Brewster angle and reflected light

AI Thread Summary
When light polarized at 45° to the plane of incidence strikes a smooth dielectric surface at the Brewster angle, the reflected beam is polarized perpendicular to the plane of incidence. The discussion clarifies that regardless of whether the incident light is polarized or unpolarized, reflected light at Brewster's angle will always be polarized perpendicularly. If the incident light is polarized in the plane of incidence, no reflection occurs. The confusion around the 45° + 90° concept is addressed, confirming that reflected light is indeed polarized 90° relative to the plane of incidence. Understanding these principles is essential for applying Brewster's angle in optics.
notnewton96
Messages
10
Reaction score
0

Homework Statement



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).

Homework Equations



Brewster Angle -
tan θ = n2/n1

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 :)
 
Physics news on Phys.org
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?

notNewton96 said:
Is the effect on the light the same as un-polarized light? So does it simply become horizontally polarized and therefore 45° + 90°?

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°.
 
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?
 
notNewton96 said:
So what you're saying is that the reflected light is simply polarized 90° relative to the plane of incidence?

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.
 
Ahh ok. That makes sense. Thank you very much for the help :)
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top