JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Geometrical Optics

At the top face Snell's law gives $\sin \theta =\mu \sin {\theta }^{\prime }$ with $\mu =\frac{4}{3}$, where ${\theta }^{\prime }$ is measured from the normal to that face. The adjacent face is perpendicular to the top face, so its normal is perpendicular to the first normal and the ray meets it at ${\theta }^{\prime \prime }=90^{\circ }-{\theta }^{\prime }$. Total internal reflection there requires ${\theta }^{\prime \prime }\ge {\theta }_c$, where

Since ${\theta }^{\prime \prime }$ shrinks as ${\theta }^{\prime }$ grows, the largest admissible $\theta$ corresponds to ${\theta }^{\prime \prime }={\theta }_c$, giving $\cos {\theta }^{\prime }=\frac{3}{4}$ and $\sin {\theta }^{\prime }=\frac{\sqrt{7}}{4}$. Substituting in Snell's law,

The maximum value of the angle is ${\sin }^{-1}\frac{\sqrt{7}}{3}$.