Why do things the hard way? The magic formulas needed are published in, among other places, Lefkowitz, Lester. 1979. The Manual of Close-Up Photography. Amphoto. Garden City, NY. 272 pp. ISBN 0-8174-2456-3 (hardbound) and 0-8174-2130-0 (softbound).

If all you want to do is calculate effective aperture given aperture set, pupillary magnification, lens orientation, and magnification use Lefkowitz' formulas. If you're using Nikkors for 35 mm Nikon SLRs, Nikon has published exposure factor curves (exposure factor given magnification and lens' orientation) for their lenses, I have the set of curves they packed with PB-4 bellows.

Well, i don't think it is about doing it the hard way or making live easy. (In my book, by the way, there is no hard way. The exact way is easy enough.)
The question was a simple one: whether the inverse square law applies.
It does, of course.

The excursion into the matter of focal length vs exit pupil position was a bit unfortunate, i think, because whatever you pick, the answer about the inverse square law remains the same.

Having said that...
Using a pocket calculator, the procedure to get to the absolute correct result requires no more than 13 key strokes, plus the input of two variables (where it says "extension" you need to enter the added extension, i.e. over the extension required for infinity focus. Where it says "focal length", the exit pupil position is required).
Figuring out the exit pupil position (one value) for each of your lenses is not much work. And with that, you can forgo any rough and ready guesstimates and always be spot on.
But i too am lazy, so my 'easy way' is to use tables. You only have to compile those once, and they'll be good forever.

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 well-known. 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.

The math is absolutely correct, Dann.
Have you checked whether you get different results using Emmanuel's math?
(And if so, Emmanuel may try to point out the errors. Which will be in the application of what Emmanuel says you should do, for i'm sure there are none in the maths i use.)

What matters in this all is how much the 'light source' is moved away from the film, relative to where it was (at infinity - to provide a good starting point). That light source is the exit pupil of the lens.

The pupil magnification is a measure for how much the exit pupil to film distance differs from the focal length. So it can be - as i have described - be used to calculate the exit pupil position. (Made easier if the lens manufacturer provides the diameters of both pupils).

From then on, it's plain sailing, easy application of the usual, well-known formulae, using the exit pupil to film distance instead of the focal length (the use of the focal length is based upon the thin lens assumption, and that only to simplify things, at the cost of accuracy).

And these formulae are simply an application of the inverse square law.

Last edited by Q.G.; 09-03-2010 at 12:52 PM. Click to view previous post history.
Reason: I should really take time to learn to use a keyboard one of these days

Your math is correct, and doing it mathematically is just fine, but some people prefer a graph such as the one attached.

You're right, Ralph.
I use tables, a somewhat less graphic way, but still a form of a 'look-up' solution.

Your graph also assumes symmetrical lenses.
Which is the only way to present such data, as long as we don't know in advance what sort of lens will be used, and how it's exit pupil position differs from it's focal length.

... Your graph also assumes symmetrical lenses.
Which is the only way to present such data, as long as we don't know in advance what sort of lens will be used, and how it's exit pupil position differs from it's focal length.

What you can't calculate, but have to understand (so no equations), is that you have to use the distance between exit pupil and film, instead of the focal length.
You don't appear ready to accept that, but you could of course also ask Emmanuel. I'm sure he will explain.

Re: "translation of axes". It works perfectly fine. But you have to remember that when you flip a lens, the entrance pupil turns into the exit pupil, the exit pupil into the entrance pupil. And with that, the f-stops of the lens change too (remember that, they are focal length/diameter entrance pupil.)
Maybe that's where your results go wrong?
(Maybe too that i don't get what you mean when you say "translation of axes"?)

I always looked at it as an increase in aperture, numerically. I'll explain. It's geometric. An aperture of f/ 1 would be twice the geometric area of an aperture of f/1.4. An aperture of f/1 would be four times the geometric area of an aperture of f/2. So, in using a tube that would extended the focal length of 2 you would be reducing the light reaching the film to one quarter the original amount. If one use, say, a 1.4 teleconverter, you would reducing the light reaching the film by one half. It would have the same light reducing affect using tubes/teleconverters as stopping down the same factor in aperture settings. The effect is in functional aperture as shutter speeds remain unaffected.

QG, thanks for the link to your site. We are at cross purposes. Or perhaps we are looking at the problem from very different directions.

If one does the calculations in terms of magnification, then a lens with pupillary magnification < 1 facing normally doesn't seem to obey the inverse square law. Wehn magnification = 1, the diameter of the circle covered is twice that of the circle covered at infinity; in this case, by the inverse squares law illumination in the circle should fall by a factor of 4, i.e., by 2 stops. This is the standard result for a lens with pupillary magnification =1. But with, say, a lens with pupillary magnification = 0.5, at 1:1, illumination falls by 3 stops. And that's why I don't agree with you.

By the way, your explanations would be clearer if you said "distance from the film plane to the exit pupil with the lens focused at infinity" instead of "position of the exit pupil". This because for unit focusing lenses, the exit pupil's position is fixed relative to the lens and as the lens is focused closer it moves relative to the film plane.