Prism, Dispersion & Scattering question from NEET 2012 (AIPMT, Mains)

Answer: (b)

1. At minimum deviation $\mu = \dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}}$. With $\delta_m = A$: $\mu = \dfrac{\sin A}{\sin\frac{A}{2}} = 2\cos\frac{A}{2}$.
2. Here $i = \frac{A+\delta_m}{2} = A$, and $i \le 90^\circ$, so $0 < A \le 90^\circ$.
3. $A \to 0$ gives $\mu \to 2$; $A = 90^\circ$ gives $\mu = 2\cos 45^\circ = \sqrt{2}$.
4. So the answer is (b): $\mu$ lies between $\sqrt{2}$ and 2.