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.
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
You're right, Ralph.
Originally Posted by RalphLambrecht
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.
That is correct.
Originally Posted by Q.G.
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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.
"Wubba, wubba, wubba. Bing, bang, bong. Yuck, yuck, yuck and a fiddle-dee-dee." - The Yeti
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.
Originally Posted by Dan Fromm
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.)