Q127 / 160 · AP EAMCET 2003 · Physics
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$\sin ^{-1} \sqrt{\frac{\mu-1}{2}}$
- Correct answerC$2 \cos ^{-1}\left(\frac{\mu}{2}\right)$
- D$\cos ^{-1}\left(\frac{\mu}{2}\right)$
Solution
Refractive index of material of prism
$\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}$
but $\quad \delta_m=A$
$\mu=\frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}}$
$\mu=\frac{\sin A}{\sin A / 2}$
$\mu=\frac{2 \sin A / 2 \cos A / 2}{\sin A / 2}$
$\cos \frac{A}{2}=\frac{\mu}{2}$
$\frac{A}{2}=\cos ^{-1}\left(\frac{\mu}{2}\right)$
$A=2 \cos ^{-1}\left(\frac{\mu}{2}\right)$
Hence, the correct answer is option: 'C'.