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Physics Question on Ray optics and optical instruments

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∘−2A180^\circ -2 A

B

90∘−A90^\circ - A

C

180∘+2A180^\circ +2 A

D

180∘−3A180^\circ -3 A

Answer

180∘−2A180^\circ -2 A

Explanation

Solution

n=sin⁡(A+dm2)sin⁡(A2)n=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}
⇒cot⁡(A2)=sin⁡(A+dm2)sin⁡(A2)\Rightarrow \cot \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}
⇒cos⁡(A2)sin⁡(A2)=sin⁡(A+dm2)sin⁡(A2)\Rightarrow \frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}
⇒sin⁡(90−A2)=sin⁡(A+dm2)\Rightarrow \sin \left(90-\frac{A}{2}\right)=\sin \left(\frac{A+d_{m}}{2}\right)
⇒90−A2=A+dm2\Rightarrow 90-\frac{A}{2}=\frac{A+d_{m}}{2}
⇒180−2A=dm\Rightarrow 180-2 A=d_{m}