Understanding Refraction of Light: Photon Absorption and Beam Direction

In summary, the refraction of light is not considered to be a photon absorption-emission process, but rather a phenomenon related to the speed of light in a given medium. From a classical perspective, the direction of the reflected and refracted photons can be explained by the conservation of momentum and the shape of the boundary between two media. From a quantum perspective, the incoming photon stimulates the electron at the same frequency, causing it to emit a reflected photon. However, this is not considered absorption as it does not change the energy level of the electron. Instead, the energy of the incoming photon can be transferred to a phonon or plasmon mode in the solid. The process of refraction does not always result in a change of direction or
  • #1
what_are_electrons
Is the refraction of light a photon absorption-emission process?
If so, why doesn't the beam scatter in all directions?
 
Physics news on Phys.org
  • #2
Classically, it is not considered as absorption. It is n and related to the speed of light in the media, while absoption is kappa. the relation is N = n + ikappa. This is a from the 100% wave point point of view.

Quantum mechanically, I would say that the fact that there is a plane boundary between the two media is important, and that the specular direction is closely related to the shape of the boundary. (This would also be true using the classical/plane wave point of view.).
 
  • #3
The process is most definitely not absorption, the incoming photon stimulates (or drives) the electron at the frequency of the photon, which in turn generates another photon (the reflected photon). Note that this process is not absorption as it does not promote the electron to a higher energy state.

The direction of the reflected photon is due to the parallel component of the momentum of the photon being conserved.

Claude.
 
  • #4
Sorry, I wrote reflection instead of refraction.

The same argument can be applied to refraction. The presence of a higher refractive index causes the parallel component of the photon's k vector to reduce by a factor of n (Where n is the refractive index of the glass). This is embodied in Snell's law.
 
  • #5
Can someone explain this to me in English?
 
  • #6
Claude Bile said:
The process is most definitely not absorption, the incoming photon stimulates (or drives) the electron at the frequency of the photon, which in turn generates another photon (the reflected photon). Note that this process is not absorption as it does not promote the electron to a higher energy state.Claude.

I agree with the first sentence, this is classical optics. But when the electron is "stimulated/driven", doesn't it gain energy? From a quantum point of view, doesn't it have to change energy level? Why isn't this absorption?
 
  • #7
Why should the electron have to change an energy level from a quantum point of view?

The energy from the incoming photon goes into emitting a new photon, not the electron.
 
  • #8
Because it seems to have done so from a classical point of view. It was relaxing, a photon came, it oscillated, a photon went, it relaxed again. During the seemingly brief time it oscillated out the next photon, it seems to have been affected somehow or done something, since the before and after photons are not the same.
 
  • #9
I understand your confusion, but this is very difficult to explain without the aid of a diagram.

The picture that of 'relaxed' and 'excited' electrons and that they spontaneously transform from one into the other is an oversimplification. If you are interested, look up pertubation theory, as it gives a better insight into what happens when a photon is incident on an electron.

Ask yourself the following questions -

How does light propagate through glass?
What happens if an incoming photon has an energy that does not correspond to an allowed energy transition (and thus cannot be absorbed), does the photon simply ignore the electron?

Claude.
 
  • #10
Excellent, thanks.
 
  • #11
What happens to the incoming photon when it "stimulates" the electron? If it is not absorbed, then you will end up with 2 photons for every 1 that goes in. The only way this can happen is if those 2 have half the frequency of the first. We know this can't be true because when you shine a certain colored light into glass, the same color comes out.
 
  • #12
Just keep in mind that you don't really have distinct energy levels in solids like you do in individual atoms or molecules. You have bands. In a conductor or semi-conductor, the photon can excite the electron into the conduction band - though this is not required - but remember that most transmitters of light are insulators or dielectrics. But if the photon can not excite an electron, it can still excite a phonon or plasmon mode.

I'm sure you'll find a decent explanation in Johnson or even Griffiths (perhaps) - or whatever is the standard E&M text where you are.

"Why doesn't the beam scatter ?" - short answer : to conserve momentum.
 
Last edited:
  • #13
Gokul43201 said:
it can still excite a phonon or plasmon mode

hmmm...thats over my head. Is that supposed to answer my question?
 
  • #14
ArmoSkater87 said:
hmmm...thats over my head. Is that supposed to answer my question?

A mode is a way to vibrate (a note on a guitar string, or a harmonic of that note, 2f, 3f, etc.). In a crystal, the atoms can vibrate in an organized manner and also have modes. When they vibrate in such a mode, we say a phonon is propagating. Gokul43201 is saying that if the energy of an incoming photon is not sufficient to change the energy (frequency) of an electron, it may still change the energy (frequency) of vibration of the atoms (associated "particle" : phonon, in analogy to photon, but for vibrating matter = sound = phono, phonograph, phonetics etc.). (Plasmon : not very familiar with this but probably the same for a plasma, or free electron gas in a a solid lattice).

I don't see why you're saying that "If it is not absorbed, then you will end up with 2 photons for every 1 that goes in." This is not necessary, nor true IMO, except in non-linear frequency doubling.
 
  • #15
Gonzolo, you are describing a photon/phonon interaction. This type of interaction usually involves some momentum transfer and thus a direction and/or wavelength shift of the incident photon.

