Ray Optics

JEE Main level, single correct, medium

249 of 295

Question 249 of 295

If the refractive index of the material of a prism is cot⁡(A2)\cot \left(\dfrac{A}{2}\right), where AA is the angle of prism then the angle of minimum deviation will be

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Answer: (A) π−2A\pi -2A

The formula to calculate the refractive index (μ)\left(\mu \right) of the material of the prism is given by
μ=sin⁡(A+δm2)sin⁡(A2)...(1)\mu =\dfrac{\sin \left(\dfrac{A+{\delta }_{m}}{2}\right)}{\sin \left(\dfrac{A}{2}\right)}...\left(1\right)
From equation (1), it follows that
cot⁡A2=sin⁡(A+δm2)sin⁡A2\cot \dfrac{A}{2}=\dfrac{\sin \left(\dfrac{A+{\delta }_{m}}{2}\right)}{\sin \dfrac{A}{2}}
⇒cos⁡A2=sin⁡(A+δm2)⇒sin⁡(π2−A2)=sin⁡(A+δm2)\Rightarrow \cos \dfrac{A}{2}=\sin \left(\dfrac{A+{\delta }_{m}}{2}\right) \Rightarrow \sin \left(\dfrac{\pi }{2}-\dfrac{A}{2}\right)=\sin \left(\dfrac{A+{\delta }_{m}}{2}\right)
⇒A+δm2=π2−A2\Rightarrow \dfrac{A+{\delta }_{m}}{2}=\dfrac{\pi }{2}-\dfrac{A}{2}
⇒δm=π−2A\Rightarrow {\delta }_{m}=\pi -2A

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