Graph of inverse square law for radiation intensity

[Luke1121]

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

 

[cepheid]

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 $\ln I = \ln(S/4\pi) - 2 \ln r$, then yes, this is correct, because

ln(ab^c) = ln(a) + ln(b^c) (the log of a product equals the sum of the logs of the individual factors in the product).

and

ln(b^c) = cln(b)

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]

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

 

[Luke1121]

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

 

[rwisz]

Yes, it would be correct to write the inverse square law equation in terms of logs as lnI=(lns/4pi)-2.lnr. This can also be related to the equation y=mx+c, where y=lnI, x=lnr, m=-2, and c=lns/4pi. This is because the inverse square law shows a linear relationship between the logarithm of intensity and the logarithm of the distance. So, when plotted on a graph, the data would form a straight line with a slope of -2 and a y-intercept of ln(s/4pi). Therefore, your interpretation is correct.