Question

Angle of minimum deviation for a prism of refractive index $1.5$ is equal to the angle of prism of given prism. Then, the angle of prism is ____ $(\sin {48}^{\circ }{36}^{\prime }=0.75)$.

• A. ${80}^{\circ }$
• B. ${41}^{\circ }{24}^{\prime }$
• C. ${60}^{\circ }$
• D. ${82}^{\circ }{48}^{\prime }$

Answer 1

The correct option is D ${82}^{\circ }{48}^{\prime }$

Given, $\mu =1.5$
As refractive index of a prism, $\mu =\frac{\sin \frac{A+\delta }{2}}{\sin \frac{A}{2}}$
where, $\delta =$ angle of minimum deviation
and $A=$ refracting angle of prism.
As per question $\delta =A$, then
$\mu =\frac{\sin \frac{A+A}{2}}{\sin A/2}$
$=\frac{\sin A}{\sin A/2}=\frac{2\sin A/2\cdot \cos A/2}{\sin A/2}$
$=2\cos A/2$
$\Rightarrow \frac{15}{2}=\cos A/2=\sin ({90}^{\circ }-A/2)$
$\Rightarrow {90}^{\circ }-A/2={\sin }^{-1}\left(0.75\right)={48}^{\circ }{36}^{\prime }$
$\Rightarrow {90}^{\circ }-{48}^{\circ }{36}^{\prime }=\frac{A}{2}$
$\Rightarrow {41}^{\circ }{24}^{\prime }=A/2$
$\therefore A={82}^{\circ }{48}^{\prime }$.