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  1. #41
    AgX
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    Quote Originally Posted by Prof_Pixel View Post
    As the distance approaches zero, the intensity asymptotically approaches infinity. You can never reach zero distance.
    But still this contradicts the idea of a limited intensity. The problem seems to be in the definition of intensity.
    Last edited by AgX; 03-07-2014 at 01:16 PM. Click to view previous post history.

  2. #42
    MattKing's Avatar
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    Quote Originally Posted by StoneNYC View Post
    He didn't really give a detailed answer... This was his reply....

    There is always a physical limitation to a light source, which itself has an associated radius. So in practice this mathematical situation never arises. But theoretically, yes.


    "Stone wrote:
    So, on the photo forum, this question arose....

    This is confusing me

    according to the inverse square lawB=I/d^2,theIllumination from a light sourcequadruples every time the distance from subject to light source is cut in half.Inconsequence doesn't that mean that the light source approaches infinite intensitywhen the distance to the light source approaches '0'?Hoew can this be?is there a flaw in the inverse square lawor is it limited to certain conditions? "
    My god, the clarity of that answer makes me flashback to first year physics!!!
    Matt

    “Photography is a complex and fluid medium, and its many factors are not applied in simple sequence. Rather, the process may be likened to the art of the juggler in keeping many balls in the air at one time!”

    Ansel Adams, from the introduction to The Negative - The New Ansel Adams Photography Series / Book 2

  3. #43
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    Quote Originally Posted by AgX View Post
    But still this contradicts the idea of a limited intensity.
    The law is good for true point sources-- in the 'perfect physics world' a true point source would occupy no space, so therefore, at the source, the light *would* be infinite. This can't happen in real life, but the law still applies, as the margin of error is small enough at reasonable distances, for normal everyday light sources. Just like distance can't be zero, as the sensor would occupy the same space, which according to the formula, would be 0, as would be the size of the area of measurement.

    Just because the perfect parameters are impossible in practice, does not mean the law doesn't apply to average, everyday light sources.

    How many people commenting have a background in Physics? Just curious.
    New-ish convert to film.
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  4. #44
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    Quote Originally Posted by ntenny View Post
    Disclaimer: I'm trained as a mathematician, not a physicist; though if I'd had an undergrad minor it would have been physics.

    -NT
    Hehe:

    http://xkcd.com/435/
    New-ish convert to film.
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  5. #45
    StoneNYC's Avatar
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    Quote Originally Posted by fretlessdavis View Post
    ~Stone | "...of course, that's just my opinion. I could be wrong." ~Dennis Miller

  6. #46
    StoneNYC's Avatar
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    "That's one of the problems in physics, the electron is supposedly a point particle, and the charge self repulsion energy is greater than the mass of the particle itself, whenever you deal with point entities and forces or fields you can get infinities which make no physical sense, in other words physical solutions always have a radius".

    He additionally said...

    "The 1 / R^2 rule does not apply when one goes down to single photon interactions."

    Was his second response ...
    ~Stone | "...of course, that's just my opinion. I could be wrong." ~Dennis Miller

  7. #47
    fretlessdavis's Avatar
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    Quote Originally Posted by StoneNYC View Post
    "That's one of the problems in physics, the electron is supposedly a point particle, and the charge self repulsion energy is greater than the mass of the particle itself, whenever you deal with point entities and forces or fields you can get infinities which make no physical sense, in other words physical solutions always have a radius".

    He additionally said...

    "The 1 / R^2 rule does not apply when one goes down to single photon interactions."

    Was his second response ...
    Heh. Pretty much none of the laws apply when you get to the subatomic level. It's a whole new world, with it's own set of rules and laws down there...
    New-ish convert to film.
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  8. #48

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    Referring back to the OP - it is a good question , relevant for macro flash.
    I think the Gauss Law still applies, but you have to consider the source with real dimensions.
    This can be done by the integral calculus -conceptually, by applying Gauss law to many small illuminated points.

    I hope my scribble is legible. It is for a 2 dimensional "flash tube" - approximately OK for a thin rectangular flash
    and could be expanded to a 3D flash reflector.
    Attached Thumbnails Attached Thumbnails Flas_CloseGauss003.jpg  
    Last edited by wombat2go; 03-07-2014 at 02:20 PM. Click to view previous post history. Reason: last line of integral

  9. #49
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    Quote Originally Posted by wombat2go View Post
    Referring back to the OP - it is a good question , relevant for macro flash.
    I think the Gauss Law still applies, but you have to consider the source with real dimensions.
    This can be done by the integral calculus -conceptually, by applying Gauss law to many small illuminated points.

    I hope my scribble is legible. It is for a 2 dimensional "flash tube" - approximately OK for a thin rectangular flash
    and could be expanded to a 3D flash reflector.
    On the phone with Dad, can't read him your written stuff as I don't understand it, but he said he agress with what you typed.
    ~Stone | "...of course, that's just my opinion. I could be wrong." ~Dennis Miller

  10. #50

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    Quote Originally Posted by AgX View Post
    But still this contradicts the idea of a limited intensity. The problem seems to be in the definition of intensity.
    Exactly! Intensity is in terms of power/area. As you approach zero distance the area approaches zero meaning the 'intensity' value approaches infinity. (...and not that the power value approaches infinity.)

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