Verified Solution • 2025–26 Session moderate

QJEEM24Ph73c6: JEE Main 2024 Physics - Class 12 JEEM24Ph73c6 Solution: The refractive index of a prism with apex angle is . The angle…

The refractive index of a prism with apex angle AA is cot⁡A/2\cot A / 2 . The angle of minimum deviation is :

Multiple Choice Options

  • A δm=180∘−3 A\delta_m=180^{\circ}-3 \mathrm{~A}
  • B δm=180∘−4A\delta_m=180^{\circ}-4 A
  • C δm=180∘−2A\delta_m=180^{\circ}-2 A
  • D δm=180∘−A\delta_m=180^{\circ}-A

Step-by-Step Verified Solution

amp;μ=sin⁡(A+δm2)sin⁡A2amp;cos⁡A2sin⁡A2=sin⁡(A+δm2)sin⁡A2amp;sin⁡(π2−A2)=sin⁡(A+δm2)amp;π2−A2=A2+δm2amp;δm=π−2A\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}

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