Question

Derive an expression for brewster law ?

Answer 1

See the diagram.

 

   (David) Brewster’s law is related to polarization of light rays. Brewster angle θ_B is the angle of incidence for a light ray polarized in a certain way p and of given wavelength, for which there is no reflection of light from the boundary.  All the light is perfectly transmitted through the transparent second medium. Brewster angle is also called the polarization angle.

 

    Usually the refracted ray is partially polarized in the electric field. The reflected ray is polarized too. If an Unpolarised light ray is incident on the interface boundary (planar) of medium 1 and medium 2 at this angle $\theta</span>_B$, then the reflected light is perfectly polarized.

 

   We prove Brewster’s law by simple qualitative method. Assume light rays are produced by oscillating electrical dipoles. They are absorbed by molecules (oscillating dipoles) at the surface of the boundary. They are then reflected (re-radiated) and retransmitted to pass through by these dipoles.  Light rays are polarized in a direction perpendicular to the direction of propagation.  Let us say incident rays are p polarized.

 

    We know that the transmitted light rays are polarized in the same direction as the incident light.  Dipoles do not radiate any light rays in the direction of the dipole moment. If the angle i and r are such that the angle BOC is 90 deg, then refracted light is in the direction perpendicular to transmitted ray. As transmitted ray is p polarized.  The direction of orientation of dipoles (dipole moment) will be in the direction expected reflected ray OB. Since dipoles are also oriented along OB, there will not be any radiation in that direction now.

 

Finally, the condition is:  i + r = 90°,  or   $\theta_B$ + r = 90°.

According to Snell’s law of refraction,  Sin i / Sin r = μ2 / μ1.

So  $Sin \theta_</span>B / Cos \theta_B = Tan \theta_B$ = μ2 / μ1.

      $\theta_B$ = Tan ⁻¹ ( μ2 / μ1)

 

The reflected rays of unpolarised light incident at $\theta_B$ are perfectly polarised in plane perpendicular to the incidence plane (and transmitted plane). There is no reflection for p polarised light incident at $\theta_B$.

 

Uses: 

   Polarised sunglasses and Brewster windows are the applications of Brewster’s law. 

Answer 2

above answer given is correct