Originally Posted by polyglot
Yes. The inverse square law is still the thing; my point was that you can no longer use the attenuation = ((extension+focal length) / focal length) ^ 2 formula because the focal length is no longer a good approximation for the exit pupil distance. Does anyone here know the true position of the exit pupils of their non-normal lenses? I for sure don't and I'd have to be at about 6 sigmas for nerdiness and the gathering of otherwise-useless technical facts so I suspect most others don't either.

That makes it a little more difficult to calculate bellows factor for non-normal lenses, as I was saying.
Zeiss publishes the pupil positions in their lens data sheets.
So i know the exit pupil position of all my Zeiss lenses.

There however is an quick and easy way to meassure the thing: the pupil magnification.
1 - Meassure the diameter of the aperture from both the rear and the front.
(Use a ruler across the front and rear of the lens, as close as you can get it to the glass itself. Read the scale holding the lens and ruler as far away as you can, without having trouble reading the scale. At arm's length would be good.)
2 - Divide the rear diameter by the front diameter (i.e. RD/FD. In technical parlance, RD is the size of the exit pupil, FD that of the entrance pupil.)
3 - Multiply the focal length by the result of step 2, and the result will be the distance between exit pupil and the film.

The result is not exact, and the precision will vary depending on how close you can get the ruler to the lens, etc.
And it helps, of course, if you know and use the exact focal length of the lens, and not just the nominal.

But it will be more than precise enough for exposure compensation calculations.