Q. Angle of minimum deviation for a prism of refractive index $1.5$ is equal to the angle of the prism. The angle of the prism is (given $\cos 41^{\circ}-24'-36''=0.75$ )
J & K CETJ & K CET 2003
Solution:
Let A be the angle of prism, and $ \delta $ the angle of minimum deviation, then refractive index of the medium of prism is given by
$\mu=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Given, $\delta=A, \,\,\, \mu=1.5 $
$\therefore 1.5=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Also, $\sin 2 \theta=2 \sin \theta \cos \theta$
$\therefore 1.5=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} $
$\Rightarrow \cos \frac{A}{2}=\frac{1.5}{2}=0.75$
$\Rightarrow \frac{A}{2}=41^{\circ}-24'-36''$
$\Rightarrow A=82^{\circ}-49'-12''$
