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Question: The refracting angle of prism is *A* and refractive index of material of prism is \(\cot \frac { A }...

The refracting angle of prism is A and refractive index of material of prism is cot⁡A2\cot \frac { A } { 2 }. The angle of minimum deviation is

A

180∘−3A180 ^ { \circ } - 3 A

B

180∘+2A180 ^ { \circ } + 2 A

C

90∘−A90 ^ { \circ } - A

D

180∘−2A180 ^ { \circ } - 2 A

Answer

180∘−2A180 ^ { \circ } - 2 A

Explanation

Solution

By using μ=sin⁡A+δm2sin⁡A2\mu = \frac { \sin \frac { A + \delta _ { m } } { 2 } } { \sin \frac { A } { 2 } }

⇒\Rightarrow cot⁡A2=sin⁡A+δm2sin⁡A2⇒cos⁡A2sin⁡A2=sin⁡A+δm2sin⁡A2\cot \frac { A } { 2 } = \frac { \sin \frac { A + \delta _ { m } } { 2 } } { \sin \frac { A } { 2 } } \Rightarrow \frac { \cos \frac { A } { 2 } } { \sin \frac { A } { 2 } } = \frac { \sin \frac { A + \delta _ { m } } { 2 } } { \sin \frac { A } { 2 } }

⇒δm=180−2A\Rightarrow \delta _ { m } = 180 - 2 A