Question

A ray PQ incident on the refracting face BA is refracted in the prism BAC as shown in the figure and emerges from the other refracting face AC as RS such that AQ = AR. If the angle of prism $A={60}^o$ and refractive index of material of prism is $\sqrt{3}$ . Calculate angle $\theta$.

Answer 1

As $AQ=AR$ it means ray $QR$ is parallel to the base $BC$ , hence the prism is in the position of minimum deviation ,

now $\mu =\frac{\sin (A+\theta )/2}{\sin A/2}$

given $\mu =\sqrt{3},A={60}^o$

therefore $\sqrt{3}=\frac{\sin (60+\theta )/2}{\sin 60/2}$

or $\sin (60+\theta )/2=\sqrt{3}\times \sin 30$

or $\sin (60+\theta )/2=\sqrt{3}/2=\sin 60$

or $(60+\theta )=120$

or $\theta ={60}^o$