What colour will be an object that reflects some wavelengths while emitting some others?

Question

Often when you make some researches online on the reasons why objects have a specific colour the answer seems to be unique: The colour is given by the wavelengths they(their atom) reflect.
All the others wavelngth are absorbed.
Reading about Black-bodies I've also learnt that objects always emit electromagnetic radiation and that the wavelength emitted is a function of the temperature and the material.
I was wondering what happens in the case of an object ( which is not a black body) that emits some light in the vistble spectrum and reflects some other visible wavelengths coming from a different light source?
Is this possible? If the answer yes, will the light be a combination of the phenomena?

Answer 1

It is possible. The light reaching your eyes would be a mix of the reflected and emmited light. So its color would be whatever that mixture produces.

Answer 2

The color of an object is always a combination of:

1. elastic scattering (metals), mirror

2. inelastic scattering, but most of these photons are deeper in the material and are non-visible wavelength

3. absorption, re-emission, like non-shiny, diffuse surface objects

Now with a metal, most of the photons get reflected, elastically scattered, that keeps the energy and phases of the photons, that is how you get a mirror image. This is called specular reflection. In this case the relative angle of the photons is kept too. Some of the photons are absorbed and re-emitted, but only a little ratio of the whole.

Now with a diffuse reflection, the relative angle of the photons is not kept, it is random. This is for example a white wall. Most of the photons in this case are absorbed and re-emitted, and their energy levels are not kept either. The incident photons' energy level only partially determine the color of the white wall. If you shine yellow color on the white wall, the re-emitted photons will have yellow-ish light. Now in this case the ratio of reflected (elastically scattered) photons is very little. Most of the visible wavelength photons are absorbed, and the atoms in the lattice structure of the wall re-emit white light, that is a combination of all colors. Now if you shine yellow light (combination of different wavelength photons) on the white wall, most of the photons will be absorbed, and the atoms will re-emit white light again, that is the combination of all wavelength photons, but it will be dominated by yellow light.

If you shine blue light on the white wall, it will seem bluish, because most of the visible wavelength photon will be absorbed and the atoms in the lattice structure will re-emit white light (combination of all wavelength photons) that is dominated by blue wavelength photons.

So the answer is that it is always the combination of reflection and re-emission, the ratio changes as per QM depending on the lattice structure.