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Refraction And Dispersion Of Light Through A Prism - Practice Questions & MCQ

Refraction And Dispersion Of Light Through A Prism - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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Quick Facts

Refraction Through A Prism 1, Dispersion Of Light 1 is considered one of the most asked concept.
35 Questions around this concept.

Solve by difficulty

A ray of light is incident at an angle of $60^{\circ}$ on one face of a prism of angle $30^{\circ}$ . The emergent ray of light makes an angle of $30^{\circ}$ with the incident ray. The angle made by the emergent ray with the second face of the prism will be:

A

$0^{\circ}$

Correct Answer

Wrong Answer

B

$90^{\circ}$

Correct Answer

Wrong Answer

C

$45^{\circ}$

Correct Answer

Wrong Answer

D

$30^{\circ}$

Correct Answer

Wrong Answer

When a glass prism of refracting angle $\mathrm{60^{\circ}}$ is immersed in a liquid its angle of minimum deviation is $\mathrm{30^{\circ}}$ . The critical angle of glass with respect to the liquid medium is:

A

$\mathrm{42^{\circ}}$

Correct Answer

Wrong Answer

B

$\mathrm{45^{\circ}}$

Correct Answer

Wrong Answer

C

$\mathrm{50^{\circ}}$

Correct Answer

Wrong Answer

D

$\mathrm{52^{\circ}}$

Correct Answer

Wrong Answer

. The refracting angle of a prism is A and the refractive index of the material of the prism is $\mathrm{cot(A/2)}$ . The angle of minimum deviation of the prism is:

A

$\mathrm{\pi +2A}$

Correct Answer

Wrong Answer

B

$\mathrm{\pi -2A}$

Correct Answer

Wrong Answer

C

$\mathrm{\frac{\pi}{2}+A}$

Correct Answer

Wrong Answer

D

$\frac{\pi}{2}\mathrm{-A}$

Correct Answer

Wrong Answer

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In an experiment for determination of the refractive index of glass of a prism by $i\, -\, \delta$ , plot, it was found that a ray incident at angle 35⁰, suffers a deviation of 40⁰ and that it emerges at angle 79⁰⋅ Ιn that case which of the following is closest to the maximum possible value of the refractive index?

A

1.5

Correct Answer

Wrong Answer

B

1.6

Correct Answer

Wrong Answer

C

1.7

Correct Answer

Wrong Answer

D

1.8

Correct Answer

Wrong Answer

A ray of light is incident normally on one of the faces of a prism of apex angle $30^{\circ}$ and refractive index $\sqrt{2}$ . The angle of deviation of the ray is:

A

$0^{\circ}$

Correct Answer

Wrong Answer

B

$12.5^{\circ}$

Correct Answer

Wrong Answer

C

$\mathrm{15^{\circ}}$

Correct Answer

Wrong Answer

D

$22.5^{\circ}$

Correct Answer

Wrong Answer

A ray of light is incident normally on one face of a right angled isosceles prism. It then emerges parallel to this face. The refractive index of the material of the prism is:

A

$1.33$

Correct Answer

Wrong Answer

B

$1.414$

Correct Answer

Wrong Answer

C

$1.5$

Correct Answer

Wrong Answer

D

$1.732$

Correct Answer

Wrong Answer

A ray $\mathrm{PQ}$ incident on the refracting face $\mathrm{BA}$ is refracted in the prism $\mathrm{BAC}$ as shown in the figure and emerges from the other refracting face $\mathrm{AC}$ as $\mathrm{RS}$ such that $\mathrm{AQ = AR}$ . If the angle of prism $\mathrm{A = 60^{\circ}}$ and refractive index of the material of prism is $\mathrm{\sqrt{3}}$ , then the angle of deviation of the ray is:

A

$60^{\circ}$

Correct Answer

Wrong Answer

B

$45^{\circ}$

Correct Answer

Wrong Answer

C

$30^{\circ}$

Correct Answer

Wrong Answer

D

None of these

Correct Answer

Wrong Answer

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A thin prism, $\mathrm{P_{1}}$ with an angle $6^{0}$ and made of glass with a refractive index of 1.54 is combined with another prism $\mathrm{P}_{2}$ made from glass with a refractive index of 1.72 to produce dispersion without average deviation. The angle of prism $\mathrm{P}_{2}$ is

