# Light coming off a mirror ?

1. Apr 15, 2010

### cragar

How come when light hits a mirror and then gets re-emitted , that the light comes off at the same angle it hits, when the light hits the glass or what-ever , when it interacts with the electrons in the atom , these electrons are orbiting so why cant they re-emit the light at some other angle .

2. Apr 15, 2010

### mathman

I suggest Feynman's book (QED) for a complete discussion.

3. Apr 15, 2010

### cragar

ok thanks

4. Apr 15, 2010

### sophiecentaur

You are basing your ideas on the simple, isolated, Hydrogen Atom model and assuming each photon interacts with just one atom. The light doesn't react with individual electrons, in this case; it interacts with the bulk of the reflecting surface - which is metallic in a mirror - and the result produces a reflected wave that obeys the laws of reflection.

5. Apr 16, 2010

### cragar

why does it obey the law of reflection .

6. Apr 16, 2010

### sophiecentaur

A guy called Huygens came up with an almost perfect description of this in the 17th Century!
If you imagine every part of the reflecting surface receiving its own bit of an advancing plane wavefront. It will absorb and then re-emit a wave with a very slight phase delay, which is common to all the other surrounding points on the surface. This behaves like a point source (a Huygens 'secondary wavelet') from which waves radiate, backwards, over a hemisphere. All the neighbouring points on the surface are doing the same thing. The result, if you add up all the secondary wavelets, is to produce just one plane wavefront in a direction which corresponds to the symmetrical rule of the law of reflection. Waves in any other direction don't exist because the secondary wavelets all cancel each other. When the surface is not perfectly smooth, of course, the geometry means that this cancellation doesn't occur and you will get light in other directions.
They use what is, effect, Huygens Principle, even today, to solve problems of diffraction reflection and refraction. Not a bad shelf life!

7. Apr 16, 2010

### Andy Resnick

It doesn't have to at all: look up the BRDF, Lambertian surfaces, or any rough surface scattering report.

You are asking (presumably) about planar surfaces with certain boundary conditions (i.e. corresponding to a perfect conductor). Then Maxwell's equations predict the field is reflected specularly. Relax those assumptions, and you get diffuse scattering.

The light is not 're-emitted', it is *elastically scattered*, unless you want to discuss Raman processes, 4-wave mixing (retroreflection), a surface painted with fluorescent ink, etc.

8. Apr 16, 2010

### sophiecentaur

@Andy Resnick
You are using a sort of hybrid classical / quantum approach for this. Fair enough. Your 'elastic scattering' still has to be, deep down, a quantum effect, of course. But this is ok if you think of the photons interacting with the material as a whole and not with individual electrons-in-atoms, which only applies in a gas, in any case. Maxwell assumes classical behaviour.

I think the OP was concerned that the individual photon - electron interaction would not support specular reflection. And it doesn't. Light entering a gas may be absorbed and re-emitted randomly - hence absorption spectroscopy - but that's gases, not condensed matter surfaces..

9. Apr 16, 2010

### cragar

10. Apr 16, 2010

What the others said can also are true, but Albert Einstein suggested Photoelectric effect. Which one of the theories that i like very much:)

i didn't read the other theories, I'll do it now.Thank you Sophiecentaur, Andy Resnick and Mathman.

You can find more at : http://en.wikipedia.org/wiki/Photoelectric_effect

11. Apr 16, 2010

### Andy Resnick

Since the OP specifically mentioned 'a mirror', I think it unwise to approximate the device as a gas.

12. Apr 16, 2010

### sophiecentaur

Of course - in which case "an electron orbiting an atom" is not a suitable model to explain what happens and leads to a wrong conclusion.

13. Apr 16, 2010

### Antiphon

These answers are all great and show there's more than one approach. Here's the simplest one I can think of:

The angles of incidence and reflection are equal for the same reason as for a (perfect point) ping-pong ball- conservation of momentum.

14. Apr 17, 2010