Ray Optics

JEE Main level, single correct, medium

248 of 295

Question 248 of 295

The refractive index of a prism with apex angle AA is cotA2cot\dfrac{A}{2}. The angle of minimum deviation is :

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Answer: (D) δm=180∘−2A{\delta }_{m}=180^{\circ}-2A

As we know, μ=sin⁡(A+δm2)sin⁡A2\mu =\dfrac{\sin \left(\dfrac{A+{\delta }_{m}}{2}\right)}{\sin \dfrac{A}{2}}
⇒cos⁡A2sin⁡A2=sin⁡(A+δm2)sin⁡A2\Rightarrow \dfrac{\cos \dfrac{A}{2}}{\sin \dfrac{A}{2}}=\dfrac{\sin \left(\dfrac{A+{\delta }_{m}}{2}\right)}{\sin \dfrac{A}{2}}
⇒sin⁡(π2−A2)=sin⁡(A+δm2)\Rightarrow \sin \left(\dfrac{\pi }{2}-\dfrac{A}{2}\right)=\sin \left(\dfrac{A+{\delta }_{m}}{2}\right)
⇒π2−A2=A2+δm2\Rightarrow \dfrac{\pi }{2}-\dfrac{A}{2}=\dfrac{A}{2}+\dfrac{{\delta }_{m}}{2}
⇒δm=π−2A=180∘−2A\Rightarrow {\delta }_{m}=\pi -2A=180^{\circ}-2A

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