Why is the Law of Reflection so?

Question

So the law of reflection states that light incident on a surface should reflect away at the same angle to the normal as it came in at. But why should this be the case? Why, if it came in at 30 degrees to the normal, can't it leave at 58 degrees or something? And as far as I have been taught, reflection would just be a photon increasing the energy of an electron, which then loses that energy re-emitting the photon. So why does the photon have to be emitted at these specific angles?

Answer 1

What you're thinking of is photo-excitation of and re-emission from an atom's bound electron. Light reflection is the result of conduction band electrons that move freely in a thin aluminum coating applied to glass to make it a mirror. These electrons are accelerated by the electric field of the incident light and redistribute themselves until the electric field is zero at the surface. This all occurs fast enough to keep up with the fact that the electromagnetic field is oscillating at high frequency (at least well into the ultraviolet). From the principle of superposition, this corresponds to the sum of the electric fields of the incident and a reflected wave of equal amplitude transmitted by the conduction electrons' current destructively interfering at the surface. A little math (see previous Jackson reference) and you get the angle of incidence = angle of reflection.

Answer 2

You can show the law of reflection is a consequence of either classical E-M theory or quantum electrodynamics. This takes more analasys than I'll put into an answer. See Jackson E-M or Feynman QED if you really want it explained.

A sort of half-assed proof (that has a grain of truth in QED) is to appeal to the principal that light exremizes travel time from point A to point B. If the light reflected from any other angle, it would have a longer travel time than the actual path it takes (where the reflected angle = the incident angle).