Question

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. $\sin ^{-1}\left(\sqrt{\frac{\mu-1}{2}}\right)$
• C. $2 \cos ^{-1}\left(\frac{\mu}{2}\right)$
• D. $\cos ^{-1}\left(\frac{\mu}{2}\right)$

Answer 1

Answer: (C)

$\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Here, $\delta_{m}=A$
$\therefore \mu =\frac{\sin A}{\sin \left(\frac{A}{2}\right)}=\frac{2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=2 \cos \left(\frac{A}{2}\right)$
$A =2 \cos ^{-1}\left(\frac{\mu}{2}\right)$