# Light transmission, reflexion, refraction and diffraction at a atomic level

1. Jun 25, 2007

### SciencePF

How to explain transmission, reflexion, refraction and diffraction at a atomic level.
For example:
1 - how does light passes through a transparent material?
2 - what happens in the boundary of separation of two materials for light to be: partially reflected, partially refracted ad partially transmitted?

2. Jun 25, 2007

### cesiumfrog

It has to do (see https://www.physicsforums.com/showthread.php?t=173745") with the way, as an electromagnetic wave passes by, the electrons are driven by it. This produces a small new electromagnetic wave, which constitutes the reflection on one side and also overlaps with the original wave (the process of interference, from which diffraction is arguably not distinct) adding up to slower transmission (i.e., refraction) on the other side.

Last edited by a moderator: Apr 22, 2017
3. Jun 25, 2007

### olgranpappy

Here's my two cents:

The way I like to discuss this problem is via Fermi's golden rule. One can start off from an atomic level and use the golden rule (see e.g. Sakurai) to calculate the absorption cross-section for photons. But this cross-section is related to the macroscopic dielectric function of continouous electrodynamics (it's roughly the imaginary part). Thus by KK-transforms one can obtain the whole dielectric function and all related quantities (e.g. the index of refraction,reflection, etc.)

4. Jun 25, 2007

### cesiumfrog

If I've understood:

- Interference/diffraction explained by the wave theory of light. Reflection/refraction explained by dielectric susceptibility (that is, the charges in the media can be driven by the EM wave, to some degree).

- Dispersion (the fact that prisms produce rainbows) requires that the dielectric susceptibility depends on the driving frequency.

- the fact that a media can sometimes absorb light (not only reflect/transmit) can be expressed by further writing the dielectric susceptibility as complex. The real part mostly* determines the refractive index and the imaginary part mostly* determines the rate of (exponential) attenuation. (* these parts can only be disentangled this easily if the media is weakly absorbing, e.g. translucent.)

- Using basic QM we can calculate the energy levels of the (simple) atom (and predict which few specific frequencies of light it will absorb).

- Using quantum perturbation theory ("Fermi's golden rule"), we can determine the rate at which some driving EM wave will (or will not) cause atoms to be excited (which we interpret as absorption of a quantum of energy from that EM wave).

- From this we deduce (an approximation to) the imaginary part of the dielectric susceptibility (which, with respect to frequency, will look like a discrete series of very isolated but Doppler-broadened peaks).

- Due to the differential equations of classical dynamics conveniently satisfying certain conditions (like causality), we can use the KK-relation to determine (an approximation to) the real part of the dielectric susceptibility (and therefore also the refractive index, etc, for all wavelengths) directly from just the (approximate) imaginary part.

Did I miss anything? (For one thing, it seems this line of argument would make certain predictions, like that the refractive index will change in a specific/trivial way as the density of a gas changes. Has this all confirmed by experiment?)

Last edited: Jun 25, 2007
5. Jun 25, 2007

### olgranpappy

No, but why stop with gases of atoms? One can also calculate the optical properties of solids and liquids and all sorts of things. Just use the same formalism as for the atomic problem but replace the sum over the "Z" electrons of the atom with a sum over the "N" electrons of the solid... Of course Z is only on the order of 10 whereas N is on the order of 100000000000000000000000... so... well... it takes a little more work but one can still calculate optical properties for a solid without too much hard labor.

6. Jun 26, 2007

### Claude Bile

1 - A handy classsical picture of transmission through a solid is as follows. Imagine a linear chain of atoms (so we are only working in 1 dimension). Now imagine an light wave incident on the 1st atom in the chain, and let us further imagine the light wave to be sinusoidal. The incoming EM field will act as a driving force on the 1st atom, causing it to polarise. Since the driving force (EM field) varies as a sinusoid, then the position of the electron cloud must also vary as a sinusoid, essentially acting as a driven harmonic osciallator.

Now the motion of the 1st atom will act as a driving force for the second atom. The second atom will act as a driving force for the 3rd atom and so on. No absorption occurs because we are not exciting the atoms at an atomic resonance. The wave slows down by a factor of n, due to the accumulated phase difference between the drive force and the motion of the electron cloud.

A fairly layman description I admit, but I feel that classically at least it paints a decent picture as to what is happening on an atomic level.

2 - Partial reflection and transmission (transmission and refraction are really the same thing here since the transmitted wave is invariably refracted) will occur at every interface in the general case. One will almost always get reflection, the amount of light reflected is governed by Fresnel equations. The only exception to this is polarised light that is incident upon a surface at Brewster's angle.

Transmission can be suppressed when moving from a higher refractive index to a lower refractive index if the angle of incidence is high enough - this is total internal reflection.

So unless you have total internal reflection or you are using polarised light incident at Brewster's angle, you will get partial reflection and transmission.

Claude.