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I am a relative novice in terms of mathematics, but I am trying to understand how to approach a problem I have.

I have an area light source that is 0.5 m² that produces 3500 luminous flux.

As I understand it, light has a falloff that follows the inverse square law.

How would I approach finding the luminous flux of a smaller version of the light that is 0.2 m² without remeasuring the output?

Is there a way to scale the original light source while taking into account the inverse square law?

Any nudge in the right direction would be appreciated.

Niall McKenna
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  • It seems to me (if I understand your question correctly—let me know if I'm missing it) that your concern lies mostly outside the inverse square law, but is really about how luminous a light is at various sizes. If the smaller light emits the same amount of radiation per unit area, then it should produce $\frac{0.2}{0.5} \cdot 3500 = 1400$ lumen, and you can proceed identically after that. – Brian Tung Aug 27 '19 at 21:28
  • I should have been more clear, the luminous flux measurement of each light is from a set distance. For example each measurement is from a distance of 5m, which is why I am thinking that the inverse square law may come into play rather than applying a linear interpolation. – Niall McKenna Aug 27 '19 at 21:31
  • It's inverse square with distance, yes, but it's still linear with the surface area of the light source. I'm suggesting that if you're measuring $3500$ lux at $5$ meters with the larger light, you'll measure $1400$ lux at $5$ meters with the smaller light (assuming that the radiation per unit area of the source is the same—ETA: luminance, I think—I never remember which term means what). – Brian Tung Aug 27 '19 at 21:37
  • That makes sense, thanks for pointing that out. – Niall McKenna Aug 27 '19 at 21:41

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