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As I learned today in school, my teacher told me that when light enters a glass slab it slows down due to the change in density and it speeds up as it goes out of the glass slab. This causes a lateral shift and the light emerges out from the point different than that from where it should have actually emerged from.

Okay so what I mean to ask is, when light enters point A on glass slab and emerges from point C why does the light speed up? Where does it get the energy it has lost when it entered the glass slab?

P.S.: Also, if I place a very very very large glass slab and make a beam of light pass through it will the light never come out as all the energy was lost in place of heat?

Amey Shukla
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    There is no energy lost in the slab. Part of the energy in the electromagnetic field is stored in the electric polarization of the atoms in the material. When the light leaves the glass slab, that energy is transferred back into the vacuum/air. – CuriousOne Dec 18 '14 at 10:32
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    Re the equations, see this post. – John Rennie Dec 18 '14 at 10:44
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    For a matter of intuition: The energy could be the same, and the speed increases simply as resistance decreases. Consider when you're spinning on a chair. If you push your legs out, you'll slow down. When you pull them in, you speed back up but you didn't gain any energy. – Cruncher Dec 19 '14 at 09:34
  • What speeds up photons when they leave a emitting atom? :=) – Georg Jan 15 '15 at 20:57
  • @Georg Those the best answer is the most perfect answer one could give. – Amey Shukla Jan 22 '15 at 10:13
  • Is it possible that photons do not slow down but instead take longer paths through the atoms and molecular structure. Could it be tested or even calculated what the true distance would be if a photon were forced back and forth between the atoms as it traveled to the other side of the transparent material. Different materials and their unique atomic structures would produce paths with different lengths or different refractive indexes. Just wondering – Bill Alsept Nov 30 '16 at 00:40
  • Published less than a month ago (June 2017), this new result is very relevant: "momentum paradox of light" https://phys.org/news/2017-06-atomic-mass-photon-momentum-paradox.html – stafusa Jul 27 '17 at 16:19

3 Answers3

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When light is propagating in glass or other medium, it isn't really true, pure light. It is what (you'll learn about this later) we call a quantum superposition of excited matter states and pure photons, and the latter always move at the speed of light $c$.

You can think, for a rough mind picture, of light propagating through a medium as somewhat like a game of Chinese Whispers. A photon is absorbed by one of the dielectric molecules, so, for a fantastically fleeting moment, it is gone. The absorbing molecule lingers for of the order of $10^{-15}{\rm s}$ in its excited state, then emits a new photon. The new photon travels a short distance before being absorbed and re-emitted again, and so the cycle repeats. Each cycle is lossless: the emitted photon has precisely the same energy, momentum and phase as the absorbed one. Unless the material is birefringent, angular momentum is perfectly conserved too. For birefringent mediums, the photon stream exerts a small torque on the medium.

Free photons always travel at $c$, never at any other speed. It is the fact that the energy spends a short time each cycle absorbed, and thus effectively still, that makes the process have a net velocity less than $c$.

So the photon, on leaving the medium, isn't so much being accelerated but replaced.


Answer to a Comment Question:

But how the ray of light maintain its direction? After it is absorbed by first atom, how does it later knows where to shot new photon again? Where is this information is preserved?

A very good question. This happens by conservation of momentum. The interaction is so short that the absorber interacts with nothing else, so the emitted photon must bear the same momentum as the incident one. Also take heed that we're NOT a full absorption in the sense of forcing a transition between bound states of the atom (which gives the sharp spectral notches typical of the phenomenon), which is what David Richerby is talking about. It is a transition between virtual states - the kind of thing that enables two-photon absorption, for example - and these can be essentially anywhere, not at the strict, bound state levels. As I said, this is a rough analogy: it originated with Richard Feynman and is the best I can do for a high school student who likely has not dealt with quantum superposition before. The absorption and free propagation happen in quantum superposition, not strictly in sequence, so information is not being lost and when you write down the superposition of free photon states and excited matter states, you get something equivalent to Maxwell's equations (in the sense I describe in my answer here or here) and the phase and group velocities naturally drop out of these.

Another way of qualitatively saying my last sentence is that the absorber can indeed emit in any direction, but because the whole lot is in superposition, the amplitude for this to happen in superposition with free photons is very small unless the emission direction closely matches the free photon direction, because the phases of amplitudes the two processes only interfere constructively when they are near to in-phase, i.e. the emission is in the same direction as the incoming light.

All this is to be contrasted with fluorescence, where the absorption lasts much longer, and both momentum and energy is transferred to the medium, so there is a distribution of propagation directions and the wavelength is shifted.


Another comment:

There was a book which said mass of photon increases when it enters glass... I think that book was badly misleading.

If you are careful, the book's comment may have some validity. We're talking about a superposition of photon and excited matter states when the light is propagating in the slab, and this superposition can indeed be construed to have a nonzero rest mass, because it propagates at less than $c$. The free photons themselves always propagate at $c$ and always have zero rest mass. You actually touch on something quite controversial: these ideas lead into the unresolved Abraham-Minkowsky Controversy.

