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Inverse Square Law concerning Light

  1. Nov 23, 2009 #1
    [a]1. The problem statement, all variables and given/known data[/b]
    How does the distance from a point source of light affect the visibility of that light from the said source?


    2. Relevant equations
    This is a bit tricky without sub points, but here goes:

    I1 x d12=I2 x d22
    where:
    I=intensity d2=distance per sq. unit

    [c]3. The attempt at a solution[/b]

    By using the ISL formula above, I plan to find the intensity of light at predefined intervals (i.e. every 10 cm) by substituting variables in the equation. After which, I will use a voltmeter and CdS cell to verify my calculations. I'm pretty sure I'm doing everything right, but some suggestions would be nice. :biggrin:
     
    Last edited: Nov 23, 2009
  2. jcsd
  3. Nov 23, 2009 #2

    mgb_phys

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    You might want to be careful defining "visibility"
    What you are measuring is how the power received by a fixed area (the lightmeter) = the irradiance of the light varies with distance.
     
  4. Nov 23, 2009 #3
    did i say visibility? I meant intensity. Ty.
     
  5. Nov 23, 2009 #4

    mgb_phys

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    intensity is the power of the source
    the correct term is irradiance = power/area
     
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