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Graph of inverse square law for radiation intensity

by Luke1121
Tags: graph, intensity, inverse, radiation, square
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Luke1121
#1
Jun14-11, 05:26 PM
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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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cepheid
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Jun14-11, 09:36 PM
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Quote Quote by Luke1121 View Post
If I=s/4pir2 would It be correct to write this in terms of logs like this:. lnI=(lns/4pi)-2.lnr
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).

and

ln(bc) = cln(b)

Quote Quote by Luke1121 View Post
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
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.
ehild
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Jun14-11, 09:42 PM
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You wrote -m=-2 which results m=2, a positive slope of the ln(I)-lnr graph. Is it right?


ehild

Luke1121
#4
Jun15-11, 02:27 AM
P: 13
Graph of inverse square law for radiation intensity

Ah of course it's not -m, seems like I confused myself. Thank you


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