Question

For a prism, its refractive index is $\cot \frac{A}{2}$. Then the minimum angle of deviation is-

• A. ${180}^{\circ }-A$
• B. ${180}^{\circ }-2A$
• C. ${90}^{\circ }-A$
• D. $\frac{A}{2}$

Answer 1

The correct option is B ${180}^{\circ }-2A$

Given, $\mu =\cot \frac{A}{2}$

The refractive index of prism is given by,

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$

$\cot \frac{A}{2}=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$

$\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$

$\sin (90-\frac{A}{2})=\sin \left(\frac{A+{\delta }_m}{2}\right)$

$\Rightarrow \frac{\pi }{2}-\frac{A}{2}=\frac{A}{2}+\frac{{\delta }_m}{2}$

$\Rightarrow \frac{\pi }{2}-\frac{2A}{2}=\frac{{\delta }_m}{2}$

$\Rightarrow {\delta }_m=\pi -2A={180}^{\circ }-2A$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.