Two equilateral-triangular prisms $\mathrm{P}_1$ and $\mathrm{P}_2$ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism $P_1$ at an angle of incidence $\theta$ such that the outgoing ray undergoes minimum deviation in prism $P_2$. If the respective refractive indices of $\mathrm{P}_1$ and $\mathrm{P}_2$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$, then $\theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{\beta}\right)\right]$, where the value of $\beta$ is _______ .

Solution

At surface BC $\begin{aligned} & \sqrt{\frac{3}{2}} \sin \mathrm{r}_2=\sqrt{3} \sin 30 \\ & \sqrt{\frac{3}{2}} \sin \mathrm{r}_2=\frac{\sqrt{3}}{2} \\ & \operatorname{sinr}_2=\frac{1}{\sqrt{2}} \\ & \mathrm{r}_2=45^{\circ} \\ & \mathrm{r}_1=60^{\circ}-45^{\circ}=15^{\circ}\end{aligned}$ At surface $A B$ $\begin{array}{ll} 1 \sin \theta=\sqrt{\frac{3}{2}} \sin 15^{\circ} \\ \theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \frac{\pi}{12}\right] \\ \beta=12\end{array}$

Hence, the correct answer is 12.