JEE Advanced 2024 — Paper 2 — PHY-III solution

JEE Advanced2024Paper 2PHY-III

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 $3$ , $θ = sin^{- 1} [\frac{3}{2} sin (\frac{π}{β})]$ , where the value of $β$ is ______.

Your answer

Solution

Use optical reversibility. In prism $P_{2}$ at minimum deviation, the ray inside $P_{2}$ is parallel to the base, so $r_{1}^{'} = r_{2}^{'} = A /2 = 3 0^{\circ}$ .

Applying Snell's law at the entry face of $P_{2}$ :

$1 \cdot sin i_{2} = 3 \cdot sin 3 0^{\circ} = \frac{3}{2} \Rightarrow i_{2} = 6 0^{\circ} .$

This ray, before entering $P_{2}$ , travelled through air after exiting $P_{1}$ at $6 0^{\circ}$ on the exit face of $P_{1}$ . Working back through $P_{1}$ , Snell's law at the exit face gives

$1 \cdot sin 6 0^{\circ} = \frac{3}{2} \cdot sin r_{1} \Rightarrow sin r_{1} = \frac{1}{2} \Rightarrow r_{1} = 4 5^{\circ} .$

Since $r_{1} + r_{2} = A = 6 0^{\circ}$ , we get $r_{2} = 1 5^{\circ} .$

Finally, Snell's law at the entrance face of $P_{1}$ :

$1 \cdot sin θ = \frac{3}{2} \cdot sin 1 5^{\circ} .$

Writing $1 5^{\circ}$ as $π /12$ radians,

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

$∴ β = 12.$

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