JEE Main 2026 — Ray Optics Question with Solution

Let the refractive index of the prism be $n_1=n$ and the refractive index of the coating on the exit surface be $n_2=\frac{n}{2}$.
At the condition of minimum deviation, the angle of refraction at the first surface $r_1$ and the angle of incidence at the second surface $r_2$ are equal, i.e., $r_1=r_2=\frac{A}{2}$, where $A$ is the prism angle.
The problem states that at the exit surface, the light meets the condition of critical angle. Therefore, $r_2={\theta }_c$.
The critical angle ${\theta }_c$ for the interface between the prism and the coating is given by $\sin {\theta }_c=\frac{n_2}{n_1}$.
Substituting the values, $\sin {\theta }_c=\frac{n/2}{n}=\frac{1}{2}$.
Thus, ${\theta }_c=30^{\circ }$.
Since $r_2={\theta }_c$, we have $r_2=30^{\circ }$.
Using the relation $r_1=r_2=\frac{A}{2}$ for minimum deviation, we get $\frac{A}{2}=30^{\circ }$.
Therefore, $A=60^{\circ }$.