Ray Optics Question 10
Question 10 - 2024 (31 Jan Shift 1)
The refractive index of a prism with apex angle $\mathrm{A}$ is $\cot \mathrm{A} / 2$. The angle of minimum deviation is :
(1) $\delta_{\mathrm{m}}=180^{\circ}-\mathrm{A}$
(2) $\delta_{\mathrm{m}}=180^{\circ}-3 \mathrm{~A}$
(3) $\delta_{\mathrm{m}}=180^{\circ}-4 \mathrm{~A}$
(4) $\delta_{m}=180^{\circ}-2 \mathrm{~A}$
Show Answer
Answer: (4)
Solution:
$\mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}}$
$\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}}$
$\sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_{m}}{2}\right)$
$\frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta m}{2}$
$\delta_{m}=\pi-2 A$
