Question

A beam of natural light is incident on an air-glass

interface(nti=1.5) at 40 degrees . Compute the degree
ofpolarization of the reflected light.

Answer 1

The degree of polarization of the reflected light
`74.6^(\circ)`

Solution:

As per the question:

Refractive index,
`\mu = 1.5`

Angle of incidence,
`\angle i = 40^(\circ)`

Now,

To calculate the degree of polarization:

We know according to the laws of reflection, angle of incidence and angle of reflection are equal.

`\angle i = \angle r = 40^(\circ)`

Also, from Brewster law:

`\mu = (sin p)/(sin r)`

`\musin r = sin p`

`p = sin^(- 1){1.5* sin 40^(\circ)} = 74.6^(\circ)`