JEE Advanced PYQ Physics – Prism and Dispersion of Light

Q 1 :

A parallel beam of light is incident from air at an angle $α$ on the side PQ of a right angled triangular prism of refractive index $n = \sqrt{2}$ . Light undergoes total internal reflection in the prism at the face PR when $α$ has a minimum value of $45 °$ . The angle $θ$ of the prism is [2016]

$15 °$
$22.5 °$
$30 °$
$45 °$

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(1)

Applying Snell's law at A

$1 \times \sin 45 ° = \sqrt{2} \times \sin r_{1}$ $∴$ $r_{1} = 30 °$

$\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}$

$∴$ $C = 45 °$

In $∆ A M B,$

$90 ° + θ + r_{1} + (90 ° - C) = 180 °$

$\Rightarrow 90 ° + θ + 30 ° + 90 ° - 45 ° = 180 °$

$∴$ $θ = 15 °$

Q 2 :

The graph shows relationship between object distance and image distance for an equiconvex lens. Then, focal length of the lens is [2006]

0.50 ± 0.05 cm
0.50 ± 0.10 cm
5.00 ± 0.05 cm
5.00 ± 0.10 cm

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(3)

From the graph, $u = - 10 cm, v = 10 cm, Δ u = Δ v = 0.1$

From the lens formula, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{10} - \frac{1}{- 10}$

$∴$ $f = 5 cm$

Differentiating the lens formula,

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ $\Rightarrow \frac{Δ f}{f^{2}} = \frac{Δ v}{v^{2}} + \frac{Δ u}{u^{2}}$ $(for maximum error in f)$

$\Rightarrow \frac{Δ f}{25} = \frac{0.1}{{(10)}^{2}} + \frac{0.1}{{(10)}^{2}}$

$\Rightarrow Δ f = 25 \times 0.1 \times 2 \times 0.01 = 0.05$

$∴$ $Focal length, f \pm Δ f = (5.00 \pm 0.05) cm .$

Q 3 :

Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be [2008]

30° for both the colours
Greater for the violet colour
Greater for the red colour
Equal but not 30° for both the colours

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(1)

For minimum deviation, the ray in the prism is parallel to the base of the prism. This condition does not depend on the colour (or wavelength) of incident radiation. Therefore, in both the cases, for both the colours by geometry, angle of refraction is r = 30°.

Q 4 :

An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation [2004]

PQ is horizontal
QR is horizontal
RS is horizontal
Any one will be horizontal

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(2)

For minimum deviation, incident angle is equal to emerging angle. And ray QR inside the equilateral prism is parallel to base.

Q 5 :

A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer [2001]

greater deviation
no deviation
same deviation as before
total internal reflection

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(3)

There will be no refraction from P to Q and then from Q to R, all being identical made of the same material. Hence, the ray will now have the same deviation.

Q 6 :

Two equilateral-triangular prisms $P_{1}$ and $P_{2}$ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism $P_{1}$ at an angle of incidence $θ$ such that the outgoing ray undergoes minimum deviation in prism $P_{2}$ . If the respective refractive indices of $P_{1}$ and $P_{2}$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$ , then $θ = \sin^{- 1} [\sqrt{\frac{3}{2}} \sin (\frac{π}{β})],$ where the value of $β$ is _______. [2024]

(12)

By using optical reversibility principle

For prism $P_{2}$ : Minimum deviation using Snell's law,

$1 \times \sin θ_{1} = \sqrt{3} \sin r$

$r_{1} = r_{2} = \frac{A}{2} = \frac{60 °}{2} = 30 °$

$\sin θ_{1} = \sqrt{3} \times \frac{1}{2}$

$∴$ $i = e = 60 °$

For prism $P_{1}$ : Incident angle $= 60 °$

$1 \times \sin 60 ° = \frac{\sqrt{3}}{\sqrt{2}} \sin r_{1}$

or, $\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{2}} \sin r_{1}$

$r_{1} + r_{2} = 60 °$

$\sin r_{1} = \frac{1}{\sqrt{2}}$ $∴$ $r_{1} = 45 °$ and $r_{2} = 15 °$

$\frac{\sqrt{3}}{\sqrt{2}} \sin (45 °) = 1 \times \sin θ$

$15 ° = \frac{π \times 15}{180} rad = \frac{π}{12} rad$

$θ = \sin^{- 1} [\frac{\sqrt{3}}{\sqrt{2}} \sin (\frac{π}{12})]$

$∴$ $β = 12$

Q 7 :

