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 √(3/2) and √(3) , then θ=sin -1 [√(3/2) sin (π/β ) ] , where the value of β is

Physics · Geometrical Optics

JEE Advanced 2024 — Paper 2 — Question 22

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 $\theta$ 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 $\theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{\beta}\right)\right]$ , where the value of $\beta$ is

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

At C, $1 \sin \alpha=\sqrt{3} \sin 30^{\circ} \Rightarrow \alpha=60^{\circ}$ At $B, \sqrt{\frac{3}{2}} \sin r_{2}=1 \sin \alpha \Rightarrow r_{2}=45^{\circ}$ At $A, 1 \sin \theta=\sqrt{\frac{3}{2}} \sin 15^{\circ}$ $\Rightarrow \theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{12}\right)\right]$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Physics
Chapter: Geometrical Optics
Topic: Apparent Depth, Glass Slab, Prism and Dispersion