


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.

Originally Posted by Dan Fromm
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.
Agree.
The bit that concerns the light loss question is that about where the exit pupil is.
Originally Posted by Dan Fromm
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)?
Yes.
Originally Posted by Dan Fromm
There is a convention, and I think you follow it. PM = exit/entrance.
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?

Originally Posted by Dan Fromm
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?
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.
Last edited by Q.G.; 09062010 at 10:59 AM. Click to view previous post history.

QG, thanks for pushing back. You were right, I was mistaken.

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Gentlemen,
I have read through this thread slowly, deliberately, with great interest.
I thank each of you sincerely for responding to my request for help in better understanding the at play of the inverse square law as it pertains to closeup imaging. I am confident now that the original idea I held was wellplaced in the basic idea at least.
This thread has turned out to be quite educational, thanks to the many seasoned imagemakers and experts among you who have taken an interest in it.
The dialogue that has resulted has rendered a wealth of additional points of understanding with respect to the various nuances, such as exit pupil, entrance pupil, actual location of the effective center of the optic and so on. I had assumed thin, single element lenses in my question, but am pleased to read about the more complex considerations. In the final analysis, and for practical purposes, it appears that the ISL works fine enough for practical purposes.
I appreciate also the math and various associated diagrams, links, and explanations.
Thanks again.
Last edited by T42; 10102010 at 08:57 PM. Click to view previous post history.
Henry
A Certified Dinosaur
Nikons F, F2, D700, Leica M3, & Kiev 4a