A monochromatic light is incident from air on a refracting surface of a prism of angle $75 °$ and refractive index $n_{0} = \sqrt{3}$ . The other refracting surface of the prism is coated by a thin film of material of refractive index $n$ , as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of $θ \leq 60 °$ . The value of $n^{2}$ is _______. [2019]

(1.50)

In $∆ X Y Z,$ $90 ° \times r + 90 ° - C + 75 ° = 180 °$

$∴$ $r + C = 75 °$ $\Rightarrow r = 75 ° - C$ $\dots (i)$

Applying Snell's law at $Z$ ,

$\sqrt{3} \sin C = n \sin 90 °$

$\sqrt{3} \sin C = n$ $\dots (i i)$

Applying Snell's law at $Y$ ,

$1 \times \sin θ = \sqrt{3} \sin r$

$= \sqrt{3} \sin (75 ° - C)$

From eq. (i),

For $θ = 60 °$

$\frac{\sqrt{3}}{2} = \sqrt{3} \sin (75 ° - C)$ $∴$ $C = 45 °$

From eq. (ii), $n = \sqrt{3} \sin 45 ° = \frac{\sqrt{3}}{\sqrt{2}}$

$∴$ $n^{2} = 1.50$

Q 8 :

The monochromatic beam of light is incident at $60 °$ on one face of an equilateral prism of refractive index $n$ and emerges from the opposite face making an angle $θ (n)$ with the normal (see the figure). For $n = \sqrt{3}$ the value of $θ$ is $60 °$ and $\frac{d θ}{d n} = m .$ The value of $m$ is _______ [2015]

(2)

Applying Snell's law at $A$ ,

$\sin 60 ° = n \sin r$ $\dots (i)$

Differentiating w.r.t. $' n'$ , we get

$0 = \sin r + n \cos r \frac{d r}{d n}$ $\dots (i i)$

Again, applying Snell's law at $B$ ,

$\sin θ = n \sin (60 ° - r)$ $\dots (i i i)$

Differentiating w.r.t. $' n'$ , we get

$\cos θ \frac{d θ}{d n} = \sin (60 ° - r) + n \cos (60 ° - r) (- \frac{d r}{d n})$

$∴$ $\cos θ \frac{d θ}{d n} = \sin (60 ° - r) - n \cos (60 ° - r) (- \frac{\tan r}{n})$ $[from (ii)]$

$∴$ $\frac{d θ}{d n} = \frac{1}{\cos θ} [\sin (60 ° - r) + \cos (60 ° - r) \tan r]$ $\dots (i v)$

From eq. (i), for $n = \sqrt{3}$ , we get $r = 30 °$

From eq. (iii), for $n = \sqrt{3}$ and $r = 30 °$ , we get $θ = 60 °$

Substituting the values of $r$ and $θ$ in eq. (iv), we get

$\frac{d θ}{d n} = \frac{1}{\cos 60 °} [\sin 30 ° + \cos 30 ° \tan 30 °] = 2 (\frac{1}{2} + \frac{1}{2}) = 2$

Q 9 :

For a prism of prism angle $θ = 60 °$ , the refractive indices of the left half and the right half are, respectively, $n_{1}$ and $n_{2}$ $(n_{2} \geq n_{1})$ , as shown in the figure. The angle of incidence $i$ is chosen such that the incident light rays will have minimum deviation if $n_{1} = n_{2} = n = 1.5$ . For the case of unequal refractive indices $n_{1} = n$ and $n_{2} = n + Δ n$ (where $Δ n ≪ n$ ), the angle of emergence $e = i + Δ e$ . Which of the following statement(s) is (are) correct? [2021]

