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
$\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin(A/2)}(A=60^{\circ }$for an equilateral prism)
$\therefore \sqrt{2}=\frac{sin\left(\frac{60^{\circ }+{\delta }_m}{2}\right)}{sin\left(\frac{60^{\circ }}{2}\right)}$
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.