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Why some photons are absorbed while propagating through glass

  1. Aug 15, 2013 #1
    Hi guys, I'm new to this forum...
    I was just reading the link given below,posted by ZapperZ, describing as how photons propagate in the form of phonons in medium like glass....
    https://www.physicsforums.com/showthread.php?p=899393#post899393[/COLOR] [Broken] and a question it struck to me,

    In the last paragraph ZapperZ wrote that, "On the other hand, if a photon has an energy beyond the phonon spectrum, then while it can still cause a disturbance of the lattice ions, the solid cannot sustain this vibration, because the phonon mode isn't available. This is similar to trying to oscillate something at a different frequency than the resonance frequency. So the lattice does not absorb this photon and it is re-emitted but with a very slight delay. This, naively, is the origin of the apparent slowdown of the light speed in the material."

    Now my question is if we use a highly monochromatic light of frequency more than the phonon spectrum of glass then, light would transmit through glass by the process of to absorbtion of photons by the lattice and again re-emission and we must get intensity at the end of glass slab equal to the intensity incident.But in reality this does not happen , some energy is absorbed by the slab and we don't get intensity of wave emerging out of slab equal to incident intensity.Why it is so???? Why some photons do get permanently absorbed by the glass plate,even if all of them had energy beyond the phonon spectrum of glass???? Ideally, all the identical photons having energy beyond the phonon spectrum of glass should have been emitted as told by ZapperZ that "if the photon has an energy beyond the phonon spectrum, then while it can still cause a disturbance of the lattice ions, the solid cannot sustain this vibration, because the phonon mode isn't available" and none of them(photons) should have been absorbed permanently by glass slab and we should get 100% intensity on the other side of the slab.But in reality,this doesn't happens,Why?????

    I think that we cannot explain this absorption of photons with the help of phonon and quantum theory.It can only be explained by classical physics,considering light to have wave nature and not particle(photon) nature.If anyone can clear this doubt of mine with the phonon theory and quantum mechanics,it will be really pleasant and exotic.This question is really nagging my brain and as you see, I am not able to get it on the track with phonon mechanism.Please help by trying to answer my question.

    Waiting for a sensible reply.:smile:
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 15, 2013 #2
    do you accept that?
    If so then you can expect some heating of the material via the energy of the incident light.
     
  4. Aug 15, 2013 #3
    So I just skimmed Zapper's material and see he says the same thing:

    This seems analogous, but not identical, to the absorption and reemissions of photons when abaorbed by orbital electrons in individual atoms.
     
  5. Aug 15, 2013 #4
    What do you think does happen?

    edit: I would think the incident light energy would equal the emitted light energy plus the heat energy
    emitted after the slab returns to its original ambient temperature.
     
    Last edited: Aug 15, 2013
  6. Aug 15, 2013 #5
    As an example, Wikipedia seems able to do it here:

    http://en.wikipedia.org/wiki/Phonons

    by the way: Kudo's on reading a technical description then digging in with questions!!
     
  7. Aug 15, 2013 #6
    I think that some photons are absorbed by the slab and their energy is converted to kinetic energy of phonons,which we commonly call as heat energy,and if you have a very sensitve thermometer then you will even notice a temperature rise of a very few degrees due to this. But this experimental reality observed here just contradicts Zapper's material that "if photon has an energy beyond the phonon spectrum(which our photons,the ones we made incident on the slab had energy beyond phonon spectrum), then while it can still cause a disturbance of the lattice ions, the solid cannot sustain this vibration and hence forth the lattice does not absorb this photon and it is re-emitted". So if all the incident photons are re emitted in this way then, we would have got 100% intensity back at the end of the slab.But as such we don't get 100% intensity back,It is a contradiction to Zapper's material.

    So, I think we ought to think this problem in a new way,by some other principles, as quantum mechanics is not proving much helpful to solve this simple yet a grave problem of physics.Or if any one can solve this problem by quantum principles,it will be so intelligent of him/her.
     
  8. Aug 15, 2013 #7
    I read that even before you told me.It gives an exotic description of what phonons are. But that's not an answer to my question. That is what phonons are, my question is different.
     
  9. Aug 15, 2013 #8

    Cthugha

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    I do not see how one can draw that conclusion. First, phonons are here just placeholders for any possible excitation which can occur in a material. You can have excitons, plasmons or whatever as well. So you need to stay clear of any resonance to get good transmission. Second, you obviously need to make sure you also include reflections from the surfaces in your energy balance. Some light will not even enter the glass because it gets reflected at the surface.

    If you do all of that, the remaining tiny amount of light not making it through glass is caused by impurities. State of the art technical glasses like suprasil have a residual absorption of less than 1ppm/cm (http://optik.heraeus-quarzglas.de/media/webmedia_local/downloads/Suprasil_and_Infrasil__Material_Grades_for_the_Infrared_Spectrum.pdf).

    Here metal impurities are one factor determining the residual absorption, while the OH concentration is the other. OH has some nasty vibrational or rotational absorption bands extending into the visible regime.
     
  10. Aug 16, 2013 #9
    First of all a heartiest thanks for replying and giving a nice reference. I am 90% clear now.But still,I can't fully interpret whether the residual impurities you talked about is only the factor responsible for absorption of light propagating through glass. If so,let's keep aside glass for a moment as it is composed of many compounds and many different kinds of atoms. Instead, let us take a pure diamond slab, containing only carbon atoms and having no impurities present in it.Let the dimensions of the slab be 15cmχ15cmχ15cm.

