Question

Trace the path of a ray of light passing through a glass prism (ABC) as shown in the figure. If the refractive index of glass is `sqrt3`, find out the value of the angle of emergence from the prism.

Answer 1

Refractive index of glass `n_g =sqrt3`

Since i = 0

At the interface AC, we have according to shell’s law

`sin i/sinr = n_g/n_a`

But sin i = sin 0 = 0

Thus sin `r = (n_a sin i)/n_g  =0`

Hence r = 0

This ray pass unrefracted at AC interface and reaches AB interface. Here we can see angle of incidence becomes 30°.

Thus, applying Snell’s law

`(sin30°)/(sin e) = n_a/n_g = 1/sqrt3`

`sin e = sqrt3 xx sin 30° = sqrt3/2`

Thus e = 60°

Hence, angle of emergence is 60°.