# Brewsters Law Formula

Brewster’s Law Brewster derived a relation amongst the polarizing angle ip and the refractive index mu which is well known as Brewster’s law. This says that the tangent of the polarizing angle is numerically equivalent to the refractive index of the medium.

A polarizing angle is created when at a certain angle of incidence, the reflected light is completely polarized, and this specific value of the angle of incidence is identified as polarizing angle. The polarizing angle ip hinges on on the refractive index mu of the transparent material.

The relation is articulated by “µ = tan ip”

When unpolarized light is found to be incident on a transparent medium at any polarizing angle then the transmitted and reflected rays are vertical to each other.

$\mu\,&space;=\,&space;tan\,&space;ip$

$since,\,&space;\mu\,&space;=\,&space;\frac{\left&space;(&space;sin\,&space;ip&space;\right&space;)}{\left&space;(&space;sin\,&space;ir&space;\right&space;)}$

$Hence,\,&space;tan\,&space;ip&space;=\,&space;\frac{\left&space;(&space;sin\,&space;ip&space;\right&space;)}{\left&space;(&space;sin\,&space;ir&space;\right&space;)}$

Examples

Some numerical on Brewster’s Law are mentioned below: Solved Samples

Problem 1: Refractive index of a polarizer is found to be 1.9128. Find the polarization angle and angle of refraction?

Refractive index of the polarizer = 1.9128

The Brewster’s law is μ = tan ip

Or, ip = tan−1tan−1 (1.9128)

Or, ip = 62o 24’

Now, Angle of refraction

It is given that ip + ir = 90 degrees

Thus, angle of refraction or ir = 90 – 62o 24’

Angle of refraction = 27.6 o

Problem 2: Compute the Brewster’s angle of light that is traveling from water (n = 1.33) into air ?

$Brewster's\,&space;angle\,&space;=\,&space;tan^{-1\left&space;(&space;\frac{n_{2}}{n_{1}}&space;\right&space;)}$
$Brewster's\,&space;angle\,&space;=\,&space;tan^{-1}\,&space;\left&space;(&space;\frac{1.5}{1.33}&space;\right&space;)$
$Brewster's\,&space;angle\,&space;=\,&space;48.4^{o}$