# Is there some analog way to measure light without blocking it?

Asked by Zyx (4142) July 24th, 2010

I want to create some high tech stuff with no previous education whatsoever. Some light needs to comply to certain specifications like not being laser light before I let it out of the system.

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Do you need to measure it, or just filter it? Many substances will filter light with certain properties (generally, its wavelength or polarity). You could also measure by blocking some portion of the light, and/or measure and reproduce what you measure.

Zaku (16992)

You can use a splitter to pass half the light the way you want and use the other half for measurement. I use Edmund Optics If they don’t have what you need, it does not exist.
It’s also fun to just browse the catalog.
Enjoy.

LuckyGuy (33340)

I had already considered a splitter so I’ll continue to look for something that uses less of the light. But thanks for the answer and the site @worriedguy

Zyx (4142)

I don’t know if this would work:

If you know the albedo of a surface, and you know the size of the a room, you could always shine the light source and measure the brightness of a pre-determined area of the wall.

You could, in principle, calculate the brightness of the source. I think the calculation would involve the inverse square law.

the100thmonkey (11220)

Yeah, but that would destroy my precious light.
It might be useful to mention this light source is not static and this measurement has to be integrated. I suppose I could settle for a way to measure the light souce efficiency, and then just use that in combination with the power I’m putting into it.

Zyx (4142)

@Zyx the light wouldn’t be destroyed, only converted into a different form of energy and/or re-radiated at different wavelengths

I don’t think it’s possible to measure the intensity of light directly without absorbing some of it. You can certainly calculate its theoretical brightness based on the specification of the components you use and the energy you put into the system, but that’s no guarantee that the calculation will accurately reflect the reality of the system.

the100thmonkey (11220)

Couldn’t you affect the light with a very subtle radiation source? That’s how lasers work too right?

Zyx (4142)

or