Question

A prism of refractive index $\sqrt{2}$ has refracting angle ${60}^0$. In order that a ray suffers minimum deviation it should be incident at an angle:

• A. ${45}^0$
• B. ${90}^0$
• C. ${30}^0$
• D. none

Answer 1

The correct option is A ${45}^0$

Given: A prism of refractive index $\sqrt{2}$ has refracting angle ${60}^{\circ }$.

To find the angle of incidence in order that a ray suffers minimum deviation

Solution:

As per the given criteria,

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

Angle of the prism, $A={60}^{\circ }$

For minimum angle of deviation we have angle of incidence is equal to angle of emergence, i.e., $i=e$

Hence, $i=\frac{A+{\delta }_m}{2}$, where ${\delta }_m$ is the minimum deviation angle.

We know,

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}⟹\sqrt{2}=\frac{\sin i}{\sin \frac{60}{2}}⟹\sin i=\sqrt{2}\times \sin \left(30\right)⟹\sin i=\sqrt{2}\times \frac{1}{2}$

Multiply and divide by $\sqrt{2}$, we get

$\sin i=\frac{1}{\sqrt{2}}⟹i={45}^{\circ }$

In order that a ray suffers minimum deviation it should be incident at an angle ${45}^{\circ }$