  • There was a book which said mass of photon increases when it enters glass...i think that book was badly misleading. – Paul Dec 18 '14 at 11:14
  • How is that the angular momentum is conserved instead of, say, scattered randomly throughout the medium? – vaitrafra Dec 18 '14 at 11:20
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    Photons don't have a mass but energy and momentum, and a book that talks about effective mass for photons is not a very good book, I guess... Well... strictly speaking "photons" are nothing but changes in the quantum numbers of a quantum field, so they don't really have an independent existence except as accounting tools for our purposes. The glass slab is the same quantum field as the vacuum outside of it, so the only thing that really changes is how fast the changes of the quantum numbers propagate from one coordinate of the field to the next... but we are unnecessarily confusing the OP. :-) – CuriousOne Dec 18 '14 at 11:21
  • Hey Rod nice summary. I always found the various reasons for light bending when entering a medium, to say the least, unsatisfactory. I would love to hear your input on the matter if you have the time. – PhotonBoom Dec 18 '14 at 12:20
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    Unless I've misunderstood you, this is incorrect. If the photons were being absorbed by the glass, they'd be re-radiated in random directions. Further, atoms tend to be pretty picky about what frequencies they absorb, which is the reason that different things are different colours. There's a Sixty Symbols video about this. – David Richerby Dec 18 '14 at 13:37
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    But how the ray of light maintain its direction? After it is absorbed by first atom, how does it later knows where to shot new photon again? Where is this information is preserved? – Andrey Dec 18 '14 at 19:16
  • @WetSavannaAnimalakaRodVance I really didn't understand a single word. Virtual states huh? I am not HS student, I had physics for 2 years at university :( I know only full absorption when electron goes to higher energy states and later emits spontaneously (the spectrum lines and stuff). But this stuff seriously is beyond me. – Andrey Dec 18 '14 at 22:09
  • @Andrey Sorry. The explanation was originally for a high school student (the OP) so that's what I meant. Think about a Feynman diagram. You have the zeroth order process, which is simply the straight through, noninteracting photon. Then, in quantum superposition with this, you have the first order processes where the photon combines with absorber to become a "raised absorber" and then reverts. You also have second and higher order terms where several transitions and reversions happen. You sum all their amplitudes coherently.The momentum and other conserved quantum numbers belong to the .... – Selene Routley Dec 18 '14 at 22:30
  • ... to the superposition as a whole. It's a little weird I know, but that is the nature of quantum superpositions. To a certain extent this is all question begging: we talk about a superposition as a whole and assume that it doesn't interact with anything, so by definition momentum is conserved. By definition the light knows where to go: if it doesn't, then we're not talking about dielectric propagation. We're talking about something else: for example: sometimes this falls apart and we have an open quantum system: probability "leaks" into continuums of states in lattice vibrations .... – Selene Routley Dec 18 '14 at 22:35
  • @Andrey .. and there are strong interactions with the lattice: energy and momentum are "lost" (to the lattice) and we have fluorescent processes instead, for example. – Selene Routley Dec 18 '14 at 22:45
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    @Andrey Another take on this is that the absorber can indeed emit in any direction, but because the whole lot is in superposition, the amplitude for this to happen in superposition with free photons is very small, because the phases of amplitudes the two processes only interfere constructively when they are near to in-phase, i.e. the emission is in the same direction as the incoming light. – Selene Routley Dec 18 '14 at 22:53
  • @WetSavannaAnimalakaRodVance Dear Rod, thanks for this wonderful answer. Are there any famous papers (that you'd recommend) on the QM treatment of light propagation in such mediums? i.e. showing some of these calculations involving phonon-polaritons and exciton-polaritons. Thanks a lot in advance. – user929304 Jan 06 '15 at 09:18
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    Yeh,but this doesnt explain the fact that light also bends when it enters glass? – Paul Apr 10 '15 at 17:11
  • @Paul No, what I have written does not. To explain that, you need to look carefully at the quantum superposition. The superposition is of a free photon and many excited atoms. The re-emitted photon is therefore effectively transmitted by a phased array: there is a unique direction where the phases line up. Other directions are improbable, by dint of destructive interference. When you do this analysis carefully, you get Maxwell's equations for the medium as the propagation equation for the quantum state. Then Snell's law and all the rest apply. – Selene Routley Apr 11 '15 at 00:05
  • @Paul You may be interested in my answer here. – Selene Routley Apr 11 '15 at 00:06
  • So sad! All incoming photons die :-) Isn't there a chance a small percentage will survive (say, when the glass is thin)? Could a finite-length beam of light have its intensity profile changed this way (e.g. some of the leading-edge photons are delayed and some not)? – Peter Mortensen Sep 27 '15 at 10:46
  • @PeterMortensen Yes indeed, with exponentially decreasing probability with thickness. The quantum superposition that results from the incidence of a one-photon state has a nonzero amplitude for the "straight through" propagation. I'm not sure whether the experiment has been done, but in principle if you shone a very bright source at a thin slip of glass, and could still resolve one-photon events, you would see events beginning after time $t/c$, with peak intensity after time $n,t/c$ where $t$ is the slab thickness – Selene Routley Sep 27 '15 at 11:48
  • @DavidRicherby You seem to be getting confused with absorption by bound electron states, such as a happens in fluorescence or when an electron shifts orbitals. Such interactions happen over nanosecond scales, which is more than enough time for the absorber's momentum / energy states to be changed by its interaction with its neighbors. In other words, the quantum system interacts with the outside world, coherence is lost and we're no longer dealing with pure quantum states when we look at the re-emitted light. The reason for the random re-emission direction in the case you are ... – Selene Routley Jan 01 '19 at 07:07
  • @DavidRicherby ... speaking of is the momentum that comes from the outside world through these "spurious" interactions. When light interacts with matter through virtual states in the way I am describing, the timescales are six or seven orders of magnitude shorter, there are no interactions between absorbers and only direct interactions between light and matter. Therefore, quantum states stay pure and no outside momentum or energy comes into play, therefore no color or direction change. – Selene Routley Jan 01 '19 at 07:10
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    @WetSavannaAnimal Various sources online seem to suggest that the absorption-re-emission explanation is wrong. e.g., https://www.youtube.com/watch?v=CUjt36SD3h8 Can you please comment? – A_P Oct 14 '19 at 21:24
  • the absorption re-emission explanation is completely incorrect – MC9000 Dec 24 '23 at 09:27
  • @MC9000 It is a mind picture, as I stated, that is used when one wants to avoid the notion of a quantum superposition of one photon and excited matter states. "Completely incorrect" is a prescriptive arrogance that has no place in a science that continually builds on itself, and, moreover, leaves the suspicion that the writer actually understands less than they think – Selene Routley Jan 11 '24 at 08:04
  • No, really, it's proven completely incorrect. We would see discreet emission lines and scattering, yet, we do not. As a former spectroscopist, specializing in AE and AA of elements and many different ionizations as well as molecular lattice absorption/emission (amorphous and crystalline). The writer does not understand this topic. – MC9000 Jan 13 '24 at 10:14
  • @MC9000 No really. There would be no dicreeet emission in an utterly amorphous structure like glass. I understand quite a bit more than you arrogantly assume. Especially when one looks at quantum superposition instead. Read up on virtual states. There's no line emission from a continuum of states. – Selene Routley Jan 14 '24 at 11:03
  • That's what I said. This is why light is NOT absorbed and re-emitted by the crystal lattice. If it were, there would be discreet emission and only particular lines would be absorbed and emitted. It's NOT discreet, hence it is absolutely incorrect to say this is the mechanism. This and the direction of emission would necessarily be in the same direction (it is not - atomic and molecular emission is in all directions and unpredictable). This is also not the case. – MC9000 Jan 16 '24 at 04:18
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A classical explanation to supplement Rod's excellent quantum mechanical one:

