Extension Tubes and the Inverse Square Law
I have been in a discussion with a friend about close up imaging with tubes and bellows. A question has arisen from that, and with which I need some help.
It is my understanding that when extending a lens with tubes or bellows, that its light at the film plane falls off with extension in lock step with the inverse square law.
For example, suppose we have a 50mm lens set to its infinity setting. Its focal length is 50mm. Now suppose that we extend that with a bellows or tubes to include another 50mm, now totaling 100mm. Suppose also that the aperture of the lens stayed at what is marked as f4 on the lens barrel.
It is my understanding that while the physical diameter of the aperture remains constant, the effective aperture changes from f4 to f8, two stops. And the light available at the film plane (because of diverging transmitted light rays coming from the open aperture area of the lens) is one fourth as much. This, I understand, agrees with the inverse square law.
I have been told that in close up imaging, explaining the light at the film plane in terms of the inverse square law is wrong to do, and oversimplified. I'm puzzled as to what is wrong about the way I am thinking about it.
Doesn't the inverse square law apply between lens and film plane just as it applies outside the camera? What am I missing in this?
Thanks for your insight in this.
A Certified Dinosaur
Nikons F, F2, D700, Leica M3, & Kiev 4a
In short, yes. You are exactly correct.
Although your example thinks in term sof extension tubes, we usually encounter this with large format cameras where it is called, bellows factor.
bellows factor = (image distance / focal length) ^2
image distance is, roughly speaking, lens to film plane distance.
apply the bellows factor like you would a filter factor. That is, multiple the factor by the metered shutter speed.
an exposure factor (either bellows factor or filter factor) can be converted to stops by...
log(factor) / log(2)
so, for your example,
factor = (image distance / focal length) ^2
= (100mm / 50mm) ^2
= (2)^2 = 2*2
and, log(4) / log(2) = 2 stops
so, if your meter reading suggests f/8 for 1/500 second you can either
open up two stops or, multiply 1/500 by four and get
4/500 = 1/125
Of course, if you're using an SLR with through the lens metering, this is all taken care of by the meter and you really need not concern yourself with it much.
Last edited by BradS; 09-02-2010 at 01:39 AM. Click to view previous post history.
Reason: add caveat at end
You are absolutely correct.
The reduction in light level is entirely due to the wider spread of the light projected by the lens, which is entirely caused by moving the lens away from the film, and is described by the inverse square law.
One could argue that the size of the light source (the exit pupil) isn't small enough (compared to the distances involved and to the film format) to be ignored, complicating things a bit, i.e. needing a bit more complex formula/law to describe what happens.
But unnecessarily so. The inverse square law works perfectly.
You are entirely correct for the symmetric thin lens approximation. If there are asymmetric lenses (e.g. with a pupil magnification ratio other than 1, i.e. telephoto or retrofocus lenses) then it gets a little messier because the lens when focused at infinity is not its focal length away from the film plane.
Edit: more importantly, its exit pupil is not the focal length away from the film.
Last edited by polyglot; 09-02-2010 at 02:28 AM. Click to view previous post history.
That's true, insofar as that the focal length isn't important at all. It's always the distance between exit pupil and film that matters.
Originally Posted by polyglot
But that doesn't really make it messier: the inverse square law is the thing for hypothetical thin lenses, symmetric lenses, asymmetric lenses, etc.
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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.
Zeiss publishes the pupil positions in their lens data sheets.
Originally Posted by polyglot
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.
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).
Also see this http://www.largeformatphotography.in...ad.php?t=65951 recent discussion on the US LF forum.
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.
I agree entirely with your understanding except that being English I would say a quarter instead of one fourth!
Originally Posted by T42
The f stop is easily explained by thinking of it in its simplest terms. i.e. focal length divided by aperture diameter. When you double the focal length and keep the aperture diameter the same then the f No. will double.
Last edited by Steve Smith; 09-03-2010 at 11:32 AM. Click to view previous post history.
"People who say things won't work are a dime a dozen. People who figure out how to make things work are worth a fortune" - Dave Rat.
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.)
Originally Posted by Dan Fromm
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.