WBCHSE Class 12 Physics Notes For Dispersion Of Light

WBCHSE Class 12 Physics Notes

Dispersion Of Light

Dispersion Of Light Definition

The phenomenon of splitting up polychromatic light into different colours is called dispersion of light.

Sir Isaac Newton observed for the first time that white rays such as sunlight is a mixture of different colours. He had observed that when a ray of white light is refracted through a prism it is separated into seven colours forming a band.

Experiment of Dispersion of Light:

It can be shown by an experiment that white light consists of seven colours i.e., it is polychromatic. A ray of white light passing through a narrow slit S is incident on a refracting face of a glass prism P. On passing through the prism the constituent colours are separated and emerge as a band of seven colours on the white screen placed on the other side of the prism.

This band of colours on the screen is called spectrum. From the bottom upwards, the colours of the band are violet, indigo, blue, green, yellow, orange, and red. This band of seven colours is commonly called VIBGYOR, the word being formed by the initial letters of the colours in the spectrum arranged in the order in which they have been written

If we observe the spectrum, we see that the deviation of different rays is different. The deviation of the violet ray is maximum and that of the red ray is minimum. For this, it is said that the refractivity of light for different colours is different.

Read and Learn More Class 12 Physics Notes

The magnitude of deviation of yellow rays is almost an average of those for red and violet rays. So yellow light is called the mean ray. Again deviation also depends on the refractive index of the material The more the refractive index of the prism, the more is the devi¬ ation. The refractive index of a medium is minimum for red light and maximum for violet light

The order of refractive indices of a medium for red, orange, yellow, green, blue, indigo and violet light is,

⇒ μ_r<μ_o<μ_y<μ_g<μ_b<μ_i<μ_v

WBCHSE class 12 physics notes Cause of dispersion of light:

1. In a vacuum or in air, all the different coloured rays travel at the same speed. But through any other medium, they travel at different speeds.
2. According to wave theory of light refractive index of a medium is given by the ratio of the speed of light in a vacuum to the speed of light in that medium.
3. The velocity of the red ray, through glass, is greater than violet ray.
4. So a medium has different refractive indices for different colours.
5. The refractive index of a red ray for a medium is less than other rays for the same medium.
6. We know that, if the refractive index of a medium decreases then the deviation of the ray also decreases.
7. That’s why violet ray is deviated most and red ray is least deviated

So, in short, it can be said that dispersion of polychromatic light takes place due to differences in the velocities of the components of the light in a medium. The medium in which the dispersion of light takes place is called a dispersive medium. Dispersion of white light takes place in glass. So glass is a dispersive medium. But vacuum or air is not a dispersive medium.

WBBSE Class 12 Dispersion of Light Notes

Dispersion Of Light Experiments Related To Dispersion Of Light

A prism does not create colour; it just splits white light into different colours:

Through the narrow slit S, white light is incident on the prism P₁ and after dispersion, it creates a VR spectrum on-screen C₁

There is a narrow slit S₁ on C₁. Displacing the screen up and down, a particular light(let yellow light) of the spectrum is sent through the slit and incident on prism P₂. The ray emerging from the second prism is incident on a screen C₂. It is observed that the ray is deviated towards the base, it means that the ray has suffered deviation but the ray is not splitting into different colours i.e., no spectrum is visible. The same incident happens for other colours.

So if colour has been created by a prism then the spectrum would be observed after refraction from the second prism. From this experiment it is further proved that the colours present in white light are pure, hence no dispersion is possible for these colours.

As we get seven different colours by the dispersion of white light, so we can get white light by the recombination of the seven colours because white light is just a mixture of the different constituents.

By two similar prisms:

Two exactly similar prisms of the same material P1 and P2 are placed side by side. Similar refracting surfaces of the two prisms are parallel (A₁ B₁  || A₂B₂ and A₁ C₁ || A₂ C₂) and their bases are on opposite sides. Through the narrow slit S white light is incident on the prism P1 and after dispersion it forms a spectrum/

The different colours of die spectrum after passing through the prism P2 recombine and the emergent beam received on a screen appears white.

By Newton’s colour disc:

Newton’s colour disc is a circular cardboard disc. It is divided usually into four quadrants. Each quadrant is painted with the VIBGYOR colours in the proportion in which they are present in white light. The disc is now rotated quickly it appears white. Visual impression persists for about \(\frac{1}{10}\) th of a second even after the stimulus is removed. The phenomenon is called the persistence of vision. So when the disc is rotated quickly the disc appears white.

Dispersion Of Light Sensation Of Colour

Colour may be defined as the visual sensation produced by a particular wavelength. Radiations of numerous -wavelengths are emitted from white light. Visible radiations are limited within a definite range of wavelength, from 4000A° to 8000A°. This range is called the visible range of spectrum or visible spectrum. The approximate wavelengths of seven elementary colours of the visible spectrum are given below in angstrom(A) unit.[1 A = 10^-10] m

Sensation of colours:

Violet ≈ 4000 A°

Green ≈ 5300 A°

Red ≈ 8000 A°

Indigo ≈ 4800 A°

Yellow ≈ 5800 A°

Blue ≈ 5050 A°

Orange ≈ 6300 A°

The sensitivity of the eye is different for different colours. A nor¬ mal human eye is most sensitive to the yellow region in a visible spectrum. Sensitivity to other colours situated on either side of the yellow region gradually diminishes.

Dispersion of light class 12 notes 

Dispersion Of Light Angular Dispersion And Dispersive Power

Deviation in a thin prism of refracting angle A is given by

δ = (μ- 1)A

When a ray of white light passes through the prism, it splits up into its constituent colours. The deviation of the mean ray,

δ  = (μ- 1)A ……………………………….(1)

Let μ_r and μ_v be the refractive indices of the prism for red and violet rays and δ_r and δ_v be the corresponding deviation.

So,

δ_r = (μ- 1)A ……………………………….(2)

δ_v = (μ- 1)A ……………………………….(3)

δ_r – δ_v = (μ_r-μ_v) A ……………………………….(4)

equation

This difference ( δ_v – δ_r) is called angular dispersion for the
two colours

Now, δ_v – δ_r= (μ_r-μ_v) A

= \(\frac{\mu_v-\mu_r}{\mu-1} \times(\mu-1) A\)

= \(\frac{\mu_v-\mu_t}{\mu-1} \times \delta\) [using equation (1)] ……………………………….(5)

So, \(\frac{\delta_v-\delta_1}{\delta}=\frac{\mu_v-\mu_t}{\mu-1}\) ……………………………….(6)

= ω

ω = [Disperative power of the refracting medium]

Thus,  δ_v – δ_r =  ω δ

Dispersive power Definition:

The dispersive power of a medium is defined as the angular dispersion for violet and red rays per unit deviation of the mean ray in the medium.

