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Why must an integrating sphere be a sphere?

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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

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user dartheize

asked Apr 6, 2014 in Experimental Physics by dartheize (25 points) [ no revision ]
retagged Apr 13, 2014

Your question is very difficult to understand, as it is unclear what you're asking. Can you phrase your question more clearly?

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user Draksis

commented Apr 6, 2014 by Draksis (0 points) [ no revision ]

@ChrisMueller: Touche. I see your edit now and that makes a bit more sense. I've deleted my completely irrelevant answer :(

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user Kyle Kanos

commented Apr 6, 2014 by anonymous [ no revision ]

@Draksis: Sorry for my unclear question

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user dartheize

commented Apr 6, 2014 by dartheize (25 points) [ no revision ]

@ChrisMueller: Thanks for editing my question

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user dartheize

commented Apr 6, 2014 by dartheize (25 points) [ no revision ]

@user40847 You're welcome. Welcome to phys.SE!

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user Chris Mueller

commented Apr 6, 2014 by Chris Mueller (0 points) [ no revision ]

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2 Answers

Surface coating of an integrating sphere is optimized for low losses. This white coating (barium sulfate or PTFE) 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 decresease 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.

Ressources

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

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user Stefan Bischof

answered Apr 6, 2014 by Stefan Bischof (60 points) [ no revision ]

Thank you very much, you enlighten me!

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user dartheize

commented Apr 6, 2014 by dartheize (25 points) [ no revision ]

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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.

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user George G

answered Apr 6, 2014 by George G (30 points) [ no revision ]

Your answer is straight forward and also right. But I prefer another answer because it's more detailed. But thanks anyway! :)

This post imported from StackExchange Physics at 2014-04-13 14:42 (UCT), posted by SE-user dartheize

commented Apr 7, 2014 by dartheize (25 points) [ no revision ]

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