A

$1.3^{\circ}$

Correct Answer

Wrong Answer

B

$6^{\circ}$

Correct Answer

Wrong Answer

C

$4.5^{\circ}$

Correct Answer

Wrong Answer

D

$7.8^{\circ}$

Correct Answer

Wrong Answer

Light of wavelength $\mathrm{ 2 \times 10^{-3} \mathrm{~m}}$ falls on a slit of width $\mathrm{ 4 \times 10^{-3} \mathrm{~m}. }$ The angular dispersion of the central maximum will be:

A

$30^{\circ}$

Correct Answer

Wrong Answer

B

$60^{\circ}$

Correct Answer

Wrong Answer

C

$90^{\circ}$

Correct Answer

Wrong Answer

D

$180^{\circ}$

Correct Answer

Wrong Answer

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Concepts Covered - 4

Refraction Through A Prism 1

A prism is a transparent medium whose refracting surfaces are not parallel but are inclined to each other at an angle A which is also known as angle of the prism.

The angle of deviation $(\delta)$ -It is the angle between the emergent and the incident ray.

For the above figure $\begin{array}{l}{\delta=\left(i-r_{1}\right)+\left(e-r_{2}\right) \text { or } \delta=i+e-\left(r_{1}+r_{2}\right)} \end{array}$

and using $Using \ \ A=r_1+r_2 \ \ we \ \ get \ \ \delta =i+e-A$

Note-From the above formula, we can say that if we interchange i and e then also we will get the same value of $\delta$ .

The plot of $\large \delta \ \ vs \ \ i$

As shown in the above figure The graph is a parabola.

If we vary i between $0^0 \ \ to \ \ 90 ^0$

then $for \ \ 0 < i < e$ the value of $\delta$ decreases

and $for \ \ e < i < 90$ the value of $\delta$ increases

And when $i=e \ \ then \ \ \delta =\delta _{min}$

and when $i=90^0 \ \ or \ \ e=90^0 \ \ \ \ then \ \ \delta =\delta _{max}$

Grazing Incidence-When i = 90°, the incident ray grazes along the surface of the prism. This is known as grazing incidence.
Grazing Emergence- When e = 90°, the emergent ray grazes along the prism surface. This is known as grazing emergence.

This happens when the light ray strikes the second face of the prism at the critical angle for glass - air.

$I.e \ \ when \ \ \ \ r _2=\theta _c \ \ then \ \ \ e=90^0$

I.e For the prism of refractive index $\mu$ places in the air.

then $i=\sin ^{-1}[\sqrt{\mu ^{2}-1} \sin A-\cos A]$ then e = 90°

Refraction Through A Prism 2

Refractive index of prism $(\mu )$ in case of minimum deviation condition-

As we learned The angle of deviation $(\delta)$ for the prism is given as $\ \delta =i+e-A$

and from The plot of $\delta \ \ vs \ \ i$ we get $i=e \ \ then \ \ \delta =\delta _{min}$

$i.e \ \ \delta_{min} =i+e-A=i+i-A=2i-A\\\Rightarrow i=\frac{A+\delta _{min}}{2}$

For the prism of refractive index $\mu$ places in the air.

For the first surface, we can write $1*sini=\mu sinr_1$

similarly For the second surface, we can write $\mu sinr_2=1* sine$

using i=e we get $r_1=r_2$

$A=r_1+r_2=2r_1\\ \rightarrow r_1=\frac{A}{2}$

So $1*sini=\mu sinr_1$ will give us

$\Rightarrow 1*sin(\frac{A+\delta _{min}}{2})=\mu sin(\frac{A}{2})\\ \Rightarrow \mu =\frac{sin(\frac{A+\delta _{min}}{2})}{ sin(\frac{A}{2})}$

For thin films (i.e $A \ \ and \ \ \delta _{min} \ are \ small$ )

Then $sin(\frac{A+\delta _{min}}{2})=\frac{A+\delta _{min}}{2}$

and $sin(\frac{A}{2})=\frac{A }{2}$

So we get

$\mu = \frac{A+\delta _{min}}{A}\\ \Rightarrow \delta _{min}=A(\mu -1)$

condition of no emergence-

i.e A ray of light incidence on a prism of angle A & Refractive index $\mu$ will not emerge out of a prism

This will happen when $A> 2\theta _{c}$

where $\theta _{c}=$ critical angle

Dispersion Of Light 1

Dispersion of light -The splitting of white light into its constituent colors or wavelengths is called dispersion of light.

or

angular splitting of a ray of white light into a number of components and spreading in different directions is called diversion of light.

This phenomenon arises due to the fact that the refractive index varies with wavelength.

When white light is incident on the prism it will split itself into its constituent colors as shown in the below figure.

