Explanation

(c) Here, angle of prism $A = 60{^\circ}$
Refractive index, $\mu = \sqrt{3}$
At the minimum deviation $\delta_{m}$, the refracted ray inside the prism becomes parallel to its base
$\mu = \frac{\sin\left( \frac{A + \delta_{m}}{2} \right)}{\sin\left( \frac{A}{2} \right)}$
$\sqrt{3} = \frac{\sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)}{\sin\left( \frac{60{^\circ}}{2} \right)} \Rightarrow \sqrt{3}\sin{30{^\circ}} = \sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)$
$\frac{\sqrt{3}}{2} = \sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right) \Rightarrow \sin{60{^\circ}} = \sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)$
$60{^\circ} = \frac{60{^\circ} + \delta_{m}}{2} \Rightarrow \delta_{m} = 60{^\circ}$
As $\delta_{m} = 2i - A,$ where $i$ is the angle of incidence
Hence, $i = \theta$
$\therefore\ \theta = \left( \frac{\delta_{m} + A}{2} \right) = \frac{60{^\circ} + 60{^\circ}}{2} = 60{^\circ}$