If you make a Huygens construction of wave propagation (I assume you know how to do that) then every point on the wave front is treated as the source of a new wave of the same frequency and phase. How that wave propagates depends on the medium it encounters. So the Huygens wavelets generated at the exit face of the glass, which "see" only empty space in front of them, simply propagate with the velocity that is appropriate for them - just like the wavelets at the entrance face see a medium of higher refractive index and thus slower propagation (and refraction for non-normal incidence) so the ones at the exit face see the opposite.

As for "heat loss" - if there are loss mechanisms inside the glass photons will get absorbed but there is always a (very small) probability that a photon will make it through - in practice that probability can become so small that you can assume no light will be detected but that is not the same thing as saying "no light can make it through this slab". Likelihood vs. certainty.

steveOw
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Floris
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Your two questions are based on the erroneous notion that light looses energy in going through a glass slab. Light has a propagation speed which depends on the density of the medium. When a light beam goes from vacuum (air) into glass, the only thing that happens is that the wave gets delayed (takes more time to travel the same distance, because of the higher density). Since v = d/t, if t gets bigger v gets smaller (for same d). What this means is that the propagation speed of light in glass gets slowed down. Once it passes the glass, the delay is gone so light resumes its previous propagation speed in air.
Although there is a small loss due to some photons striking the "cores" of the glass molecules, the major loss is due to the light "spreading out" (not coherent light). With the proper light (coherent), one can use "miles" of glass fibers and still have light come out the far end.

Guill
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    Coherence as used by you in the last paragraph means monochromatic, which means (for mono mode fiber) no dispersion. You seem to imply it also means no absorption. Absorption is a function of wavelength, not of coherence. – Floris Aug 12 '15 at 11:19