The value of $Δ e$ (in radians) is greater than that of $Δ n$
$Δ e$ is proportional to $Δ n$
$Δ e$ lies between 2.0 and 3.0 milliradians, if $Δ n = 2.8 \times 10^{- 3}$
$Δ e$ lies between 1.0 and 1.6 milliradians, if $Δ n = 2.8 \times 10^{- 3}$

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Select one or more options

(2, 3)

Given angle of prism $A = 60 °$

For minimum deviation,

$r_{1} = r_{2} = \frac{A}{2} = 30 °$

and from Snell's law, $n \times \sin i = n_{1} \sin r_{1}$

$\Rightarrow 1 \times \sin i = n_{1} \sin \frac{A}{2}$

$∴$ $\sin i = \frac{3}{2} \times \sin 30 ° = \frac{3}{2} \times \frac{1}{2}$ $\Rightarrow \sin i = \frac{3}{4}$

At another face of prism,

$n_{1} \sin 30 ° = 1 \times \sin (e)$

On differentiating both sides, $Δ n \sin 30 ° = Δ e \cos (e)$

$Δ e = \frac{Δ n}{2 \cos (e)}$

or, $Δ e = \frac{Δ n}{2 \sqrt{1 - \frac{9}{16}}} = \frac{2}{\sqrt{7}} Δ n$

$Δ e = \frac{2}{\sqrt{7}} Δ n \Rightarrow Δ e < Δ n$ and $Δ e \propto Δ n$

$Δ n = 2.8 \times 10^{- 3}$

$∴$ $Δ e = \frac{2.8 \times 10^{- 3} \times 2}{\sqrt{7}} = 2.11 \times 10^{- 3} rad = 2.11 mrad$

Q 10 :

For an isosceles prism of angle A and refractive index $μ$ , it is found that the angle of minimum deviation is $δ_{m} = A .$ .

Which of the following options is/are correct? [2017]

For the angle of incidence $i_{1} = A$ , the ray inside the prism is parallel to the base of the prism.
For this prism, the refractive index $μ$ and the angle of prism A are related as $A = \frac{1}{2} \cos^{- 1} (\frac{μ}{2})$
At minimum deviation, the incident angle $i_{1}$ and the refracting angle $r_{1}$ at the first refracting surface are related by $r_{1} = \frac{i_{1}}{2}$
For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_{1} = \sin^{- 1} [\sin A \sqrt{4 \cos^{2} (\frac{A}{2}) - 1} - \cos A]$

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Select one or more options

(1, 3, 4)

For minimum deviation,

$i_{1} = e$

$r_{1} = r_{2} = r (say) = \frac{A}{2}$

$δ_{m} = 2 i_{1} - A$

Here, $δ_{m} = A$ $\Rightarrow 2 i_{1} - A = A$

$∴$ $i_{1} = A \Rightarrow e = A$

$∴$ $r_{1} = \frac{i_{1}}{2}$

Relation between $μ$ and $A$

$μ = \frac{\sin i_{1}}{\sin r_{1}} = \frac{\sin A}{\sin (\frac{A}{2})} = \frac{2 \sin (\frac{A}{2}) \cos (\frac{A}{2})}{\sin (\frac{A}{2})} = 2 \cos (\frac{A}{2})$

When emergent ray is tangential to the surface,

$μ = \frac{\sin 90 °}{\sin r_{2}} = \frac{1}{\sin r_{2}}$ $\Rightarrow r_{2} = \sin^{- 1} (\frac{1}{μ})$

But, $r_{1} + r_{2} = A$ $∴$ $r_{1} = A - r_{2}$

$∴$ $r_{1} = A - \sin^{- 1} (\frac{1}{μ})$

Applying Snell's law at P,

$μ = \frac{\sin i_{1}}{\sin r_{1}}$ $∴$ $i_{1} = \sin^{- 1} [μ \sin (A - \sin^{- 1} (\frac{1}{μ}))]$

For minimum deviation through an isosceles prism, if $∠ B = ∠ C, P Q ∥ B C .$

Q 11 :

A right angled prism of refractive index $μ_{1}$ is placed in a rectangular block of refractive index $μ_{2}$ , which is surrounded by a medium of refractive index $μ_{3}$ , as shown in the figure. A ray of light $' e'$ enters the rectangular block at normal incidence. Depending upon the relationships between $μ_{1}$ , $μ_{2}$ and $μ_{3}$ , it takes one of the four possible paths 'ef', 'eg', 'eh' or 'ei'.

