How does the Ocean polarize light?

Question

I know that the light gets polarized in the plane of the ocean, but why is that? It seems to me that the molecules are free to be in any position or location, so why is it that the 'vertical' polarization gets ignored?

Answer 1

I'm going to focus on Chris White's rewording of your question

"if the photon is only hitting a few molecules in random orientations, how does it know what direction to be polarized?"

because, unless you can delve into Maxwell's equations, you're not going to get much joy from anything else. So let's just accept that, as on the Wikipedia page, there are the Fresnel equations coming from the solution of Maxwell's equations and that they say that light of different polarisations behaves differently when reflected from a dielectric interface. All of these equations assume that the mediums involved are continuums, with bulk, extrinsic refractive indices. So, given this context, how indeed do continuum ideas square with a photon hitting a few molecules in random orientations?

The answer is that the photon does not just hit a few molecules - it's not like a billiard ball bouncing off things. A photon propagates following Maxwell's equations. When light enters a medium, the propagating wave interacts with ALL of the molecules in its field of influence. In a medium, we don't just have pure photons, we have a quantum superposition of free photons and exited matter states. And ALL of the molecules contribute to that quantum superposition. Indeed, an optical photon's wavelength is about three orders of magnitude longer than the molecular size, so the quantum superposition will involve of the order of $10^9$ to $10^{11}$ molecules even if the light field is focused to its diffraction limited spot. So the photon really does see effectively a continuum.

The particle side of the photon's nature does not show itself until it is "observed" by being permanently absorbed, e.g. in a photodetector or camera film. Until then, even at the one photon at a time level, the propagation is perfectly well described by the classical Maxwell equations.

Answer 2

From Wikipedia's article on Brewster's angle:

When light encounters a boundary between two media with different refractive indices, some of it is usually reflected as shown in the figure above. The fraction that is reflected is described by the Fresnel equations, and is dependent upon the incoming light's polarization and angle of incidence. The Fresnel equations predict that light with the p polarization (electric field polarized in the same plane as the incident ray and the surface normal) will not be reflected if the angle of incidence is

$\theta_\mathrm B = \arctan \left( \frac{n_2}{n_1} \right),$

where $n_1$ is the refractive index of the initial medium through which the light propagates (the "incident medium"), and $n_2$ is the index of the other medium. This equation is known as '''Brewster's law''', and the angle defined by it is Brewster's angle.

The physical mechanism for this can be qualitatively understood from the manner in which electric dipoles in the media respond to $p$-polarized light. One can imagine that light incident on the surface is absorbed, and then re-radiated by oscillating electric dipoles at the interface between the two media. The refracted light is emitted perpendicular to the direction of the dipole moment; no energy can be radiated in the direction of the dipole moment. That is, if the oscillating dipoles are aligned along the supposed direction of the reflection, no light is reflected at all. In this case, all the light would be refracted in the direction perpendicular to the direction of the dipoles. Thus, if $\theta_1$ is the angle of supposed reflection (which is equal in magnitude to the angle of incidence), the angle of refraction $\theta_2$ would be equal to (90° - $\theta_1$).

The above geometric condition can be expressed as

$\theta_1 + \theta_2 =90^\circ$

where $\theta_1$ is the angle of reflection (or incidence) and $\theta_2$ is the angle of refraction.

Using Snell's law,

$n_1 \sin \left( \theta_1 \right) =n_2 \sin \left( \theta_2 > \right),$

one can calculate the incident angle $\theta_1 = \theta_\mathrm{B}$ at which no light is reflected:

$n_1 \sin \left( \theta_\mathrm B \right) =n_2 \sin \left( > 90^\circ - \theta_\mathrm B \right)=n_2 \cos \left( \theta_\mathrm B > \right)$

Solving for $\theta_\mathrm{B}$ gives

$\theta_\mathrm B = \arctan \left( \frac{n_2}{n_1} \right).$

Answer 3

As others have pointed out, the exact formula of the reflected light should be derived with Maxwell's Equation. If you are simply asking for the qualitative reason for the asymmetry of the polarization, it can be attributed to the asymmetry of the boundary geometry relative to the path of the incoming light. The size of the effect is determined by the wavelength of the light relative to the size and separation of the molecules. WetSavannaAnimal aka Ro has addressed why the material is treated as a continuum but has not mentioned the geometric asymmetry of the boundary (interface). There is no phase change in the horizontal line in the reflecting surface perpendicular to the incoming light path, while there is a large phase shift in the intersection line of the vertical plane with the interface.