Light is incident from air onto the surface of a liquid.

Question

Light is incident from air onto the surface of a liquid. The angle of incidence is 53.0 degrees, and the angle of refraction is 34.0 degrees. At what angle of incidence would the reflected light be 100% polarized?

I'm not exactly sure as to what this question is asking. When light becomes polarized it reflects of the surface at the same angle as the angle of incidence or Brewster's angle according to my notes. I tried using Brewster's law but I got 53.123 degrees and the answer is 55 degrees according to my professor.

Please help me! I really don't understand.

Answer 1

Hi Olivia!

 

The key here is that the light needs to be reflected at Brewster's angle, which occurs when the angle between the reflected ray and the refracted ray is 90^o (this can be demonstrated using the Fresnel equations mathematically, or by a carefully drawn diagram, but we usually take it as a given at this level).  With some geometry, it can be shown that:

 

θ_B = tan^-1(n₂/n₁)

 

where

 

θ_B = Brewster's angle

n₁ = incident medium's index of refraction

n₂ = refracted medium's index of refraction

 

We are told the incident medium is air, which is usually taken to have n = 1 (actually, about n = 1.003).  So what we need to solve for Brewster's angle is a knowledge of n₂, which is what he information about the refraction for a given incidence angle tells us.  From that, via Snell's law, we know:

 

n₁sin(θ_i) = n₂sin(θ_r) --> (1)sin(53^o) = n₂sin(34^o)

 

Use this to solve for n₂, then plug into the Brewster's angle expression.  Doing so, I obtained your teacher's answer.  If you need any more help with this, or have to derive the Brewster's angle expression in terms of the indices of refraction, just let me know.