# Inverse square law - brightness and distance [closed]

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, WolpertingerAug 18 '17 at 14:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, M. Enns, Jon Custer, honeste_vivere, Wolpertinger
If this question can be reworded to fit the rules in the help center, please edit the question.

• Where do these numbers come from? – ccorbella Aug 18 '17 at 9:21
• 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. – MrBrushy Aug 18 '17 at 9:22
• They are from the recorded data i have been given – Ryan Aug 18 '17 at 9:22
• Edited units sorry – Ryan Aug 18 '17 at 9:23
• 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. – 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?