Is a photoresistor or photodiode sensitive enough to measure the difference between a volume with air and vacuum accurately?
Maybe you can enlighten on the connection between light (photons) and pressure.
I missed that class.
Hi,
Um, What optical technique would you use??
Is air darker than vacuum ? (just wondering... 
)
mhouellemont:
Is a photoresistor or photodiode sensitive enough to measure the difference between a volume with air and vacuum accurately?
I think not no.
What difference would you expect in terms of opacity?
How big is this volume?
What effect are you looking for?
Is it back scatter?
What indeed do you actually want to do that you think a measurement with a photo transducer is the answer?
Refraction?
A pressure sensor would be a better choice.
I was thinking some of the light would be refracted so at vacuum it would be receiving slightly more light. But I'm not sure if the difference between the two would be measurable.
I was thinking some of the light would be refracted so at vacuum it would be receiving slightly more light.
No it would not be measurable.
Well, it could be measurable because of the refraction thing. You could have a small rectangular glass box. Have a laser beam hitting it from a 45 degree angle. The beam will refract when it goes into the glass box. The thinner the air, the more it refracts.
it could be measurable ....
Big difference be could and can.
The refractive index of air is 1.000277, while it could be measured it would be rather silly to try unless you had some very accurate equipment.
I never knew the difference was so small. A mirage happens between different air layers. Though a mirage is about reflection, not refraction, I thought it still needed a difference in the refractive indices of the layers.
How about doing it the other way around? Have a beam starting inside the glass, hitting the surface at the angle that barely refracts the escaping beam, while a part of the beam reflects back into the glass. This reflected beam has always the same angle and its intensity could be measured. When the air gets thinner, at some point a total reflection happens and the intensity jumps up. I believe it's more or less an on/off thing, when the total reflection happens. But due to inaccuracies in the glass surface and the beam and to the width of the beam, one could get a linear change in the readings from a photodiode or an ldr, when the refractive index changes from 1.00028 to 1. It all would require lots of precision placing the laser in the exact angle.
Though a mirage is about reflection, not refraction
No, a mirage is about refraction.
You're right! Refraction it is.
Here's a link to an article bout total reflection. The part about evanescent wave might have something in it, too.
The difference between the conditions is pressure. So use a pressure sensor. You could even rig a makeshift one up with a simple rubber membrane or a sealed piston that bumps a switch since all you want is yes/no. Depending on pressures involved.
Using light and these ideas of linear changes are going to get you nowhere. You will never achieve a true empty-space vacuum. You will just be looking at 'some air' and 'less air'. Arbitrary points, and in any case even slighter effect than anything mentioned since you will never have a true vacuum devoid of any mass.
mhouellemont:
I was thinking some of the light would be refracted so at vacuum it would be receiving slightly more light. But I'm not sure if the difference between the two would be measurable.
Refraction merely changes the direction the light travels, not its intensity. Air is very transparent until
you have to pass light through 10's of km of it, in which case dust and Rayleigh scattering will make
measurable changes to light intensity.
If the light path is short you won't be able to sense anything. The refractive index of air is only about 1.0003
or so anyway.
I presume the intent is to remotely determine if some transparent container is a vacuum or not? More details,
there's probably another way to do this (strain guage on the container perhaps?)