Total Internal Reflection question from JEE Main 2021 (25 Jul, Shift 2)

Answer: (a)

1. At the top face: $\sin\theta = \frac{4}{3}\sin\theta'$.
2. The side face is perpendicular to the top face, so the angle of incidence there is $\theta'' = 90^\circ - \theta'$.
3. TIR needs $\sin\theta'' \ge \frac{3}{4}$, i.e. $\cos\theta' \ge \frac{3}{4} \Rightarrow \sin\theta' \le \frac{\sqrt{7}}{4}$.
4. $\sin\theta \le \frac{4}{3}\cdot\frac{\sqrt{7}}{4} = \frac{\sqrt{7}}{3}$.
5. So the answer is (a) $\theta_{max} = \sin^{-1}\frac{\sqrt{7}}{3}$.