Question about slowdown of the speed of light in solid medium

  • #26
In this case you can forget energy: there is no energy absorption, no energy transitions; they oscillate without the need of any energy.
So they oscillate in presence of electromagnetic wave, right?
 
  • #28
And what do you think, is this http://science.howstuffworks.com/light10.htm"?
The atoms in some materials hold on to their electrons loosely. In other words, the materials contain many free electrons that can jump readily from one atom to another within the material. When the electrons in this type of material absorb energy from an incoming light wave, they do not pass that energy on to other atoms. The energized electrons merely vibrate and then send the energy back out of the object as a light wave with the same frequency as the incoming wave. The overall effect is that the light wave does not penetrate deeply into the material.
 
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  • #29
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Yes, this can be slightly confusing if not understood well; it means that the atoms don't make a transition between different energy levels but that they simply "resonate" according to the em radiation hitting on them and then they re-emits the same em wave (but with a slight delay). Think of a sort of "reflection" of a light beam on a mirror: if the reflection is perfect, the mirror doesn't absorb any energy from the beam, even if the em wave do interact with the mirror's atoms.
"Absorbed" in that contest means exactly this, that is, "energy is not absorbed from the em wave"; it doesn't mean that "there is no interaction"; so "re-emitted" about photons in this case means that they do interact but without energy absorption.

So what happens to the original photon after the interaction ?
 
  • #30
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  • #31
No. The article means to talk about a perfect reflection (no energy absorption at all), so it's confusing to say "When the electrons in this type of material absorb energy from an incoming light wave".
I think that some materials, like some metals absorb that energy of the wave. Like in the photoelectric effect. And the glass electrons have enough energy, so they don't absorb it, they are just letting pass through the material, that's why it is transparent. But when the light interacts with the glass surface the EM wave make them vibrate at same frequency as the light passes. But still I can't understand why air can't reflect light or it reflects the light very poorly.
 
  • #33
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I think that some materials, like some metals absorb that energy of the wave. Like in the photoelectric effect.
They have already explained you the concept of energy bands. A smooth metal's surface like Al or Ag (mirrors are made covering a glass surface with those metals) reflects quite well visible light without absorbing it appreciately, but if you increase the frequency and so you go into UV, you will, at the end, find frequencies at which the metals become opaque as is a piece of carbon for visible em radiation (= light).
So, you can't have photoelectric effect with a metal which doesn't absorb energy at the frequency of light you are hitting it. (Actually there is always a slight absorption, no material is perfect, however this doesn't affect the essential concept). For this reason photocathods inside photomultipliers are covered with alkali metals or other special materials that absorbs well in the visible spectrum releasing electrons. So, photoelectric effect has nothing to do with reflection.
And the glass electrons have enough energy, so they don't absorb it, they are just letting pass through the material, that's why it is transparent.
I haven't understood what you said.
But when the light interacts with the glass surface the EM wave make them vibrate at same frequency as the light passes. But still I can't understand why air can't reflect light or it reflects the light very poorly.
Imagine to be inside water, 100 m down the surface: would you say that water reflects light there? Probably not. But, on the other hand, you know that light is reflected by a water surface in air. Try to think about it. (You have to begin thinking to yourself, from now on. :smile:)
 
  • #34
They have already explained you the concept of energy bands. A smooth metal's surface like Al or Ag (mirrors are made covering a glass surface with those metals) reflects quite well visible light without absorbing it appreciately, but if you increase the frequency and so you go into UV, you will, at the end, find frequencies at which the metals become opaque as is a piece of carbon for visible em radiation (= light).
So, you can't have photoelectric effect with a metal which doesn't absorb energy at the frequency of light you are hitting it. (Actually there is always a slight absorption, no material is perfect, however this doesn't affect the essential concept). For this reason photocathods inside photomultipliers are covered with alkali metals or other special materials that absorbs well in the visible spectrum releasing electrons. So, photoelectric effect has nothing to do with reflection.
I haven't understood what you said.Imagine to be inside water, 100 m down the surface: would you say that water reflects light there? Probably not. But, on the other hand, you know that light is reflected by a water surface in air. Try to think about it. (You have to begin thinking to yourself, from now on. :smile:)
Ok, thank you very much for the help. When we are at the air, I have 2 simulations for refraction of light, which are connected with slowing the speed of light will passing through medium. http://www.walter-fendt.de/ph14e/refraction.htm" [Broken]. My question was, why air have 1.003 index of refraction?
 
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  • #35
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Doesn't anybody on this thread have an EM textbook?
 
  • #36
ZapperZ
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Doesn't anybody on this thread have an EM textbook?

No I don't. I have several. Now what's your point?

Zz.
 
  • #37
So what happens to the original photon after the interaction ?

Yes, what happens to the original photon after the interaction?
 
  • #38
Is there any analogy between the way that the phonon mass behavior of atoms in a transparent solid transmits light and the way a metal conducts electrical current?
 
  • #39
ZapperZ
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Is there any analogy between the way that the phonon mass behavior of atoms in a transparent solid transmits light and the way a metal conducts electrical current?

Not exactly, because there is no "carrier" for optical transport in solids, where as there is one for electrical transport. One can typically use a Boltzmann transport equation to describe charge and even heat transport, whereas I don't recall the same thing being used for optical transport.

Zz.
 
  • #40
Does the photon still exists?
 
  • #41
Physicsissuef, the way that you keep making posts that add nothing new but reiterate your previous questions give the appearance of self-centeredly clamoring for attention. I'm not saying that you're necessarily being childish, just that this approach may actually discourage people from answering you.
 
  • #42
Physicsissuef, the way that you keep making posts that add nothing new but reiterate your previous questions give the appearance of self-centeredly clamoring for attention. I'm not saying that you're necessarily being childish, just that this approach may actually discourage people from answering you.

I think my questions are very simplified, so everyone can understand. So my question was, what happens with the light after the interaction with the surface?
 
  • #43
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Yes, what happens to the original photon after the interaction?

Nobody knows ?
 
  • #44
pam
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No I don't. I have several. Now what's your point?

Zz.
Read the section on index of refraction.
 
  • #45
ZapperZ
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Read the section on index of refraction.

I have. Can you tell me how E&M phenomenology explains why carbon atoms arranged in different configuration can give different index of refraction? I've gone through Jackson, and he certainly says nothing about it.

And if you somehow forgotten why this is relevant, note that you took exception to the info quoted from the FAQ, in which I had tried to explain why "photons" moving through a medium interact with the structure of the material, and why the active phonon modes of the material dictates the nature of the optical transport in that material.

Zz.
 

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