Question 11

For equilateral prism $P_2$, the refracting angle is $A = 60^\circ$. For minimum deviation, the internal refracted angle is $r_2 = A/2 = 30^\circ$.

The angle of incidence on $P_2$ from vacuum is $i_2$, given by $\sin i_2 = n_2 \sin r_2 = \sqrt{3} \sin 30^\circ = \frac{\sqrt{3}}{2}$. Thus, $i_2 = 60^\circ$.

Since the sides of $P_1$ and $P_2$ are parallel and the space between them is vacuum, the angle of emergence $e_1$ from $P_1$ must equal the angle of incidence $i_2$ on $P_2$. So, $e_1 = 60^\circ$.

Applying Snell's law at the second surface of $P_1$: $n_1 \sin r'_1 = \sin e_1 \Rightarrow \sqrt{\frac{3}{2}} \sin r'_1 = \sin 60^\circ = \frac{\sqrt{3}}{2}$.

$\sin r'_1 = \frac{\sqrt{3}}{2} \cdot \sqrt{\frac{2}{3}} = \frac{1}{\sqrt{2}}$, which gives $r'_1 = 45^\circ$.

For prism $P_1$, $r_1 + r'_1 = A \Rightarrow r_1 = 60^\circ - 45^\circ = 15^\circ$.

Applying Snell's law at the first surface of $P_1$: $\sin \theta = n_1 \sin r_1 = \sqrt{\frac{3}{2}} \sin(15^\circ) = \sqrt{\frac{3}{2}} \sin \left( \frac{\pi}{12} \right)$.

Comparing with $\theta = \sin^{-1} \left[ \sqrt{\frac{3}{2}} \sin \left( \frac{\pi}{\beta} \right) \right]$, we get $\beta = 12$.