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  1. #31
    keithwms's Avatar
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    Quote Originally Posted by Maris View Post
    I'm the source. The calculation is not particularly difficult although it requires some knowledge of basic physics. The number is of an "order of magnitude" accuracy because it involves recasting photometric quantities into energy units and assumes some values for a "typical" film. Annoyingly, I can't find a reference to anyone else having done the sums. I'd appreciate it if you or someone else on APUG would run the numbers and check my answer.
    Haha sounds like something I'd inflict on my poor students!

    First of all, let's be clear: photons don't have mass so the argument that they deposit mass would be incorrect.

    However, photons do carry a momentum, namely p=E/c, and so you could pick ~532 nm as your center wavelength and that is, what, 2.3 eV energy, roughly.

    So then the argument would be that the impulse F*t equals the change in momentum imparted by the photons. Each photon contributes a momentum of roughly 2.3 eV/c or 4x10^(-19) kg m^2/s^2 divided by 3x10^8 m/s, so let's say ~10^(-27) kg m/s . Then in one second you get an effective force of, what, 10^(-27) Newtons. You set that "weight" equal to m*g and use g= 10 m/s^2 to get a mass of something like 10^(-26) kg. Then you say you get a thousand photons deposited on your film in a second, and you arrive at 10^(-23) kg.
    "Only dead fish follow the stream"

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  2. #32
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    I dare say that a thousand photons is a bit shy of reality in a typical exposure.

    I didn't check your math, though I have no reason to doubt the accuracy.

    But I would expect the number of photons to be closer to 10^7. That's just a wild guess on my part.
    Michael Batchelor
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    www.industrialinformatics.com

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  3. #33
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    Now this is just getting amusing - what were you all talking about? :-)

    Another option would be to caclutale the total change in energy of the the atomic bonds on a pre and post exposure film and then relate that to its mass/energy equivalent.

    Just don't forget to include the weight of the soul you captured.......
    n
    "There is no such thing as objective reality in a photograph"

    My flickr and (gasp!) dpug photos - take a look if you like.

  4. #34
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    Quote Originally Posted by Maris View Post
    I'm the source. The calculation is not particularly difficult although it requires some knowledge of basic physics. The number is of an "order of magnitude" accuracy because it involves recasting photometric quantities into energy units and assumes some values for a "typical" film. Annoyingly, I can't find a reference to anyone else having done the sums. I'd appreciate it if you or someone else on APUG would run the numbers and check my answer.
    Did you do it on your bathroom scales?

  5. #35
    keithwms's Avatar
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    Oh I don't know how many photons hit the film offhand, Michael, I was just using whatever I remember. To be more correct, one would need to specify the light source intensity, aperture and bellows factor etc.

    But the estimate is just a light-hearted joke (I hope); there is no mass gained by exposure to photons. Whatever mass change actually results is a loss of mass during development.
    "Only dead fish follow the stream"

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