Polarization of Light at Boundary: Exploring the Brewster Angle and Absorption

[arish]

when light incident at the boundary of another medium with brewester angle why does only parallel component absorb and perpendicular component is reflected. WHY?
we say that whenever incident light frequency matches with that of medium(with motion of atoms ) ,light is absorbed.so why not the frequency of perpendicular component of incident light match with that of medium?

 

[tyco05]

I'm not exactly sure what you are trying to say here, but this might help:


The optical properties (or any others for that matter) of a material are not necessarily the same in all directions.

Take the example of the electrical conductivity of graphite. It will conduct along the x or y directions, but not the z direction.

 

[EMAAN]

I'm not exactly sure what you are trying to say here, but this might help:


The optical properties (or any others for that matter) of a material are not necessarily the same in all directions.

Take the example of the electrical conductivity of graphite. It will conduct along the x or y directions, but not the z direction.

but why always perpendicular component is reflected?what is special in it

 

[tyco05]

http://scienceworld.wolfram.com/physics/FresnelEquations.htmlmight help

 

[TGarzarella]

To understand it, you'll need to study Maxwell's equations, and the boundary conditions it predicts for field components normal and perpendicular to the boundary.

 

[Claude Bile]

The Parallel component does not get absorbed - it gets transmitted. The theory only takes into account the refractive index of the medium - any additional material properties (such as absorption) are ignored.

Claude.

 

[arish]

sorry guys, i m still confused

 

[Claude Bile]

Okay, you are asking why the perpendicular component gets absorbed by the material when incident at Brewster's angle, even if the material does not absorb at whatever wavelength we are dealing with.

My answer to you is this - the medium does NOT absorb the wave, it transmits it.

The question then is why is the perpendicular component 100% transmitted at this rather unremarkable angle of incidence? For that I do not have a good explanation except to say that this behaviour is consistant with Maxwell's equations. The numbers just happen to work out as they do. For better insight you need to do what TGarzarella suggested and study the equations themselves.

 

[daniel_i_l]

First of all, I'm not sure that I understood you but when light is reflected off a medium at Brewsters angle the light is polarized PARALLEL to the surface of the medium (I'll always be talking about the electric component), if the medium is a table and the light hits it from left to right, then the light that's reflected will radiate only along the "in out" axis = from your side of the table to the other side.
The reason for this is simple, the EM field of light always propagates perpendicular to the direction of the light. Now let's split it up into two components, the "in out" (normally symbollized by a dot) and a second component perpendicular both to the "in out" and to the direction of the light (if the light was going to the right, this component would point up). Now, when light hits a medium it's electric field oscillates the electrons in the medium which generate another EM field in the direction of the oscillation, this new EM field is the reflected light.
When the light comes at Brewsters angle, the reflected light always makes a right angle with the incident ray(this is one way to calculate BA with Snells equation), because of this, the direction of the reflected ray will be the same as the second component of the incident light described above. Since they're in the same direction, the electrons that this component oscillates cannot contribute to the new reflected ray because, as I mentioned above, none of the EM is in the direction of the ray! So only the "in out" component can generate the reflected ray and that means the this ray will be polarized parallel to the surface. This Wiki describes it well.
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Please tell me if that wasn't clear!

 

[Claude Bile]

When the light comes at Brewsters angle, the reflected light always makes a right angle with the incident ray(this is one way to calculate BA with Snells equation).

It is the reflected ray and the transmitted ray that occur at right angles.

Your explanation is a little confusing though, i needed to refer to Wiki to figure out what you were trying to say. The overall argument though appears sound.

Thanks for posting that, I was at a loss to explain why Brewster's angle occurs at the angle it does using physical arguments.

Claude.

 

[oldschoolprof]

My scientific skepticism has been piqued by a recent posting concerning the release of a new high tech movie. James Cameron’s Avatar is slated for December release and is supposed to shake the foundations of the 3-D cinema experience. One of the technologic breakthroughs involved is the use of polarized glasses (no surprise here). The innovation in this scenario is the use of “untinted” polarized lenses. This seems to me to violate my concepts of polarization. If this proves to be true, indeed this would be a quantum improvement in the current crop of glasses which have huge attenuation of the light reaching the viewer’s eyes. Is it theoretically possible to produce an “untinted” polarized lens, or is this just hype?

 

[Born2bwire]

Take a look at the Disney Digital 3D and Dolby 3D. They now use circularly polarized light which will allow all the light projected and reflected through for the purposes of the movie.

http://en.wikipedia.org/wiki/Disney_Digital_3D

EDIT: Quick clarification. The old linear polarization scheme reduces the projected light by 50%. This is not by the glasses that the audience wears. Yes, any polarizer will reduce the unpolarized light by 50%, but since the projected light is already polarized, the glasses do not affect the brightness of what is seen on screen. However, to create the linearly polarized light for the projection, you have to pass the light from the projector's lamp through a polarizer, this is where the light is reduced by 50%. I guess you could compensate by installing a light that is twice as bright but I doubt this would be common practice (if it is practical at all). Circularly polarized light though will not reduce the intensity of the light after it passes through the filter, it still let's through both horizontal and vertical linear polarized light, just that the two linear components are given a temporal displacement. So the circular polarizer used for the more modern 3D systems will not reduce the intensiy of the projector's lamp like the older linearly polarized scheme.