Physics - Geometrical Optics Question with Solution | TestHub

The relation for refractive index of prism is $\mu =\frac{\sin i}{\sin r}$ $\dots (1)$ The condition for minimum deviation is $r=\frac{A}{2}=\frac{60^{\circ }}{2}=30^{\circ }$ Putting the given value of $\mu =\sqrt{2}$ and $r=30^{\circ }$ in eq (1), we get $\begin{array}{rl}\sqrt{2}& =\frac{\sin i}{\sin 30^{\circ }}\\ \text{ So, }\sin i& =\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\\ \sin i& =\sin 45^{\circ }\\ i& =45^{\circ }\end{array}$