The deviation is given as $\delta =(\mu -1)A$

$Since \ \mu _{violet}>\mu _{red} \\ So \ \ \delta _{violet}>\delta _{red}$

Angular dispersion ( $\theta$ )- Angular separation between extreme colors

i.e $\boldsymbol{\theta}=\boldsymbol{\delta}_{V}-\boldsymbol{\delta}_{R}=\left(\boldsymbol{\mu}_{V}-\boldsymbol{\mu}_{R}\right) \boldsymbol{A} .$

It depends upon $\mu$ and A.

Dispersive power (ω)- Ratio of angular dispersion to mean deviation.

i.e $\omega=\frac{\delta_{v}-\delta_{r}}{\delta}$

where where $\delta$ is deviation of mean ray (especially yellow)

$\begin{aligned} using \ \ \delta _{v} &=\left(\mu_{v}-1\right) A, \delta _{r}=\left(\mu_{r}-1\right) A \\ we \ get \ \ \omega &=\frac{\mu_{v}-\mu_{r}}{\mu_{y}-1} \quad \text { where } \quad \mu_{y}=\frac{\mu_{v}-\mu_{r}}{2} \end{aligned}$

where

$\mu _{v}$ = Refractive index of violet

$\mu _{r}=$ Refractive index of red

$\mu _{y}=$ Refractive index of yellow

Dispersion Of Light 2

Condition for deviation without dispersion-

This means an achromatic combination of two prisms in which net(or) resultant dispersion is 0, but and deviation is produced as shown in the below figure.

$\begin{array}{ll}{\text { For the two prisms, }} \\ {\qquad \begin{aligned} \theta _{net}=0\Rightarrow \theta _1+\theta _2=0 \\ \left(\mu_{v}-\mu_{r}\right) A+\left(\mu'_{v}-\mu'_{r}\right) A '&=0 \\ \Rightarrow \quad A^{\prime}=\frac{\left(\mu_{v}-\mu_{r}\right) A}{\mu'_{v}-\mu'_{r}} & \end{aligned}}\end{array}$

where

$\mu _{v}$ = Refractive index of violet ( prism 1)

$\mu _{r}=$ Refractive index of red ( prism 1)

$\mu{}' _{v}$ = Refractive index of violet ( prism 2)

$\mu{}' _{r}=$ Refractive index of red ( prism 2)

Similarly

$\begin{array}{ll}{\text { For the two prisms, }} \\ {\qquad \begin{aligned} \theta _{net}=0\Rightarrow \theta _1+\theta _2=0 \\ & \end{aligned}}\end{array}$

$\omega \delta+\omega^{\prime} \delta^{\prime}=0\\ \Rightarrow \delta '=-\delta \frac{\omega }{\omega '}\\ \Rightarrow \delta_{net}=\delta +\delta '=\delta\left[1-\frac{\omega}{\omega^{\prime}}\right]$

where $\omega \ \ and \ \ \omega^{\prime}$ are the dispersive powers of the two prisms and their corresponding mean deviations are $\delta \ \ and \ \ \delta^{\prime}$ .

Condition for Dispersion without deviation-

A combination of two prisms in which deviation produced for the mean ray by the first prism is equal and opposite to that produced
by the second prism will give a dispersion of light without deviation.

This combination of two prisms is also called a direct vision prism.

i.e $\delta _{net}=0 \ \ while \ \ \theta _{net}\neq0$

As shown in the above figure as emergent rays from the second prism is parallel to the incident white ray of prism 1.

this will give $\delta _{net}=0$ .

$\begin{aligned} \text { For zero deviation ,} \\ i.e \ \ \delta _{net}=0 \left( i.e \ \ \delta+\delta^{\prime}=0\right) \\ \Rightarrow \quad(\mu_y-1) A+\left(\mu_y^{\prime}-1\right) A'=0 \\ \Rightarrow \quad A^{\prime}=\frac{(\mu_y-1) A}{\left(\mu_y^{\prime}-1\right)} \end{aligned}$

and the Angular dispersion is given as

$\theta_{\text {net }}=\theta _1+\theta _2=\left(\omega \delta+\omega^{\prime} \delta^{\prime}\right)=\left(\omega \delta-\omega^{\prime} \delta\right)=\theta\left(1-\frac{\omega^{\prime}}{\omega}\right)$

Study it with Videos

Refraction Through A Prism 1

Refraction Through A Prism 2

Dispersion Of Light 1

Dispersion Of Light 2

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