A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to $60^{\circ}$. The refractive index of the prism material is $\sqrt{2}$. The angle of incidence (in degrees) is__________

<p>To find the angle of incidence when a ray of light experiences minimum deviation in a prism, we use the formula for the refractive index $\mu$:</p> <p>$ \mu = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} $</p> <p>Given:</p> <p><p>The angle of the prism, $A = 60^{\circ}$</p></p> <p><p>The refractive index of the prism material, $\mu = \sqrt{2}$</p></p> <p><p>Minimum deviation, $\delta_m = 30^{\circ}$</p></p> <p>At minimum deviation, the angle of incidence $i$ equals the angle of emergence $e$. Therefore, the relation between the angle of minimum deviation and the angle of the prism is represented as:</p> <p>$ \delta_m = 2i - A $</p> <p>Solving for $i$:</p> <p>$ \delta_m = 2i - A \implies 30^{\circ} = 2i - 60^{\circ} \implies 2i = 90^{\circ} \implies i = 45^{\circ} $</p> <p>Thus, the angle of incidence is $45^{\circ}$.</p>