# Brewster angle and reflected light

1. May 18, 2013

### notnewton96

1. The problem statement, all variables and given/known data

Light polarised at 45° to the plane of incidence is incident on a smooth dielectric surface at
the Brewster angle. State the polarisation of the reflected beam (relative to the plane of
incidence).

2. Relevant equations

Brewster Angle -
tan θ = n2/n1

3. The attempt at a solution

I simply don't understand the question. If the light is polarized and incident at the Brewster angle shouldn't there be no reflection as in brewster windows? I can't actually seem to find much material that covers already polarized light and the brewster angle. Is the effect on the light the same as un-polarized light? So does it simply become horizontally polarized and therefore 45° + 90°?

Any help would be appreciated :)

2. May 18, 2013

### TSny

Light that is polarized at 45o to the plane of incidence can be treated as a superposition of two components, with one component polarized in the plane of incidence and the other component polarized perpendicular to the plane of incidence. What happens to each of those components?

That is essentially right, if by horizontally polarized you mean polarized perpendicular to the plane of incidence. But I don't understand the 45° + 90°.

3. May 18, 2013

### notnewton96

I thought that it would become polarized perpendicular to its incidence polarization. So what you're saying is that the reflected light is simply polarized 90° relative to the plane of incidence?

4. May 18, 2013

### TSny

Yes, that's right. It doesn't matter whether or not the incident light is polarized or unpolarized. If the angle of incidence is Brewster's angle, then any reflected light will be polarized perpendicularly to the plane of incidence. If the incident light happens to be polarized in the plane of incidence, then no light will be reflected.

5. May 18, 2013

### notnewton96

Ahh ok. That makes sense. Thank you very much for the help :)