Ray Optics

JEE Main level, single correct, medium

254 of 295

Question 254 of 295

The refracting angle of a prism is AA and refractive index of the material of the prism is cot⁡(A2)\cot \left(\dfrac{A}{2}\right). Then the angle of minimum deviation will be

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Answer: (A) 180∘−2A180^{\circ}-2A

As we know, angle of minimum deviation is given by, ⇒μ=sin⁡(δm+A)2sin⁡A2\Rightarrow \mu =\dfrac{\sin \dfrac{\left({\delta }_{m}+A\right)}{2}}{\sin \dfrac{A}{2}}
⇒cot⁡(A2)=sin⁡(δm+A)2sin⁡A2\Rightarrow \cot \left(\dfrac{A}{2}\right)=\dfrac{\sin \dfrac{\left({\delta }_{m}+A\right)}{2}}{\sin \dfrac{A}{2}}
⇒cos⁡(A2)sin⁡(A2)=sin⁡(δm+A)2sin⁡A2\Rightarrow \dfrac{\cos \left(\dfrac{A}{2}\right)}{\sin \left(\dfrac{A}{2}\right)}=\dfrac{\sin \dfrac{\left({\delta }_{m}+A\right)}{2}}{\sin \dfrac{A}{2}}
⇒sin⁡(90∘−A2)=sin⁡(δm+A2)\Rightarrow \sin \left(90^{\circ}-\dfrac{A}{2}\right)=\sin \left(\dfrac{{\delta }_{m}+A}{2}\right)
Therefore, 90∘−A2=δm+A290^{\circ}-\dfrac{A}{2}=\dfrac{{\delta }_{m}+A}{2} ⇒δm=180∘−2A\Rightarrow {\delta }_{m}=180^{\circ}-2A

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