Q 12-09-041JEE MainJEE Main 2026 (6 Apr, Shift 2)Easy
Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is ______
Answer: (C) $\sqrt{2}$
Concept. At minimum deviation $\mu=\dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2}$.
Step 1. Values. $A=60^\circ$, $\delta_m=30^\circ$: $\mu=\dfrac{\sin45^\circ}{\sin30^\circ}=\sqrt2$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
