JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Geometrical Optics

For a prism of prism angle

θ = 6 0^{\circ}

, 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 ?

(A)The value of $Δ e$ (in radians) is greater than that of $Δ n$
(B) $Δ e$ is proportional to $Δ n$ ✓ Correct
(C) $Δ e$ lies between 2.0 and 3.0 milliradians, if $Δ n = 2.8 \times 1 0^{- 3}$ ✓ Correct
(D) $Δ e$ lies between 1.0 and 1.6 milliradians, if $Δ n = 2.8 \times 1 0^{- 3}$

Answer: (B), (C)

Solution

At minimum deviation for a symmetric prism with

A = 6 0^{\circ}

, the ray passes symmetrically:

sin i = n sin (A /2) = 1.5 \times sin 3 0^{\circ} = 1.5 \times 0.5 = 0.75

i = sin^{- 1} (0.75)

At the exit face, applying Snell's law:

n_{1} sin 3 0^{\circ} = 1 \times sin e

sin e = \frac{n _{1}}{2}

Differentiating:

d n sin 3 0^{\circ} = d e cos e

d e = \frac{d n}{2 cos e}

For

n = 1.5

sin e = 0.75

, so

cos e = 1 - 0.7 5^{2} = 7 /4

d e = \frac{d n}{2 \times 7 /4} = \frac{2 d n}{7}

Since

\frac{2}{7} \approx 0.756 < 1

, we have

d e < d n

, so (A) is incorrect.

Clearly

d e \propto d n

, so (B) is correct.

For

Δ n = 2.8 \times 1 0^{- 3}

Δ e = \frac{2 \times 2.8 \times 1 0 ^{- 3}}{7} = \frac{5.6 \times 1 0 ^{- 3}}{2.646} \approx 2.12 \times 1 0^{- 3} rad = 2.12 mrad

This lies between 2.0 and 3.0 mrad, so (C) is correct and (D) is incorrect.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · One or more correct options