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I am working on a research project that needs to project a small UV laser dot at a 3D object. Because of the geometry of the sculpture, the projection distance varies between 500mm to 1000mm as the dot scans. I would like this dot diameter to be somewhere between 0.2 to 0.5mm.

I believe by collimating a laser beam and telescope it to a small diameter (0.2 to 0.5mm), I can avoid focusing the beam to an ever changing distance (500 to 1000mm). Is it realistic to create such a narrow beam from a cheap diode laser source? I plan to purchase a 20X and 40X microscope objective for changing the diameter of the beam, with a pin hole in the middle that acts as a spatial filter.

Money is a limiting factor and I currently have a 405nm laser source 500mW that look like this: enter image description here

Victor Leung
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    You can't maintain a narrow beam waist over 500mm. – Jon Custer Aug 14 '17 at 17:50
  • It would be helpful if you explained why @JonCuster – boyfarrell Aug 14 '17 at 23:58
  • How narrow is too narrow? Or what is a practically maintainable diameter over 500mm? Thanks. – Victor Leung Aug 15 '17 at 02:38
  • The best you can do is what Gaussian beam optics dictates see: https://physics.stackexchange.com/questions/189160/gaussian-beam-shape. However, that assumes that you have a Gaussian beam. Diode lasers often have severely aberrated beams, which makes things worse. – flippiefanus Aug 16 '17 at 05:47
  • Be extremely careful with these light sources. They will hurt your eyes badly and you will hardly know it - because 405 nm is right at the edge of your sensitivity curve. It won't hurt, but suddenly you have a black dot in your field of view where you used to have sight... Trying to align these sources requires looking at the spots - and therein lies the catch. BE CAREFUL. – Floris Aug 17 '17 at 14:02
  • @JonCuster This is not a narrow beam over these distances. The Rayleigh length is 150 meters. – Selene Routley Aug 17 '17 at 22:09
  • @Floris Thank you for the information. I will be very careful. – Victor Leung Aug 19 '17 at 14:35
  • @WetSavannaAnimalakaRodVance Thank you for confirming this calculation. – Victor Leung Aug 19 '17 at 14:36

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A 0.2mm UV beam can easily be kept narrow up to a distance of 1000mm (a meter).

The Wikipedia Gaussian beam page tells you pretty much everything you need to know to think about these kinds of issues. For example, two useful parameters are the Rayleigh distance $z_R$ and beam divergence half angle $eta$:

$$z_R = \frac{\pi\,\mu^2}{2\,\lambda}$$

where $\mu$ is the $1/e^2$ beam diameter at the focus / laser output and

$$\eta = \frac{2\,\lambda}{\pi\,\mu}$$

is the vertex half angle of the cone defined by the $1/e^2$ diameter of the asymptotic farfield.

The Rayleigh distance is the distance the beam must propagate from the focus (laser output) to expand to twice its diameter. For your parameters, assuming a wavelength of 400nm, I make the Rayleigh distance to be 150 meters! So, if your beam is anything approaching a Gaussian beam, you are all set to go.

As in the comments, sometimes cheap lasers can have a great deal of beam aberration, particularly astigmatism. Depending on your application, a way around this is to use two infinite conjugate objectives to focus your laser onto a subresolvable pinhole and recollimated the output. This will give you a perfect, aberration free beam, but you will lose power: the more aberrated the beam, the more you lose. You may have enough power, or you could compensate by choosing a higher power laser. Be mindful of laser safety considerations. You calculate the pinhole size needed as follows. The numerical aperture of the focussing beam is:

$$\eta = \frac{\mu_L}{2\,f}$$

where $f$ is the focal length of the objective and $\mu_L$ the $1/e^2$ laser beam diameter. Don't be distracted by the numerical aperture printed on the objective: the laser sets how full the objective is, so the effective focal length is the pertinent specification. Now, given the numerical aperture, the resolvable diameter is:

$$\mu = \frac{2\,\lambda}{\pi\,\eta} = \frac{4\,\lambda\,f}{\pi\,\mu_L}$$

Choose a pinhole of about 2/3 of this diameter, and use the arrangement I have described to purify the beam.

Selene Routley
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  • This. Maybe worth mentioning that the addition of a chopper and a lock in amplifier-like arrangement allow you to work with much lower light levels and still get a decent result (because it can act as a very effective narrow band filter for the signal of interest). – Floris Aug 17 '17 at 14:00
  • Thank you. I refer to your explanation about how spatial filter works and now I understand more. https://physics.stackexchange.com/questions/276254/how-does-an-optical-spatial-filter-work/276396 – Victor Leung Aug 19 '17 at 14:43
  • I'm not sure how to find the value of μL? (the 1/e^2 laser beam diameter). Is that the diameter that I want to achieve after collimation (in this case 0.2mm)? Or is that the diameter of the beam when it exit the laser module before being focused on the pinhole (in this case it is a oval 4mm x 5mm) ? Thank you. – Victor Leung Aug 28 '17 at 10:19
  • $\mu_L$ is the same as the $1/e^2$ intensity diameter. I'm also referring to the collimated beam. The output from the laser should be collimated: try using a Galilean telescope to focus to your wished-for diameter. If it propagates without divergence over 1m after this, then you won't need to use the pinhole at all. I simply mention the pinhole as a backup if the laser beam quality is too poor for your purposes. – Selene Routley Aug 28 '17 at 10:32
  • I plug the numbers into your formula and get a Rayleigh range of 150 mm, not 150 m. – The Photon Mar 18 '20 at 22:52