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Question about the brightness of light and the inverse square law.

  1. Feb 27, 2010 #1
    Okay, I know that the brightness of light dims by 1 / r^2. I just need a little help about brightness and so forth. What exactly is the unit of brightness ? Also, when using 1 / r^2

    Say I am 5 meters away from a light bulb and find the brightness, then I step back 47 meters. Would I just find the brightness by 47 - 5 = 42.

    Then take 1 / 42^2 = 1/1764

    So, then if I move 42 meters away it would be 1/1764 less bright. Then it would be from my original position?
     
  2. jcsd
  3. Feb 27, 2010 #2
    Hi zero..you have the right concept, but the wrong math:

    http://en.wikipedia.org/wiki/Brightness

    The radial lines of light, like those of gravitvy and electromagnetism from a point charge, decrease according to the area of a concentric sphere....4pi r2....

    So you want the ratio of spherical areas at the two distances to determine the relative luminance....
    1/(5)2/ 1/(47)2....which is NOT your calculated answer...

    What would be the ratio in two dimensions? In one dimension??
     
  4. Feb 28, 2010 #3
    The lumen (symbol: lm) is the SI unit of luminous flux, a measure of the power of light perceived by the human eye. Luminous flux differs from radiant flux, the measure of the total power of light emitted, in that luminous flux is adjusted to reflect the varying sensitivity of the human eye to different wavelengths of light. The lumen is defined in relation to the candela by

    1 lm = 1 cd·sr

    That is, a light source that uniformly radiates one candela in all directions radiates a total of 4π lumens. If the source were partially covered by an ideal absorbing hemisphere, that system would radiate half as much luminous flux—only 2π lumens. The luminous intensity would still be one candela in those directions that are not obscured.
     
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