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

Solution :

For prism at minimum deviation :

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

Given :

$$A = 60^\circ$$

$$\mu = \sqrt2$$

Therefore,

$$\sqrt2 = \frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{\sin30^\circ}$$

$$\sqrt2 = \frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{1/2}$$

$$\sin\left(\frac{60^\circ+\delta_m}{2}\right) = \frac{1}{\sqrt2}$$

$$= \sin45^\circ$$

Therefore,

$$\frac{60^\circ+\delta_m}{2} = 45^\circ$$

$$60^\circ+\delta_m = 90^\circ$$

$$\delta_m = 30^\circ$$

At minimum deviation :

$$i = e$$

and

$$\delta_m = i+e-A$$

$$= 2i-A$$

Therefore,

$$30^\circ = 2i-60^\circ$$

$$2i = 90^\circ$$

$$i = 45^\circ$$

Final Answer :

$$45^\circ$$