But for film, you can go smaller film for a given pupil size ie DoF. This increases apparent brightness though not improving detail or gathering more light. Better still, change film.
There's this equivalence thingy in photography. It yields two important conclusions, namely no free photons and no free hyper-focus.
Let me begin with the second ...
I'll publish another answer since the question was updated significantly with numbers and after updating, it doesn't seem to be about the different exposure but rather why the exposure difference is larger than expected.
You have to remember that the focal length of a lens changes as it is focused.
Normally a lens that is twice as long gives an image twice ...
Is this law only an approximation for object distances much bigger than focal length (which is of course the "regular practical usage"...)?
Yes. The simple formula assumes that magnification approaches zero as object distance approaches infinity. As object distances grow smaller, the Magnification factor increases and light is lost due to the ...
If you keep f fixed and vary R, you vary the aperture number N by your equation. You also vary the area of the lens. For a circular lens, A = π R^2. Assume that you allow the same light flux to the lens. Flux is a measure of photons arriving per area per time. Your film at the other end becomes exposed (developed) according to the number of photons that ...
These ratios hold for exposure:
A1/A2 = (f2/f1)^2 = SS2/SS1
An = area of the aperture, commonly a circle = πr^2
fn = f-number (f-stop) = focal_length/aperture_diameter
SSn = shutter speed in seconds
In your image, you are using an aperture radius that is fixed. The two lenses do not have the same aperture number. Since N=f/R, the longer lens has twice the aperture number. Thus, its exposure is one fourth of the exposure of the shorter lens.