The refractive index of a prism whose angle $A= 60°$ is $\sqrt{2}$. Then the angle of minimum deviation ${\delta }_{\mathit{m}}$ will be 

Refractive index, $\mu = \frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin {\displaystyle \frac{A}{2}}}$

Given, $A=60°$, $\mu =\sqrt{2}$, angle of minimum deviation ${\delta }_m$ = ?

$\Rightarrow \mu = \frac{\sin \left(\frac{60°+{\delta }_m}{2}\right)}{\sin {\displaystyle \left(\frac{60°}{2}\right)}}$

$\Rightarrow \sqrt{2} = \frac{\sin \left(30°+\frac{{\delta }_m}{2}\right)}{\sin {\displaystyle \left(30°\right)}}$

$\Rightarrow \frac{1}{\sqrt{2}}= \sin \left(30°+\frac{{\delta }_m}{2}\right)=\sin {45}^0$

$\therefore {\delta }_m = 30°$