Dispersive power depends on the nature of the medium. It is a unitless dimensionless quantity.

From equation(6) we get,

ω = \(\frac{\mu_v-\mu_r}{\mu-1}=\frac{d \mu}{\mu-1}\)

Angular dispersion

δ_v – δ_r =  ω × δ

= Dispersive power × Deviation of middle ray

Desperation power of flint glass is greater than that of crown glass

Desperation power of any material medium is always positive (since μ_v -μ_r and μ>1)

Dispersion of light class 12 notes 

Dispersion Of Light Prism Combination

We have already seen that polychromatic light gets dispersed as well as deviated while refracted through a prism.

Two prisms are used to get deviation without dispersion or dispersion without deviation. The prisms are placed side by side with their refracting angles in opposite directions.

Deviation without dispersion:

The materials and the refractive angles of two prisms are so chosen that the dispersion due to one of the prisms can be cancelled by the other. It means the white ray after passing through the prism gets refracted without any dispersion. But deviation may happen. Such type of combination of prisms is called achromatic combination

Let the refractive indices of the prisms be μ and μ’ and their retractive angles are A and A‘ respectively Vertices of prisms are opposite and placed back to back.

If the  combination is  not to produce a net dispersion, then the angular dispersion of one prism I* equal ami opposite to another prism

δ_v – δ_r =δ’_v – δ’_r……………………….. (1)

Or, \(\left(\mu_v-\mu_r\right) A=\left(\mu_v^{\prime}-\mu_r^{\prime}\right) A^{\prime}\)

Or, \(\frac{\left(\mu_v-\mu_r\right)(\mu-1) A}{(\mu-1)}=\frac{\left(\mu_v^{\prime}-\mu_r^{\prime}\right)\left(\mu^{\prime}-1\right) A^{\prime}}{\left(\mu^{\prime}-1\right)}\)

Or, ωδ= ω’δ’ ……………………….. (2)

This is the condition for no dispersion:

Here  ω and ω’  are the dispersive powers of die prisms and δ and δ’ are die mean deviations Iry die prisms respectively

Total deviation: As the two deviations are each other, the total deviation,

δ – δ’ = (μ – 1)A – (μ’ – 1)A’

Now considering

⇒ \((\mu-1) A-\left(\mu^{\prime}-1\right) \cdot \frac{d \mu}{d \mu^{\prime}} \cdot A\)

= \((\mu-1) A\left[1-\frac{\left(\mu^{\prime}-1\right) d \mu}{(\mu-1) d \mu^{\prime}}\right]\)

= \((\mu-1) A\left[1-\frac{d \mu /(\mu-1)}{d \mu^{\prime} /\left(\mu^{\prime}-1\right)}\right]\)

= \(\delta\left(1-\frac{\omega}{\omega^{\prime}}\right)\) ……………………..(3)

Although the dispersion of white light is acceptable during the experiment with the spectrum of light, it is a real problem in different optical instruments.

The dispersion may cause the images formed by such instruments coloured and blurred which is unwanted. It is therefore necessary to deviate the light without dispersing it and prisms(as well as lenses) that do this are called achromatic(Without colour’).

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Short Notes on Prism and Dispersion

Dispersion without mean deviation:

In this case, the polychromatic ray gets dispersed in different colours. Only the mean ray (i.e., yellow-ray) remains parallel to the incident ray. Let the refractive indices of the prisms for yellow rays be μ and for and their refracting μ’ angles are A and A’ respectively.

The prisms are placed, Here μ’>μ.

Let the deviations of the mean ray for two different prisms used to be δ and δ’ respectively. For no mean deviation by this combination

δ-δ’ = 0

Or, (μ-1)A = ((μ’-1)A’

Or, \(\frac{A}{A^{\prime}}=\frac{\mu^{\prime}-1}{\mu-1}\) …………………………. (4)

This is the condition for no mean deviation. In this case, though no deviation occurs for the yellow ray, all the other rays deviate and hence a spectrum is formed.

Not dispersion:

As angular dispersion due to one prism takes place in the opposite direction of the angular dispersion due to the other prism, so net dispersion:

= (δ_v– δ_r) –  δ’_v– δ’_r = (μ_v-μ_r)A – (μ’_v-μ’_r)A’

= dμ.A-dμ’.A’

Since

= μ_v-μ_r = dμ, μ’_v-μ’_r= d’μ,

= \(\frac{d \mu}{\mu-1} \cdot(\mu-1) A-\frac{d \mu^{\prime}}{\mu^{\prime}-1}\)

= ωδ- ω’δ’

Dispersion of light class 12 notes 

Dispersion Of Light Numerical Examples

Examples of Applications of Light Dispersion

Example 1: Aprism of 6° angle is made of crown glass. The material of the prism for red and blue light are 1.514 and 1.532 respectively. Find the angular dispersion produced by the prism. Also, calculate the dispersive power of the material of the prism
Solution:

Angular dispersion

δ_b – δ_r= (μ_b-μ_r ) × A

= (1.532- 1.514) × 6°

= 0.018 × 6°= 0.108°

Mean refractive index = \(\frac{\mu_b+\mu_r}{2}=\frac{1.532+1.514}{2}\)

= 1.523

∴ Dispersive power of the material of the prism,

= \(\omega\frac{\mu_b-\mu_r}{\mu-1}=\frac{1.532-1.514}{1.523-1}\)

= \(\frac{0.018}{0.523}\)

= 0.034(approx).

Example 2. The- refractive indices of quartz relative to air in the wavelengths 4500 A° and 4600 A° are 1.4725 and 1.4650, respectively. What is the angular dispersion in degree angstrom‾¹ unit? Suppose, the angle of incidence =45°.
Solution:

μ₁ = 1.4725, μ₂ and= 1.4650.

