A prism of refractive index $\sqrt{2}$ has refracting angle $60^0$. In order that a ray suffers minimum deviation it should be incident at an angle:

Given: A prism of refractive index $\sqrt{2}$ has refracting angle $60^{\circ }$. 

To find the angle of incidence in order that a ray suffers minimum deviation 

Solution:

As per the given criteria,

refractive index of the prism, $\mu =\sqrt{2}$

Angle of the prism, $A=60^{\circ }$

For minimum angle of deviation we have angle of incidence is equal to angle of emergence, i.e., $i=e$

Hence, $i=\frac{A+{\delta }_m}{2}$, where ${\delta }_m$ is the minimum deviation angle.

We know,

$$\begin{gathered} \mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}} \\ ⟹\sqrt{2}=\frac{\sin i}{\sin \frac{60}{2}} \\ ⟹\sin i=\sqrt{2}\times \sin (30) \\ ⟹\sin i=\sqrt{2}\times \frac{1}{2} \end{gathered}$$

Multiply and divide by $\sqrt{2}$, we get

$$\begin{gathered} \sin i=\frac{1}{\sqrt{2}} \\ ⟹i=45^{\circ } \end{gathered}$$

In order that a ray suffers minimum deviation it should be incident at an angle $45^{\circ }$