Explanation of Reflection and Refraction

Question

I've studied Huygens wave theory. But Huygens assumed that light is a wave in 'ether' which we know doesn't exist. Light is basically a electromagnetic wave. So I want to understand how light actually interacts with matter both as a photon and as a wave to produce Reflection and Refraction. I want to understand why the angle of incidence and angle of reflection are same. Why does speed of light decrease in an optically denser medium? And what does it have to do with the change in angle of refraction?

I know that when a photon is incident upon an atom, it exicites an electron to a higher energy orbital and when the electron again falls down on a lower energy orbital, it releases a photon. But why the direction of the incoming photon and the released photon have to be same? What happens in the case of refraction?

I'm not a graduate yet. So I think I'll have to wait until I'm able to read Quantum Electrodynamics. — Apoorv Potnis, Sep 08 '16 at 03:09
Then you can at least read the shorter introductory book on QED by feynman himself. It hardly has any rigorous mathematics, but quite a few explanations. — Lelouch, Sep 08 '16 at 03:11
Basically he has this technique he found known as 'feynmans sum over histories', which considers all possible paths a photon can take between A and B, and shows that there is greatest probability for the photon to move via paths, which most of our classical models predict. — Lelouch, Sep 08 '16 at 03:14
So, a photon has largest probability of getting reflected and reflected along the angles which laws of Reflection and refraction predict. — Apoorv Potnis, Sep 08 '16 at 03:18
Yes. But its best to wait for another answer here. Its not wise to take every word of a high schooler, especially when it comes to QED. — Lelouch, Sep 08 '16 at 03:19
To show all these things, Maxwell used the help of the entire vector calculus. If you have a better understanding on vector calculus, then I prefer reading Introduction to Electrodynamics by Griffiths for starters. — UKH, Sep 08 '16 at 04:24
I think these are way too many questions here. You should try to raise a single specific problem per post, otherwise the question becomes too broad. — glS, Sep 08 '16 at 09:00

score 2 · Answer 1 · answered Sep 08 '16 at 08:22

Propogation of light in matter is usually treated as an electric wave in a medium with a dielectric constant other than that of vacuum. That's the wave picture, and it doesn't depend on 'luminiferous ether'. The reflection condition at a flat surface comes from the polarization in the medium giving rise to two outgoing wave solutions, one inside the medium and one outside. If the dielectric constant is very high (like, in a metal below the plasma frequency), the energy in the reflection is 100% of the incoming wave.

The particle picture, is that photons going through a chunk of glass are scattered at a multiplicity of sites (every atom), and the scattering is mainly forward-directed, but with a time delay (or, if you prefer, a phase shift that creates a group delay). When the photons are long wavelength compared to atom size, the scattering centers don't matter (many atoms overlap each single photon), only the delay. For X-rays, the scattering centers give rise to diffraction peaks, not just 'refraction', of course. It's not clear to me how diffraction can be modeled with photons as particles.

Never think a photon 'excites an electron to a higher energy orbital' in transparent media, it is an E-field that changes the orbitals inside an atom. This is generally called the Stark effect.
Polarizing the atoms, yes; electrons doing orbital hopping, no. I'll disregard Zeeman effect (magnetic), because significant magnetic interaction is rare in transparent materials.

Reflection and refraction angles and coefficients are, in the particle picture, determined by energy and momentum conservation, with the photon-in-the-medium having some mass. The angle of incidence equalling the angle of reflection is a subtle consequence of time-reversability and symmetry.

In both pictures, the answers that come out are the same: light slows, because the material includes charges that can be displaced, atoms that can become dipoles, and mainly the work done on that displacement or polarization does not get absorbed, but after a slight delay is returned to the wave.

I'm not accepting any answer as I've not learnt physics that much as understanding QED. I didn't get the "actual" and exact answer but anyway thanks for the answer :-) — Apoorv Potnis, Sep 11 '16 at 07:19

HolgerFiedler · Answer 2 · 2016-09-08T06:19:52.313

In the case of reflection with equal incomming and outgoing angles we are tolking about statistical processes. It is obvious that for elastic scattering the energy and the momentum of the involved photons lead to this reflection angle. The real behaviour of single photons is an other thing for some reasons:

they reach the mirror with different phase of their electric and magnetic field components
the absorbing and re-emitting molecule is in different position and rotation to the ideal point of the incident, and it is in vibration too.

Perhaps this are not all boundary conditions. More important is the statement that this law depends from the material and cleanness of the mirrors surface and the wavelength of the light, that has to be reflected. For different wavelengths or other materials or not polished surfaces one does not get the awaited result.

To say it shorter, sometimes the reflection is ideal in a statistical meaning and behaves like an unelastic scattering.

There is a case the equal angles are not holds more. Under some angle the outgoing light is polarised and this is a asymmetry and the unelastic scattering can't be more the precise model.

I'm not accepting any answer as I've not learnt physics that much as understanding QED. I didn't get the "actual" and exact answer but anyway thanks for the answer :-) — Apoorv Potnis, Sep 11 '16 at 07:20

Explanation of Reflection and Refraction

2 Answers2