Here's a thought experiment:

Suppose you have a light source and a camera in a box that that has no exit or entry points. Nothing can get in or out. We observe the activity within the box via the camera, but for this purpose, no light can escape through the camera. Just humor me here.

If the light goes out, one would expect for the interior of the box to go dark, right? But why? If nothing can get in or out of the box, the photons should still be in the box, and thus remain lit regardless of the source of light being on or off. The question is, therefore, where did the photons go?

This is a rhetorical question, and hopefully get some insights to satisfy my curiosity.

Cheers!

Absorbed by the material from which the box was constructed.

  The law of entropy:

   All energy ultimately degrades to heat energy.

  (As zachsdad said...)

Absorbed? Doesn't light either reflect or refract when in contact with matter? When the light is on, the fact that we can see the walls at all indicates that photons are reflecting off the walls. What if the walls were made from a purely reflective surface like that of a mirror? When the light goes out the box would still be dark.

So where do the photons go in that case?

even the best mirrors do not return all of the energy that hits them.  eventually all of the light energy would be absorbed by the mirror's material, and could be detected in the form of heat on the walls of the box.

   I am not a quantum physicists,but simply put, light is a form of energy in a particular state.  Let's assume it is light in the visible range of the spectrum (of a certain wavelength).  When the light is turned off in this mirrored box (actually it doesn't really matter what the box is made of), the light energy is dispersed through several means:  heating the air, refraction by the air, heating the mirror, refraction off the mirror--etc... so it very rapidly losses its energic state, falling from visible light to a lower energic state.  If we had infrared ( and extremely sensitive) eyes, we would see the residue of this energy dispersion- for a short time.

   Then that heat is further dispersed...

  Now no more silly questions.

  Think of light as energy in a particular state, not a "thing" having permanence of some sort.

They are still absorbed. When a photon strikes an atom, there are two cases:

• The electrons in the atom are at energy levels that allow them to absorb the energy of the photon completely, by rising to higher levels.
• The electrons in the atom are at higher levels, which means any absorbed photon's energy must be re-radiated (by emitting a "new" photon).

So, the photons will bounce around in the mirrored box for awhile until all instances of these two cases have been satisfied. The box will warm slightly, and then cool as the electrons in the affected atoms have settled into their "normal" state. As time passes, the energy is transferred through the material that forms the walls. Eventually, equilibrium is restored.

Thank you everyone for putting up with my "silly" questions. So light photons are not things that can be contained but excess energy released when energetic states of matter thresholds ar reached.

The idea that photons are emitted in the first place though says that photons are stored in the contaner from which they are emitted. Energy in this form, E, is contained in mass, refering to E=mc^2. So mass then equals E/c^2. It is a pointless quandary however.

Thanks again.

Not precisely. It isn't the photons themselves that are stored. Photons are defined to be the particle that represents the energy for transmission through space.

Consider the transmission of a "ray" of light through a sheet of glass: it is neither a "solid" beam of light nor a "connected" stream of photons passing through the glass without interacting with the atoms in the glass. Instead, it is a series of photon/electron transitions, wherein each successive photon is absorbed and re-emitted in turn.

According to quantum electrodynamic theory (QED), the re-emitted photons are actually emitted in all possible directions (including back along the path from which they entered), with a higher probability of transmission along the direction of travel (assuming the sheet of glass is perfectly flat and sufficiently thin).

If the glass is replaced by, say, a similar thickness of graphite or other opaque material, the situation is the same except that the index of refraction is very much greater and the energy of the photons is absorbed rather than transmitted. There is, however, a very small probability that one incident photon will trigger a chain of absorption/remission events from atom to atom and its energy will make it "through" the plate of material.

The photons are the means of transmitting the quanta of energy. The transfer of energy from a photon to an electron is the means of absorbing or reflecting (reflection being absorption followed by re-emission within the "space" of a single atom). The same kinds of actions are happening in both transparent and opaque materials.

So, where is the quandary?

Only in my brain apparently...

So this is a chain reaction, and not a stream. In that case, light emitted from a source hits an atom of material, like glass, and gets absorbed into the atom, the extra energy the atom now posesses releases a new photon, and transfers its energy into the next atom, and so on, just like in a thermonuclear detonation, except there a neutron is the trigger and not a photon.

Only in my brain apparently...

So this is a chain reaction, and not a stream. In that case, light emitted from a source hits an atom of material, like glass, and gets absorbed into the atom, the extra energy the atom now posesses releases a new photon, and transfers its energy into the next atom, and so on, just like in a thermonuclear detonation, except there a neutron is the trigger and not a photon.

Yep. The trick here is (as cyberpatzer indicated) not to think of photons and electrons as "things" but as states of energy.

In an atom, an electron represents a quantum of energy at an average radius from the nucleus that corresponds to the potential that keeps the electron at that radius (distance). If you hit the atom with a photon, it represents an incoming energy level that is greater than that currently in the electron's orbit. If the atom can absorb the energy, then it does absorb the energy, causing the electron's orbital radius to increase. That is, the electron jumps to the new energy level represented by the energy added by the photon being absorbed. If this increase in energy makes the electron's orbit unstable, then it re-emits the added energy (one photon's quantity of energy) and the electron settles back to its original energy level (orbital radius).

In stimulated emission (as Einstein postulated and lasers deliver) there is a strong preference for the forward direction. That is the extreme case, where a solid (in the laser's case, crystalline) material exhibits an architecture that prefers the forward direction and can actually amplify the re-emission (assuming it's fed by a separate power source). The opposite case is represented by absorption by a solid which has no preferential direction of transmission. An easy case to think about is a fog, wherein a gas or liquid suspension of particles offers no easier forward path than sideways path, so the photons are re-emitted in random directions, leading to dissipation of the "forward" light. The extreme case of this is an absolutely opaque medium where all the incoming photons' energies are absorbed and none can be remitted.

In between, you have other media which both absorb and re-emit, or re-emit at non-visible wavelengths (such as infrared) as mentioned earlier.

They go to the same place as my left socks. 

Photons wonder in the Universe unil they hit something that absorbs them.

Well to answer this question I am going to go M theory on you.  :p

 If photons are open ended strings as the theory suggests, then the longer thier exsistence the more energy they will lose by interacting with other matter and energy.  So this would mean over variable time, that if they are not converted to another partical through some interaction, that the photons frequency and amplitude will decrease, shifting it from say a gamma ray to a visible light ray to say a radio wave as it interacts with other constitutes.

 So now matter and energy is never created or destroyed so we know that the mass of a photon will always be converted into other constitutes of equalling mass, if the photon interacts correctly, it will cease to exsist but it's mass and energy will "live on" in other interactions or particles/strings.

At least thats what I think...  Not that it says much.  :p