Prism, Dispersion & Scattering question from NEET 2021

Answer: (b)

1. The ray falls normally on the hypotenuse, so it enters undeviated: $r_1 = 0$.
2. Prism angle between the hypotenuse and the vertical face is $90^\circ - 60^\circ = 30^\circ$, so $r_2 = A - r_1 = 30^\circ$ at the vertical face.
3. Critical angle $\sin C = \frac{1}{\sqrt{3}} \Rightarrow C \approx 35^\circ > 30^\circ$, so the ray emerges.
4. $\sin e = \sqrt{3} \sin 30^\circ = \frac{\sqrt{3}}{2} \Rightarrow e = 60^\circ$
5. So the answer is (b) $60^\circ$.