Suppose, the angle of refraction in case of light of wavelength 4500  A°  (= λ₁) = r₁ and angle of refraction in case of light of wavelength 4600 A° (= λ₂) = r₂

Therefore angular dispersion between the given wavelength range,  δ = r₁-r₂

So, angular dispersion for the unit difference of wavelength

⇒ \(\frac{\delta}{\Delta \lambda}=\frac{r_2-r_1}{\lambda_2-\lambda_1}\)

According to the

sin i =  μ₁ sin r₁

Or, sin r₁ = \(\frac{\sin 45^{\circ}}{1.4725}\)

= 0.4802

∴ r₁ = sin^-1(0.4802)

= 28.7°

Similarly in sin r₂ = \(\frac{\sin 45^{\circ}}{1.4650}\) = 0.4827

r₂ = sin^-1 (0.4827)= 28.86°

∴ δ = r₂-r₁ = 28.86°- 28.7° = 0.16°

∴ \(\frac{\delta}{\Delta \lambda}=\frac{0.16}{4600-4500}\)

= 0.0016°  A^-1

Example 3. The refractive Indices of crown glass for red and blue light are 1.517 and 1.523 respectively. Calculate the dispersive power of crown glass with respect to the two colours.
Solution:

Dispersive power of crown glass, ω = \(\frac{\mu_b-\mu_r}{\mu-1}\)

= \(\frac{\mu_b+\mu_r}{2}\)

Here,μ_b = 1.523, μ_r = 1.517

∴ \(\frac{\mu_b+\mu_r}{2}\)

= \(\frac{1.523+1.517}{2}\)

= 1.520

ω =  \(\frac{1.523-1.517}{1.520-1}\)

= \(\frac{0.006}{0.520}\)

= 0.0115

Practice Problems on Dispersion and Refraction

Example 4. The refractive indices of crown and flint glass for red light are 1.515 and 1.644. Again, the refractive indices of crown and flint glass for violet light are 1.532 and 1.685 respectively. For making the lens of spectacles, which glass would be more suitable and why?
Solution:

Dispersive power of crown glass

ω_C = \(\omega_C=\frac{1.532-1.515}{\left(\frac{1.532+1.515}{2}\right)-1}\)

= \(\frac{0.017}{0.523}\)

= 0.0325

Dispersive power of flint glass,

ω_F= \(\frac{1.685-1.644}{\left(\frac{1.685+1.644}{2}\right)-1}\)

= \(\frac{0.041}{0.6645}\)

= 0.0617

For spectacles, the material of the lens should have minimum dispersive power so as to minimise chromatic aberration. Hence crown glass is more suitable than flint glass as the material for spectacles.

Second method:

Angular dispersion for crown glass,

(μ_r – μ_v )A = (1.532 -1.515)A = 0.017A

Angular dispersion for flint glass,

(μ_v – μ_r )A = (1.685- 1.644)A = 0.041A

If It Is considered that the lens of a spectacle is the combination of prisms then A will be the refracting angle of any such prism. Angular dispersion should be minimal for the spectacle lens. As crown glass has less angular dispersion than flint glass, for the preparation of spectacle lens crown glass is more preferable.

Example 5. Write down the expression- for dispersive power.— The itcfractivc index of crown glass for violet and red colour is 1.523 and 1,513 respectively. Calculate the dispersive power of crown glass
Solution:

The refractive index of crown glass for violet light,

Hv = 1.523; refractive index of crown glass for red light

Hr = 1.513

So, average refractive index, μ= \(\frac{1.523+1.513}{2}\)

= 1.518

Dispersive power, ω= \(\frac{\mu_v-\mu_r}{\mu-1}\)

= \(\frac{1.523-1.513}{1.518-1}\)

= 0.0193

Class 12 Physics Dispersion Notes

Dispersion Of Light Impure And Pure Spectrum Definition

Impure spectrum:

An impure spectrum Is a spectrum In which the constituent colours overlap each other and hence cannot he distinctly separated from one another.

Pure Spectrum:

Pure Spectrum Is a spectrum in which all the constituent colours occupy different and distinct positions and do not overlap one another

Impure And Pure Spectrum:

If it was possible to isolate a single ray of white light and pass it through a glass prism it would split up into seven distinct and separate single colours producing what could be called a pure spectrum. In actual practice, it is not possible to isolate a single ray. Even if a narrow pencil is taken, it will contain a large number of rays and if the pencil is divergent, each ray in the pencil

Passing through a prism will produce a spectrum of Its own on the screen and these different spectra will partially overlap. No colour can he separately Identified, As a result, the spectrum becomes Impure, as Illustrated.

The spectra produced by the rays between SA and SB lie In the region if R₁V₂. So the colours overlap each other producing an impure spectrum,

Class 12 physics dispersion notes 

Dispersion Of Light Method of Producing Pure Spectrum

A source of white light Illuminates the narrow vertical slit S.  The slit is placed at the principal. the focus of a convergent lens L₁.

Which renders the emergent rays parallel. The emergent rays are received by the prism P placed In (be the position of minimum deviation for yellow rays. The dispersed rays come out of the prism and proceed In parallel directions.

A second convergent lens .L₂ placed beyond the prism brings the different groups of parallel rays to different foci, violet at V and red at R on-screen M, the foci of other colours lying In between them. Thus a real pure spectrum is formed on the screen M which is placed at the focal plane of the lens L₂

Conditions for the formation of pure spectrum:

For producing pure spectrum the following conditions should be
satisfied

1. The slit should be narrow. Otherwise, many light rays being
   incident on the prism will produce an impure spectrum
2. A convex lens should ho placed between the silt to find the prim so that  the slit In at the principal focus of the lend and a parallel emergent heart) may fall on the prism.
3. The prism should be placed In the position of minimum deviation for the mean rays (l.e. yellow light) so that all other rays emerge with minimum deviation.
4. The refracting edge of the prism should be parallel to the length of the slit.
5. A convex lens with a suitable focal length should be placed beyond the prism so as to converge all the parallel rays of Identical colour emerging from the prism to a single point on the screen.
6. The screen M is to be placed at the focal plane of the lens l
7. All the lenses used must be achromatic

Important Definitions in Dispersion of Light

Dispersion Of Light Rainbow

An Interesting natural phenomenon of the dispersion of light is the rainbow. It is formed by the splitting of sunlight Into different colours . The formation of a rainbow is due to the refraction and reflection of sunlight by water droplets. After rain, huge droplets of water remain suspended in the sky. An observer standing with his back towards the sun can see the rainbow.

Sometimes a little above the primary rainbow, another rainbow called the secondary rainbow of lesser brightness can be seen The rainbow can be seen in the form of a circular arc spread across the sky the primary rainbow has a red colour on the outside of the arc and violet colour at the inside. In the secondary rainbow, the colour spectrum is arranged in reverse order.

Formation of the Primary (First Order) Rainbow:

Let a sun ray AB be incident on a suspended water droplet at the point B. The refracted ray, inside the water droplet is BC. From point C, the droplet to D and then refracted to air through DE to reach the eyes of the observer.

