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

Given: A ray of light undergoes deviation of ${30}^{\circ }$ when incident on an equilateral prism of refractive index $\sqrt{2}$.

To find the angle made by the ray inside the prism with the base of the prism.

Solution:

As per the given criteria,

angle of deviation, $\delta ={30}^{\circ }$

refractive index of the prism, $\mu =\sqrt{2}$

As the prism is equilateral, the refractive angle, $A={60}^{\circ }$

And we know,

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$

Substituting the values, we get

$\sqrt{2}=\frac{\sin \left(\frac{60+{\delta }_m}{2}\right)}{\sin \frac{60}{2}}⟹\sqrt{2}=\frac{\sin \left(\frac{60+{\delta }_m}{2}\right)}{\sin 30}⟹\sqrt{2}\times \frac{1}{2}=\sin \left(\frac{60+{\delta }_m}{2}\right)⟹\frac{60+{\delta }_m}{2}={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)⟹\frac{60+{\delta }_m}{2}=45⟹{\delta }_m=90-60={30}^{\circ }$

Hence the minimum deviation angle is equal to the deviation angle, hence at minimum deviation condition the ray travels parallel to the base inside the prism.

The angle between the ray and base inside the prism is zero.