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Microscope magnification with camera

  1. Oct 4, 2014 #1
    I have two lenses for my microscope. The object lens which magnifies 63 X. Then there is my eyepiece lens with f=100 mm which focuses the incoming light at an camera. How can I calculate the total magnification? All calculations so far for the eyepiece involve the 25 cm of the human eye, but now I focus it at an camera.
     
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  3. Oct 6, 2014 #2

    Andy Resnick

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    Can you provide some additional details? For example, is the objective infinity-corrected or not? Also, there should not be an eyepiece associated with the camera path.

    The easiest approach is to image a scale bar/known object and determine how many pixels = 1 mm (or um or whatever else is convenient)
     
  4. Oct 6, 2014 #3
    It's an optical tweezer setup made by Thorlabs. Part 9 is a 100 mm biconvex lens, part 8 is a 2" long tube and 7 is the camera which has no lens. part 12 is the objective with 63X magnification.

    Optical Tweezer.png
    Your suggestion is what I ended up doing, and from that I was able to calculate back to the total magnification of the system. However, I want to be able to calculate the magnification in the normal way.
     
  5. Oct 7, 2014 #4

    Andy Resnick

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    If the microscope objective is infinity corrected, the focal length is given by the magnification written on the lens and IIRC assumes a 160mm tube length: for example, a 63X objective has f = 2.5 mm. Nikon assumes 210mm tube length and Leica 170mm, tho (again, IIRC). Then your 'system' magnification is the ratio of focal lengths: 40X using the 100mm 'tube lens'. Does this agree with a reference object?
     
  6. Oct 10, 2014 #5
    Not entirely. I used micron sized polysterene as a reference and moved these 0,01 mm. Then calculating back from that we got a magnification of 0,6 for for the eyepiece.
     
  7. Oct 10, 2014 #6

    Andy Resnick

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    I don't understand- you are using a 63X objective, a 100mm focal length biconvex lens *and* an eyepiece, and you measured a magnification of only 0.6X?
     
  8. Oct 11, 2014 #7
    Almost, the 100 mm biconvex lens here is the eyepiece. What I did was the following:
    I know the pixel size (3,8 micron). I push an object in its view 0,01 mm. Then I look at the distance it covered. I see that a certain point has moved 80 pixels (or something like that)
    That must mean it has moved 80*3,8 micron on the cameralens. However, since the distance is magnified, the real distance is still that 0,01 mm.
    So I have this information:

    It moved 80*3,8 micron on the image
    in reality it moved 0,01 mm

    So from that I can find the total magnification. Which was something like 38X. But the magnification of the objective was 63X. That means that the magnification of the eyepiece (the 100 mm lens) must be something around 0,6X.

    That is however not really helping for me, because I want to be able to calculate forwards, not backwards if see what I mean
     
  9. Oct 13, 2014 #8

    Andy Resnick

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    Now I'm totally confused- I thought you wanted to calculate the *system* magnification- and 38X is close enough to 40X, IMO.

    Note that the contributions of the individual elements to the overall system spec (magnification, aberration, length, etc.) will vary depending on the spacing of the elements. Presumably your two lenses are spaced like a simple two-lens telescope (lens spacing = f1 + f2), and then the overall magnification is the ratio of focal lengths.

    Maybe I'm not sure about your goal here- a full analysis of the system? A re-design to meet a certain system spec? etc. etc.
     
  10. Oct 13, 2014 #9
    I do want to calculate the total system magnification, but I want to do so using my own known or calculated values.
    I know the magnification of the objective
    I do not know the magnification of the eyepiece.
    Hence, I want to know the latter so I can calculate the former.

    So I know the following
    M(obj) * M (eye) = Mtotal
    63 X * M(eye) = Mtotal

    Mtotal has been estimated to be around 38X, which means M(eye) must be around 0,6
    So I want to find M(eye) so I can calculate Mtotal.
    I have estimates of both thanks due to my pixelmoving trick, but that's just that. An estimate. I want to calculate it using geometrical optics.
     
  11. Oct 14, 2014 #10

    Andy Resnick

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    Sorry, I still don't understand your goal(s). If your measurement has so much uncertainty that you can only call it an 'estimate', then I suggest re-doing the measurement.
     
  12. Oct 18, 2014 #11
    that's would go by calculating lens optics what have gotten out. not sure how to make about that the camerapiece is in shorter distance than the 100 mm focal point.
    http://en.wikipedia.org/wiki/Lens_(optics)#Imaging_properties

    http://en.wikipedia.org/wiki/Magnification#Calculating_the_magnification_of_optical_systems

    those make an approximation to my understanding that you have air or vacuum inside the telescope. it's lasers so probably vacuum...
    not sure if those give you directly the needed ones

    http://en.wikipedia.org/wiki/Lens_(optics)#mediaviewer/File:Thin_lens_images.svg

    in that's picture it's maybe e that's you are looking for. for thin biconvex lenses. they should be approximately the same at least for others.

    but back to business.. you need basically measure manually the distances to images and becos 63X for lower is known no need to sort out parameters if it's oil/water filament althought that would probably take those lasers out by heating or something or not. for the eye would use 100 mm instead of 25 cm becos of closed apparatus.

    sorry if mess up. kinda newbie.
     
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