Why does the light of this infrared laser become visible after reflection? We use a 780 nm laser in our lab, and that makes it in the near infrared (IR) range. The majority of people can not see this wavelength of light. However, when the beam reflects off of an object (see image), the light becomes visible. This image has been taken with an iPhone camera which has a poor (or non existent) IR filter, though the light is visible with the eye.
Some questions have been asked which are related. The answer to one of them suggests that when an object is stationary, the reflected beam should lose energy. So why is the reflected beam experiencing an increase in energy here?

 A: You can never see any light beam from the side. You only see light (of whatever wavelength) propagating directly into your eye. When laser beams sometimes appear as a visible line through the air, what is happening is that dust (and to some extent molecules too) in the air are scattering the light, sending some of it towards your eye. When the beam hits a solid object, then unless the surface is extremely flat (such as a precise and clean mirror) there will be scattering at all angles, so some will go towards your eye. It is this light which you are seeing.
In the case of infra-red radiation, the human eye sensitivity does not drop off immediately for wavelengths above 700 nm; it is low but non-zero, and the scattered radiation from a laser beam is often bright enough to be seen (obviously it depends on the intensity of the original beam). I have in this way seen 852 nm, for example. However, when you can see a wavelength such as this, you should take care: the radiation entering your eye is brighter than you may think, because your eye's sensitivity is low but you are seeing it. Eye protection is for this reason especially important with wavelengths outside the normal visible range.
A: I am sure that no conversion of photon energy is happening here.
Especially you need an up-conversion in energy which is very unlikely. Normal flourescence cannot be the cause here. There are detector cards for up-conversion of laser light, but they need to be "charged" by sunlight before they can be used . And this is very special material.
More likely the laser is relatively strong and the sensitivity of the eye is still sufficiently high.
For example: While I was working with 762 nm laser (Oxygen A-band) I, and all my colleagues, were able to clearly see the beam (even though 762 nm is already classified as IR). The laser had a power of ~ 300 µW and the collimated beam was clearly visible on a sheet of white paper at daylight conditions. When spread over 1 cm area the beam was very visible with light of.
Even though 780 nm is certainly further in the IR than 760 nm, your laser might be more powerful and the eye is still able to see the beam.
But the beam will probably be much more powerful than the perceived brightness suggests.
A: The other answers are perfectly correct, assuming specular (mirror like) reflection, that is, elastic scattering (leaving the energy level of photons almost unchanged).
But there is another case which I would like you to consider, that is diffuse reflection, and absorption re-emission.

Diffuse reflection is the reflection of light or other waves or particles from a surface such that a ray incident on the surface is scattered at many angles rather than at just one angle as in the case of specular reflection.
But the above scheme continues to be valid in the case that the material is absorbent. In this case, diffused rays will lose some wavelengths during their walk in the material, and will emerge colored.


https://en.wikipedia.org/wiki/Diffuse_reflection
Now the most important thing about your case, is that the surface on the picture is not only causing specular reflection, but diffuse too. This means, that:

*

*it reflects some photons in random directions


*it not only elastically scatters, but absorbs some photons, and re-emits them at different (in your case visible) wavelength. This is the answer to your question. Yes, some of the photons can actually gain energy, and from the incident IR wavelength, they are re-emitted as visible wavelength, and those are the photons that you see with the naked eye.

In some cases under intense illumination it is possible for one electron to absorb two photons allowing for the emission of radiation of a higher photon energy (shorter wavelength) than the absorbed radiation

https://en.wikipedia.org/wiki/Fluorescence
The question is an interesting one, and the only way to test this is to do this with different objects. If the laser itself is invisible to the naked eye (is truly IR and not in the visible range), but if you shine the laser on the wall or other objects, the dot might become visible, because of diffuse reflection and absorption re-emission, where some photons are re-emitted in the visible range (you can even do fluorescence).
A: You see the light because its frequency, or wavelength for that matter, is visible for your eye, not because it somehow, magically, increases its energy.
That the ray hits objects, reflects or scatters and is coming to your eye creates an impression of light. It also proves that the frequency used is (still) visible for the human eye although probably less than the "regular, visible" spectrum.
