A ray of light undergoes deviation of 30^o when incident on an equilateral prism of refractive index $\sqrt{2}$. The angle made by the ray inside the prism with the base of the prism is ___ (in degrees) 

$\begin{array}{l}\mu =\frac{\sin \left(\frac{\mathrm{A}+\delta \mathrm{m}}{2}\right)}{\sin \mathrm{A}/2},\therefore \sqrt{2}=\frac{\sin \left(\frac{60+\mathrm{m}}{2}\right)}{\sin 60/2}\\ \left(\mathrm{\angle }\mathrm{A}={60}^{\circ }\text{ for an equilateral triangle }\right)\Rightarrow \frac{60+\delta \mathrm{m}}{2}={45}^{\circ }\Rightarrow \delta \mathrm{m}={30}^{\circ }\end{array}$

The condition is for minimum deviation. 

In this case the ray inside the prism becomes parallel to base. 

Therefore the angle made by the ray inside the prism with the base of the prism is 0^o