Generally at point C, inside the water droplet, partial reflection of the ray takes place and a part of it gets refracted into the air. So, the brightness of emerging ray DE becomes less than that of the incident ray AB. shows only one reflection of the refracted ray, inside the water droplet at, point C. For different angles of incidence, more than one reflection inside the curved surface of the water droplet is possible.

More the reflection less is the brightness of the rays reaching the eyes

The  deviation of the Lightray:

According to the angle of incidence at B = i

The angle of refraction = r

Deviation of the ray =(i-r)

Angle of Incidence at C = Angle of reflection r

Deviation of the ray =- 180° – 2r

The angle of Incidence at D= r

The angle of refraction = i

The deviation of the ray = i-r

So. the total deviation of the Incident ray with respect to the observer,

δ =(i-r) +(180°-2r)+(i-r)

δ =  2i-4r+180°……………………………. (1)

Condition of deviation angle to be minimum:

Differentiating equation(1) with respect to i,

⇒ \(\frac{d \delta}{d i}=2-4 \frac{d r}{d i}\)

Or, \(\frac{d r}{d i}=\frac{1}{2}\) ……………………………. (2)

For, the deviation angle ti to be minimum

⇒ \(\frac{d \delta}{d i}=0\)

So, \(2-4 \frac{d r}{d i}=0\)

Or,

⇒ \(\frac{d r}{d i}=\frac{1}{2}\)

Again, the refractive index of water,

μ = \(\frac{\sin i}{\sin r}\) ‘

Or, \(\frac{1}{\mu} \sin i\)

In this case, differentiating w.r.t. l, wo get

Or, \(\cos r \frac{d r}{d l}=\frac{1}{\mu} \cos 1\)

= \(\frac{\cos l}{\mu \cos r}\) ………………….. (3)

Comparing equations (2)and (3) , we get

½ = \(\frac{\cos i}{\mu \cos r}\)

Or, 2 cos i = cos r

4 cos²t= μ²cos²r = μ²(1-sin²r)

= μ² – μ² sin²r

=μ² –

4 cos² i+ sin² i = 3 cos² i+ 1

μ² = 4cos² i + sin² i=3cos² i+1

⇒ \(\sqrt{\frac{\mu^2-1}{3}}\)

So, If the value of μ  is known, the value of i can be calculated from equation (4).

Then, putting the values of i and μ In the equation \(\mu=\frac{\sin l}{\sin r}\) = the value of r can ho calculated.

From equation(1), with the help of values, μ and i, the value of the minimum angle of deviation can be obtained.

Now, the refractive Index of the colour red in water is 1.331. Minimum angle of deviation of colour red, due to one-time reflection. Inside the water droplet = 138°. When parallel red rays coming from the sun reach the eyes of the observer with minimum devia¬ tion, the inclination angle of these rays with respect to sun rays, becomes (180°- 138°) = 42°.

An Important point to note Is that only in minimum deviation, many light rays remain In overlapping conditions which increases the brightness considerably.

As the angle of minimum deviation of other colours is not 138°, hence with an inclination angle 42°, the tire red colour appears to be most prominent, whereas other colours remain almost obscure. One can find a rainbow’s bright spectrum of red colour along an arc-subtending angle 42° at the eye,.making eye as the centre.

Similarly, the minimum angle of deviation for a violet ray is 140°. So, the colour seen by an observer with an inclination angle (180°- 140°) or 40° is deep violet.

Clearly, the other colours are seen as a bright spectrum between red and violet. In this way, by one reflection inside the water droplet, a first-order rainbow is formed with red colour at the top (outside) and violet colour at the bottom(inside).

Formation of Second Order Rainbow

Sunrays can also reach the eyes of the observer after having two reflections on the inside surface of the water droplet.

Calculations show that the inclination angle of the red ray at the eye of observation is due to reflection at a minimum deviation angle is 52°. For violet ray, this inclination angle is 55°. Hence the other rain-white bow, which forms at an angular interval of 3° (from 52° to 55°) is called the second-order rainbow

The inclination of the rainbow can easily be shown

The inclination of the rainbow can easily be shown with the help of a figure like. At an inclination angle of 40° to 42°, the rainbow seen in the direction of R₁ V₁ is the first-order rainbow. A little above it, the second-order rainbow is seen at an inclination angle of 52°-55°

Points to note:

1. At each reflection, inside the planes of the water droplet, a portion of the light ray is absorbed or refracted. Hence, its brightness also gets reduced.

For this reason, the second-order rainbow is of much lesser brightness than the first one. For more reflections, the higher order rainbows can be formed, but their brightness would be so low that they are almost invisible.

2. Rainbow is an example of an almost pure spectrum. Due to the dispersion of sunrays, its seven colours can be seen almost distinctly. But some impurity still creeps in. Due to the finite size of the sun, the rays do not remain perfectly parallel.

As a result the inclination angles for minimum deviation also vary. So there is bound to be some overlapping1 bf; colours in the rainbow, and consequently the specfruriT must incur some

3. Two observers, standing near each other, do not see them! same rainbow because the circular arcs drawn, taking their eyes as the centre, are always different. This is why the two of them see two different rainbows at the same time. More finely, it can be said, one single observer sees two different rainbows in his two eyes.

4. Sometimes, an observer, looking straight sees a rainbow in the sky and at the same time a reflection of the rainbowin a pond or in a reflector on the earth’s surface.

It is very clear from the that, this image is not the image of the one seen by the observer. Looking straight, the observer sees the rainbow at P₁ position in the sky, on the other hand, the image seen through the reflector is the image of the rainbow which is at position P₂ in the sky.

A comparatively darker zone lying in between the primary and secondary rainbow is called Alexander’s dark band. In 200AD, the Greek philosopher Alexander explained the phenomenon. The dark band is formed due to differences in the angles of deviation of the primary and secondary rainbows.

The appearance of this dark band to a person standing at a particular place means that no light is being dispersed to his eyes from the region of the dark band. Just like the rainbow, the position of the dark band is not fixed. Light dispersed from this region may reach the eyes of another person standing at a different position. So, he may see a rainbow in the same region and a dark band at some other region

Atomic therapy helps to explain the formation of various spectra. [Discussion of this topic Indetail will be discussed later].

In spectroscopy, the different kinds of spectra obtained  from different sources of light are mainly of two types

1. Emission spectrum and
2.  Absorption spectrum

Emission Spectrum:

Anybody can be excited in such a way that It emits light. The nature of this light depends on the temperature and material of the body. The spectrum produced by Ibis light In the spectro¬ scope Is called the emission spectrum.

