Q 12-09-090JEE MainJEE Main 2024 (27 Jan, Shift 1)Medium
If the refractive index of the material of a prism is $\cot\left(\dfrac{A}{2}\right)$, where $A$ is the angle of prism, then the angle of minimum deviation will be:
Answer: (A) $\pi - 2A$
$$\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\frac A2} = \cot\frac A2 = \frac{\cos\frac A2}{\sin\frac A2}$$
So $\sin\left(\dfrac{A+\delta_m}{2}\right) = \cos\dfrac A2 = \sin\left(\dfrac\pi2 - \dfrac A2\right)$, giving
$$\frac{A+\delta_m}{2} = \frac\pi2 - \frac A2 \;\Rightarrow\; \delta_m = \pi - 2A$$
Solution by Sreeraj P, M.Sc Physics