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 (in degree)?

Answer 1

Answer: 0.00

For prism,
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$
$\Rightarrow \sqrt{2}=\frac{\sin \left(\frac{60^{\circ }+{\delta }_m}{2}\right)}{\sin \frac{60^{\circ }}{2}}$
$\sqrt{2}\times \frac{1}{2}=\sin \left(30^{\circ }+\frac{{\delta }_m}{2}\right)$
$\frac{1}{\sqrt{2}}=\sin \left(30^{\circ }+\frac{{\delta }_m}{2}\right)$
$\Rightarrow 45^{\circ }=30^{\circ }+\frac{{\delta }_m}{2}$
$\Rightarrow {\delta }_m=30^{\circ }$
Hence, clearly $30^{\circ }$ is the angle for minimum deviation. Thus the given ray passes through prism with minimum deviation.
Hence in this case ray passes parallel to base of prism or angle made with base of prism is $0^{\circ }$.