JEE Main · 2009 · 12 Apr 2009 · Offline · Geometrical Optics

For the ray to skim along the curved wall it must strike that wall at exactly the critical angle $C$, where $\sin C=\frac{1}{\mu }=\frac{\sqrt{3}}{2}$, giving $C=60^{\circ }$. Inside the rod the ray that does this makes an angle $r=90^{\circ }-C=30^{\circ }$ with the normal at the flat entry face, so $\sin r=\frac{1}{2}$. Applying Snell's law at that entry face, $\sin \theta =\mu \sin r=\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}$, so the incident angle is ${\sin }^{-1}\frac{1}{\sqrt{3}}$.