Thank you for reporting, we will resolve it shortly
Q.
A ray of light passes through an equilateral prism such that the angle of incidence is equal to the angle of emergence and the latter is equal to $ \frac{3}{4} $ the angle of prism. The angle of deviation is
It is observed if $\angle i=\angle e$ deviation produced is minimum.
and $ i=\frac{A+\delta_{m}}{2}$
Here $A=60^{\circ}$
and $\angle i=\angle e=\frac{3}{4} \angle A$
$\therefore \delta_{m}=2 \times \frac{3}{4} \times 60^{\circ}-60^{\circ}=30^{\circ}$