Prism, Dispersion & Scattering question from JEE Main 2024 (31 Jan, Shift 1)

Answer: (c)

1. $\mu = \dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}} = \cot\frac{A}{2} = \dfrac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$
2. $\sin\frac{A+\delta_m}{2} = \cos\frac{A}{2} = \sin\left(90^\circ - \frac{A}{2}\right)$
3. $\frac{A+\delta_m}{2} = 90^\circ - \frac{A}{2} \Rightarrow \delta_m = 180^\circ - 2A$
4. So the answer is (c).