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Lamps that produce visible light can light up a whole room because the ceiling and the walls diffuse light. This way, when there is a single lamp in a room, something that is in the shadows is not completely dark. If I set up an infrared lamp in a room, will the whole room be lit by infrared light like with a regular lamp, or will it be absorbed by the walls ? If it is not, will the wavelengths be the same as the ones emitted by the lamp ?

Thanks !

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  • $\begingroup$ What's your guess? Why do you expect the walls do behave differently for infrared light? $\endgroup$
    – elcojon
    Jan 11, 2013 at 22:15
  • $\begingroup$ @elcojon Because dielectric properties of materials are a function of wavelength, and at 20 µm it may be vastly different compared to 0.5 µm. $\endgroup$
    – gerrit
    Jan 11, 2013 at 23:04

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There are two aspects to be considered here:

  • Does the wall reflect or absorb the radiation?
  • If the wall reflects, is this specular or diffuse reflection?

For both of these sub-questions, the answer depends on the material that the wall is made of and on the wavelength of the infrared radiation. At wavelengths close to the visible (for example, at 0.8 µm), the behaviour is likely similar to the visible — so similar, in fact, that there are meteorologists that refer to any radiation up to 1 µm as "visible", even though we clearly can't see such radiation. At 20 µm, however, behaviour may be vastly different than at 0.5 µm.

The exact answer depends on too many factors, so the answer is: it depends. If you can specify exactly what material the wall is made of, and in what wavelength you are interested, a more precise answer can be given.

For example, for the walls of the Ice Hotel, mid-infrared radiation (10–20 µm or so) is certainly absorbed a lot more than visible light!

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  • $\begingroup$ Thanks ! I just wanted to know whether it was possible to "illuminate" a whole room with infrared light, so I guess it is possible if wavelengths around 1µm are used. $\endgroup$ Jan 18, 2013 at 18:35
  • $\begingroup$ @personne3000 At 1 µm, yes. At far infrared, say 100 µm, I guess so but I'm not sure. $\endgroup$
    – gerrit
    Jan 18, 2013 at 21:44
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The diffuse reflection of light off a boundary will still occur at wavelengths outside of the visible spectrum provided that the absorptivity of the light in the IR is similar to the visible wavelengths.

Say the bulk structure (the wall) did not interact strongly with that wavelength, then the light would continue to propagate and thus not scatter diffusely like the visible light did in the room.

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The basic answer is, infrared is similar to visible. The light will diffusely reflect off normal walls.

For a more detailed answer, maybe you want to know "Exactly what percentage of infrared light will diffusely reflect off a white-painted wall?" (for example). You'll have to be very lucky to find someone who can give you a specific answer. It certainly depends (at least to some extent) on the exact brand of white paint and the exact wavelength of infrared light.

But I would guess that it's unlikely that an ordinary white paint behaves drastically different for near-infrared light than it does for visible light.

(There are very specialized paints that DO have drastically different reflectance for near-infrared light versus visible ... these kinds of paints are designed for outdoor use, related to the heat management of buildings. Again, ordinary paints are unlikely to have that property.)

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  • $\begingroup$ Infrared is not necessarily similar to visible. For example, clouds absorb infrared radiation, but reflect visible radiation. $\endgroup$
    – gerrit
    Jan 11, 2013 at 23:03
  • $\begingroup$ What you stated above is actually only true for near IR (NIR). Once you get into the mid to long wave you are actually measuring a lot of emissivity due to heat and not all reflection anymore... $\endgroup$
    – daaxix
    Jan 12, 2013 at 3:42
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In first approximation we can say that the walls will diffusely reflect most of the infrared light. Some of it will be absorbed on the wall and re-emitted at a different wavelength. Because there may be meta-stable energy levels in the wall paint, where the excited states can only lose the energy via non-radiative energy exchange, for example collision. Some of the light will just be absorbed and not re-emitted.

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