Question

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

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

Answer 1

Answer: (D)

${180}^o-2A$

Using prism formula,
$\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$ ........ (i)
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 eq. (i)
$\frac{cos\left(\frac{A}{2}\right)}{sin\left(\frac{A}{2}\right)}=\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$
$\Rightarrow sin(\frac{\pi }{2}-\frac{A}{2})=sin(\frac{A}{2}+\frac{{\delta }_m}{2})$
$\Rightarrow {\delta }_m=\pi -2A={180}^o-2A$