Explanation

(c) For minimum deviation through a prism, the refractive index of material of prism is given by
$\mu = \frac{\sin\left( \frac{A + \delta_{m}}{2} \right)}{\sin\left( \frac{A}{2} \right)}$
$given,\ A = 60{^\circ},\ \mu = \sqrt{2}$
$\therefore\ \sqrt{2} = \ \frac{\sin\left( \frac{A + \delta_{m}}{2} \right)}{\sin\left( \frac{60{^\circ}}{2} \right)}$
$or\ \sin\left( \frac{A + \delta_{m}}{2} \right) = \ \sqrt{2}\sin{30{^\circ}}$
$or\ \sin\left( \frac{A + \delta m}{2} \right) = \ \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$
$or\ \sin\left( \frac{A + \delta_{m}}{2} \right) = \sin{45{^\circ}}$
$or\ \frac{A + \delta_{m}}{2} = 45{^\circ}$
But we know angle of incidence
$i = \ \frac{A + \delta_{m}}{2} = 45{^\circ}$