As above… Surely it must go somewhere?
My child asked me this question and it was a difficult one to answer.
[Physics] Where does the light go when I turn off the light switch
everyday-lifevisible-light
Related Solutions
1) Yes, in the real world, only a tiny portion of the light scatters by the Rayleigh scattering. This may be reinterpreted as the simple fact that generic places in the blue sky are far less bright than the Sun. It means that the generic places of the sky become blue but the Sun itself remains white. For the same reason, distant mountains keep their color. Also, the distant mountains don't increase the amount of blue light from other directions much simply because the intensity of light reflected from distant mountains into our eyes is vastly smaller than the intensity of light coming directly (or just with Rayleigh scattering) from the Sun to our eyes. And even if it were not smaller, e.g. when the Sun is right below the horizon and the mountains are needed, we won't be able to easily distinguish that the blue sky actually depends on the mountains.
Rayleigh scattering is caused by particles much smaller than the wavelength, i.e. individual atoms and molecules, so it doesn't really matter which of them they are. The rate of Rayleigh scattering is therefore more or less proportional to the air density which means that a vast majority of it occurs in the troposphere, especially the part closer to the surface.
2) The changing atmospheric density only impacts the angle of the propagation of the sunlight substantially if the atmospheric density changes at distance scales comparable to the wavelength. If the length scale at which the density changes is much longer than that, the impact on the direction of light is negligible and calculable by Snell's law.
If you ever watch Formula 1 races, you may see some fuzzy waving water-like illusion near the hot asphalt. This is indeed caused by density fluctuations caused by the variable heat near the asphalt (well, a campfire could have been enough instead of Formula 1). However, in this case the direction of light only changes slightly because the regions of hot and cold air are still much longer than the wavelength (half a micron or so).
If you think about ways how to get density fluctuations comparable to the wavelength which is really short, you will see that the source is in statistical physics and the naturally fluctuating air density due to statistical physics is actually nothing else than an equivalent macroscopic description of the Rayleigh scattering! When you calculate the Rayleigh scattering, you may either add the effect of individual air molecules; or you may directly calculate with a distribution of many air molecules and the source of the effect is that their density isn't really constant but fluctuates. So these two calculations are really equivalent. They are the microscopic and macroscopic description of the same thing, like statistical physics and thermodynamics.
If the blue light manages to come from a direction that differs from the direction of the source of light, the Sun, then – assuming that the atmosphere doesn't emit blue light by itself, and it doesn't (at least not a detectable amount of it) – it is scattering by definition. To get a substantial change of the direction, you need small particles, and that's by definition Rayleigh scattering. So there's no other source of the blue sky than the Rayleigh scattering – although the Rayleigh scattering may be described in several ways (microscopic, macroscopic etc.).
Well, there's also the Mie scattering – from particles much larger than the wavelength, especially spherical ones, like water droplets. However, for the Mie scattering to be substantial, you need a substantial change of the index of refraction $n$ inside the spheres, which is OK for water. Also, the Mie scattering is much less frequency-dependent (because $n$ only slightly depends on the frequency, nothing like the fourth power here) than the Rayleigh scattering so it doesn't influence the overall color much. Not only during the sunset, some grey-vs-white strips on the clouds near the horizon are caused by the Mie scattering. The Rayleigh scattering really has a monopoly on the substantial change of the color.
Different molecules and different crystalline structures have frequency dependent absorption/reflection/transmission properties. In general, light in the human visible range can travel with little absorption through glass, but not through brick. UV can travel well through plastic, but not through silicate-based glass. Radio waves can travel through brick and glass, but not well through a metal box. Each of these differences has a slightly different answer, but each answer is based on molecular resonance or crystalline structure (or lack thereof) or electrical conductivity.
Bottom line: There isn't one general answer for why $\lambda_A$ goes through material X but $\lambda_B$ doesn't.
Best Answer
Light travels at 300,000,000 meters per second. There is a very small period of time after switching it off where there still are photons from the bulb in the room. But these get absorbed/scattered by the wall and thus you don't see them. This what happens to all of the light that came out of the bulb significantly before you turned it off as well -- the photons (or EM radiation, take your pick) are no longer present in the room in their initial form. When the light hits the wall, some of it is absorbed, which heats up the wall ever so slightly (just how sunlight can heat things. For that matter, you can warm your hand even with a flashlight if you hold it in place fo a couple of minutes. You can do the same by placing your hand near a lightbulb). This heat dissipates through the wall. Also, some of it is scattered back with a shift in wavelength -- the electromagnetic radiation is still present in the room, but it is no longer visible. It could be radio waves/infrared waves or even ultraviolet waves. It's harmless, though. Eventually, it all leaves the room in the form of heat.
Remember, a light bulb doesn't "contain" light. It contains a filament, which glows when you pass electricity through it. This exactly what happens when you heat a metal -- the electrons get excited and start emitting visible light. Here, you are supplying energy to the bulb (in the form of electricity), and it gets heated, giving off light. For CFL bulbs the system is more complicated, but it still involves the excitation of electrons.