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JEE MAINS 2024
31.01.24 S1
Question
The refractive index of a prism with apex angle A is cot⁡A/2. The angle of minimum deviation is :
Select the correct option:
A
$\delta_m=180^{\circ}-A$
B
$\delta_{\mathrm{m}}=180^{\circ}-3 \mathrm{~A}$
C
$\delta_{\mathrm{m}}=180^{\circ}-4 \mathrm{~A}$
D
$\delta_m=180^{\circ}-2 \mathrm{~A}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. $$ \begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\ & \frac{\pi}{2}-\frac{\mathrm{A}}{2}=\frac{\mathrm{A}}{2}+\frac{\delta \mathrm{m}}{2} \\ & \delta_{\mathrm{m}}=\pi-2 \mathrm{~A} \end{aligned} $$
Question Tags
JEE Main
Physics
Easy
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