1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
  2. jcsd
  3. Jun 14, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Graph of inverse square law for radiation intensity
  1. The Inverse Square Law (Replies: 2)