Question

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

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

Answer 1

The correct option is A ${45}^{\circ }$

Refractive index $\mu =\sqrt{2}$

Angle of prism $={60}^{\circ }$

For minimum deviation $i=e$

So,$A+{\delta }_m=t+v$ or $i=\frac{A+{\delta }_m}{2}$ for minimum ,i=r

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

$\sin i=\mu \sin \left(\frac{A}{2}\right)$

$\sin i=\mu \sin \left(\frac{60}{2}\right)$

$=\sqrt{2}\sin 30=\frac{1}{\sqrt{2}}$

$i={45}^{\circ }$