A ray of light is incident at \(60^{\circ}\) on one face of a prism of angle \(30^{\circ}\) and the emergent ray makes \(30^{\circ}\) with the incident ray. The refractive index of the prism is \(\left(\sin 30^{\circ}=0 \cdot 5, \sin 60^{\circ}=\sqrt{3} / 2\right)\)

Given \(\mathrm{A}=30^{\circ}, \mathrm{i}_1=30^{\circ}\) and \(\delta=30^{\circ}\)
For a prism,
\(\begin{array}{ll}
& \delta=\left(i_1+i_2\right)-\left(r_1+r_2\right) \\
& \delta=\left(i_1+i_2\right)-A \\
& 30=60+i_2-30^{\circ} \\
\therefore \quad & i_2=0
\end{array}\)
This means the emergent ray is perpendicular to the face from which emerges out.
As \(\mathrm{i}_2=0, \mathrm{r}_2=0\)
But \(\mathrm{r}_1+\mathrm{r}_2=\mathrm{A}\)
\(\mathrm{r}_1+0=30^{\circ}\)
\(\therefore \quad \mathrm{r}_1=30^{\circ}\)
Using Snell's law at the first face of the prism,
\(\begin{aligned}
& \mu_1 \sin i_1=\mu_2 \sin r_1 \\
& 1 \sin 60=\mu_2 \sin 30
\end{aligned}\)
\(\begin{aligned} & \frac{\sqrt{3}}{2}=\frac{\mu_2}{2} \\ \therefore \quad & \mu_2=\sqrt{3}=1.732\end{aligned}\)