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 $\mu$:

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

Given:

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

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

Minimum deviation, $\delta_m = 30^{\circ}$

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:

$ \delta_m = 2i - A $

Solving for $i$:

$ \delta_m = 2i - A \implies 30^{\circ} = 2i - 60^{\circ} \implies 2i = 90^{\circ} \implies i = 45^{\circ} $

Thus, the angle of incidence is $45^{\circ}$.