Total Internal Reflection question from JEE Main 2009 (AIEEE)

Answer: (c)

1. Grazing along the wall means the ray meets the wall at the critical angle: $\sin C = \frac{\sqrt{3}}{2} \Rightarrow C = 60^\circ$.
2. The wall is perpendicular to the end face, so the refraction angle at the end is $r = 90^\circ - 60^\circ = 30^\circ$.
3. $\sin\theta = \frac{2}{\sqrt{3}}\sin30^\circ = \frac{1}{\sqrt{3}}$.
4. So the answer is (c) $\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$.