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Why must an integrating sphere be a sphere? Why can't it be an integrating cube? What is the difference? Could I use a cube to measure total illuminance like an integrating sphere does?

enter image description here

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Your question is very difficult to understand, as it is unclear what you're asking. Can you phrase your question more clearly? – Shivam Sarodia Apr 6 '14 at 15:47
@ChrisMueller: Touche. I see your edit now and that makes a bit more sense. I've deleted my completely irrelevant answer :( – Kyle Kanos Apr 6 '14 at 16:34
@Draksis: Sorry for my unclear question – dartheize Apr 7 '14 at 0:46
@ChrisMueller: Thanks for editing my question – dartheize Apr 7 '14 at 0:47
@user40847 You're welcome. Welcome to phys.SE! – Chris Mueller Apr 7 '14 at 1:01
up vote 6 down vote accepted

Surface coating of an integrating sphere is optimized for low losses. This white coating (barium sulfate or PTFE/Teflon) acts like an ideal lambertian scatterer.

Lambertian scatterer

  • all light is scattered (Ok, not 100%, but a very high percentage like 99,5%. See ressources)
  • it is emitted in the hemisphere following the cosine law: perpendicular to the surface it's highest. Intensity decrease follows a cosine law.

First generation stray light (blue in OP's picture) shows this light cone. Imagine this cone at the corner of a cube: some light will hit a wall again and suffers tiny losses. Detector port in cubic geometry hat a lower propability to to be hit with the ray of highest energy. With a sphere however all surface normal vectors point to its center. Remember, that these rays "carry more energy" according to Lambert's cosine law. It will have lower losses than a measurement head with a cube geometry. A spherical geometry reduces the necessary number of stray events.


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Thank you very much, you enlighten me! – dartheize Apr 7 '14 at 0:45

In a sphere, any light emitted from the center will reflect off the sides at normal incidence come back to the center. In a cube, some rays never return to the center, so you aren't measuring all of the light emitted, which defeats the purpose of the device.

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Your answer is straight forward and also right. But I prefer another answer because it's more detailed. But thanks anyway! :) – dartheize Apr 7 '14 at 8:19

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