Polarizer Polarizing Filter
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Questions about light polarization, scattering and reflection?
I'm very interested in how light behaves, and I have a few questions about it.
1.) What is the difference between a particle scattering light and a particle reflecting light diffusely?
2.) When light is polarized by being reflected, why exactly is it that it is only polarized at/near the Brewster angle? How does this filter out all the other components of light?
3.) In diagrams for polarization by reflection that I've seen, it showed only teh electric component of light being reflected when it was polarized, is this what really happens? How can it still be light if it's just an electric field?
4.) When light passes through a polarizer like sunglasses, what happens to all the light that doesn't get through? Is it absorbed, reflected, and if absorbed, then why don't sunglasses seem to get really hot from absorbing all that light?
1) reflection is properly a macroscopic phenomenon, arising when light impinges on a many-atoms system (like a solid). It comes from the sum of the scattering of light by every single atom of the target, so you can say that a particle scatters light but saying that a particle reflects light has little meaning in my point of view 
3) (I'm answering to this before n.2) by convention, the polarization of light is the direction of the electric field. It is sufficient to give one field (either E or B) for an em wave since the other one can be calculated easily via Maxwell's equations.
2) you can always define a degree of polarization (DOP) for a wave: DOP = 0 means unpolarized light, DOP = 1 means fully polarized light. it is only near the Brewster angle that the reflected light has a DOP near 1, but even for an unpolarized incident wave the reflected light (practically) never has a DOP = 0.
There's a simple way to understand why the Bewster angle "polarizes" light: imagine the target surface as made of atoms with one electron; each electron oscillates along the E field of the wave creating an oscillating dipole. It is known that an oscillating dipole irradiates the maximum power in the plane normal to the direction of oscillation, and does not irradiate in the direction of oscillation.
The Brewster angle is the angle for which the direction of oscillation (and then the direction of the refracted E field, inside the medium) is parallel to the direction of the reflected ray: this means that the dipoles do not emit in the direction of the reflected wave.
This mechanism only works if the incident light is polarized on the plane of incidence (p-polarization), so that the E field lies in the plane fo the incident, reflected and refracted rays. If the incident light is polarized normally to the plane of incidence (s-polarization) the E field is always normal to the reflected ray.
i understand that a picture would make it much easier to get, so you may find some illuminating pictures and explanations here:
http://www.google.it/imgres?imgurl=http://nothingnerdy.wikispaces.com/file/view/polarisation_brewster.gif/98302927/polarisation_brewster.gif&imgrefurl=http://nothingnerdy.wikispaces.com/11.5%2BPolarisation&usg=__O9BIrr2RkKx-4RO5YRuJ_zE6bUU=&h=288&w=360&sz=7&hl=it&start=6&sig2=K_x3nYp_qv9nkn7sOx6_JQ&zoom=1&tbnid=E8qJH3LyM6fagM:&tbnh=97&tbnw=121&ei=Ixd-TczcA8fIswac24T6Bg&prev=/images%3Fq%3Dbrewster%2Bangle%26um%3D1%26hl%3Dit%26safe%3Doff%26client%3Dfirefox-a%26sa%3DN%26rls%3Dorg.mozilla:it:official%26biw%3D1280%26bih%3D595%26tbs%3Disch:10%2C912&um=1&itbs=1&biw=1280&bih=595
http://en.wikipedia.org/wiki/Brewster%27s_angle
4) the light that is not transmitted is slightly absorbed, but the major part is reflected or, in general, re-emitted (for example as black body radiation, which is mostly in the infrared region). the filter will heat up a bit but not nearly enough to melt or similar beacuse there's an equilibrium between absorbed and re-emitted light power.
i don't know if i was clear but i tried to xD
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