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

Question

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

hmmmmm. I try solving it and see what I get

anonymous · Answer

k thanks !

anonymous · Answer

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

anonymous · Answer

Is it correct?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

cool I'll have a look at it

anonymous · Answer

i just posted it