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?
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3$\begingroup$ Your question is very difficult to understand, as it is unclear what you're asking. Can you phrase your question more clearly? $\endgroup$ – Shivam Sarodia Apr 6 '14 at 15:47
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1$\begingroup$ @ChrisMueller: Touche. I see your edit now and that makes a bit more sense. I've deleted my completely irrelevant answer :( $\endgroup$ – Kyle Kanos Apr 6 '14 at 16:34
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$\begingroup$ @Draksis: Sorry for my unclear question $\endgroup$ – dartheize Apr 7 '14 at 0:46
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$\begingroup$ @user40847 You're welcome. Welcome to phys.SE! $\endgroup$ – Chris Mueller Apr 7 '14 at 1:01
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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.
- 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.
Resources
- Labsphere Spectralon data sheet: 99,5% hemispherical reflectance value: So 0,5% loss.
- Spherical geometry is more expensive than cube geometry: Stellarnet
- There also is a cylindrical geometry: ILX Lightwave manuals
answered Apr 6 '14 at 17:22
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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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$\begingroup$ Your answer is straight forward and also right. But I prefer another answer because it's more detailed. But thanks anyway! :) $\endgroup$ – dartheize Apr 7 '14 at 8:19
