Question

A ray of light suffers minimum deviation when incident at $60^{\circ }$ prism of refractive index $\sqrt{2}$ The angle of incidence is :

• A. ${\sin }^{-1}(0.8)$
• B. $60^{\circ }$
• C. $45^{\circ }$
• D. $30^{\circ }$

Answer 1

Answer: (C)

For minimum deviation through a prism, the refractive index of material of prism is given by
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Given, $A=60^{\circ },\mu =\sqrt{2}$
$\therefore \sqrt{2}=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{60^{\circ }}{2}\right)}$
or $\sin \left(\frac{A+{\delta }_m}{2}\right)=\sqrt{2}\sin 30^{\circ }$
or $\sin \left(\frac{A+{\delta }_m}{2}\right)$
$=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}$
or $\sin \left(\frac{A+{\delta }_m}{2}\right)=\sin 45^{\circ }$
or $\frac{A+{\delta }_m}{2}=45^{\circ }$
but we know angle of incidence
$i=\frac{A+{\delta }_m}{2}=45^{\circ }$