Question

The refractive index for a prism is given as $\mu =cot\frac{A}{2}$ . Then, the angle of minimum deviation in terms of angle of the prism is

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

Answer 1

Answer: (D)

The refractive index of a prism is given by
$\mu =\frac{sin\left(\frac{A+5_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$
Where, A = angle of prism
${\delta }_m=$ angle of minimum deviation
Given, $\mu =cot\left(\frac{A}{2}\right)=\frac{cos\left(\frac{A}{2}\right)}{sin\left(\frac{A}{2}\right)}$
So, from equation (i)
$\frac{cos\left(\frac{A}{2}\right)}{sin\left(\frac{A}{2}\right)}=\frac{sin\left(\frac{A+{\left(\delta \right)}_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$
$\Rightarrow sin\left(\frac{\pi }{2}-\frac{A}{2}\right)=sin\left(\frac{A}{2}+\frac{{\left(\delta \right)}_m}{2}\right)$
$\Rightarrow {\delta }_m=\pi -2A$
$\Rightarrow {\delta }_m=180^{\circ }-2A$