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  1. #21

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    Quote Originally Posted by Dan Fromm View Post
    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,

    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.)

  2. #22

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    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.

  3. #23

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    Quote Originally Posted by Dan Fromm View Post
    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.

    Quote Originally Posted by Dan Fromm View Post
    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.

    Quote Originally Posted by Dan Fromm View Post
    There is a convention, and I think you follow it. PM = exit/entrance.
    Good to hear that!

  4. #24

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    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?

  5. #25

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    Quote Originally Posted by Dan Fromm View Post
    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.; 09-06-2010 at 11:59 AM. Click to view previous post history.

  6. #26

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

  7. #27
    T42
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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 close-up imaging. I am confident now that the original idea I held was well-placed 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; 10-10-2010 at 09:57 PM. Click to view previous post history.
    Henry
    A Certified Dinosaur
    Nikons F, F2, D700, Leica M3, & Kiev 4a

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