OpenStudy (anonymous):

The intensity of light at a distance r from a source is given by L = Ir^(−2), where I is the illumination at the source. Starting with the values I = 20, r = 20, suppose we increase the distance by 4 and the illumination by 4. By (approximately) how much does the intensity of light change? dL= ??? and its not -0.0083

6 years ago
OpenStudy (anonymous):

\[\Delta L \approx dL=L _{l}dl+L _{r}dr\]

6 years ago
OpenStudy (anonymous):

i know that but when i solve it i get -0.0083 and its wrong !

6 years ago
OpenStudy (anonymous):

hmmmmm. I try solving it and see what I get

6 years ago
OpenStudy (anonymous):

k thanks !

6 years ago
OpenStudy (anonymous):

\[L _{l}=\frac{1}{r^2}=\frac{1}{400}\]\[dl=\Delta l=4\] and,\[L _{r}=\frac{-2l}{r^3}=\frac{-2(20)}{(20)^3}=\frac{-40}{8000}=\frac{-1}{200}\]\[dr=\Delta r=4\]Thus,\[dl=\frac{1}{100}-\frac{4}{200}=\frac{-2}{200}=\frac{-1}{100}=-0.01\]

6 years ago
OpenStudy (anonymous):

Is it correct?

6 years ago
OpenStudy (anonymous):

yeah thanks so much ! i must have done some arithmetic wrong then !

6 years ago
OpenStudy (anonymous):

i have another question that is not answered yet i'm posting it now !

6 years ago
OpenStudy (anonymous):

what threw me is that your answer is so close to the one I got

6 years ago
OpenStudy (anonymous):

cool I'll have a look at it

6 years ago
OpenStudy (anonymous):

i just posted it

6 years ago