Q. If the refractive index of a material of equilateral prism is $\sqrt{3}$, then angle of minimum deviation of the prism is
A) $30{^\circ}$
B) $45{^\circ}$
C) $60{^\circ}$
D) $75{^\circ}$
Explanation
(c) $\mu = \frac{\sin\left( \frac{A + \delta_{m}}{2} \right)}{\sin\frac{A}{2}} \Rightarrow \sqrt{3} = \frac{\sin\left( \frac{60{^\circ} + \delta_{m}}{2} \right)}{\sin\frac{60{^\circ}}{2}}$
$\Rightarrow \frac{\sqrt{3}}{2} = \sin\left( 30{^\circ} + \frac{\delta_{m}}{2} \right) \Rightarrow \delta_{m} = 60{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.