Why is light reflected off road and water surfaces horizontally polarised?

[jsmith613]

light reflected off road surfaces or water surfaces is HORIZONTALLY polarised. Why is this?

I would presume therefore that if reflected light is horizontally polarised, light reflected by objects beneath the water is vertically polarised (assuming you see it from under water)

 

[Andy Resnick]

light reflected off road surfaces or water surfaces is HORIZONTALLY polarised. Why is this?

From the difference in the Fresnel reflection coefficients:

http://hyperphysics.phy-astr.gsu.edu/HBASE/phyopt/freseq.html

As to your second paragraph, I'm not sure that's correct- you would need to run the numbers.

 

[Naty1]

Some interesting descriptions here:

http://en.wikipedia.org/wiki/Polarized_light#Polarization_effects_in_everyday_life

note that the polarization you describe is partial...

 

[jsmith613]

This did not really help.

Is there a basic explanation for an AS level student of physics?

thanks

 

[Andy Resnick]

I don't know- what is an AS level of knowledge?

 

[Dr Lots-o'watts]

This did not really help.

Is there a basic explanation for an AS level student of physics?

thanks

Yeah, if a horizontal (-ly polarized) beam hits the road, it makes electrons move horizontally. The electron motion stays on the horizontal surface of the road, so the reflected light they give off can reach your eyes.

However, if a vertical (-ly polarized) beam hits the road, it makes electrons move vertically, but since this movement is below the surface of the road, the energy is absorbed by the depth. The depth absorbs the would-be reflected beam, so nothing bounces (is reflected) back.

Unpolarized light is thus horizontally polarized when reflected because its vertical component is absorbed by the vertical movement of electrons underneath the surface.

In summary, waves emitted by electrons moving horizontally on a surface, are free to propagate away from the bulk matter. Waves that are emitted by electron moving down through a surface are absorbed by the bulk matter.

 

[Bob S]

The effect is caused by external reflection of light at or near Brewster's angle. Fishermen who wear polarizing sunglasses can see the fish better. See

http://en.wikipedia.org/wiki/Brewster's_angle

It also occurs for internal reflection (like under water), below angles where total internal reflection occurs..

Bob S.

 

[Andy Resnick]

Yeah, if a horizontal (-ly polarized) beam hits the road, it makes electrons move horizontally. The electron motion stays on the horizontal surface of the road, so the reflected light they give off can reach your eyes.

However, if a vertical (-ly polarized) beam hits the road, it makes electrons move vertically, but since this movement is below the surface of the road, the energy is absorbed by the depth. The depth absorbs the would-be reflected beam, so nothing bounces (is reflected) back.

Unpolarized light is thus horizontally polarized when reflected because its vertical component is absorbed by the vertical movement of electrons underneath the surface.

In summary, waves emitted by electrons moving horizontally on a surface, are free to propagate away from the bulk matter. Waves that are emitted by electron moving down through a surface are absorbed by the bulk matter.

Ummm...be careful. If the electrons are free to move, they will absorb the light. Polarizers, if you think of them as 'picket fences', transmit light with a polarization *perpendicular* to the lines.

 

[sophiecentaur]

Ummm...be careful. If the electrons are free to move, they will absorb the light. Polarizers, if you think of them as 'picket fences', transmit light with a polarization *perpendicular* to the lines.

Electrons in a metal rod (very free to move) don't absorb radio waves - they re-radiate. It depends upon the conductive part of the complex refractive index i think.

 

[Andy Resnick]

... in any case, the transmitted polarization state is perpendicular to the metal rods.

 

[sophiecentaur]

The radiated E field is parallel to the rods. Polarisation is defined as the E field plane. My comment would also apply to a plane metal reflector. No / low loss is involved due to electron movement in the metal surface.

 

[Dr Lots-o'watts]

Ummm...be careful. If the electrons are free to move, they will absorb the light. Polarizers, if you think of them as 'picket fences', transmit light with a polarization *perpendicular* to the lines.

The electrons here are bounded (dipoles). The model is called the electron-oscillator model.

 

[sophiecentaur]

A few points to remove possible confusions:
The question was originally about polarisation by reflection. Polarisers ('polaroid'), which are mostly used these days are transmission devices. A reflection polariser can be made for 'RF' wavelengths using a grid of parallel wires and that will produce a polarisation that is parallel with the wires because of the currents induced in the wires. The remaining energy (the transmitted beam) polarisation will be at right angles to the wires.
Polaroid also uses the fact that currents are induced in the direction of the aligned particles but they are not 'good conductors' (in the RF sense) and they absorb the energy in one polarisation rather than reflecting it. So they are not really relevant to the OP - although several links seem to produce confusion when they consider the fact that polaroids can be used to observe the polarisation at reflective surfaces and don't make a point that the glasses work quite differently.Polarisation by reflection at a surface is complete at the Brewster angle, which is when the VP cannot propagate in the reflected direction. The reason that it can't propagate is that the E vector would be along the line of the reflected ray - which cannot happen for a space wave, which needs a component of E at right angles to the direction of propagation.
See http://webphysics.davidson.edu/physlet_resources/bu_semester2/c27_brewster.html" for a convincing animation.

Explanations of polarisation due to electron flow in the surface are ok but it must be realized that the actual distances that you can consider the electrons moving are extremely small (think of the frequency of vibration and the resulting acceleration that would be required for a large amplitude vibration). They don't disappear down into the material and dissipate their energy, as was implied in an earlier post because they don't get that far. Actually, you need to consider the interaction with the material as a whole, rather than with individual electrons to get a good picture of what's going on.

 

[Dr Lots-o'watts]

They don't disappear down into the material and dissipate their energy, as was implied in an earlier post because they don't get that far.

The waves yes (if vertical polarization), not the electrons.

 

[Andy Resnick]

Yes, the polarizing nature of reflection is a different mechanism than that of transmission.

The Fresnel reflection coefficients are usually calculated for dielectrics, not conductors- electron motility does not apply to dielectrics- there is no induced current.

The boundary conditions account for the difference in reflectivity between the two components:

http://electron9.phys.utk.edu/optics507/modules/m3/plane_wave.htm

I suppose the animation sophiecentaur posted is a graphical version of the boundary conditions.

 

[sophiecentaur]

The waves yes (if vertical polarization), not the electrons.

I'm not sure that makes sense. After all, there will be a transmitted wave, through the glass, which doesn't get attenuated so why should this apply to a wave that 'isn't reflected'?

 

[Bob S]

light reflected off road surfaces or water surfaces is HORIZONTALLY polarised. Why is this?

The skylight incident on road or water surfaces is often very polarized in the VERTICAL direction, which enhances the perception of the polarization of the reflected light.

Blue skylight, due to Rayleigh scattering on individual air molecules, when scattered at 90 degrees from the incident sunlight, is nearly completely polarized in the plane perpendicular to the plane of Rayleigh scattering, which would be VERTICALLY polarized (near sunrise or sunset) when reflecting off water. This is the polarization blocked by reflection at Brewster's angle. Skylight at 90 degrees from the Sun is completely polarized only when the air is completely absent of any air pollution, smog, fog, clouds, etc.

Bob S

 

[jsmith613]

I suppose the animation sophiecentaur posted is a graphical version of the boundary conditions.

I did not quite get this animation. It seemed to me that both types of waves had arrows in the same direction just different magnitudes?