Note: Why objects reflect light of their color is already answered here:
Physics of color: why do visible colors absorb all colors except the color itself?
My question is whether my following theory of why this happens is also valid (stop me anywhere that I go wrong):
My thinking was that a red object has red light-waves. When a light-wave hits it, the red portion of the light-wave is in phase with the object's red light-waves, and so due to constructive interference the red light is amplified (which is why we see it most).
Conversely, most other color portions of the light-wave that hits the object are out of the phase with the object's red light-waves, and due to destructive interference we don't see these colors.
Is this theory correct?
Best Answer
First, in the article you mention, it says that biological color perception is different from physical color wavelength associations.
When you say a red object, it can reflect many wavelengths of EM waves, which altogether our eye and brain perceive as red.
But let us just talk about an object that would reflect only red wavelength EM waves, so in a physical way it would be red.
What happens with that object is, that the molecular structure of the object is so that the atoms' electrons energy levels (as per QM) have available levels so the atoms can be excited and so some wavelength photons will be absorbed. Other photons will not be absorbed (because they do not fit the available energy level's differences) and so those will be reflected.
The physical color of the object, so the EM waves that come from the object are reflected and emitted both. Some of the emitted are not even visible wavelengths.
So the physical color of the object comes from:
1.reflection of red wavelength photons
2.emission of red wavelength photons
So basically the object will absorb and re-emit some photons, and will reflect some others. All of those photons together will be the physical color of the object (the ones that are in the visible wavelength). So some of the EM waves coming from the object will not even be in the visible wavelength. But you are talking about the visible red wavelength ones so I am going to talk about those wavelengths.
The photons that are reflected are usually elastically scattered, so their wavelength is the same, their energy level is the same, only their direction is changed.
The photons that are absorbed and re-emitted are sometimes not emitted at the same wavelength (some of them). It is because when an atom gets excited, and the electron moves to a higher energy level (as per QM), the electron will come down to the lower level again. But the electron can come down in multiple steps. And in that case it is emitted multiple photons with different wavelengths then the originally absorbed photon. So if a photon gets absorbed, sometimes it is re-emitted with the same wavelength, in that case the electron came down to the lower energy level in only one step.
But sometimes the electron will come down to the lower energy level in multiple steps, and sometimes the electron will be excited so (when absorbing a photon) that it will move to a higher energy level of more then one level away, so it will go up more then one level at one step (with one absorption).
So basically one absorption can have multiple re-emissions, and in that case the absorbed wavelength, in your case red, will have nothing to do with the re-emitted wavelength.
In your case, the reason the object is red (physically again, not biologically) is because:
it is reflecting and emitting red wavelength photons.
And because it is absorbing other wavelengths, non-reds.
And it is only re-emitting red wavelength photons.
Even if it is absorbing non-red wavelengths, it will still re-emit red wavelength photons.