Ray Optics and Optical Instruments

Let the prism angles at B and C be $B$ and $C$ with $B+C=90^{\circ }$. The ray inside is parallel to BC, so its angle of refraction at AB is $r=90^{\circ }-B$ and its angle of incidence at AC is $90^{\circ }-C=B$. Emerging along AC means $B$ is the critical angle: $n\sin B=1$. At P, $\sin 30^{\circ }=n\sin r=n\cos B$, so $n\cos B=\frac{1}{2}$. Squaring and adding: $n^2=1+\frac{1}{4}\Rightarrow n=\frac{\sqrt{5}}{2}$.