JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 10)
The refractive index of a prism with apex angle $$A$$ is $$\cot A / 2$$. The angle of minimum deviation is :
$$\delta_m=180^{\circ}-3 \mathrm{~A}$$
$$\delta_m=180^{\circ}-4 A$$
$$\delta_m=180^{\circ}-2 A$$
$$\delta_m=180^{\circ}-A$$
توضیح
$$\begin{aligned}
& \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\
& \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\
& \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\
& \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta m}{2} \\
& \delta_m=\pi-2 A
\end{aligned}$$