JEE Advanced 2024 Paper 2 · Q11 · Geometrical Optics

Question

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 $	heta$ such that the outgoing ray undergoes minimum deviation in prism $P_2$. If the respective refractive indices of $P_1$ and $P_2$ are $\sqrt{\dfrac{3}{2}}$ and $\sqrt{3}$, then $	heta = \sin^{-1}\left[\sqrt{\dfrac{3}{2}}\sin\left(\dfrac{\pi}{\beta}ight)ight]$, where the value of $\beta$ is _____.

SolveKar AI · Accepted Answer

For minimum deviation in $P_2$ (equilateral, $A = 60°$): the ray inside $P_2$ is symmetric, so the angle of refraction at each face is $A/2 = 30°$.

Using Snell at the $P_2$ entry face: $\sin r_2 = n_2 \sin 30° / 1$ — wait, let me redo. At entry to $P_2$ (from vacuum): $1\cdot\sin\alpha = n_2 \sin 30°$ where $\alpha$ is angle of incidence into $P_2$. So $\sin\alpha = \sqrt{3}\cdot(1/2) = \sqrt{3}/2$ ⇒ $\alpha = 60°$.

The ray emerging from $P_1$ enters $P_2$ at angle $\alpha = 60°$. Geometry of parallel-sided equilateral prisms means the ray must emerge from $P_1$ at angle 60° (measured from normal of $P_2$'s entry face, which is parallel to $P_1$'s exit face's normal). So angle of refraction at exit face of $P_1$ in vacuum is $60°$, giving angle inside $P_1$ at the exit face: $\sin r_1' \cdot n_1 = \sin 60°$, so $\sin r_1' = \dfrac{\sin 60°}{\sqrt{3/2}} = \dfrac{\sqrt{3}/2}{\sqrt{3/2}} = \dfrac{\sqrt{3}/2}{\sqrt{3}/\sqrt{2}} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$, so $r_1' = 45°$.

In $P_1$ (also equilateral, $A = 60°$), $r_1 + r_1' = 60°$ ⇒ $r_1 = 60° - 45° = 15°$.

At entry face of $P_1$ from vacuum: $\sin	heta = n_1 \sin r_1 = \sqrt{3/2}\,\sin 15°$.

So $	he…[truncated]

JEE Advanced 2024 Paper 2 · Q11 · Geometrical Optics

Solution