
The relationship between extension (lens’ rear node to film) and magnification is e = f*(m + 1) where e is extension, f is the lens’ focal length, * is the multiplication operator and m is magnification. This is true for all lenses, regardless of pupillary magnification. When the lens is focused at infinity, m = 0.
For a symmetrical lens, with entrance and exit pupils the same size, i.e., pupillary magnification = 1, effective f/ number is f/ set * (m + 1) and the exposure increase factor is (m + 1)^2. Read f/ as “f over.” ^ is the exponentiation operator. So with a lens whose aperture is set to, say, f/4 the effective aperture when magnification = 1 is f/8 and, equivalently, the exposure increase factor is 4.
For a lens with pupillary magnification <> 1 mounted front forward, effective f/ number is f/ number set * ((m/p) + 1 ) where p is the pupillary magnification. With the lens reversed, effective f/ number is f/ number set * (1/p)*(1 + pm).
All of these relationships are wellknown. Emmanuel Bigler, professor of optics and microtechniques at École Nationale Supérieure de Mécanique et des Microtechniques de Besançon recently derived them from first principles just to check. See the LF format link I posted earlier in this thread.
I don’t see how a lens with pupillary magnification <> 1 can behave as all of you assert. Please present your derivations or admit you were mistaken.