The energy of a photon of light, or more generally, electro-magnetic radiation, is determined by its frequency.

E = hf
E is energy, f is frequency, and h is a constant, planks constant.

so

f = E/h

so yo could say that a photon's frequency is determined by its energy. So the more energy you give it, the higher its frequency will be.

Blue light has more energy than red light.

The electrons around the nucleus of an atom can not just do any old thing they want. There are discrete, different orbitals that they can occupy. Each of these orbitals has a different energy associated with it. When light strikes an electron, the electron may transition from one orbital to another, absorbing the photon. It might reflect the photon instead.

Right now, I'm looking at a yellow post-it note. White light, comprised of many frequencies, hits the post-it. Most of the non-yellow light is absorbed, most of the yellow light is reflected, do, I see it as yellow.

There are a lot of complications that I left out, but this is the jist of it.

Njorl

Could anyone tell me what is happening on an atomic or subatomic level when materials fade when exposed to light? From njorl's explaination I guess it has something to to with the electrons.

Now for non mirrors the atomic structure dictates which photons will be adsorbed. If there is no energy state availble which corresponds to the energy of the photon it will simply pass the atom by. Often a photon will be adsorbed which jumps an electron past several possible energy states, now when the electron decays to it ground state it may take a different path this means that several lower energy photons may be emitted in place of single one adsorbed. This is why objects have color. All adsorbed visible light excites an electron to some state, from the excited state there is a prefered path which all electrons must take, this path will include a transition which corresponds to the color of the object.

Njorl

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Originally posted by JackHoyes:
Could anyone tell me what is happening on an atomic or subatomic level when materials fade when exposed to light? From njorl's explaination I guess it has something to to with the electrons.

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To put it very simple, visible light gives the electrons some energy, but not enought to leave the atom/molecule. While UV light (where the photons have more energy) is able to kick electrons out of the atom/molecule, so triggering chemical reactions which destroy molecules.

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Originally posted by Integral:
The reason we have the "angle of reflection = angle of incidence" law is because that is the mimimun time path between your eye and the mirror.

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In reflection, the path of least time is also the path of least distance. Draw a line representing a reflective surface. Make two dots on the same side of the line, like this:

point 1 ---------------->.

point 2------------------------------------->.

reflective surface_____________________________________________

Now, there are many paths you could draw from one point to another, bouncing off the line, but the one that will be shortest is the one with angle of incidence equalling angle of reflection.

Something similar can be done with refration, but it is more complicated. Since in refraction, the two media have different speeds for light, the calculation is more difficult. The minimum path distance would be a straight ine, but the minimum path for time has a bend at the refracting surface, which shortens the path in the slow medium, but lengthens the path in the fast medium.

Njorl

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Originally posted by njorl:
It is Fermat's principle of least time.

In reflection, the path of least time is also the path of least distance. Draw a line representing a reflective surface. Make two dots on the same side of the line, like this:

point 1 ---------------->.

point 2------------------------------------->.

reflective surface_____________________________________________

Now, there are many paths you could draw from one point to another, bouncing off the line, but the one that will be shortest is the one with angle of incidence equalling angle of reflection.

Something similar can be done with refration, but it is more complicated. Since in refraction, the two media have different speeds for light, the calculation is more difficult. The minimum path distance would be a straight ine, but the minimum path for time has a bend at the refracting surface, which shortens the path in the slow medium, but lengthens the path in the fast medium.

Njorl

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Originally posted by Integral:... The states available are determined by the atomic structure, in a mirror there are many states very closely spaced, so all visible light photons can be adsorbed, but as soon as they are adsorbed they are reemitted. Emitten photons travel in a completely random direction, some will be readsorbed instantly others will leave the surface of the mirror, but in no perfered direction. The reason we have the "angle of reflection = angle of incidence" law is because that is the mimimun time path between your eye and the mirror. If that law were true at the atomic level then you would only be able to see objects in the mirror from specific view points.

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Hold on, a metal mirror is made of FREE electrons, thus they all are moving IN PHASE with the incident wave. In this case their coherent motion produces secondary oscillating electric field which thus cancels everywhere except one specific direction in which it is seen in phase, and we then call this direction "mirror angle of reflection".

It's same with c=f

The wrong thing here is that f as frequency expresses in rote over sec (not even rote PER sec(you cannot even think about one over sec or Hz)).Therefore it wold have to be ct=n, but then the equation is invalid.

n=rotes;
c=LSpeed
=wave length
t=time

What is the right answer I don't know at the moment.