How Does the Speed of Light Reflect?

[gonzo]

Since the reflection of a photon is just really it's absorbtion and reemission, I was wondering how long that took on average? Does it cause problems in very accurate calculations of the speed/time that light goes around a path that involves mirrors? Is this an error that would accumulate in the classic "light clock" involving photons bouncing back and forth between two mirrors?

 

[Photongod]

Interesting thought! One thing I remember from a Feynman lecture is that the absorbtion and re-emission can take place at all points within the thickness of the glass. It does not all happen at one place. Some reflections must take longer than others, due to the fact that some photons have to travel furthur through glass than others, and since light is slower in glass than in say air, it seems like some sort of error could be introduced. Photongod

 

[kanato]

I don't think reflection is absorption and reemission... I don't think there'd be any reason in that case for it to go a certain direction or even have the same energy.. especially in an insulator, when the energy of the photon may be less than the band gap, but you still get reflections.

 

[dextercioby]

I don't think reflection is absorption and reemission... I don't think there'd be any reason in that case for it to go a certain direction or even have the same energy.. especially in an insulator, when the energy of the photon may be less than the band gap, but you still get reflections.

Then what is REFLECTION...??

Daniel.

 

[masudr]

Kanato, reflection is most definitely absorption and emission of a photon. If a material can do so to a significant range of the visible wavelengths then we call that material reflective.

Masud.

 

[kirovman]

Just a question on this:

Doesn't emission after absorption happen in an arbitrary direction, so why the preference to emit at the same angle of reflection as that of incidence?

Sorry, done an optics module last year, and doing a laser physics and light matter interaction module this year, but interestingly this hasn't cropped up (or I haven't been paying attention in lectures)

 

[kanato]

I would think it would be better described as scattering.. what about reflection of an insulating material with a large bandgap? absorbtion cannot occur in such a system.

 

[Photongod]

Kirovman asked:
Just a question on this:
Doesn't emission after absorption happen in an arbitrary direction, so why the preference to emit at the same angle of reflection as that of incidence?

The best answer is found in "QED" by Feynman--he explains it there with some good illustrations and text--much better than I could do here. It is a matter of various photon paths cancelling each other out, with the incident angle being the path of least time.
Photongod

 

[Barryg]

Think of a phase conjugated array

If I'm standing outdoors with a mirror, I can project an image of the sun onto the wall of my neighboor's house. I've heard it said that this projected beam is the result of the constructive interference of the re-emitted photons... much the same way that military radar antennas work. (That sounds OK but what if we consider an experiment where only one photon is reflected?)

 

[vanesch]

Since the reflection of a photon is just really it's absorbtion and reemission, I was wondering how long that took on average? Does it cause problems in very accurate calculations of the speed/time that light goes around a path that involves mirrors?

This is slippery and hand-waving, but one could at first sight say that reflection where you get a 180 degrees phaseshift could be seen as a "delay" of about half a period, while the reflections that are "in phase" are "immediate". But as I said, that's very handwaving, and in fact not correct, because this is not the group delay (which is 0: phase shift independent of frequency). So better consider it "immediate".

cheers,
Patrick.

 

[ZapperZ]

Keep in mind one very important thing. When we say "reflection" done by ordinary mirror, it is a reflection of a smooth METALLIC surface. You may have a sheet of glass on top of the metallic surface to protect it from being scratched, but the reflection is still done predominantly by the metallic surface.

The reason why metals are good reflector of light, especially within the visible range, is due to the conduction electrons. These conduction electrons forms what we call "plasmons". The oscillating electric field from light causes the conduction electrons to oscillate. This, in turn, causes an EM radiation of the same frequency, but 180 degrees out of phase to be reradiated. This is the "reflected" light.

We already had a thread on here on why the angle of incidence is equal to the angle of reflection, so I won't go into that again.

Zz.

 

[marlon]

I really need to make some remarks here because i have been reading some quite confusing things about plasmons here. When an oscillating E-field is incident to a metal for example, the conduction electrons will feel this E-field and as a result they start to oscillate themselves. The elecrons will screen (ie block off)the E-field. Due to these induced conduction electron oscillations, you get local differences in charge density throughout the medium. These socalled density-waves (ie the gradients of charge density can be represented by a wave) can be quantized alla QFT. The associated particle is the plasmon. So a plasmon is not just a bunch of electrons, it represents a charge-density-wave (or more specifically the particle associated with it). If you are familiar with the concepts of QFT, this is not difficult the understand (i mean, a particle being associated to a wave). What this means is that if a wave goes from one energy level to another (ie : excitation), the associated energy change dE can be seen as the energy dE of some particle with certain mass m and so on...What this m will be, is determined by the Einstein energy relation and the principles of QFT.

If the incident EM-wave has a frequency below the socalled plasma frequency, the light will be reflected by the metal. The reason being, the electrons in the metal screening the electric field of the light. Light of frequency above the plasma frequency is transmitted, because the electrons cannot respond fast enough to screen it. They cannot start to oscillate as fast as the incident wave, if you will.

In most metals, the plasma frequency is in the ultraviolet, making them shiny in the visible range.

regards

marlon

Here is a nice site http://home.hccnet.nl/ja.marquart/BasicSPR/BasicSpr01.htm

 

[MeJennifer]

Very interesting topic!
So does anybody know the average delay between the absorbtion followed by another emission of a photon for a mirror?

Suppose we have an experiment where we measure that a photon at A was emitted in the directIon of a mirror at B and, some small time later, we measure that a photon was absorbed at A.

It seems that we could we not say definitively that the absorbed photon came from the miror, since an absorbtion followed by an emission seems just one possible superposition.

But it seems we could (in principle) measure the time differential between the emission and the absorbtion. But that would mean we could infer the path taken without making a measurement!
Where is the error in this reasoning?

 

[Mentz114]

Agreed that this is an interesting topic. I'd like to add that if one assumes a delay in reflection, of a radar beam from a planet say, then by redrawing the space-time diagram it is possible to derive the Schwarzschild metric from SR calculations.
Sounds crackpot but there's a paper in the XArchiv that shows it.

If I remember correctly, the delay revises the observers calculations of time and distance so it's as if she was in a curved space-time. I think it's more of a lucky coincidence than anything real.

Let me know if this off-topic and I'll delete it.

[I can't find the paper but I'll keep looking]