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)
$\frac{cos \left(\frac{A}{2}\right)}{sin ⁡ \left(\frac{A}{2}\right)}=\frac{sin ⁡ \left(\frac{A + \left(\delta\right)_{m}}{2}\right)}{sin ⁡ \left(\frac{A}{2}\right)}$
$\Rightarrow \, sin \left(\frac{\pi }{2} - \frac{A}{2}\right)=sin ⁡ \left(\frac{A}{2} + \frac{\left(\delta\right)_{m}}{2}\right)$
$\Rightarrow \, \delta_{m}=\pi -2A$
$\Rightarrow \, \, \delta_{m}=180^\circ -2A$