question_answer

A refracting angle of a prism is A and the \[A\] Refractive index of the prism is cot\[\left( \frac{A}{2} \right)\]. Then, Angle of minimum deviation is:

A)  \[{{180}^{\circ }}-2A\]  

B)  \[{{90}^{\circ }}-A\]

C)  \[{{180}^{\circ }}+2A\]                 

D)  \[{{180}^{\circ }}-3A\]

Correct Answer: A

Solution :

In the prism ABC, \[\delta \] is the angle of minimum deviation, A is angle of prism. The refractive index of the material of the prism is given by \[\mu =\frac{\sin \,\left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \,\frac{A}{2}}\] Given, \[\mu =\cot \frac{A}{2},\]                                 \[\cot \frac{A}{2}=\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[\Rightarrow \]               \[\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] \[\Rightarrow \]               \[\cos \frac{A}{2}=\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)\] Using     \[\sin \left( \frac{\pi }{2}-\theta  \right)=\cos \,\,\theta ,\] we have \[\Rightarrow \]               \[\sin \left( \frac{\pi }{2}-\frac{A}{2} \right)=\sin \frac{A+{{\delta }_{m}}}{2}\] \[\Rightarrow \]               \[{{\delta }_{m}}=\pi -2A={{180}^{\circ }}-2A\] Note: For a prism there is one and only one angle of minimum deviation.