Q. Refractive index of the material of a prism is 1.5. If $\delta_{m} = A,$ what will be a value of angle of the given prism? $\left( where\delta_{m} = minimum\ deviation;A = angle\ of\ prism \right)$
A) $82.8{^\circ}$
B) $41.4{^\circ}$
C) $48.6{^\circ}$
D) $90{^\circ}$
Explanation
(a) $\mu = 1.5$
$\delta_{m} = A$
We know that
$\mu = \frac{\sin\left( \frac{A\ + \ \delta_{m}}{2} \right)}{{sin\ }\frac{A}{2}}$
$1.5 = \ \frac{\sin\left( \frac{A\ + \ A}{2} \right)}{\sin\frac{A}{2}} = \frac{\sin A}{\sin\frac{A}{2}}$
$1.5 = \ \frac{{2\ sin}{\frac{A}{2}\ .\cos{\ \frac{A}{2}}}}{\sin\frac{A}{2}}$
$1.5 = 2\cos\frac{A}{2}$
$\cos{\frac{A}{2} = \frac{1.5}{2} = 0.75}$
$cos41.4 = 0.75$
$\frac{A}{2} = 41.4$
$A = 82.8$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.