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

Answer 1

Answer: (D)

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