If you had a container that had perfect white walls on the inside and you somehow filled it with light, would the light always be reflecting inside the container because the color white does not absorb any light rays?
Yes, a perfect reflector container could trap light. There are a couple of problems, of course, that make this impractical: there are no perfect reflectors, and there's no way to INSERT light (or observe it) if the container is closed.
A clean silver surface might reflect 99% of all light that strikes it, but a 1 liter container (circa 0.1m in size) implies that in a single second, there are three billion collisions of light with wall. That leaves about zero light in the container.
The only exception would be if your container were to rise in temperature and emit its own light. Odd as this sounds, if it's not a perfect reflector, a container WILL come into thermal equilibrium with its internal trapped light. The 'perfect reflector' hypothetical situation implies that no thermal contact, and no thermal equilibrium, occurs.
There are two concepts:
- First, if you refer to a perfectly reflecting surface (=a perfect mirror) then the light would always get reflected by assumption.
- Second, if you refer to a system where the walls are in thermal equilibrium with its environment then the amount of emitted light is equal to amount of absorbs light. However, the "photons" are not reflected, however, I guess the color of the paint is not important. It just shifts the amount of absorbed and emitted light. See here.