Question

The angle of minimum deviation for a prism of refractive index $\sqrt 3$ is equal to the angle of the prism. Then the angle of the prism is

• A. $30^{\circ}$
• B. $45^{\circ}$
• C. $60^{\circ}$
• D. $90^{\circ}$

Answer 1

Answer: (C)

Here, $\mu = \sqrt{3},\delta_{m} = A$
where $A$ is the angle of prism
$ \because\mu = \frac{sin\left(\frac{A+\delta_{m}}{2}\right)}{sin \left(\frac{A}{2}\right)}$
$ \sqrt{3} = \frac{sin\left(\frac{A+A}{2}\right)}{sin\left(\frac{A}{2}\right)}$
$\sqrt{3} = \frac{sin\left(A\right)}{sin\left(\frac{A}{2}\right)} $
$ =\frac{2 sin\left(\frac{A}{2}\right) cos \left(\frac{A}{2}\right)}{sin\left(\frac{A}{2}\right)} $
$ = 2 cos \left(\frac{A}{2}\right) $
$cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2} $
$ \frac{A}{2} = cos^{-1}\left(\frac{\sqrt{3}}{2}\right) $
$ \frac{A}{2} = 30^{\circ} $
$ A = 60^{\circ}$