The angle of minimum deviation of an equilateral prism is equal to it…

The correct answer is Option (4) → 1.732

Reasoning:

For a prism at minimum deviation:

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

Given: equilateral prism $⇒A = 60^\circ$

Also given: $\delta_m = A = 60^\circ$

Substitute:

$\mu = \frac{\sin \left( \frac{60^\circ + 60^\circ}{2} \right)}{\sin \left( \frac{60^\circ}{2} \right)} = \frac{\sin 60^\circ}{\sin 30^\circ}$

$= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} = 1.732$