Total Internal Reflection question from JEE Main 2013 (23 Apr, Shift 1)

Answer: (d)

1. Top face: $\sin45^\circ = \mu\sin r \Rightarrow \sin r = \frac{1}{\sqrt{2}\mu}$.
2. Angle of incidence at the vertical face is $90^\circ - r$; TIR needs $\cos r \ge \frac{1}{\mu}$.
3. $1 - \frac{1}{2\mu^2} \ge \frac{1}{\mu^2} \Rightarrow \mu \ge \sqrt{\frac{3}{2}}$.
4. So the answer is (d) $\sqrt{\frac{3}{2}}$.