Question

A prism is made up of material of refractive index $\sqrt{3}$. The angle of prism is $A$. If the angle of minimum deviation is equal to the angle of the prism, then the value of $A$ is

• A. $30^{\circ }$
• B. $45^{\circ }$
• C. $60^{\circ }$
• D. $75^{\circ }$

Answer 1

Answer: (C)

Given $M=\sqrt{3}$ and ${\delta }_m=A$
Now, $\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin A}{\sin \frac{A}{2}}$
$=\frac{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$=2\cos \left(\frac{A}{2}\right)$
Now. $2\cos \left(\frac{4}{2}\right)=\mu =\sqrt{3}$
or $\cos \left(\frac{A}{2}\right)\Rightarrow \frac{\sqrt{3}}{2}$
$=\frac{A}{2}=30^{\circ }$
or $A=60^{\circ }$