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Why can you see stars (1/r or 1/r^2 dropoff of power)?

  1. Sep 21, 2008 #1
    I read somewhere that people can see stars because an electromagnetic wave drops off by 1/r, therefore the power delivered by the electromagnetic wave stays strong enough to activate the receptors in your eye. I believe, this 1/r relation was realized by Maxwell when he was analyzing and combining the Maxwell equations. But, even still, the intensity or flux of light waves should be dropping off by 1/r^2 because (according to Gauss' law) the light from that distant star all passes through a notional sphere that gets bigger and bigger the farther one gets away from that distant star. The sphere expands by r^2 for an increase of distance r. So why doesn't the intensity of the light drop off by 1/r^2 or does it? Even though the light itself decreases by 1/r, the intensity still has to follow this 1/r^2 relation.

    Your eye being triggered depends on the power delivered by that electromagnetic radiation right? If this is true, what would be the equation for this? Dependent on 1/r or 1/r^2?

    If someone could just generally shed some illumination on this overall situation and what is going on here with 1/r versus 1/r^2 and when one or the other matters, I would greatly appreciate it. Thanks! -xerxes73
     
  2. jcsd
  3. Sep 21, 2008 #2

    olgranpappy

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    Homework Helper

    The intensity is the square of the field. So if the field goes like 1/r the intensity goes like 1/r^2.

    This is the case for light radiating from a localized source (like a star); the electric and magnetic fields both depend on the distance from the star as
    [tex]
    E\sim\frac{1}{r}e^{ikr}\;,
    [/tex]
    where ck is the angular frequency of the light.

    As for the intensity:
    [tex]
    I\sim |E|^2\sim \frac{1}{r^2}\;.
    [/tex]
     
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