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"Reflected light is polarized". The following photo demonstrates this:

enter image description here(source)

When unpolarized light is incident at this angle, the light that is reflected from the surface is therefore perfectly polarized (Wikipedia).

But if at Brewster's angle only a particular polarization (p-polarized) is perfectly transmitted, then I'd expect the reflected light to be randomly polarized except this one missing polarization that was perfectly transmitted.

In short, I understand that the transmitted light is perfectly polarized (p-polarization), but I do not understand why the reflected light is perfectly polarized as well (s-polarized), if the incident light is randomly polarized.

A particle point of view would be appreciated.

Community
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Sparkler
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    What is "randomly polarized" light? Is that the same as "unpolarized"? – CuriousOne Jun 27 '16 at 02:19
  • @CuriousOne yes – Sparkler Jun 27 '16 at 02:24
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    Unpolarized light is simply a mixture of both polarizations. You can separate them and then you have two polarized light sources. – CuriousOne Jun 27 '16 at 03:20
  • @CuriousOne are you saying light coming from the sun, for example, has only two kinds of polarisation? (I thought polarisation angle can be anything, hence "random") – Sparkler Jun 27 '16 at 03:27
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    Electromagnetic waves have two independent polarization components that you can express in an infinity of linear combinations. That has nothing to do with their source. – CuriousOne Jun 27 '16 at 03:33
  • @CuriousOne the components depend on the axes you choose and different polarization angles will result in different projections, whose magnitudes will depend I the axes you choose. Do I have a conceptual error here? – Sparkler Jun 27 '16 at 03:40
  • @CuriousOne "unpolarized light is simply a mixture of two polarizations" it is not just 'a' mixture, it is a very specific one, namely the one where the density matrix is invariant under rotations (or if you like to think classically, the state of coherence). That is not just two polarizations added together, you also need the off-diagonal terms. Since the state of coherence is an ensemble average of the randomly fluctuating field I think "randomly polarized" hits it a lot better than what you said. But you probably mean the same thing. I also fail to see how this is relevant to the question. – Wolpertinger Jun 27 '16 at 08:28
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    @Numrok: Unless you have an instrument that can correlate the two polarizations at $10^{15}$ samples/second, it doesn't actually matter. Maxwell is perfectly linear, so which operation you perform first, polarization averaging or time averaging, is completely irrelevant. Please look at my other comment that makes it clear that the polarization components are a feature of electromagnetic waves rather than their sources. That is at the heart of the matter, the coherence problem is a side-show. – CuriousOne Jun 27 '16 at 08:41
  • @CuriousOne that is of course if you want perfectly unpolarized light down to the photon level. Instead you can just make a classical unpolarized wavefield though (i.e. unpolarized in the many photon limit). And the off-diagonal terms are not a "side-show" then at all. I agree, it is a feature of the wave, but the correlation terms matter since the field is not fully characterized by diagonal terms alone (how would you get interference otherwise). but now we are in this technical discussion and like i said: not relevant to the question... – Wolpertinger Jun 27 '16 at 08:48
  • @Numrok: I think we are looking at a window illuminated by sunlight... I kind of doubt that this particular radiation source has given us entangled quantum pairs... I also kind of doubt that the OP has fully understood the classical theory, yet, so no need to push QED on him, just yet. :-) – CuriousOne Jun 27 '16 at 08:58

2 Answers2

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In your final two paragraphs you have it backwards. At Brewster's angle the reflected light is totally polarized, but the total polarization of the transmitted light is usually rather weak. Compare reflection coefficients $r$ and transmission coefficients $t$ from the Fresnel equations:

enter image description here

Reflected light is completely polarized at Brewster's angle because the direction of propagation $\vec S = \vec E \times \vec B$ is perpendicular to the electric field $\vec E$. A hand-wavy way to describe Brewster's angle is to observe that there, the angle between transmitted and reflected light is 90$^\circ$. Therefore your EM oscillations in the transmitted beam can't contribute evanescently to the reflected wave, except where $\vec E$ is parallel to the surface.

rob
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You are correct in asserting that unpolarized light contains a mixture of many polarizations. However, each of these polarizations can be expressed as a combination of horizontally and vertically polarized light. Diagonally polarized light can thus be seen as containing both horizontally as well as vertically polarized light.

When a horizontal beam of diagonanally polarized light strikes a vertical window under the brewster angle, the horizontally polarized component will be perfectly transmitted. The vertically polarized component of the diagonal beam will be partially reflected and partially transmitted. The reflected light is thus fully vertically polarized, while the transmitted light still contains both polarizations. However, since part of the vertically polarized light has been reflected, the polarization of the transmitted light will be somewhere between diagonal and horizontal.

For unpolarized light this means that for every polarization only the vertically polarized components will be partially reflected. The transmitted beam will thus still be a mixture of polarizations, but on average it will be somewhat horizontally polarized.

Note that here I used horizontal and vertical polarization as the two components of the light, but one might just as well have chosen to express the light as a combination of two components polarized along the diagonals. In reflection experiments, one typically chooses to use a component in the same plane as the direction of the beam and the surface normal (p-polarized) and a component perpendicular to this (s-polarized).

Physically, the reason for this effect is that the reflection of light from a glass plane can be seen as the absorption and re-emission of light. The dipoles in the glass start to oscillate when they absorb a photon. They can then again emit this photon, which can be seen as the reflected light. Note that as the electric field is always perpendicular to the propagation direction of the light, the dipoles also oscillate perpendicularly to this propagation direction. Since the electric field of the emitted light also has to be perpendicular to the propagation direction of the emitted light, the dipoles can only emit photons in a direction perpendicular to the direction of oscillation.

For the Brewster angle, the direction of propagation inside the glass (thus after refraction) makes a 90 degree angle with the direction of the reflected light. This means that when the dipoles in the glass are excited by p-polarized light, they cannot emit in the direction of reflection, since the oscillation direction and the direction of reflection are aligned. At the Brewster angle, there is thus no reflection possible of p-polarized light.

Note that in the picture sketched here, the reflection occurs everywhere in the glass, not just at the interface. The resulting reflectivity and transmitivity are however identical to models which model the reflection to happen only at the interface. For a detailed explanation on how these models are equivalent Feynmans book "QED" is an excellent introduction. It explains the principles of QED with hardly any math.

Crimson
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  • so both refracted and reflected light is polarized now? 2) can you please also explain from a particle point of view?
    • – Sparkler Jun 27 '16 at 12:01
  • The transmitted light is not fully polarized. I corrected the mistake in my answer. I also added a physical explanation – Crimson Jun 28 '16 at 08:38
  • "in the picture sketched here" - which one? – Sparkler Jun 28 '16 at 12:36