Question

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Answer 1

Answer:

The refractive index of the prism is 1.53

Minimum angle of deviation when the prism is immersed in water is 10.9°

Explanation:

Given

Refracting angle of the prism

tex]A=60^\circ[/tex]

Minimum angle of deviation

$\delta_m=40^\circ$

The refractive index of the material of the prism is given by

$\boxed{\mu=\frac{\sin{(A+\delta_m)/2}}{\sin(A/2)}}$

$\implies \mu=\frac{\sin(60^\circ+40^\circ)/2}{(\sin 60^\circ /2)}$

$\implies \mu=\frac{\sin 50^\circ}{\sin 30^\circ}$

$\implies \mu=\frac{0.766}{0.5}$

$\implies \mu=1.53$

When the prism is placed in water, the refractive index of the prism w.r.t water

$\mu_{g/w}=\frac{\mu_g}{\mu_w}$

$\implies \mu_{g/w}=\frac{1.53}{1.33}$

$\implies \mu_{g/w}=1.16$

If minimum angle of deviation is $\delta_m$ then

$1.16=\frac{\sin(60^\circ+\delta_m)/2}{(\sin 60^\circ /2)}$

$\implies 1.16=2\sin(60^\circ+\delta_m)/2$

$\implies \sin(60^\circ+\delta_m)/2=0.58$

$\implies (60^\circ+\delta_m)/2=\sin^{-1}0.58$

$\implies (60^\circ+\delta_m)/2=35.45^\circ$

$\implies \delta_m=70.90^\circ-60^\circ$

$\implies \delta_m=10.90^\circ$

Hope this helps.