A ray of light undergoes deviation of ${30}^{\circ }$ when incident on as an equilateral prism of refractive index $\sqrt{2}$ . Find the angle made by the ray inside the prism with the base of the prism.

$\mu =\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

here, D=Minimum deviation

 $$\begin{gathered} \sqrt{2}=\frac{\sin \left(\frac{{60}^0+\mathrm{D}}{2}\right)}{\sin \left(\frac{{60}^0}{2}\right)} \\ \Rightarrow \mathrm{D}={30}^0 \end{gathered}$$

$$\begin{gathered} \text{ In this case the ray inside the prism becomes parallel to base.} \\ \text{Therefore the angle made by the ray inside the prism with the} \\ \text{base of the prism is zero. } \end{gathered}$$