Question

A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is $\frac{3}{4}$th of the angle of prism. Calculate the speed of light in the prism.

Answer 1

Here, Angle of prism, $A={60}^o$

Angle of incidence, i=$\frac{3}{4}\times 60$

i+e=A+$\delta$

e is emergence angle

where $\delta$ is deviation angle

At minimum deviation $\delta ={\delta }_{min}$

i=e

2i=A+${\delta }_{min}$

$2\times 45=60+{\delta }_{min}$

${\delta }_{min}=30$

By prism law,

$\mu =\frac{\sin \frac{A+{\delta }_{min}}{2}}{\sin \frac{A}{2}}$

$\mu =\frac{\sin \frac{60+30}{2}}{\sin \frac{60}{2}}=\frac{2}{\sqrt{2}}=1.41$

Let v be velocity of light in prism

$v=\frac{c}{\mu }=\frac{3\times {10}^8}{1.41}$=$2\times {10}^8$ m/s