Question

A ray of light undergoes deviation of $30^\circ$ when incident on an equilateral prism of refractive index $\sqrt{2}$ The angle made by the ray inside the prism with the base of the prism is.......

Answer 1

Let $\delta_m$be the angle of minimum deviation. Then
$\mu=\frac{sin\Big(\frac{A+\delta_m}{2}\Big)}{sin(A/2)} (A = 60^\circ$for an equilateral prism)
$\therefore \sqrt{2}=\frac{sin\Big(\frac{60^\circ+\delta_m}{2}\Big)}{sin\Big(\frac{60^\circ}{2}\Big)}$
Solving this we get $\delta_m=30^\circ$
The given deviation is also $30^\circ (i.e. \delta_m)$
Under minimum deviation, the ray inside the prism is parallel
to base for an equilateral prism.