Q. The refractive index of a prism for a monochromatic wave is $\sqrt{2}$ and its refracting angle is $60{^\circ}.\ $For minimum deviation, the angle of incidence will be
A) $30{^\circ}$
B) $45{^\circ}$
C) $60{^\circ}$
D) $75{^\circ}$
Explanation
(b) $\mu = \frac{\sin\left( \frac{A + \delta_{m}}{2} \right)}{\sin\frac{A}{2}}$
$\sqrt{2} = \frac{\sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)}{\sin\frac{60{^\circ}}{2}}$
$\frac{1}{\sqrt{2}} = \sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)$
Or ${\ sin}{45{^\circ} = \sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)}$
$\delta_{m} = 30{^\circ}$
Or $i = \frac{A + \delta_{m}}{2}$
$= \frac{60 + 30}{2} = \frac{90}{2} = 45{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.