Q. A prism of refractive index $\mu$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is $A$, then the value of $A$ in terms of $\mu$ is
A) $\sin^{- 1}\left( \frac{\mu}{2} \right)$
B) $\sin^{- 1}\sqrt{\frac{\mu - 1}{2}}$
C) $2\cos^{- 1}\left( \frac{\mu}{2} \right)$
D) $\cos^{- 1}\left( \frac{\mu}{2} \right)$
Explanation
(c) Given $\delta_{m} = A,$ as $\mu = \frac{\sin\left( \frac{A + \delta_{m}}{2} \right)}{\sin\left( \frac{A}{2} \right)}$
$\Rightarrow \mu = \frac{\sin\left( \frac{A + A}{2} \right)}{\sin\left( \frac{A}{2} \right)} = 2\cos\frac{A}{2} \Rightarrow A = 2\cos^{- 1}\left( \frac{\mu}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.