It is entirely possible (probable in some cases) that the photon will simply pass through the electron cloud. The E-field of the photon interacts with the atom by polarising it. This polarisation oscillates in phase with the E-field of the photon and propagates through the medium as a polarisation wave. The photon then emerges from the medium, unchanged in direction, wavelength or polarisation (Assuming that exotic effects such as the Faraday effect, or nonlinear effects are assumed not to be present).

Claude.
 
  • #16
Claude Bile said:
It is entirely possible (probable in some cases) that the photon will simply pass through the electron cloud. The E-field of the photon interacts with the atom by polarising it. This polarisation oscillates in phase with the E-field of the photon and propagates through the medium as a polarisation wave. The photon then emerges from the medium, unchanged in direction, wavelength or polarisation (Assuming that exotic effects such as the Faraday effect, or nonlinear effects are assumed not to be present).

Claude.

Humm... However, when this occur, doesn't the polarization itself take in energy from the field? If you have a bunch of randomly-oriented dipoles, and you pass a field through it that will cause a net torque on each dipole to cause a specific orientation, this requires that work being down by the field. So each time the field oscillates, the act of polarization will use up some energy out of the field. So if what you said is the mechanism, there will be energy loss after passing through the cloud.

I'm guessing that this scenario is similar to the beam-loading effect in particle accelerators. If you have a large amount of charge particles (electrons) in a part of the beamline, the RF field that you put in isn't the same as the RF field that is transmitted, because the charge particles literally suck up some of the RF field that is causing it to oscillate in the RF cavity. It is why dark currents in accelerating structures and photoinjectors are not desirable.

Zz.
 
  • #17
Claude Bile said:
Gonzolo, you are describing a photon/phonon interaction. This type of interaction usually involves some momentum transfer and thus a direction and/or wavelength shift of the incident photon.

Could this change in momentum, direction and wavelength be linked to what a change in n does at each interface? Varying index of refraction do change a beam's direction and wavelength.

Claude Bile said:
It is entirely possible (probable in some cases) that the photon will simply pass through the electron cloud. The E-field of the photon interacts with the atom by polarising it. This polarisation oscillates in phase with the E-field of the photon and propagates through the medium as a polarisation wave. The photon then emerges from the medium, unchanged in direction, wavelength or polarisation (Assuming that exotic effects such as the Faraday effect, or nonlinear effects are assumed not to be present).

When polarisation occurs, doesn't anything change quantum mechanically? It also seems to me that the electron/atom/dipole/primitive cell did gain energy, because when the photon is through, the dipole will spring back to its original state if the photon polarisation stretched it. Also, photon beams are not always polarized.

Nonlinear effects are related to saturation effects when very high fields (many many photons) or "sensitive" materials are involved. The question here as I understand it is only about one photon or the common linear regime in general.
 
Last edited by a moderator:
  • #18
The direction change at an interface is due to continuity conditions imposed by Maxwell's equations at a boundary.

Quantum mechanically, the photon does perturb the electron slightly, but the perturbation is over a very short time scale (femtoseconds or less). It is entirely possible that the electron experiences a change in energy over this time frame, but this energy is quickly removed as it propagates through the medium.

And no, photon beams are not always polarised (in fact, obtaining perfect polarisation is quite impossible), a single photon will posess a polarisation.

ZapperZ, I am not sure the exact details, as this is beginning to move outside my field of knowledge. Your question is a good one, I have a thought;

the situation you describe would seem to indicate that a photon passing through a medium would leave a trail of aligned dipoles in its wake. This (to my knowledge) is not observed in practice. Perhaps there is a quantum mechanical explanation?

Claude.
 
  • #19
What I understand from perturbation theory is that it is a change in the ground energy. Like when the classic mass-on-a-spring is in a G field. If this is the case, I'm almost convinced that polarizing an electron cloud doesn't necessarily change the electron's energy level.

Just like the mass-on-a-spring doesn't change energy when it is switched from one G field to another.
 

Related to Understanding Refraction of Light: Photon Absorption and Beam Direction

1. What is refraction of light?

Refraction of light is the bending of light as it passes through different mediums, such as air, water, or glass. This occurs because light travels at different speeds in different mediums, causing the light rays to change direction.

2. How does photon absorption affect refraction of light?

Photon absorption is the process in which light particles, or photons, are absorbed by atoms or molecules in a material. When light passes through a material, some photons may be absorbed, causing the light to lose some of its energy and change direction. This can affect the overall direction and intensity of the refracted light.

3. What determines the amount of refraction in a material?

The amount of refraction in a material is determined by the speed of light in that material, also known as the refractive index. The higher the refractive index, the greater the change in direction of the light as it passes through the material. The refractive index is dependent on the density and composition of the material.

4. How is the direction of a refracted light beam determined?

The direction of a refracted light beam is determined by Snell's Law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speeds of light in the two mediums. This law helps to predict the amount and direction of the refraction of light as it passes through different mediums.

5. What are some real-life applications of understanding refraction of light?

Understanding refraction of light is important in many fields, including optics, engineering, and medicine. Some real-life applications include the design of lenses for glasses and cameras, the creation of microscopes and telescopes, and the development of medical imaging techniques such as MRI and ultrasound. Additionally, understanding refraction helps in understanding the behavior of light in the natural world, such as the formation of rainbows and mirages.

Similar threads

  • Quantum Physics
Replies
4
Views
203
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
233
Replies
1
Views
673
Replies
8
Views
943
Replies
43
Views
2K
Replies
4
Views
981
Replies
1
Views
476
Back
Top