The refractive index of a prism with apex angle $A$ is $\cot A / 2$. The angle of minimum deviation is :
Solution
<p>$$\begin{aligned}
& \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\
& \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\
& \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\
& \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta m}{2} \\
& \delta_m=\pi-2 A
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Refraction and Snell's Law
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