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Objective Back focal plane diameter

  1. May 8, 2009 #1
    Hi,

    Have a query on the back focal plane diameter of an objective.

    I have got an equation back focal diameter = 320mm X NA / M

    where NA is the numerical aperture and M is the magnification.

    Just wondering is there any explanation for it?

    Thanks
     
  2. jcsd
  3. May 9, 2009 #2

    Andy Resnick

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    I have not seen that formula before, but I can guess- it looks like someone is using 160mm tube length objectives, a formula relating the size of the Airy disc (or minimum resolution metric) to the numerical aperture, and the fact that the back pupil plane is the Fourier transform of the front focal plane. Thus, an Airy disc goes to a circle function, with a diameter given by the scaling factor 2*f*NA/M (f is the back focal length of the objective = 160 mm)

    The very simple formula obscures a lot of heavy-duty lens design, but perhaps so much the better.
     
  4. May 9, 2009 #3
    Thanks....

    I have seen this equation in use in an infinity corrected objective instead of those 160mm objectives but the tube lens used in conjunction is 160mm.

    Anyway, I guess this might kinda explain the equation
     
  5. Nov 21, 2009 #4
    The reference could come from "The Handbook of Biological Confocal Microscopy, 3rd ed", 2006, which certainly cites it in Table 9.1.

    While the handbook, and Stelzer (who I think wrote that part) are both very good, that formula is not, I believe, quite correct.

    The correct formula is:
    BFP dia = 2 * F * NA
    where f = lens focal length and NA = lens NA (n sin (theta).

    Separately, I am also unaware of any infinity-corrected micoscope objectives with 160mm tube lens focal length- Nikon, Leica, and Mitutoyo are 200, Olympus 180, and Zeiss 165mm.

    For an infinity corrected system consisting of an objective and tube lens (also known as a "4-f system or a Keplerian telescope), the lenses are spaced by the sum of their focal lengths and the magnification M = F tube lens / F objective.

    Since in a microscope the tube lens focal length is fixed, the objective focal length is determined by its magnification. E.g. for a Nikon system with F tube lens = 200mm, a 100X objective will have a 2mm focal length since 200mm / 100x = 2mm.

    For a finite-conjugate (eg 160mm tube length) objective, there is not simple way to get the focal length, since the lens is a thick lens and so the focal lengths must be measured from the principle points of the lens, which may be well separated and are generally never disclosed by microscope companies. However, since for those systems an intermediate image of magnification M is formed at 160mm behind the shoulder (near the threads) of the objective, and 160mm >> object distance to the lens (which is a few mm usually), then using the lens equation 1/f = 1/So + 1/Si we can ignore 1/Si (since it will be << 1/So) and end up with f ~ So. Since M = Si / So, we can insert that and get f ~ Si / M, and since we approximated Si = 160mm, we end up with f ~ 160mm / M. Plugging that into the BFP formula above gives BFP dia = 2 * F * NA = 320mm / M * NA, which is what's cited int he Handbook. Remember, though, unlike inifinity systems, that's an approximation, even if a pretty good one.

    Thinking about WHY the formula is BFP dia = 2 * F * NA is trickier... Drawing a ray at some angle theta through the focus of a thin lens until it hits the lens and then becomes parallel to the optics axis, you would think that the formula would be:
    height above axis = F * tan(theta)
    which would give
    BFP dia = 2 * F * tan(theta)
    which is not sin(theta) ( = NA).

    The easiest explanation for why this is not true is that... it is true. However, for microscope objectives, heroic efforts have been made to design them to behave as if they follow the small angle (paraxial) approximation even for large angle rays, so that the imaging will be good. For small angles, tan(theta) = sin(theta) and the formula becomes what I stated originally.

    Another way to think about this is that both the Abbe sine condition and general Fourier optics principles require the lens system to make linear transformations based on the sine of the angle at which rays leave the object, which is related to the spatial frequencies (sine waves in a Fourier series or transform) that make up the distribution of light at the sample and which one wants to image properly at the image plane. The linearity requirement there means that the diameter of the BFP (which is the Fourier plane to the front focal plane) must scale linearly with NA. Since it does scale like 2 * F * NA for small angles (where tan ~ sin), a good microscope objective must be designed to maintain this scaling at larger angles (where tan not equal sin (equivalently, NA)). Which is, again, just to say that people worked hard to get that relationship to be true for high-NA objectives.

    Hope that helps.
     
    Last edited: Nov 21, 2009
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