Discussion Forum

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 

Answer:

Explanation:

Using prism formula

$\mu=\frac{\sin\left(\frac{A+\delta_{m}}{2}\right)}{\sin(\frac{A}{2})}$ .....(i)

 where , A= angle of prism

 $\delta_{m}$= angle of minimum deviation

Given, $\mu=\cot(\frac{A}{2})=\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

   So, from eq.(i)

$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}=\frac{\sin\left(\frac{A+\delta_{m}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

 $\Rightarrow \sin (\frac{\pi}{2}-\frac{A}{2})=\sin\left(\frac{A}{2}+\frac{\delta_{m}}{2}\right)$

$\Rightarrow \delta_{m}=\pi-2A=180^{0}-2A$