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Graph of inverse square law for radiation intensity

  1. Jun 14, 2011 #1
    If I=s/4pir2 would It be correct to write this in terms of logs like this:. lnI=(lns/4pi)-2.lnr Also how could this relate to y=mx+c? I think it's y=lnI. X=lnr. -m= -2 and c= lns/4pi. Is this correct? Thank you
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  3. Jun 14, 2011 #2


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    If you meant [itex] \ln I = \ln(S/4\pi) - 2 \ln r [/itex], then yes, this is correct, because

    ln(abc) = ln(a) + ln(bc) (the log of a product equals the sum of the logs of the individual factors in the product).


    ln(bc) = cln(b)

    That looks right. I mean, y is the dependent variable (in this case log of intensity), x is the independent variable (in this case log of radial distance) . The slope m is the constant factor that multiplies the independent variable. The intercept c is what you get when x = 0.

    EDIT: It should be m = -2, NOT -m = -2.
    Last edited: Jun 14, 2011
  4. Jun 14, 2011 #3


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    You wrote -m=-2 which results m=2, a positive slope of the ln(I)-lnr graph. Is it right?

  5. Jun 15, 2011 #4
    Ah of course it's not -m, seems like I confused myself. Thank you
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