A Spectroscope Is an optical Instrument used for creating and analysing a spectrum. An object can In, excited In many ways; for example, by Increasing Its temperature, transforming it to a gaseous state, then by passing electric discharge through It at a low pressure, etc. A Spectrometer Is an optical Instrument with the help of which a spectrum Is created and analysed.

Dispersion Of Light Physics Class 12

Emission spectra may be divided Into three classes according to their formation:

1. Continuous spectrum,
2. Line spectrum and
3. Band spectrum

1. Continuous spectrum:

It is an unbroken band of light in which all the spectral colours from red to violet are present. There is no gap in this band from red to violet. In other words, this spectrum is not fragmented into various parts by black bands or lines.  That is why it is called a continuous spectrum.

Source of Continuous spectrum:

Solids or liquids, when heated to very high temperatures become incandescent and give rise to a continuous spectrum; for example filament of an electric bulb, white hot molten metal etc. When an iron rod is heated to a high temperature it becomes red hot. On further heating, the rod turns white hot.

White hot substances may exist in the solid or in the liquid state. The filament of an electric bulb is white hot, whereas the fila¬ ment of an electric heater is red hot. Incandescent gaseous substances at extremely high pressure also show this type of spectrum, an example being the solar spectrum.

Characteristics of Continuous spectrum:

The intensity of the yellow portion of the continuous spectrum is maximum and intensity diminishes gradually as we move towards the red and the violet extremities of the spectrum.

From a continuous spectrum, we can more or less predict the temperature of the source. However, the chemical composition of the source cannot be known from this spectrum because different materials of varying compositions produce similar continuous spectra

2. Line Spectrum :

If consists usually of a number of bright lines separated by dark spaces. Hence, ItU called line spectrum. Obviously, the light of all wavelength are not present m this spectrum.

Source of Line Spectrum :

Atoms of vap[ours or generous elementary substances, in their incandescent states produce line spectra. A bit of metallic salts such as NaCl when introduced in to a colourless bunsen flame gives rise to this kind of spectrum.

A vacuum tube containing a gas and made luminous by an electric discharge with an induction coil also gives rise to a number of isolated bright lines. Vapour – lamps also give line spectra colour as well the position of these lines differs with different gases.

Characteristics of Line Spectrum :

Line spectrum is the characteristic of an of element Any element can be identified by observing its line spectrum. The colour and position of the lines of the line spectrum of every element are definite and fixed. In the sodium line spectrum, two yellow lines occur very close to each other.

These are called the line of wavelength 5896 A° and the D₂ line of wavelength 5890 A°. It has been proved experi¬ mentally that no other element except sodium, produces a spectrum with such yellow lines. In the hydrogen line spectrum, red, blue and two violet lines are present On analysing and studying the line spectra, therefore, we can identify the substances and also gather various information regarding the atomic structures of elements.

By analyzing the spectra of a star we can know what elements are present there. Often, by observing the brightness of the lines, we can even guess the quantity of the elements.

Real-Life Scenarios in Dispersion Experiments

3. Band Spectrum:

It is a spectrum comprising a number of bands of colours, arranged one after the other with some dark gaps in between. As this spectrum consists of bands of light it is called band spectrum. It is obvious that light of all wavelengths is not present in this spectrum.

Source of Band Spectrum:

The light emitted from the molecule of descent gaseous substance produces a band spectrum that can be obtained by sending electric discharge into a discharge tube containing oxygen or nitrogen gas at low pressure. Chemical compounds such as cyanogen, and nitric oxide also give rise to band spectra.

Characteristics of Band Spectrum:

The bands in this spectrum are sharply defined on one side and get diffused on the other side. The portion of the band which is bright is distinctly demarcated. This is called the band head. However, as we move in the other direction.

On analysing these bands with a high-resolving power instrument, it is observed that each band is a collection of some discrete lines packed very close to each other. Towards the band head, the lines occur very close to each other. But, as we move towards the band tail the space between the lines goes on increasing.

A band spectrum is a light spectrum emitted from the atom of an element or compound. By analysing the band spectrum, considerable information regarding the structure of molecules can be obtained

Absorption Spectrum

When white light passes through a transparent material, light of one or more colours present in white light may be absorbed by that transparent material.

The spectrum produced in the spectrometer by the light of other remaining colours emitted from the transparent material is called the absorption spectrum. Dark bands are seen in the spectrum due to the absorption of some colours(or wavelengths).

The absorption spectrum may be divided into two types according to their formation:

1. The line absorption spectrum and
2. Band absorption spectrum

1. Line absorption spectrum:

It is a spectrum in which a number of dark lines separated by some distance are seen.

Source Line absorption spectrum:

It is the result of selective absorption of some colours (or wavelengths) by. some substances(monoatomic gas or vapour).

A substance which is capable of emitting some wavelengths absorbs the same wavelengths when light from an external source at a higher temperature is made to pass through it.

Sodium flame(about 900°C) gives a line spectrum. When light from a source (electric arc at 3500°C) at a higher temperature “is passed through it, the flame absorbs the same wavelengths, thereby producing dark lines in their place.

Characteristics of Line absorption spectrum:

The dark lines are parallel to each other and coloured region exists between two consecutive lines. Like the line emission spectrum, the positions of the dark lines in line absorption spectrum depend on the nature of the gas.

So by observing the position of dark lines in the line absorption spectrum, absorbent substances can be identified i.e., line absorption spectrum expresses the characteristics of the atoms of the absorbent substance

2. Band absorption spectrum:

It is a spectrum in which a large number of dark lines (due to absorbed wavelengths) group together and are so close to each other that it appears as though a number of dark bands cross the spectrum. The black bands are called absorption bands.

Source of Band absorption spectrum:

This is obtained when light giving a continuous spectrum, coming from a source at a higher temperature passes through polyatomic gases O₂, N₂( CO₂, etc.) at a lower temperature and pressure.

Characteristics of Band absorption spectrum:

In the absorption spectrum a large number of dark lines group together. In this spectrum, dark bands are formed at those places where colour bands are seen in the emission spectrum.

Because the colour bands, emitted (radiated) by the gas molecules in their incandescent state, are absorbed by them from the white light, this spectrum expresses the characteristics of the molecules of the absorbent substance.

