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Collimating a point source formula?

  1. Aug 18, 2010 #1
    Hi Everyone,
    I am trying to find a formula that relates numerical aperture of a lens to the beam diameter that it would collimate a point source from. I could also do this from the effective focal length if that is easier.
    Thanks for your help!
     
  2. jcsd
  3. Aug 18, 2010 #2

    Andy Resnick

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    The numerical aperture is given by:

    [tex] na = n \frac{D}{2f} [/tex]

    where n is the refractive index of the medium (not the lens), D the diameter of the lens, and f the focal length. Set the lens diameter to the (desired) beam diameter, and place the lens a distance 'f' from the source.
     
  4. Aug 18, 2010 #3
    Hi there,
    Thanks for the response. The question that I still have about that relation is that it implies that I should be able to achieve any diameter beam size for a lens of a given numerical aperture when in fact experimentally I don't find that to be the case. When I move the distance between the end of the light guide and the lens I see a definite location where the spot becomes well collimated and not fuzzy. I wonder if this is coming from the fact that I don't really have a point source but a light guide that is .125" in diameter. Is F the focal length of the lens and that in reality the only parameter for tuning the spot diameter is the na? Finally, is the focal length measured from the center of the lens or is it the back focal length? I saw that parameter thrown around the literature...
    Thanks again for the help!
     
  5. Aug 19, 2010 #4

    Andy Resnick

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    I'm not sure I understand your first question- you do have the ability to generate an arbitrary beam diameter- make the focal length of the lens equal to the diameter, for example. Then a 1" optic will make a 1" beam and a 10" optic will make a 10" beam.

    Aside from issues of cost and manufacturing, there are limits: the divergence of the beam from the light guide, the fact that the light guide is not a point source, aberrations from the lens... etc. If you provide some details of your setup and application, I may be able to give more specific information.

    The focal length is defined based on the location of the aperture stop; for a singlet this is the center line of the lens itself. For a thin lens it doesn't make a difference if it's the center line or the back surface.
     
  6. Aug 4, 2011 #5
    Hi,

    i'm having the same problem.

    I also find experimentally that using two collimating lens with the same diameter but different focal lengths (45cm, 20cm) yields two different beam sizes (about 12mm, 6mm). The lens have been positioned so that the point source is at their front focal plane.

    Alternatively, this article http://www.newport.com/Focusing-and-Collimating/141191/1033/content.aspx" suggests to use geometrical optics to compute the beam size, but my experiments disagree with such predictions as well.

    Using the resolution formula lambda*f/h, where lambda is the wavelength, f the focal length, and h the diameter of my real point source, i get a good prediction.

    Is anybody willing to enlighten me on how one should compute the beam size?
     
    Last edited by a moderator: Apr 26, 2017
  7. Aug 4, 2011 #6

    Andy Resnick

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    What's your source?
     
  8. Aug 4, 2011 #7
    Last edited by a moderator: Apr 26, 2017
  9. Aug 4, 2011 #8

    Andy Resnick

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    35um is fairly large- I would think that the light is coming out of the pinhole with a full-angle spread of 47 degrees (based on the NA of the objective). However, based on your numbers, the light is coming out of the pinhole with an effective NA of 0.03 (20cm lens) and 0.026 (45 cm lens).

    Is your pinhole properly aligned with the objective?
     
    Last edited by a moderator: Apr 26, 2017
  10. Aug 4, 2011 #9
    Ok, so the problem could be that the laser beam entering the spatial filter is too small! Not illuminating the whole aperture.

    I have a laser beam with w_0 = 0.23mm (beam radius), lambda = 632.8nm, 1.77mrad divergence. The objective legend says 20x0.40. I get the focal length doing 160mm/40 = 4mm (right?).

    Using w_f = lambda*f/(pi*w_0) = 3.5um 1/e^2 spot radius. So i need to change objective, or demagnify the laser beam! The entrance pupil of the microscope should instead be fully illuminated (correct?) but how would you ensure that?
     
  11. Aug 5, 2011 #10

    Andy Resnick

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    Ok, now we are getting somewhere.

    1) If you are not filling the back aperture of the objective lens, you are not making use of the full NA of the objective. One way to ensure the beam fills the back aperture is simply to use a diverging lens in between the laser and objective.

    2) The front focal length of a 20X objective (assuming it's a 160mm tube length objective) is 160/20 = 8 mm. Some objectives are designed using a 170mm or 200mm tube length (the tube length should be printed on the objective), so YMMV. Infinity-corrected objectives are a little different.

    3) I still don't understand why your pinhole is so large- what is the application? In general, I use a 10 um pinhole, 15um at the largest.
     
  12. Aug 5, 2011 #11
    Sorry, i meant f=8mm. The problem is that no tube length is indicated, so i don't really know f. The pinhole is _not_ properly sized, i need to make a spatial filter, so i need a smaller one/different objective!

    However i have a 1mW laser and i'd like not to lose all the energy in the spatial filter. Are there any advantages in using infinitely corrected objectives or lenses instead of a finite tube length objective?

    Thanks a lot for answering!
     
  13. Aug 5, 2011 #12

    Andy Resnick

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    If there's no tube length indicated but the objective is specified as a 'DIN standard', and/or the objective has an RMS thread (slightly smaller diameter than a c-mount) then it's most likely a 160mm tube length objective.

    There's no reason to use an infinity-corrected objective in a spatial filter instead of a finite-tube length objective. The objective doesn't have to be particularly well-corrected, either- spatial filtering will clean up any aberrations.

    When I set up a spatial filter, I tend to use a 40X objective and a 10 um pinhole, and am happy with passing 25%-50% of the incident power. 1 mW is significantly weaker than I use, tho. I'd try using what you have first to see if it's sufficient. If you are not vignetting the central peak, use a smaller pinhole. If the intensity is too low, try a higher-power objective.
     
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