question_answer

If the refractive index of a glass prism is cot  (A/2) and A is angle of prism, then angle of minimum deviation is:

A) \[\left( \frac{\pi }{2}-A \right)\]

B)  \[\left( 2\pi -\frac{A}{2} \right)\]

C)  \[\left( \frac{\pi -A}{2} \right)\]

D)  \[(\pi -2A)\]

Correct Answer: D

Solution :

 Key Idea: \[\sin ({{90}^{o}}-\theta )=cos\theta \] The refractive index \[(\mu )\] of a prism of angle A, and minimum deviation \[{{\delta }_{m}}\] is given by \[\mu -\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin A/2}\] Given, \[\mu =\cot \frac{A}{2}\] \[\therefore \] \[\cot \frac{A}{2}=\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin (A/2)}\] \[\Rightarrow \] \[\frac{\cos A/2}{\sin A/2}=\frac{\sin \frac{(A+{{\delta }_{m}})}{2}}{\sin (A/2)}\] \[\Rightarrow \] \[\cos \frac{A}{2}=\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)\] \[\therefore \] \[\sin \left( {{90}^{o}}-\frac{A}{2} \right)=\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)\] \[\Rightarrow \] \[{{90}^{o}}-\frac{A}{2}=\frac{A+{{\delta }_{m}}}{2}\] \[\Rightarrow \] \[{{180}^{o}}-A=A+{{\delta }_{m}}\] \[\Rightarrow \] \[{{\delta }_{m}}={{180}^{o}}-2A=\pi -2A\]