Dispersion of light physics class 12 

Dispersion Of Light Solar Spectrum

The spectrum formed in the spectrometer by the light coming from the sun is called the solar spectrum In the solar spectrum generally seven colours from red to violet are found in a continuous fashion. So solar spectrum is a continuous emission spectrum.

1. Fraunhofer lines:

If we carefully study the continuous spectrum of sunlight we will notice that a large number of dark lines cross the whole length of the spectrum. So, the solar spectrum is actually a line absorption spectrum. The existence of these dark lines was first observed by Wollaston in 1802.

In 1814 German scientist Fraunhofer made a systematic study of these lines. These lines are known as Fraunhofer lines. He designated the major Fraunhofer lines by several letters from A to 1C. The lines A, B, and C are in the red, D in the yellow, E and F in the green, G in the indigo, and H and K in the violet part of the spectrum. The positions of these dark lines are fixed and definite

Origin of Fraunhofer lines:

The explanation of the dark lines was first given by Kiivhhoffwho. from several experiments, came to the conclusion that the vapour of an element absorbs those wavelengths at a particular temperature which it would emit if it was incandescent at a drat temperature.

From this theory, the existence of dark lines is easily explained. The layer which we see when we look at the sun is called the photosphere. This is generally considered the surface of the tired sun. The average temperature of the photosphere is 5700 K. The substances are in a gaseous state in this layer. The temperature of the core is 15 × 10s K. The radiation from the core passes through the photosphere which is at a comparatively low temperature. So different wavelengths of light are absorbed here.

According to Kirehhbffs law. White light emitted by the sun is robbed while passing through the enveloping lavor. of diose waves which correspond to the waves that die element would emit if they wore incandescent. As a consequence of the absorption of these waves, dark lines are observed.

Conceptual Questions on Spectrum Formation

 Significance of Fraunhofer lines:

From the study of Fraunhofer lines, it has been found that many elements are present on eartii exist in the photosphere. Similarities have been found between Fraunhofer lines and some of our known elements spectra. So it can be inferred that these elements are present in sun’s photosphere. By analysing the solar spectrum the presence of about 70 elements (H₂ , Fe, Ca etc.) in the sim’s atmosphere has been confirmed. Before the existence of helium gas on the earth was discovered, its presence was predicted from the line absorption spectra.

Telluric lines: It is interesting to note that not all the dark lines across the continuous solar spectrum have been formed due to absorption by various vapours present in the enveloping layer of the sun. When sunlight passes through the atmo¬ sphere of the earth, oxygen, and water vapour eta the atmosphere absorb light of different wavelengths. This causes a few more dark lines in the solar spectrum. These dark lines are called Telluric lines. Telluric lines are not visible if the solar spectrum is observed from a place at a height of 3 km or more from the surface of die the earth, from an artificial satellite or from the moon

Dispersion Of Light Different Parts Of Electromagnetic Spectrum

It has been found on analysing the solar spectrum and the spectra of different objects that there are wavelengths of light which are greater or less than the wavelengths of visible light The properties of these radiations are similar to those of visible light

These are called electromagnetic waves. They move with the velocity of light. They also exhibit properties like reflection, refraction, interference, diffraction, polarisation etc. Different parts of an electromagnetic spectrum are discussed elaborately in the chapter‘Electromagnetic wave!

Dispersion Of Light Application Of Spectral Analysis

We can get lots of information about an element by spectral analysis. Spectral analysis is used in the following areas:

1. Identification of an element:

We can detect an element by its own characteristic spectrum even if it is present in very small quantities. Each elementary substance either in gaseous or in a vapour state under suitable stimulus produces its own peculiar spectrum. As for example, lithium gives a red line, sodium gives only two yellow lines.

2. Determination of temperature of a substance:

The nature of the continuous spectrum of a substance depends on its temperature. So by the analysis, of the continuous spectrum, we can get approximately the temperature of a substance. By studying the spectrum of light of distant star we can get an idea of its temperature.In this context, the Saha equation on thermal ionisation of elements deduced by Dr. Megnad Saha is immensely useful.

3. Determination of the chemical composition of a substance:

The atoms of gaseous elements give rise to line spectra. This line spectrum indicates the nature of an element, i.e., it can be said that line spectra are different for atoms of different elements.

So by studying the spectrum obtained from a substance, we can know what elements are present in it, the atomic and molecular structure of matter, the temperature of matter, elements present in the sun and other stars and planets and their quantities, the presence of any unknown element can be found with the help of spectral analysis.

Dispersion of light physics class 12

Dispersion Of Light Colour Of Different Bodies

Coloured objects whether opaque or transparent do not really possess any colour of their own.  The colour emitted by them depends on

• The colour of the Incident light and
• The proportion of light absorbed by them.

1. Colour of opaque body:

An opaque body appears in that colour, which is reflected by the body. A red rose appears red in white light since it absorbs all colours of white fight except red which Is reflected to the observer’s eye.

The red rose will appear bright red in red light but black in all other Jights-blue, green, yellow, and so on, because it absorbs all the colours of light except red,

It should be noted that black or white is not any special colour. A body appears black if no light is reflected from it and a body appears white in white light if it reflects all the components of white light.

The black cloth of an umbrella looks black because when white light is incident on it, all the components of it are absorbed by it i.e., no colour is reflected from the cloth of the umbrella. Again a white cloth looks white since it reflects all components of white light and absorbs none.

It may be noted as a warning that costly coloured clothes should not be purchased at night. To identify the actual colour of clothes it is to be viewed in sunlight. ‘Ifre artificial light used in shops is not perfectly white. One of the other of the components of white light remains absent.

For example, in the light of a gas lamp blue colour is generally absent. So the dress which looks blue in sunlight looks black in the light of a gas lamp.

2. Colour of a transparent body:

When white light falls on a transparent body, it absorbs certain components and transmits the remaining portions which account for its colour. Thus, a transparent plate of red glass appears red, since it. absorbs almost all the colours of white light except red which is transmitted by it and this transmitted colour falling on the eye gives the impression of the colour of the glass plate. Hence, objects like these emit that colour of white fight which they do not absorb.

If white is allowed to fall on a red glass a red glass plate, and if a blue
plate held In the path of red Jighr, it follows as no red light
can through the blue glass, the two plate? together cut off all the light. So the combination of the two glass plates looks dark back”

The colour of some transparent objects is not pure. For instance, a piece of yellow transparent glass transmits not just yellow but red and green colour light to pass through it as well. Consequently, a red or green object does not appear black when it Is seen through a yellow coloured glass,

Again, when we grounds a piece of coloured transparent body, its colour fades. In other words, In Its powdered/ground state, It appears almost white, because, the incident light Is repeatedly reflected by the different layers of minutely ground particles.

