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.
Originally Posted by Q.G.
That makes it a little more difficult to calculate bellows factor for non-normal lenses, as I was saying.