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Assuming the two lasers have identical characteristics and power output except their wavelength, which one would be more successful and easy in cutting clear glass?

In case there is doubt that you can even cut clear glass with a laser, see here these two video demonstrations:

Visible spectrum blue laser cutting glass

Infrared laser cutting glass

Secondly, what is the refractive index of glass at infrared wavelengths (~1.52 at visible white light)?

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    $\begingroup$ Refractive index database: refractiveindex.info $\endgroup$ Dec 5, 2021 at 16:11
  • $\begingroup$ @AnonymousPhysicist Great! Thanks a lot. $\endgroup$
    – Markoul11
    Dec 5, 2021 at 16:12
  • $\begingroup$ The title of the 2nd video says it's cutting sapphire, not glass. But the same (probable) 1064nm laser will be transmitted quite well through both $\endgroup$
    – Chris H
    Dec 6, 2021 at 13:19
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    $\begingroup$ As for the 1st video, it looks like it wasn't even focussed, and it barely scored the glass rather than cutting it (that could even have been heating from the charring substrate rather than direct absorption) $\endgroup$
    – Chris H
    Dec 6, 2021 at 13:21

2 Answers 2

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In industry, cutting lasers are not selected based on their wavelength. They are selected based on their cost, which is closely related to energy efficiency. Carbon dioxide lasers, which are infrared lasers, are used because they have the best energy efficiency.

"Clear" glass will absorb some portion of the infrared spectrum, as will any other solid. So some infrared radiation, depending on the wavelength, will cut the glass better than visible radiation, which is not absorbed. The exact mechanism of cutting can vary.

It is also important to consider the geometry of the beam of radiation; owing to the diffraction limit, this is not independent of the wavelength.

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  • $\begingroup$ "cutting lasers are not selected based on their wavelength" and "some infrared radiation, depending on the wavelength, will cut the glass better than visible radiation" seem to be in disagreement. Surely wavelength is important enough to be one of the factors in the decision. $\endgroup$
    – Matt
    Dec 5, 2021 at 18:05
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    $\begingroup$ The question asks about cutting glass, not etching, and ineffective lasers are a waste of money at any cost. How can you possibly rank cost above ability to do the intended job? $\endgroup$
    – Matt
    Dec 5, 2021 at 23:41
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    $\begingroup$ @Matt Your questions are not really relevant because in practice almost all commercial laser cutters are CO2 laser cutters. Wavelength is just not a consideration in the industry. $\endgroup$ Dec 6, 2021 at 0:13
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    $\begingroup$ 1064nm laser cutters also exist. 1064 (near IR) will be much less efficiently absorbed in normal glass than the 10.6µm from a CO2 laser. This doesn't necessarily make the CO2 a better choice if it's absorbed in a thin layer on the incident face $\endgroup$
    – Chris H
    Dec 6, 2021 at 13:15
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    $\begingroup$ @Markoul11 Yes, if everything else is equal. Although you probably dont care about the wavelength dependence on diffraction for most laser cutting applications. $\endgroup$
    – Matt
    Dec 6, 2021 at 21:17
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You dont define what it means to cut easier, but I will address how you select wavelength to cut faster and how you can select wavelength to be more precise in your cuts (although this precision likely does not matter for the vast majority of laser cutting applications)

For lasers that differ only in wavelength, you would typically want to select the wavelength with the highest absorption in the material you want to cut, if everything else was equal. Photons that are reflected or transmitted through the material won't transfer their energy to the matetial you are cutting and therefore wont contribute to the cut. This way you maximize cutting speed. Obviously other considerations come into play with practical lasers, with laser power being a big one.

If you want to do extremely high precision cutting (micro/nanoscale), you will tend to want a shorter wavelength. Lasers with a shorter wavelength can be focused to a smaller spot size, increasing precision. This is only important at micro or nanoscale cutting.

There are other ways a laser can cut a material than by direct absorption in the material to cut, but I will ignore those for this answer since you have control of the laser wavelength.

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  • $\begingroup$ "Photons that are reflected or transmitted through the material won't transfer their energy to the matetial you are cutting and therefore wont contribute to the cut." This is wrong for high peak power lasers. I've seen one of the most experienced laser engineers accidentally cut through a Bragg mirror (>99.99% reflectivity) using a pulsed laser. $\endgroup$ Dec 6, 2021 at 20:04
  • $\begingroup$ "This way you maximize cutting speed." In laser cutter I have used, the cutting speed is not limited by the laser. It is limited by the thermal conductivity of the object you are cutting (they catch fire if you go too fast!) or by the stepper motors. $\endgroup$ Dec 6, 2021 at 20:06
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    $\begingroup$ @Anonymous None of that is relevant to the question. It was purely about the laser aspects. If cutting at nanoscale precision we arent talking "cheap cutters". $\endgroup$
    – Matt
    Dec 6, 2021 at 21:11
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    $\begingroup$ @Anonymous I cannot believe you are claiming photons that are reflected or transmitted through a medium nevertheless transfer their energy to the medium. That just isnt true. Of course you can cut a Bragg reflector with 99.99% reflection. It is that 0.01% of absorbed photons who do the cutting. That is the entire point of my answer. All else being equal, a wavelength that is highly absorbed is more efficient. Or you are operating in one of the other laser cutting modes I said I wasnt going to address, such as ionizing the atmosphere. $\endgroup$
    – Matt
    Dec 6, 2021 at 21:14
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    $\begingroup$ I am claiming that photons that are reflected or transmitted in linear optics nevertheless transfer energy to the medium in nonlinear optics. It's a widely reproduced experiment. In the labs I've worked in, this was not negligible. $\endgroup$ Dec 6, 2021 at 21:58

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