But light can be absorbed only if it penetrates the object to some extent In a ground or powdered state, the thickness of the particles Is less so light enters from one side of the particle and exits through the other. Obviously, the more finely you ground the coloured transparent body the more white It will appear in its powdered form due to diffused reflection.

From our everyday experience, we know that ordinary deep water appears greenish but it may appear dark when the depth is comparatively large. A very thin plate of metal appears to be of different colours in reflected and transmitted light This is because of selective reflection. For e.g., an extremely thin gold plate reflects red, orange or yellow colour.

So, the fight that emerges from this plate contains a greater number of green, blue and violet light rays. Suppose, a thin gold plate is illuminated by a white fight light reflected from the plate reaches the eye it will appear orangish-yellow in colour. Again, if a fight ray that emerges from the file plate reaches the eye, then it will appear greenish-blue in colour

Primary and Complementary Colours

We know that white light is made of the seven colours of the spectrum in the right proportion. These seven colours are called pure colours. But in fact, other colours can be obtained by combining three special colours from among the seven colours of the spectrum. But these three colours cannot be prepared from any other colours.

These three colours are red, green and blue. So, these three colours are called primary colours, e.g. by mixing red and green we can obtain yellow colour. Magenta is obtained by mixing red and blue. It is to be noted that the colour obtained by mixing primary colours is not a pure colour. For example, the yellow obtained by mixing red and green is not a pure colour, because if this yellow colour is incident on a prism, red and green colour will be obtained, due to dispersion.

Physicists Maxwell, Helmholtz, Konig, et all performed various experiments to prove that more primary colours a new colour could be a white red absorbed light in nil and air molecules d light propagates In obtained. They set up and then allowed three beams of light rays with the prizes to be incident on it i.e., each beam partially overlapped the other two on incidence

The colour obtained by mixing the primary colours is: 

Red + green = yellow

Red + blue = magenta

Blue + green = peacock blue

Red + peacock blue = red + blue + green = white

Green + magenta = green + red + blue = white

Blue + yellow = blue + red + green = white

Again for the preparation of white colour, all three primary colours are not always required. By mixing any two colours white colour can be prepared. Any two spectral colours which on mixing together give the sensation of white are known as complementary colours. For example, white colour is obtained by mixing yellow and blue or green and magenta. So yellow-blue and green-magenta are complementary colours.

After washing of white clothes often blue dye is used because white clothes appear a bit yellow after many wash. Since blue and yellow are complementary colours, the clothes turn white if it is dyed in blue.

WBCHSE physics class 12 dispersion notes

Dispersion Of Light Scattering Of Light

When light wave is incident on a particle of small size in comparison to the wavelength of the incident light, the particle absorbs energy from the incident light without changing its state and radiates this energy as the wave forms in different directions. Actually, the particle here acts as a secondary source.

When Sunlight passes through the earth’s atmosphere. light isa absorbed by fine dust particles and air molecules in the atmosphere which radiates light in different directions. This phenomenon is called the scattering of light. It is illustracted A dust particle or air molecule when struck by a light wave is set into vibration.

Immediately afterwards it radiates the absorbed light in all directions Since the number of dust particles and air molecules in the atmosphere is very large , the scattered light propagates in alll directions.

Rayleigh’s taw of scattering:

The intensity of scattered light (1) varies inversely as the fourth power of Its wavelength, i.e.,\(I \propto \frac{1}{\lambda^4}\)

According to this law, the light of shorter wavelengths violet, blue, etc. is scattered much more than the light of longer lengths red, orange etc.

According to Rayleigh scattering, the intensity(I) of of scattered light is proportional to the sixth power of diameter(d) of suspended dust particles in the atmosphere. wave

The Blue of the sky

The phenomenon is due to the scattering of sunlight by the sus¬ pended dust particles, air molecules and gas molecules in the atmosphere. When rays of the sun are scattered, the intensity of the blue and violet colour is high, following Rayleigh’s inverse fourth power law. Again our eyes are more sensitive to blue light compared to violet light. So the sky appears blue.

in the absence of the atmosphere, light rays would not be tired. so the sky would appear black even during the day. moon has no atmosphere. Consequently, the moon’s sky appears black

Redness Of the Rising And The Setting Sun

During sunrise and sunset, the sun appears red. This is also due to the scattering of light. When the sun is directly overhead, sun Rays have to transverse less distance through the earth’s atmosphere than when the sun is situated near the horizon

.As the wavelength of other colours are less than that of red, they suffer more scattering and spread over a larger expanse before reaching the observed red colour suffers least scattering and so this colour reaches to us more in comparison to other colours So the sun appears red at sunrise and sunset The is also why red signals without being scattered much.

Dispersion Of Light Raman Effect

In 1928 Indian scientist Sir C V Raman observed that when a beam of monochromatic light was passed through organic liquids such as benzene and toluene, the scattered light contained other frequencies in addition to that of the incident light This phenomenon is known as Raman effect. In addition to liquids, gases and transparent solids exhibit this effect

Raman observed die scattered light by spectrometer placed at right angle to the incident light and found that in addition to the unmodified original spectral line(main line), a number of new lines were present on both sides of the main line. These lines are known as Raman lines and the spectrum produced is called Raman spectrum.

1. Features of Raman Spectra:

1.  In the Raman spectrum, some weak spectral lines of lower and higher frequencies are observed on both sides of the main line.

Spectral lines of lower frequency (or longer wavelength) higher frequency ( shorter. wavelength}are called Stokes line Stokes of lines anti-stokes lines are observed in fluorescence spectrum lines around.  anti-Stokes lines are observed only in the Raman spectrum.

The frequency of the incident radiation and scattered nidi ation is the same in the region of the original spectral lines of the Raman spectrum. This scattering of light by molecules. without change in frequency is known as Rayleigh scatter¬ ing and the spectral line is called Rayleigh line. Anti-Stokes line, Rayleigh line and Stokes line altogether are called Raman lines

2. Stokes lines and anti-Stokes lines are situated on both sides of the Rayleigh line at equal frequency intervals. The frequencies of the lines are directly related to that of the incident light.

