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Crossed polarizers and F-Stop relationship

  1. May 14, 2015 #1
    This is not homework.

    I understand that cross-polarizing (using 2 polarizing filters) will reduce the transmitted light intensity "I" to ##cos^2{\theta}##. This is well understood.

    I have played with polarizers for years. I have a set of polarizing filters and also bought an official variable ND filter for my DSLR which uses the same principle.

    What I would like to do, at least as a mental exercise, is find a way to calibrate a set of markings on the rim of a polarizing filter in units of f-number. What is unclear to me is the relationship of "I" from the above formula to F-Stop. I'm confused by transmitted intensity vs. transmittance.

    Should the intensity "I" number be squared again, square-rooted, or left as is? Will the markings ultimately be linearly spaced (after applying the trig of course) or some logarithmic scale? We know that cross polarizing by 90 degrees will in theory result in infinite F-Stop.

    We also know that no real polarizers are ideal. Also even a single polarizer cuts the light transmission by some amount - likely greater than 50%. At full cross-polarization there is still some light coming through. This light is dark blue or purple. With these additional issues at the minimum and maximum cross-polarization angles can we still manage to find a reasonably simple relationship between light transmittance and F-Stop?

    Can anyone come up with a fairly straight-forward mental exercise to come up with a solution?

    I may have to set up a test bed but I can already foresee several obstacles:
    • use accurate digital tilt meters to measure filter rotation angle
    • concoct a diffuse light source (immune to polarizing filters as much as possible)
    • read camera's exposure meter (which may be susceptible to polarized light or be influenced by the color shift) or interpret RAW image's pixel values - but this may involve trying to reverse engineer the sensor's response curves and any math that was used to linearize it.
  2. jcsd
  3. May 14, 2015 #2


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    Gold Member

    The intensity varies with the square of the f number. So if the lens is f1.8 max, then f3.5 gives about a fourth the intensity, and so with f8 etc. For the polariser, the intensity depends on cos^2 theta. So it looks as if the f number depends simply on cos theta. Make the maximum position equal to your lens, such as f1.8, and then mark the angles where cos theta is 0.25, 0.0625, 0.0125 etc. Unfortunately, this is a very uneven scale.
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