The Benzene Prism from the IPhOO

The first hint is that the problem says light leaves from the $"vertical$ $leg"$ . Thus, we get the angle which touches the table as $60^o$ . Let the angle of refraction be $r$ for the first surface. The measurements of the angles are given below:

Image

For first refraction, $\frac{sin 60^o}{sin r}= \frac{3}{2} \implies sin r= \frac{1}{\sqrt{3}} \implies cos r= \sqrt{\frac{2}{3}}$

For second refraction, $\frac{sin(60^o-r)}{sin(90^o-\gamma)}= \frac{2}{3} \implies \frac{\frac{\sqrt{3}}{2}cosr- \frac{1}{2}sinr}{cos \gamma}= \frac{2}{3} \implies cos \gamma=\frac{3}{2}(\frac{1}{\sqrt{2}}-\frac{1}{2\sqrt{3}})$

Thus, we get $\gamma =51.123^o$