3.  The frequency difference (Δf) of the Stokes and the anti-Stokes lines from the Rayleigh line does not depend on the frequency of the main line. But it depends on the nature of the scatterer. The frequency difference (Δf) is called the Raman shift. If f₀ be the frequency of the Rayleigh time, the frequencies of the Stokes line and that of the anti-Stokes line are given by.

f’ = f₀ – Δf (Stokes line)

f’ = f₀ + Δf (Anti – Stokes line)

4. The intensity of a Raman line when expressed as a fraction of the Rayleigh line is usually a few hundredths In liquids and a few thousandths in gases. No accurate data are available as regards the absolute intensities of Raman lines in liquids and gases. The Stokes lines are always more intense than the corresponding anti-Stokes lines. GD Raman lines are generally polarised

Explanation of Raman Effect on the Basis of Quantum Theory:

According to quantum theory, any radiation Is considered as the flow of photon particles, each of energy hf. When such a light photon falls on the molecules ofa solid, liquid or gas, the photon undergoes due types of collisions with the molecule.

The molecule may merely deviate the photon without absorbing its energy which will result in the appearance of an unmodified line in the scattered beam.

The Molecule may absorb part of the energy of the incident photon, giving rise to the modified Stokes Line whose frequencies will evidently be less than that of the incident radiation.

It may also happen that, the molecule itself being in an excited state, imparts some of its intrinsic energy to the incident photon and this will produce the anti-Stokes line of frequency greater than that of the incident radiation.

In the first case, an elastic collision takes place between the photon and the molecule. So the frequency of the scattered photon becomes equal to its initial frequency. It proves the existence of the Rayleigh line in the Raman spectra.

In the second and third cases be study of certain aspects of nuclear physics, such as the spin stainelastic collision takes place between them. As a result, two cases are tistics as well as the isotopic constitution of the nucleus. may arise. The frequency of the scattered photon may decrease (origin of Stokes line) or increase(origin of anti-Stokes line)

Let intrinsic energy of the molecule before collision = E_p intrinsic

The energy of it after collision = E_q

Mass of molecule = m

The velocity of the molecule before collision = v

Velocity after collision = v’

The energy of the incident photon = hf

Energy of the incident photon = hf’

From the principle of conservation of energy, we have

⇒ \(E_p+\frac{1}{2} m v^2+h f=E_q+\frac{1}{2} m v^{\prime 2}+h f^{\prime}\)…………………(1)

As the collision does not appreciably change the temperature of the surroundings, we may assume that the kinetic energy of the molecule remains practically unaltered in the process

Hence from equation( 1) we have,

⇒ \(E_n+h f=E_a+h f^{\prime} \quad \text { or, } h\left(f^{\prime}-f\right)=E_p-E_q\)

Or, \(f^{\prime}=f+\frac{E_p-E_q}{h}\)

Now, remains

If E_p – E_q . then f’ = f i.e., frequency light of So its scattered light remains identical with the incident light. So it denotes Rayleighline.

2. If Ep<Eq, then <f i.e., the frequency of the scattered light decreases. So this represents Stokes’s line.

3. If Ep>Eq, then f >f i.e., the frequency of the scattered light increases. So this represents the anti-Stokes line.

2. Applications of Raman Effect:

This effect has many applications.

1. It has been put to use in the study of the structure of molecules and crystals.
2. This effect has also been applied in the study of certain aspects of nuclear physics,such as the spin-statistics as well as the isotopic constitution of the nucleus

WBCHSE physics class 12 dispersion notes

Dispersion Of Light Conclusion

1.  White light Is composed of seven colours. These seven colours are:

1. Violet
2. Indigo.
3. Blue.
4. Green.
5. yellow,
6. Orange and
7. Red.

2. Splitting up of polychromatic (or mixed) light into its funda¬ mental colours is known as dispersion of light.

3. Dispersion of light only occurs if the speed of the different colours of light are different in a medium and that medium is called dispersive medium. Light rays of all colours travel with equal velocity In vacuum and air.

4. Prism cannot produce colour; it can only separate different colours present in white light. The visible range of the spectrum is 4000A° to 8000A°.

5. The difference of the deviations suffered by lights of two different colours due to refraction is called angular disper¬ sion with respect to those two colours.

6. The ratio of the differences of the deviations between violet and red coloured lights deviation of light mean colour Is called the dispersive power of the refractive medium.

7. The spectrum In which different coloured lights are distinctly visible as they do not overlap with each other is called spectrum.

8. Due to the scattering of sunlight sky looks blue and the sun appears rod during sunrise or sun sot.

9. If sun ray Incidents on water droplets are suspended In air, then due to dispersion, the ray splits Into several colours. Due to reflection and refraction Inside the water droplet, the dispersed rays of different colours emerge from It with u deviation. These rays form a rainbow.

10. When a beam of visible monochromatic light, passes through a transparent medium(solid/liquid/gaseous), then the scattered radiation, which takes place along the direction of the normal of the incident light, contains other radiations of lower and higher wavelengths, In addition to the radiation of original wavelength. This phenomenon Is called the Raman Effect.

11. If the refracting angle of a thin prism be A, then for refrac¬ tion of light through this prism,

The difference in the deviations of the violet and red coloured ray

δ_v– δ_r = (μ_v– μ_r)A

2. Dispersive power of the prism,

ω = \(\frac{\delta_v-\delta_r}{\delta}=\frac{\mu_v-\mu_r}{\mu^t-1}\)

[ μ_v= Refractive index for violet ray,  μ_r = Refractive  index for red ray, μ = Refractive index for yellow ray]

Dispersion Of Light Assertion-Reason Type

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

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

Question 1.

Statement 1: The blue colour of the sky Is on account of a Mattering of sun light.

Statement 2:  lit The Intensity of scattered light varies inversely as the fourth power of wavelength of light,

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

Question 2.

Statement 1: The visible spectrum consists of all colours from violet to red.

Statement 2: Visible spectrum is nothing but wave length splitting

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

Question 3.

Statement 1: Yellow light Is used as a danger signal.

Statement 2: It Is because eye Is most sensitive to yellow colour.

Answer: 4. Statement 1 is False, and statement 2 is true.

Question 4.

Statement 1: A rainbow Is formed In the sky on a rainy day,

Statement 2:  Halnbowls formed due to the dispersion of sun rays when they fall on the suspended tiny droplets of water,

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

Question 5.

Statement 1:  A prism Is not the source of colours of light,

Statement 2: A prism has different refractive Indices for different colours of light.

Answer: 4. Statement 1 is False, and statement 2 is true.

Dispersion Of Light Match The Columns

Question 1. The different types of spectrum and their sources are given in column a1 and column 2 respectively, matching the column

Answer:  1- B, 2-D, 3- A, 4- C

Question 2. 

Answer:  1 – C, 2-A, 3- D, 4- B