Question about slowdown of the speed of light in solid medium

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The discussion centers on the phenomenon of light slowing down in solid materials, specifically addressing how photons interact with lattice structures. It clarifies that photons are not absorbed in the traditional sense but rather cause atoms to resonate, leading to a slight delay in re-emission. This interaction does not involve energy transitions between electron levels but rather a collective response of multiple atoms, akin to a reflection rather than absorption. The conversation also highlights that a single photon interacts with many molecules simultaneously, challenging the simplistic view of individual absorption and re-emission. Overall, understanding the role of phonons and collective atomic behavior is crucial to grasping the complexities of light propagation in solids.
  • #31
lightarrow said:
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.
 
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  • #32
what do u think?
 
  • #33
Physicsissuef said:
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
lightarrow said:
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" . My question was, why air have 1.003 index of refraction?
 
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  • #35
Doesn't anybody on this thread have an EM textbook?
 
  • #36
pam said:
Doesn't anybody on this thread have an EM textbook?

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

Zz.
 
  • #37
GT1 said:
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
CaptainQuasar said:
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
CaptainQuasar said:
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
Physicsissuef said:
Yes, what happens to the original photon after the interaction?

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

Zz.
Read the section on index of refraction.
 
  • #45
pam said:
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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