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I shone a green laser light(~532 nm) through a glass slab and what I saw inside was that the light beam was now red. Imagine a regular refraction diagram but color the beam outside green and that on the inside red.

As frequency doesn't change upon entering another medium, effective light velocity in a medium becomes directly proportional to that of wavelength. So, as speed decreases shouldn't wavelength also do so? But in this case, it increases. Or is this logic wrong?

Also, it can't be due to light losing energy.

What is happening in here?

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  • $\begingroup$ What color is it after it exits the back side of the glass? $\endgroup$
    – garyp
    Commented Apr 1, 2016 at 16:11
  • $\begingroup$ But it's green when it comes out... And also it is not because the light is being seen through the slab... I kept the slab above the green beam, and looked from above- it was still green. Conclusion: it only occurs when beam PASSES THROUGH the slab. $\endgroup$ Commented Apr 1, 2016 at 16:11
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    $\begingroup$ Please describe the glass in detail -- what is its composition? Is it a standard optical glass, like BK7? Also what is the power of the laser beam? Some glasses can fluoresce, which could give your red color; if the laser fluence is high enough the glass may fluoresce. Also what type of laser goggles are you wearing; these can alter color perception. $\endgroup$ Commented Apr 1, 2016 at 21:21
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    $\begingroup$ @adam karlson: <1.000 mW? A green laser pointer? Or what? If a laser pointer see technologyreview.com/s/420214/… They often emit more than one wavelength. I can see 780 nm scatterred in air, so possibly you are seeing 808 nm leakage scattered in the glass. Cheap green laser pointers lack adequare filters, and should be tossed. Test with another, newer laser source. $\endgroup$ Commented Apr 2, 2016 at 11:13
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    $\begingroup$ Having just seen this, I decided to try my luck using a 532 nm green laser pointer, 535 nm interference filter (to eliminate any 808 nm light the laser might also inadvertently emit) and a glass plate 1/4” thick and about 2.5” square. All room lightsout, I can see a faint green trace for the light passing through the glass plate. No red and way too dim for me to photograph. So, a null result, but I don’t doubt your observation. Personally, I think it is fluorescence from some impurity in the particular piece of glass you have (or had). Anyway, good question to upvote. Raman seems improbable. $\endgroup$
    – Ed V
    Commented Apr 18, 2023 at 22:33

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I also noticed this happening, also with a 532nm green laser, but the beam inside was somewhat more orange than red. This only happened with one particular piece of glass that I had, the sphere from this set: https://www.amazon.com/dp/B08CZ9PCJP/?ref_=cm_wl_huc_item It is advertised as "k9 material", and other than this color change, I have no reason to believe it is a different material. it may be due to the shape, but changing the angle of incidence does not change the color. I tried to shine through different types of glass laying around my house, and water too, and I found one particular whiskey glass that does a similar (but far less noticeable) color change. After further reading, it seems to be raman scattering, a type of fluorescence, and is usually due to Ge impurities in the glass.

Garmysheva, T.Y., Shendrik, R.Y., Paklin, A.S. et al. Luminescence of Oxygen-Deficient Centers in Quartz Glasses. Glass Phys Chem 48, 232–235 (2022). https://doi.org/10.1134/S1087659622030038

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    $\begingroup$ Are you sure it's Raman scattering, given that inelastic Raman scattering is typically weak (of course the amplitudes could be strong in the specific material) – Wikipedia says: "Rayleigh scattering usually has an intensity in the range 0.1% to 0.01% relative to that of a radiation source. An even smaller fraction of the scattered photons (approximately 1 in 1 million) can be scattered inelastically, with the scattered photons having an energy different (usually lower) from those of the incident photons—these are Raman scattered photons." <en.wikipedia.org/wiki/Raman_scattering> $\endgroup$ Commented Dec 8, 2022 at 21:24

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