Condition for minimum deviation in a prism

For a prism at minimum deviation, $i=e$ and the refraction angles inside the prism are equal: r_1=r_2=rac{A}{2} where $A$ is the prism angle.

Given data

$A=60^\circ, \qquad \mu=\sqrt{2}$

Therefore, $r=\frac{A}{2}=\frac{60^\circ}{2}=30^\circ$

Apply Snell's law at first surface

Since the prism is in air, $\mu=\frac{\sin i}{\sin r}$

Substituting values: $\sqrt{2}=\frac{\sin i}{\sin 30^\circ}$

Since $\sin 30^\circ=\frac{1}{2}$

we get $\sqrt{2}=\frac{\sin i}{1/2}$

$\sin i=\sqrt{2}\cdot \frac{1}{2}=\frac{\sqrt{2}}{2}$

Find $i$

$\sin i=\frac{\sqrt{2}}{2}$ hence, $i=45^\circ$

Final answer

The angle of incidence is $\boxed{45}$