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
$\because \mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$
$\sqrt{3}=\frac{sin\left(\frac{A+A}{2}\right)}{sin\left(\frac{A}{2}\right)}$
$\sqrt{3}=\frac{sin\left(A\right)}{sin\left(\frac{A}{2}\right)}$
$=\frac{2sin\left(\frac{A}{2}\right)cos\left(\frac{A}{2}\right)}{sin\left(\frac{A}{2}\right)}$
$=2cos\left(\frac{A}{2}\right)$
$cos\left(\frac{A}{2}\right)=\frac{\sqrt{3}}{2}$
$\frac{A}{2}=co{s}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$\frac{A}{2}=30^{\circ }$
$A=60^{\circ }$