Match the paths in List I with conditions of refractive indices given in List II and select the correct answer using the codes given below the lists. [2013]

List I List II

P. $e \to f$ 1. $μ_{1} > \sqrt{2} μ_{2}$

Q. $e \to g$ 2. $μ_{2} > μ_{1} and μ_{2} > μ_{3}$

R. $e \to h$ 3. $μ_{1} = μ_{2}$

S. $e \to i$ 4. $μ_{2} < μ_{1} < \sqrt{2} μ_{2} and μ_{2} > μ_{3}$

Codes:

P-2, Q-3, R-1, S-4
P-1, Q-2, R-4, S-3
P-4, Q-1, R-2, S-3
P-2, Q-3, R-4, S-1

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(4)

$e \to f$

When the ray enters from the rectangular block to prism, the angle of incidence is greater than the angle of refraction. Hence, $μ_{2} > μ_{1} .$ The ray then moves away from the normal when it emerges out of the rectangular block. Therefore, $μ_{2} > μ_{3} .$

$e \to g$

As there is no deviation of the ray as it emerges out of the prism, $∴$ $μ_{2} = μ_{1} .$

$e \to h$

As the ray emerges out of the prism, it moves away from the normal. $∴$ $μ_{2} < μ_{1} .$ The ray moves away from the normal as it emerges out of the rectangular block. $∴$ $μ_{2} > μ_{3} .$

$e \to i$

At the prism surface, total internal reflection has taken place.

$∴$ $Critical angle 45 ° > C$ $\Rightarrow \sin 45 ° > \sin C$

$\Rightarrow \frac{1}{\sqrt{2}} > \frac{μ_{2}}{μ_{1}}$ $∴$ $μ_{1} > \sqrt{2} μ_{2}$

Topic Question Set

Q 2 :

The graph shows relationship between object distance and image distance for an equiconvex lens. Then, focal length of the lens is [2006]

Q 3 :

Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be [2008]

Q 4 :

An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation [2004]

Q 5 :

A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer [2001]

Q 10 :

For an isosceles prism of angle A and refractive index $μ$ , it is found that the angle of minimum deviation is $δ_{m} = A .$ .

Which of the following options is/are correct? [2017]

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	List I		List II
P.	$e \to f$	1.	$μ_{1} > \sqrt{2} μ_{2}$
Q.	$e \to g$	2.	$μ_{2} > μ_{1} and μ_{2} > μ_{3}$
R.	$e \to h$	3.	$μ_{1} = μ_{2}$
S.	$e \to i$	4.	$μ_{2} < μ_{1} < \sqrt{2} μ_{2} and μ_{2} > μ_{3}$

Topic Question Set

Q 1 : A parallel beam of light is incident from air at an angle α on the side PQ of a right angled triangular prism of refractive index n=2. Light undergoes total internal reflection in the prism at the face PR when α has a minimum value of 45°. The angle θ of the prism is [2016]

Q 2 : The graph shows relationship between object distance and image distance for an equiconvex lens. Then, focal length of the lens is [2006]

Q 3 : Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be [2008]

Q 4 : An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation [2004]

Q 5 : A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer [2001]

Q 8 : The monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle θ(n) with the normal (see the figure). For n=3 the value of θ is 60° and dθdn=m. The value of m is _______ [2015]

Q 10 : For an isosceles prism of angle A and refractive index μ, it is found that the angle of minimum deviation is δm=A.. Which of the following options is/are correct? [2017]

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Q 2 :

The graph shows relationship between object distance and image distance for an equiconvex lens. Then, focal length of the lens is [2006]

Q 3 :

Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be [2008]

Q 4 :

An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation [2004]

Q 5 :

A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer [2001]

Q 10 :

For an isosceles prism of angle A and refractive index $μ$ , it is found that the angle of minimum deviation is $δ_{m} = A .$ .

Which of the following options is/are correct? [2017]