Q 12-09-101JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
The refractive index of a prism with apex angle $A$ is $\cot\dfrac A2$. The angle of minimum deviation is:
Answer: (D) $\delta_m = 180^\circ - 2A$
$$\mu = \frac{\sin\frac{A + \delta_m}{2}}{\sin\frac A2} = \frac{\cos\frac A2}{\sin\frac A2} \Rightarrow \sin\frac{A + \delta_m}{2} = \cos\frac A2 = \sin\left(90^\circ - \frac A2\right)$$
$$\frac{A + \delta_m}{2} = 90^\circ - \frac A2 \Rightarrow \delta_m = 180^\circ - 2A$$
Solution by Sreeraj P, M.Sc Physics