a while ago I tried to charge a glow-in-the-dark sticker using a simple red laser pointer. It was a large sticker, of the type used to mark emergency exits and fire extinguishers here in Germany. I thought I could draw a picture this way, with glowing strokes.

To my surprise, the exact opposite happened - the sticker was already glowing a little bit and the areas I illuminated with my laser turned dark instantly, as if the charge of the sticker was removed by the laserlight.

The effect was reversible, the darkened areas would charge and glow as usual if exposed to normal light.

So I was able to draw a picture after all - however, it was a negative.

Can anyone explain why the sticker was discharged instead of charged by the laserlight?

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    $\begingroup$ I have a near-UV 405nm violet laser pointer that has the opposite effect. It makes glow-in-the-dark materials glow extremely intensely for a few seconds before returning to typical glow. $\endgroup$ Oct 6, 2013 at 21:36
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    $\begingroup$ Pics or it didn't happen :) $\endgroup$ Oct 7, 2013 at 5:09
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    $\begingroup$ @TimPietzcker: When I said "a while ago", what I actually meant was "about seven years ago" :) - my mobile had no camera at that time.. ;) I still have the same laser pointer and a similar sticker at hand though, I'll try to capture the effect in a short video as soon as possible! $\endgroup$
    – Toastor
    Oct 7, 2013 at 9:21
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    $\begingroup$ I was unable to record it yet, because the windows have no blinds or roller shutters at our office and it is too bright. However, I found somebody elses video on youtube - he is using an infrared laser and is constantly illuminating the GITD star wit UV so the dark portions get charged again right away. But the effect is visible. He even demonstrates the effect of a UV laser in contrast. I hope this will convince you that the effect is real :) But I will still try to make my own video anyway, once I have the opportunity. Here's the link: youtube.com/watch?v=kUteUH7mz0A $\endgroup$
    – Toastor
    Oct 7, 2013 at 15:19
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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/22691 $\endgroup$
    – akhmeteli
    Oct 9, 2013 at 2:39

2 Answers 2


How interesting.

Presumably the "glow-in-the-dark" effect comes from the decay of a meta-stable excited state. It gets charged by sufficiently energetic photons, and decays slowly because some selection rule prevents a direct transition without an external influence.

If this is the case, we can guess that the laser is exciting the meta-stable state to a still higher energy state which is not subject to the selection rule so that it promptly falls to the ground state1, from which the low energy photons in the laser don't have the energy to push back into the meta-stable state.

I think I'll try to make a demo out of that. Thanks.

1 Presumably you could see the light of this decay if you filtered out the intense red from the laser. Try a green or blue filter. Or a spectrograph.

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    $\begingroup$ I think this is almost certainly so. I was thinking this was simple bleaching (destruction of fluorophore - there seems to be a "mean number of photons emitted till failure" with many complex molecular fluorophores) but then I saw the REVERSIBLE bit. Very neat indeed. Great suggestion too - hope the OP does this and posts the result $\endgroup$ Oct 6, 2013 at 21:31
  • $\begingroup$ Your explanation sounds reasonable. Unfortunately I don't have access to a spectrograph or a spectroscope. But I will try to filter out the laser light as you suggested - great idea! Although I fear there may be little to see. In the video I posted in one of my comments above, an IR laser is used to achieve the effect. The laser is invisible to the camera, but the material doesn't glow either before it gets dark. Maybe when the proposed higher energy state falls to the ground level, it emits another (invisible) wavelength than the expected meta-stable state... $\endgroup$
    – Toastor
    Oct 7, 2013 at 17:17
  • $\begingroup$ This seems unlikely to me. It would require that the laser have a frequency that couples well with the higher state. I suspect that the laser is providing momentum (that the previous lack thereof prevented a more direct decay). $\endgroup$
    – Aron
    Jan 29, 2015 at 1:12

"dmckee" is probably right about the mechanism. In our lab, we often work with high-power 1064 nm NIR laser beams, and an obvious issue when dealing with them is that the light is invisible to the eye. For visualization purposes, many companies like ThorLabs and Newlight Photonics sell IR visualization cards that consist of a pink material bonded to a laminated card. The cards need to be "charged" by holding them to a light source, but after charging they do not emit any light. However, when they are struck by NIR light, they glow red brightly, and the spot that has been struck by NIR light will slowly be "depleted" until it no longer responds to NIR exposure. I strongly suspect that the material is excited to a metastable triplet state by visible light, and that NIR light promotes it to a state which has large decay channels.

In any case, if your glow-in-the-dark sticker is one of the usual glowing green types, you probably already know that they can't be excited using light which is of a longer wavelength than they emit. That's why red light will not charge glow stickers, but blue and UV will.


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