JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Geometrical Optics

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

\frac{3}{2}

and

3

, then

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

, where the value of

β

is ____.

Answer: 12

Solution

For minimum deviation in prism

P_{2}

(equilateral,

A = 60°

), the ray passes symmetrically, so the angle of refraction at entry into

P_{2}

30°

.

At the interface BC between

P_{1}

and

P_{2}

\frac{3}{2} sin r_{2} = 3 sin 30° = \frac{3}{2}

sin r_{2} = \frac{1}{2} \Rightarrow r_{2} = 45°

Inside prism

P_{1}

, using geometry of equilateral triangle:

r_{1} = 60° - 45° = 15°

At surface AB (entry into

P_{1}

sin θ = \frac{3}{2} sin 15°

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

Therefore:

β = 12

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer