Question

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 $\cos 41^{\circ }-24^{\prime }-36^{\prime \prime }=0.75$ )

• A. $82^{\circ }-49^{\prime }12^{\prime \prime }$
• B. $72^{\circ }-48^{\prime }-30^{\prime \prime }$
• C. $41^{\circ }-24^{\prime }-36^{\prime \prime }$
• D. $31^{\circ }-49^{\prime }-30^{\prime \prime }$

Answer 1

Answer: (A)

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, $\sin 2\theta =2\sin \theta \cos \theta$
$\therefore 1.5=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}$
$\Rightarrow \cos \frac{A}{2}=\frac{1.5}{2}=0.75$
$\Rightarrow \frac{A}{2}=41^{\circ }-24^{\prime }-36^{\prime \prime }$
$\Rightarrow A=82^{\circ }-49^{\prime }-12^{\prime \prime }$