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                 \[\mu =\frac{\left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin A/2}\] Given,        \[\mu =\cot \frac{A}{2}\] \[\therefore \]                      \[\cot \frac{A}{2}=\frac{\sin \frac{(A+{{\delta }_{m}})}{2}}{\sin (A/2)}\] \[\Rightarrow \]        \[\frac{\cos A/2}{\sin A/2}=\frac{\sin \frac{(A+{{\delta }_{m}})}{2}}{\sin A/2}\] \[\Rightarrow \]                   \[\cot \frac{A}{2}=\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)\] \[\therefore \] \[\sin \left( {{90}^{o}}-\frac{A}{2} \right)=\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)\] \[\Rightarrow \]               \[{{90}^{o}}-\frac{A}{2}=\frac{A+{{\delta }_{m}}}{2}\] \[\Rightarrow \]               \[{{180}^{o}}-A=A+{{\delta }_{m}}\] \[\Rightarrow \]                          \[{{\delta }_{m}}={{180}^{o}}-2A=\pi -2A\]