Change in Enlarger Head Height corresponds to Change in Exposure Time, but by how much?

Discussion in 'Darkroom Developing and Printing' started by Gregory W Blank, Jul 20, 2004.

  1. Gregory W Blank

    Dan Quinn Guest

    I read Mr. Kings remark as meaning double the area "not quite double"
    the time. As the lens nears the film plane it's speed nears it's
    maximum. A f5.6 lens is only that speed with an infinitely
    large enlargement. Dan
    Dan Quinn, Jul 23, 2004
    1. Advertisements

  2. Gregory W Blank

    f/256 Guest

    Again, I think the opposite is true, a f/5.6 lens is only that speed with
    zero enlargement, in other words, with the film carrier at an infinitely
    distance from the lens, that would place the baseboard one focal length from
    the lens. It is the distance lens to baseboard what you have to divide by
    the diameter of the aperture (entrance pupil) to find the f/stop of the
    lens. An enlarger is a camera with the "object or subject" being
    photographed inside of it and the "film" outside of the camera.

    f/256, Jul 23, 2004
    1. Advertisements

  3. Gregory W Blank

    Dan Quinn Guest

    I'm considerably less than 100% convinced. Dan
    Dan Quinn, Aug 15, 2004
  4. Gregory W Blank

    f/256 Guest

    You really gave it a good -4 weeks- thought, before deciding the % of
    convincedness!! :)

    The original msg included a description of why that is, but removed that
    paragraph before sending the msg. It read something like this: Contrary to
    your photographic camera, an enlarger is a camera with the subject/object of
    the picture inside the "camera obscura" and the photographic emulsion
    outside in the open, that is why, BTW, the need to situate the enlarger
    inside a darkroom. In essence, an enlarger takes pictures of a piece of
    film and project its image on the paper. It is the distance image plane to
    lens (exit pupil) what constitute the focal length of the lens, in other
    words, the distance lens to baseboard, and -similarly to any other regular that distance is equal to the nominal focal length of the
    lens only when the subject/object (in other words the piece of film being
    enlarged) is at an infinite distance from the lens. The marked f/stops are
    only that for when the lens is focused at infinity.

    f/256, Aug 15, 2004
  5. Gregory W Blank

    Mike King Guest

    When making large changes in magnification another variable is reciprocity

    I usually try to vary light intensity rather than exposure time for this

    A US$20 Ilford EM-10 meter is better and easier to use to calculate exposure
    than a calculator.
    Mike King, Aug 15, 2004
  6. Gregory W Blank

    Dan Quinn Guest

    OK. The speed of the lens is the lens to baseboard focal length
    divided by the diameter of the lens opening. A 50mm lens at f4 has
    an opening of 12.5mm. If the baseboard is 500mm below the lens the
    effective lens speed is f40. If the lens is stoped down to f8 the
    effective speed becomes f80, at f11, f120. Is my arithmetic
    correct? Dan
    Dan Quinn, Aug 16, 2004
  7. Gregory W Blank

    f/256 Guest


    Except for the obvious typo of f120 that should be f110, I believe you are

    To prove it: if you see the answer to question 7 in the LENS FAQ, you will see that the factor by
    which we have to multiply the nominal f-number to get the effective
    f-number, is ( 1 + M ), where M is the magnification. In the same FAQ you
    have that the Gaussian form of the lens equation is 1/Si + 1/So = 1/f
    (question 4), and also M = Si/So (question 5). So we have:

    So= ?
    1/Si + 1/So = 1/f
    1/500 + 1/So = 1/50
    1/So = 1/50 + 1/500
    1/So = (10 + 1)/500
    So = 500/9
    M = Si/So
    M = 500/(500/9)
    M = 9
    So the f-number bellows factor is ( M + 1) = ( 9 + 1) = 10
    f/4 -> f/40
    f/8 -> f/80

    Admittedly a more complicated way to look at it. Used here just to up the
    convincedness power of the answer :)

    f/256, Aug 16, 2004
  8. Gregory W Blank

    f/256 Guest

    A correction to the finding of So:

    1/Si + 1/So = 1/f
    1/500 + 1/So = 1/50
    1/So = 1/50 - 1/500 (negative sign instead of +)
    1/So = (10 - 1)/500
    So = 500/9
    f/256, Aug 17, 2004
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.