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
$i=\frac{A+\delta_{m}}{2}=45^{\circ}$