Flash vs camera distance

Discussion in 'Exposure Discussion' started by Michel Hardy-Vallée, Mar 11, 2007.

  1. Michel Hardy-Vallée

    Michel Hardy-Vallée Membership Council Council

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    Next week I will be doing simple head and shoulder portrait setups for my job, using E100G film in 35mm (using my Spotmatic), one Bowens strobe in a softbox, and a large reflecting card for the fill-in. With the strobe, I have rented a flash meter, Minolta Auto IV. Given that it's a portrait setup, I will most likely use my 135mm lens, but I'll bring the 50mm as well during my tests to see which one I like best.

    I go for simple, simple, simple. The portraits are not creative, they are to be useful. I will place my strobe high up, around 45deg from subject, and the reflector card on the opposite side. I have time to do trial and error. I will balance key and fill light using the flash meter with flat diffusor, and test for 1-2 stops of difference between short and broad side, to see what difference works best. Once my lights are balanced, I need to figure out exposure.

    Does the camera-to-subject distance matter in determining exposure with flash? I know that the flash-to-subject distance matters, because of the inverse square law, but given that I'm working only with artificial light, won't the position of camera matter in my exposure calculation as well? I was thinking about just putting my flash meter with spherical diffuser in front of the subject's face, aim toward the camera, and use that reading for exposure.
     
    Last edited by a moderator: Mar 11, 2007
  2. David A. Goldfarb

    David A. Goldfarb Moderator Staff Member Moderator

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    Camera to subject distance does not matter for this purpose, unless you need to think about bellows factor (shooting medium or large format usually, with a magnification ratio of 1:10 or greater).
     
  3. Michel Hardy-Vallée

    Michel Hardy-Vallée Membership Council Council

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    Thanks David. Is it just because of the narrow range of distances in the present case?
     
  4. David A. Goldfarb

    David A. Goldfarb Moderator Staff Member Moderator

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    Well, consider photographing a floodlit building at night. You can be across the street or two miles away, and as long as it's fairly clear outside, the exposure will be about 1/4 sec., f:2, at ISO 400. If atmospheric conditions enter into the calculation in the studio, be sure everyone is wearing a respirator.
     
  5. Michel Hardy-Vallée

    Michel Hardy-Vallée Membership Council Council

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    David, experience agrees with your example, but I am nagged by how to explain it. Here's my reasoning:

    P1) A floodlit building reflects the light that falls on it, and can thus be considered as a source of diffused light.

    P2) The more lights that falls on a film, the more it is exposed.

    P3) A light source like a flash becomes less and less intense according to distance.

    C1) There should be less and less light available to the camera as it is further away from the floodlit building.

    C2) The further you are from a floodlit building, the more you should expose your film to have similar tonal rendition.

    So why is it that the reflected light source does not behave like the flash light? :confused:

    I'm sure there's something fundamental that I'm missing, but I can't figure out what.
     
  6. MattKing

    MattKing Subscriber

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    Michel:

    A photon, is a photon, is a photon....

    If sufficient photons reflect off a surface and travel to your camera, it doesn't matter (within reason) how far they need to travel. They always travel in a straight line, and don't lose energy.

    The flash (or other light source) to subject distance only affects the intensity of the light as it hits the subject, and therefore the intensity of the light reflecting back to the camera. Once that light is on its way, it doesn't lose energy.

    This, like many observations, depends on some assumptions:

    1) there isn't too much dirt, dust or other atmospheric interference; and
    2) the surface in question isn't highly reflective (so as to serve more as a mirror).

    A flashlight emits light that spreads, and thus becomes less intense over distance.

    A laser probably doesn't lose much intensity either over a long distance.

    Matt
     
  7. Michel Hardy-Vallée

    Michel Hardy-Vallée Membership Council Council

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    I get it now, thanks a lot Matt! A corpuscular theory of light makes it much more easily understandable.
     
  8. Helen B

    Helen B Member

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    That is all correct, apart from C2. The thing that you have forgotten is that the fall-off in light reaching the aperture is exactly compensated for by the reduced area of film it has to illuminate.

    If you are twice as far from the illuminated object, the intensity of the light reaching you will be a quarter (in a given time a quarter the number of photons from the object will pass though the aperture), but the area of the image of the object will also be a quarter.

    Best,
    Helen
     
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  9. Michel Hardy-Vallée

    Michel Hardy-Vallée Membership Council Council

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    Thank you Helen, that's another good point!