Question

A refracting angle of a prism is 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. ${180}^{\circ }-3A$

Answer 1

Answer: (A)

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.