Extension Tubes and the Inverse Square Law

Hello Forum.

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.

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.
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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.

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an exposure factor (either bellows factor or filter factor) can be converted to stops by...

log(factor) / log(2)

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so, for your example,

factor = (image distance / focal length) ^2
= (100mm / 50mm) ^2
= (2)^2 = 2*2
= 4
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.

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.

polyglot said:

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.

Click to expand...

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.
But that doesn't really make it messier: the inverse square law is the thing for hypothetical thin lenses, symmetric lenses, asymmetric lenses, etc.

One way to get around the calculation is to use through the lens metering on an slr.

Steve

Q.G. said:

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.
But that doesn't really make it messier: the inverse square law is the thing for hypothetical thin lenses, symmetric lenses, asymmetric lenses, etc.

Click to expand...

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.

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

polyglot said:

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.

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

Click to expand...

Zeiss publishes the pupil positions in their lens data sheets.
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.info/forum/showthread.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.

T42 said:

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.

Click to expand...

I agree entirely with your understanding except that being English I would say a quarter instead of one fourth!

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.


Steve.

Dan Fromm said:

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.info/forum/showthread.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.

Click to expand...

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.

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

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.

RalphLambrecht said:

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

Click to expand...

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.

Q.G. said:

... 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.

Click to expand...

That is correct.

Coocoo, show me the equations. Translation of axes won't do what you assert.

The equations are simple.
You will know them.

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.

Dan Fromm said:

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.

Click to expand...

Dan,

This last bit is exactly what you need to keep in mind.
It's the exit pupil moving away from the film that is what matters.
The inverse square law applies, for lenses with a pupil magnification >1, =1 and <1.
That, with the lens facing normally, and with the lens reversed.

The magnification is used in formulae to get the amount of change, but it can only be used if the lens is absolutely symmetrical (or rather when the distance between exit pupil and film and rear focal lenght are the same.)
Else, we need to know how much, relatively, adding X amount of extension moves the exit pupil away from the film.

For that you use the same formulae that is used to calculate magnification, but using the distance of the exit pupil to the film instead of the focal length. After all, that formulae is nothing more than an expression of just that.
(But what we know then is no longer the image magnification, but the thing we need to know to find out how much less light there will be. So the pupil position can only be used in calculations aiming to find out exactly that. To find out magnification, the focal length has to be used.)

Then finally, the formulae used to calculate the shutterspeed factor is nothing else but the inverse square law. (That to calculate the aperture factor too, but in that the inverse square law is obscured by the things needed to account for the 'square root thing' aperture stops obey.)

It works perfectly with all lenses (i.e. those with pupil magn. <1, =1 and >1, so really all), simply because it is the correct thing.

When you reverse a lens, you also flip pupils (the entrance pupil becomes the exit pupil, and vice versa. You have to 'flip' pupil magnification then too, and find out the new exit pupil position for the reversed lens.

The focal length may change too when you reverse a lens, but since the focal length doesn't matter for the exposure compensation needed when adding extension, that does not matter.
The calculations using the (new) exit pupil position still yield correct results. When you apply those to f-stops, you must however remember that f-stops change when you reverse the lens.


Now to your lens, pupil magnification <1: why does it not obey the inverse square law?
You say that at 1:1, the circle it projects is twice the size of the one it projects at infinity, thus the factor should be four. Ask yourself why you think it would be, and you'll find your fault.

Think of it this way: a lens with pupil magnification of 0.5 is sitting (well... the exit pupil is. Parts of the lens too, of course. That's why they make them that way.) half as far from the film as the focal length may suggest. Add the amount of extension needed to get 1:1, i.e. as much as the focal length of the lens, and you haven't moved the lens twice as far, but 3 times as far away from the film as it was at infinity.

A lens with pupil magnification of 2 is twice as far from the film as the focal length suggests when set to infinity. Add enough extension (= focal length) to get to 1:1, and it will have moved only 1.5x further than it was at infinity.

(By the way: i don't think there is a convention prescribing how to calculate pupil magnification: either exit/entrance or entrance/exit. So if the above seems backwards, try 'flipping' the pupil magnification numbers. The principle remains the same, of course.)

QG, when a lens is focused at infinity its rear nodal plane is one focal length from the film plane. The exit pupil can be anywhere, as long as its in front of the film plane. This is one definition of focal length.

Do you agree that the relationship between magnification, focal length, and extension, defined as the distance between the lens' rear nodal plane and the film plane, is e = f * (m +1)?

There is a convention, and I think you follow it. PM = exit/entrance.

Dan Fromm said:

QG, when a lens is focused at infinity its rear nodal plane is one focal length from the film plane. The exit pupil can be anywhere, as long as its in front of the film plane. This is one definition of focal length.

Click to expand...

Agree.

The bit that concerns the light loss question is that about where the exit pupil is.

Dan Fromm said:

Do you agree that the relationship between magnification, focal length, and extension, defined as the distance between the lens' rear nodal plane and the film plane, is e = f * (m +1)?

Click to expand...

Yes.

Dan Fromm said:

There is a convention, and I think you follow it. PM = exit/entrance.

Click to expand...

Good to hear that!

QG, so far, so good.

Now, do you agree that for all lenses the diameter of the image circle at 1:1 is twice the diameter of the image circle at infinity?

Dan Fromm said:

QG, so far, so good.

Now, do you agree that for all lenses the diameter of the image circle at 1:1 is twice the diameter of the image circle at infinity?

Click to expand...

Why do you think it is?

You are probably thinking that the lens, at infinity, is at a distance from the film equal to the focal length. To get to 1:1, you need to add extension equal to the focal length. Thus twice as far away as when at infinity.
But unless the lens is a symmetrical thin lens, the lens is not (!) at a distance from the film equal to the focal length when set to infinity. So adding extension equal to the amount of the focal lenght does not (!) double the distance.
Hence your simple geometric assumption does not hold.

So: no.


The rear focal length is measured from the rear principal plane. The thing important for light loss is the position of the exit pupil. The two are rarely found in the same spot.
So considerations concerning actual image geometry, i.e. magnification, do not tell us the important bit about bellows factors and light loss.