Physics Asked by Sniggerfardimungus on October 8, 2020
Apologies if this is the wrong Stackexchange. All the laser-cutting searches led to graphicdesign.stackexchange.com. ?!?!?!?
I have access to a laser cutting rig, a 200W CO2 beast, that needs to do a bit of an unusual job – cut something underwater. We’re not looking to have the hardware submerged, but we’d like to change the focus distance from ~8mm to (wait for it) about 1m. (That’s meters, not millimeters.) If we prefocus to about 1m, traversing the boundary between air and water will bring the focus distance in to about .75m, which is where we need it. It’s just that none of us have enough optics in our blood to know how to go about finding a lens to fit our focus distance.
I’m assuming that doing this will affect our focus quality considerably. Is the patient dead, Doc? If the water itself is a deal-breaker, MAYBE we can work something else out, but we’re never going to get a cutter head into the space where we need to do the cut – so we’re stuck doing a sniper job.
Disclaimer: I know a bit about optics and laser beams. I have no specialization when it comes to laser cutters but I can make educated guesses.
The formulas for a Gaussian Beam are of critical relevance here. The laser cutter operates by focusing a laser beam to a tight focus with a certain waist size $w_0$. You could tell me how large the waist is but I would guess it is sub-mm to ensure high intensity for cutting and a small size to allow for machining small internal radii where necessary. I'll suppose the waist is 100 microns. I'll also guess that the cutter uses light with a wavelength of 10 micron.
The spatial profile (that is the radius of the beam as a function of distance) of any Gaussian beam (and the beam coming out of your cutter is certainly a guassian beam) is given by
$$ w(z) = w_0 sqrt{1 + left(frac{z}{z_R}right)^2} $$
With $z_R = frac{pi w_0^2}{lambda}$. For the beam I described we have
begin{align} w_0 =& 100 mu m z_Rapprox& 3mm end{align}
If $zgg z_R$ we can approximate $1 + left(frac{z}{z_R}right)^2approx left(frac{z}{z_R}right)^2$ which leads to:
$$ w(z) approx frac{lambda}{pi w_0} z $$
We see that away from the focus of size $w_0$ at $z=0$ the beam expands linearly as it propagates with a slope of $frac{lambda}{pi w_0}$. Importantly the smaller the focus is the faster the beam diverges. This is a general feature in optics and explains why you typically must focus at short distances to get a small focus.
In your case lets ask the question, if there is a waist of $w_0=100mu m$ at $z=0$ then how big must the beam be 1 meter away?
$$ w(z=1 m) approx 3 cm $$
The beam will have a radius of approximately 3 cm, or a diameter of a little more than 2 inches. This is a fairly reasonable size for a lens though probably much bigger than the OEM lens used in the laser cutting rig.
Without getting into too many details we can try to assess the feasibility of this. I'll assume the following optics scheme. It sounds like the rig natively creates a focused beam 8mm from the output of the cutting head. If you let this propagate some further distance the beam will begin diverging. Eventually it will reach a waist of 3 cm. At this point you can place a lens which will collimate the beam. The focal length will depend on the rate of divergence. After this if you place a lens with a focal length of 1 m you will get your 100 micron spot 1 meter away as desired.
Now what are the practical concerns. There are many. These aren't listed in any particular order
Finally, the more you can handle a larger waist the easier things will be. All of these problems will get better. However, increasing the waist decrease the intensity (intensity is inversely proportional to waist squared) so I'm sure there is a limit to how large you can make it given the laser power until it stops cutting well.
Good luck with your application. Consider consulting with the laser cutter manufacturer before attempting this mod.
Answered by jgerber on October 8, 2020
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