Question:
A prism of refractive index $\mu$ and angle of prism A is placed in the position of minimum angle of deviation. If minimum angle of deviation is also $\mathrm{A}$, then in terms of refractive index
Correct Option: 1
Solution:
$\mu=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}$
$\mu=\frac{\sin \left(\frac{\mathrm{A}+\mathrm{A}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}$
$\mu=\frac{\sin \mathrm{A}}{\sin \frac{\mathrm{A}}{2}}=2 \cos \frac{\mathrm{A}}{2}$
$A=2 \cos ^{-1}\left(\frac{\mu}{2}\right)$