JEE MAIN - Physics (2024 - 27th January Morning Shift - No. 3)
If the refractive index of the material of a prism is $$\cot \left(\frac{A}{2}\right)$$, where $$A$$ is the angle of prism then the angle of minimum deviation will be
$$\pi-2 \mathrm{~A}$$
$$\frac{\pi}{2}-2 \mathrm{~A}$$
$$\pi-\mathrm{A}$$
$$\frac{\pi}{2}-\mathrm{A}$$
คำอธิบาย
$$\begin{aligned}
& \cot \frac{\mathrm{A}}{2}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)}{\sin \frac{\mathrm{A}}{2}} \\
& \Rightarrow \cos \frac{\mathrm{A}}{2}=\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right) \\
& \frac{\mathrm{A}+\delta_{\min }}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2} \\
& \delta_{\min }=\pi-2 \mathrm{~A}
\end{aligned}%$$