Explanation

(b) Given, refractive index of prism $\mu = 1.732$
Let the angle of prism is$\ A$.
The angle of minimum deviation = The angel of prism
$\delta_{m} = A$
The refractive index of prism
$\mu = \frac{\sin\left\lbrack \frac{A\ + \ \delta_{m}}{2} \right\rbrack}{\sin\frac{A}{2}}$
$Or\ \ \mu = \frac{\sin\left\lbrack \frac{A\ + \ A}{2} \right\rbrack}{\sin\frac{A}{2}}$
$Or\ \ 1.732 = \frac{\sin A}{\sin\frac{A}{2}}$
$Or\ \sqrt{3} = \ \frac{{2\ sin}{\frac{A}{2}.\cos{\ \frac{A}{2}}}}{\sin\frac{A}{2}}$
$Or\ \ A = 60{^\circ}$