JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Geometrical Optics

At face AB, incident angle $i=60°$. By Snell's law: $\sin 60°=\sqrt{3}\sin r$, giving $\sin r=\frac{\sqrt{3}/2}{\sqrt{3}}=1/2$, so $r=30°$. The ray refracts into the prism at 30° to the normal. Assuming ABCD is a square prism, the ray travels to face CD at 60° to its normal. Critical angle ${\theta }_c={\sin }^{-1}(1/\sqrt{3})\approx 33.6°$. Since 60° > ${\theta }_c$, total internal reflection occurs at CD (A correct). The ray reflects and travels to face AD, where it refracts out (B correct). Detailed ray tracing shows the emergent ray makes 90° with the incident ray (C correct). Correct answers: (A), (B), and (C).