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Geometrical Optics question

2015 · Q126
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Geometrical Optics question

2015 · Q126

NEETPhysicsGeometrical OpticsMCQ+4 / −1
The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is
  1. A
    90o −-− A
  2. B
    180o + 2A
  3. C
    180o −-− 3A
  4. D
    180o −-− 2A
View written solutionFree

Correct answer: D

As μ\mu μ = sin⁡(A+δ2)sin⁡(A2){{\sin \left( {{{A + \delta } \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}sin(2A​)sin(2A+δ​)​

⇒\Rightarrow⇒ cot⁡A2\cot {A \over 2}cot2A​ = sin⁡(A+δ2)sin⁡(A2){{\sin \left( {{{A + \delta } \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}sin(2A​)sin(2A+δ​)​

⇒\Rightarrow⇒ cos⁡A2sin⁡A2=sin⁡(A+δ2)sin⁡(A2){{\cos {A \over 2}} \over {\sin {A \over 2}}} = {{\sin \left( {{{A + \delta } \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}sin2A​cos2A​​=sin(2A​)sin(2A+δ​)​

⇒\Rightarrow⇒ sin(π2−A2)=sin⁡(A+δ2){sin\left( {{\pi \over 2} - {A \over 2}} \right) = \sin \left( {{{A + \delta } \over 2}} \right)}sin(2π​−2A​)=sin(2A+δ​)

⇒\Rightarrow⇒ π2−A2{{\pi \over 2} - {A \over 2}}2π​−2A​ = A2+δ2{{A \over 2} + {\delta \over 2}}2A​+2δ​

⇒\Rightarrow⇒ δ\delta δ = 180o - 2A

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