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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?

Enter image description here

Peter Mortensen
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jamie1989
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    But that will just be the camera picking up the infrared. What you can see in the picture can be seen with your eye. – jamie1989 May 27 '21 at 07:11
  • What about safety? If you can see this your eyes may be damaged. – my2cts May 27 '21 at 08:50
  • Goggles are used and mpe calculations are always done. Alignment of the beam is always done with minimal power. Optical benches are covered. Inevitably, you do sometimes forget to put on the goggles whilst walking into the lab, which is how I noticed this. – jamie1989 May 27 '21 at 08:54
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    I have always been able to see 785nm. It's dangerous because it looks at least 1000x weaker than it is (compared to typical red lasers). 830nm is also visible but weaker still. – Chris H May 27 '21 at 15:55
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    "when an object is stationary, the reflected beam should lose energy" - this is only true in the sense that reflection isn't 100% efficient. The energy per photon does not change with reflection from a nonmoving object (ie the wavelength doesn't change), unless fluorescence or some nonlinear process is involved – llama May 27 '21 at 16:27
  • Side note, your iphone camera is almost certainly a CMOS camera and definitely has a pretty good IR filter on it, but the filter is designed to mimic your eye's response to light in order to take pictures accurately. You're eye is slightly sensitive to this light, so the filter would be designed to be similarly transmissive. Without any filters, pictures, especially those taken outside look very unnaturally red. – aquirdturtle May 27 '21 at 23:39

4 Answers4

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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.

Andrew Steane
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    So the OP could verify this by "puffing" some vapor into the beam path and seeing if the scattered light is visible – Carl Witthoft May 27 '21 at 14:07
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    @CarlWitthoft because it's so much weaker than visible, I suggest using a camera to do that. I was recently aligning 1064nm using a webcam as a viewer because it has far greater depth of focus than our IR viewer; I've also used my phone in the past, but my current phone has both a rubbish camera and a too-good IR filter – Chris H May 27 '21 at 15:56
  • I just bought a blue 100mW laser to use for writing on paper ... are the kind of 5 bucks red laser protection glasses (such as https://www.ebay.de/itm/402139458156) that you can get online adequate protection when working with something low power such as that laser? Also, such glasses are available in grey, green, yellow, and probably a bunch of other colors, as well. Is it important which color you choose based on the color of the laser? – Sixtyfive May 28 '21 at 20:11
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    @Sixtyfive If you really mean 100mW not microwatt then be super-careful: that is a powerful laser and needs a lot of care. I suggest you inform yourself about laser safety in general. The danger is a stray reflection off a ring or pen or watch or something like that finding its way into an eye. There is a serious risk of partial or complete blindness. You have to think about all-round protection: goggles rather than just glasses, and to ensure you actually use them you may need to think about comfort and also wavelength selectivity. – Andrew Steane May 28 '21 at 22:46
  • Yes, milli. As in, 1/10W. When you talk about wavelength selectivity, is that about the color of the goggles that I was worrying about? If so, should they be blue, since the laser is blue (405nm according to the seller's listing), or is that the exact wrong thinking? There's also googles like https://www.ebay.de/itm/353262603112 available, but at least the black ones seem like they might be completely opaque... – Sixtyfive May 28 '21 at 22:55
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    @Sixtyfive Your goggles should block the laser but allow through enough of the other wavelengths so that they allow you to see ok. You need a manufacturer's clear statement about their transmission at various wavelengths. Typically a filter that blocks blue will pass red, so will look red when you look through it, but don't rely on comments on an internet thread: get yourself a serious laser safety handbook (or equivalent online) and read it. – Andrew Steane May 28 '21 at 23:04
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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.

Andreas H.
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    Indeed, 780nm is visible to many people, even though detectability is dropping rapidly with increasing wavelength. 760nm can be seen by almost anyone, 800nm gets much more hit-or-miss. I don't think anyone in the lab could see 820nm. – Jon Custer May 27 '21 at 12:46
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    @JonCuster 300mW of 830nm onto a white card at arms length was visible in a dark room, just. The same power of 785nm would have been bright enough to collimate in room lights, and 640nm dazzling – Chris H May 27 '21 at 16:04
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    @ChrisH I once worked with a 670nm laser in a rubidium lab (780nm), which caused the older grad students to freak out at how relatively blasé I was being with so much power before I explained it was only a few mW at 670 – llama May 27 '21 at 16:24
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    @ChrisH - well, I only had laser diodes of a few mW back in the 80's. 300mW could well have been visible even to my bad eyes... – Jon Custer May 27 '21 at 17:00
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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.

enter image description here

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:

  1. it reflects some photons in random directions

  2. 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).

Árpád Szendrei
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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.

Peter Mortensen
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tansy
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