Question

If the refractive index of a glass prism is cot $\left(\frac{A}{2}\right)$ and $A$ is angle of prism, then angle of minimum deviation is

• A. $\left(\frac{\pi}{2}-A\right)$
• B. $\left(\pi - \frac{A}{2}\right)$
• C. $\left(\frac{\pi -A }{2}\right)$
• D. $(\pi - 2A)$

Answer 1

Answer: (D)

The refractive index $(\mu)$ of a prism of angle A, and minimum deviation $\delta_m$ is given by

$ \mu = \frac{\sin\, \left(\frac{A + \delta_m}{2}\right)}{\sin\, A/2}$
Given, $ \mu = \cot \frac{A}{2} $
$\therefore \cot \frac{A}{2} = \frac{\sin\, \frac{(A + \delta_m)}{2}}{\sin\, A/2}$
$ \Rightarrow \frac{\cot\, A/2}{\sin\, A/2} = \frac{\sin\, \frac{(A + \delta_m)}{2}}{\sin\, A/2}$
$ \Rightarrow \cos\frac{A}{2} = \sin \left(\frac{A + \delta_m}{2}\right)$
$ \therefore \sin \left(90^{\circ} -\frac{A }{2}\right) = \sin \left(\frac{A + \delta_m}{2}\right)$
$ \Rightarrow 90^{\circ} -\frac{A }{2} = \frac{A + \delta_m}{2}$
$ \Rightarrow 180^{\circ} - A = A + \delta_m $
$ \Rightarrow \delta_m = 180^{\circ} - 2A = \pi - 2A $