    Now let's pass a highly coherent and a monochromatic(single wavelength) light, having frequency greater than the phonon spectrum of diamond. Now we even take into account reflections at both the surfaces(Light entering and light leaving). So, at the end of the slab,we must get intensity equal to,

    Emitted light intensity = Incident intensity - intensity reflected at the surfaces [There will be no absorption as the diamond slab is free from impurities and so there will be no absorption, as you told, that is absorption will depend only on the residual impurities and not on the length of the slab]

    So, this was what you were trying to express....?????
     
  11. Aug 16, 2013 #10
    What are, generally, the mechanisms for residual absorption in otherwise clear substances?

    For example, how does water absorb blue light?
     
  12. Aug 16, 2013 #11

    Cthugha

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    That depends on what wavelength range you are thinking of. Diamond has a band gap (http://en.wikipedia.org/wiki/Band_gap) of around 5.5 eV. That means that at higher photon energies you need to consider absorption by electronic degrees of freedom in the system. As the electron dispersion is a continuous one, you will be able to get some absorption for more or less all photon energies higher than the band gap. 5.5 eV is around 225 nm, so this does not really affect the visible range. What can still happen, is a process called multi-photon absorption. Even if one photon alone does not have enough energy to create an electronic excitation, two photons (or 3,4,5,...) arriving simultaneously may be able to do it. However, the probability of such processes is usually pretty low (but obviously depends on the intensity of your light beam). At low optical powers, the effect of this process is, however, tiny, but it can occur and would depend on the length of the slab.

    Also, the number of impurities given in the system is usually given by a somewhat constant density. So as you increase the length of the slab, you will also increase the number of total impurities, which will also create increasing absorption with increasing crystal length.
     
  13. Aug 16, 2013 #12
    ns_phonon....
    well, it kinda is, at least in part.....If you read closely, you'll find valuable insights regarding your questions...
     
  14. Aug 16, 2013 #13
    I googled some reasons, as for why does absorption of light occurs while it is propagating through a medium and I found a link which explains the general cause of both reflection and absorption. The discussion in the link is based on Professor Richard P. Feynman's lectures at the California Institute of Technology. It tries to explains how light is absorbed exponentially in any transparent medium. But it seems to be a little bit old-fashioned and do not use the principles of quantum mechanics upto the mark. But it explains reflection and even absorption of light from metals and glass very sucessfully and as such we can't explain this permanent exponential absorption phenomenon of light by quantum mechanics and the phonon theory,we must accept this explanation,am I right????

    I think there is no other way, we can explain the exponential absorption phenomenon of light during propagation of light through a transparent medium. I think it is not possible to explain this exponential absorption phenomenon, as proposed by scientists Lambert and Bouguer in the name of "Lambert- Bouguer exponential law of absorption " on the basis of quantum mechanics and phonon theory of transmission of light through a transparent medium.

    The Link : http://www.madsci.org/posts/archives/1997-05/863034503.Ph.r.html

    In this link the lines in the last paragraph, "So there is a general rule that if any
    material gets to be a very good absorber at any frequency, the waves are
    strongly reflected at the surface and very little gets inside to be absorbed
    " gave me a great satisfaction.

    If anyone here has the guts to explain this "Lambert- Bouguer exponential law of absorption" on the basis of quantum mechanics and phonon theory of transmission of light through a transparent medium." then he/she is most welcomed.
     
  15. Aug 16, 2013 #14

    Cthugha

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    It uses QM. In fact, it is the same explanation Zapperz gave. You just go from the spectral region where phonon resonances are dominant (IR) to the region where electronic resonances are dominant (visible). The sole difference is that you can "escape" phonons by going to higher energies, while you cannot do that for electrons.
     
  16. Aug 22, 2013 #15
    Actually, diamond can absorb ONLY light with frequencies greater than the phonon spectrum of diamond!

    Diamond consists of atoms that all have the same electronegativity. Also the diamond lattice is symmetric so every bond has identical half-lattices at each end. Therefore, a phonon cannot possibly emit a photon, nor be created by photon absorption.

    One phonon cannot. Two or more phonons can. The interaction of two or more phonons provides the asymmetry that light can interact with.

    Accordingly, pure, type IIa diamond does absorb infrared. It possesses a fairly wide absorption band for two phonon absorption, around 1700...2600/cm, and second, weaker band for three phonon absorption around 3000...3700/cm.

    Note that these are still weak!
    In liquid water, the mainline infrared absorption, at about 3300/cm, has intensity of something like 12 000/cm - infrared with wavelength 3000 nm is absorbed in just 800 nm of water. Looking at higher overtones, the first combination band at 1950 nm still has intensity of over 100/cm, first overtone at 1450 nm is around 30/cm.

    Whereas in diamond mainline is impossible as stated, and the first overtone being the strongest line has intensity of just over 12/cm at its peak near 2000/cm. Diamond, unlike most other hard crystals (which possessing polar bonds have very high intensities of mainline single phonon absorption), is fairly but not completely clear in infrared.

    Yet if you look at more water, it is brilliant blue!

    It does NOT have only mainline/one phonon and first overtone/two phonon absorption. The water absorption bands go on - increasingly weaker but STILL significant. A 1200 nm overtone still has intensity over 1/cm, and 970 nm overtone over 40/m.

    Going on, red 740 nm shoulder has intensity 3/m. Not that there is a lot of absorption between the lines, too - the 3rd overtone looks like shoulder not maximum. Strong shoulders continue - yellow 4th overtone at 605 nm and 250/km, green 5th overtone at 515 nm and 30/km, blue 6th overtone at 455 nm and 9/km... what looks like 7th overtone is at 425 nm and 5/km or so. There are weaker combination bands in between as well - 665 nm, 555 nm...

    Do diamond and glass likewise have series of intrinsic multiphonon/overtone absorption bands and shoulders blueward of their main absorption bands?
     
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