Total Internal Reflection question from NEET 1998 (AIPMT)

Answer: (a)

1. At the end face: $\sin i = n\sin r$. At the lateral face the angle of incidence is $90^\circ - r$.
2. TIR at the lateral face needs $\cos r \ge \frac{1}{n}$ for every $i$.
3. Worst case $i = 90^\circ$: $\sin r = \frac{1}{n}$, so $1 - \frac{1}{n^2} \ge \frac{1}{n^2} \Rightarrow n^2 \ge 2$.
4. So the answer is (a) $n > \sqrt{2}$.