I am working on a lab report wherein I am given a bunch of readouts from an experiment that measures the brightness (lux) of light at set distances (cm). When I apply the inverse square law, $$\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2},$$ the answer comes out way off of the expected distance.

Example: $$\sqrt{\frac{163.1\text{ lux}·(35\text{ cm})^2}{23\text{ lux}}}=93.20\text{ cm}$$ I would be expecting 10! Am I doing something wrong here?


closed as off-topic by Kyle Kanos, M. Enns, Jon Custer, honeste_vivere, Wolpertinger Aug 18 '17 at 14:30

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  • $\begingroup$ Where do these numbers come from? $\endgroup$ – ccorbella Aug 18 '17 at 9:21
  • $\begingroup$ What are those numbers? which are lux, which are distances, and what are your units? Without further information it is unclear what you're asking. $\endgroup$ – MrBrushy Aug 18 '17 at 9:22
  • $\begingroup$ They are from the recorded data i have been given $\endgroup$ – Ryan Aug 18 '17 at 9:22
  • $\begingroup$ Edited units sorry $\endgroup$ – Ryan Aug 18 '17 at 9:23
  • $\begingroup$ The unit lux has a weighting factor associated with the response of the human eye called illuminance. I am not sure that matters too much here. Which illuminance do you know is correct? Solve for the other one assuming the second distance is 10 cm. $\endgroup$ – honeste_vivere Aug 18 '17 at 13:50

As $I_2 > I_1$ if $d_1 > d_2$, I think you just got your lux'es upside-down (inversed). What you wanted was $$35\sqrt{\frac{23}{163.1}}=13.14 $$ Is that close enough to $10$ for you?

  • $\begingroup$ Thats not what the inverse square formula says though. I1/I2=D1^2/D2^2 where I1=163.1 I2=23 d1=35 and D2=10 $\endgroup$ – Ryan Aug 18 '17 at 10:20

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