lambda=2 \pi\)

∴ I +I’+2\(\sqrt{I \cdot I}\) = 4I

∴ k = 4I

When path difference = \(\frac{\lambda}{3}\) , phase difference

Φ= \(\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{3}=\frac{2 \pi}{3}\)

I” = I+I+2\(\sqrt{I \cdot I}\). cos \(\sqrt{I \cdot I} \cdot \cos \frac{2 \pi}{3}\) = I

I” = I = \(\frac{k}{4}\)

Question 25. In a double slit experiment, the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experiment apparatus is immersed in water? hike refractive Index of water to be \(\frac{4}{3}\)

Angular width, θ = \(\frac{\lambda}{d}\) = 0.2°

When the entire experiment is done underwater,\(=\frac{\lambda}{u}\)and angular width,

= \(\frac{\lambda^{\prime}}{d}=\frac{\lambda}{\mu d}\)

= \(\frac{\theta}{\mu}=\frac{0.2}{4 / 3}\)

= 0.15

Question 26. Use Huygens’ principle to show that a point object placed in front of a plane mirror produces a virtual image at the back of the mirror whose distance is equal to the distance of the object from the mirror.
Answer:

Let A be a point source of light at a distance y from the plane mirror MM’. In the absence of the mirror let the wavefront travel A’ in time t. But due to the presence of the M mirror the reflected wavefront will reach We point A in the same time interval t.

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Distance y From The Plane Mirror MM

Thus AA’ = ct (here, c= velocity of light

i.e., AO+ OA’ = ct

Again, AO+OA = ct

OA = OA’

Question 27. Following is a list of some factors which could possibly influence the speed of wave propagation. Nature of the source, The direction of propagation, Motion of the source and or observer, Wavelength, Intensity of the wave.

On which of these factors, if any, does:

  1. The speed of light in a vacuum,
  2. The speed of light in any medium like glass or water, depends.

Answer:

  1. The speed of light in a vacuum is an absolute (universal) constant and does not depend on any factor.
  2. The speed of light in any other medium depends only on the wavelength of light that the medium

Question 28. In a double slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1 0. What is the spacing between the two slits?
Answer: 

Using, dsinθ = nλ we get

d = \(\frac{n \lambda}{\sin \theta}=\frac{1 \times 600 \times 10^{-9}}{\sin 0.1^{\circ}}\)

= \(3.43 \times 10^{-4}\)m

Question 29.

1. Using Huygens’ principle, draw the diagrams to show the nature of the wavefronts when an incident plane wavefront gets

  1. Reflected from a concave mirror,
  2. Refracted from a convex lens.

2. Draw a diagram showing the propagation of a plane wavefront from a denser to a rarer medium and verify Snell’s law of refraction
Answer:

Class 12 Physics Unit 6 Optics Chapter 6 Light Wave And Interference Of Light Incident Wavefront And Incident Wavefront

Let be the time taken by the wavefront to travel the distance BC, Thus BC = vt

Similarly AE. = v2t

For triangles ABC And AEC,

sin i = \(\frac{B C}{A C}=\frac{v_1 t}{A C}\)

And sin r \(\frac{A E}{A C}=\frac{v_2 t}{A C}\)

Where i and r are the angles of incidence and refraction respectively

\(\frac{\sin i}{\sin r}=\frac{v_1}{v_2}\)…………… (1)

If c is the speed of light vacuum, then

⇒ \(\mu_1=\frac{c}{v_1}\)

Or, \(v_1=\frac{c}{\mu_1}\)

And \(\mu_2=\frac{c}{v_2}\)

Or, \(v_2=\frac{c}{\mu_2}\)

Here, μ12 and μ2 are known “as the refractive indices of medium 1 and medium 2 respectively. From equation

⇒ \(\frac{\sin i}{\sin r}=\frac{c}{\mu_1} \times \frac{\mu_2}{c}=\frac{\mu_2}{\mu_1}\)

Or, in i = sin

This is Snell’s law of refraction.

Question 30. A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interfer¬ ence fringes in Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are sepa¬ rated by 0.28 mm, calculate the least distance from /. the central bright maximum where the bright fringes of the two wavelengths coincide
Answer:

⇒ \(\frac{n_1 \lambda_1 D}{2 d}=\frac{n_2 \lambda_2 D}{2 d}\) [2d= distance between the slits

Or, \(\frac{n_1}{n_2}=\frac{\lambda_2}{\lambda_1}\)

= \(\frac{600}{800}=\frac{3}{4}\)

∴ 3rd order of 800 nm will overlap with 4th order of 600 nm

∴  Distance of the point from central bright maximum

y = \(\frac{n_1 \lambda_1 D}{2 d}=\frac{3 \times 800 \times 10^{-9} \times 1.4}{2.8 \times 10^{-4}}\)

= 12 × 10-3 m

= 12 mm

Question 31. If one of two identical slits producing interference in Young’s experiment is covered with glass, so diet die light intensity passing through it is reduced to 50%, find the ratio ofdie maximum and minimum intensity of the fringe in the interference pattern.

What kind of fringes do you expect to observe if white light is used instead of monochromatic light?
Answer:

Let the amplitudes of the light waves passing through the slits be and a2 and the corresponding intensi¬ ties be  I and I2.

According to the die problem,

⇒ \(I_2=0.5 I_1=\frac{I_1}{2}\)

∴ \(a_2^2=\frac{a_1^2}{2}\)

Or, \(a_2=\frac{a_1}{\sqrt{2}}\)

= \(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\left(\frac{a_1+\frac{a_1}{\sqrt{2}}}{a_1-\frac{a_1}{\sqrt{2}}}\right)^2=\left(\frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}\right)^2\)

= \(\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)^2 \approx 34\).