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 _____
$(\sin {48}^o{36}^{\prime }=0.75)$

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

Answer 1

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

Given : $\mu =1.5$ ${\delta }_m=A$
Using $\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$
or $\mu =\frac{\sin A}{\sin \frac{A}{2}}$ $(\because A={\delta }_m)$
Or $\mu =\frac{2\sin \frac{A}{2}\times \cos \frac{A}{2}}{\sin \frac{A}{2}}$
$⟹2\cos \frac{A}{2}=\mu$
Or $\cos \frac{A}{2}=\frac{1.5}{2}=0.75$
We get $\cos \frac{A}{2}=\sin \left({48}^o{36}^{\prime }\right)=\cos ({90}^o-{48}^o{36}^{\prime })=\cos \left({41}^o{24}^{\prime }\right)$
$⟹\frac{A}{2}={41}^o{24}^{\prime }$
Thus $A={82}^o{48}^{\prime }$
Option D is correct.