# Using Light Meter inside Camera Obscura

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• 09-16-2012, 07:51 AM
RalphLambrecht
couldn't you use a darkroommeter instead?
• 09-16-2012, 11:49 AM
Chan Tran
I think darkroom meter makes a lot of sense. I believe this walk in camera has an electrical outlet near by just in case the darkroom meter need AC power.
• 09-17-2012, 11:44 AM
holmburgers
Quote:

Originally Posted by PanaDP
I don't think f/1.0 is equivalent to having no aperture. If that were the case, apertures larger than f/1.0 would be impossible, yet there are lenses faster than f/1 and they certainly do not magnify light.

If the diameter of the aperture is equal to the focal length (1:1), there is no attenuation of light and nearly 100% will make it through (minus absorption in the glass of course). That's effectively the same as having no aperture. To prove it; take a reading of a scene like normal, suppose the meter reads f/5.6 @ 1/250th. Now, put your lens @ f/5.6 and place the meter at the film plane. At f/1.0 the meter will tell you that you need an exposure of 1/250th.

Now, as for a f/0.95 or larger aperture actually magnifying or amplifying the light, think about this; the image formed by the lens is significantly smaller than the subject that you're photographing. With that in mind, if you can pack the same amount of light into a smaller area, you can indeed amplify the light at the film plane, relative to the subject.

No one seems to believe me on this point, but the reality is that physics proves it. The 2nd point you make about amplifying light is a little harder to wrap my head around, but I believe there's an equation out there somewhere that can show this as well. (I gotta be careful touting the "math", because if you asked me to do it, I'd have trouble...)

Here, the first two equations on the wikipedia exposure value page (http://en.wikipedia.org/wiki/Exposure_value) should be able to prove it if you feel like plugging in some numbers.
• 09-18-2012, 07:18 PM
By golly, I tried that and it does indeed work. I stand thoroughly corrected and better educated than before. I can't believe that years of school (much of it in photographic technology) and more years of working hadn't taught me that.

I'm still thinking about the apertures larger than f1.0 but I have a feeling you're also right, that it's the effect of packing the larger area of light into a smaller area. Not exactly amplifying the light but creating the same effect.
• 09-20-2012, 03:49 AM
johnielvis
I thought about this and it didn't seem right--so I started on the energy throughput analysis and had a thought "SOMEBODY'S done this before"...a little search and VOILA:

http://www.cctv-information.co.uk/i/...Through_Lenses

so f/1 gives 20%...20% is 2 1/3 stops less than what you'd meter outside. THE F/1 "trick" takes care of the f number of the lens though--it compensates for that....
• 09-20-2012, 09:27 AM
Chan Tran
Quote:

Originally Posted by holmburgers
If the diameter of the aperture is equal to the focal length (1:1), there is no attenuation of light and nearly 100% will make it through (minus absorption in the glass of course). That's effectively the same as having no aperture. To prove it; take a reading of a scene like normal, suppose the meter reads f/5.6 @ 1/250th. Now, put your lens @ f/5.6 and place the meter at the film plane. At f/1.0 the meter will tell you that you need an exposure of 1/250th.

Now, as for a f/0.95 or larger aperture actually magnifying or amplifying the light, think about this; the image formed by the lens is significantly smaller than the subject that you're photographing. With that in mind, if you can pack the same amount of light into a smaller area, you can indeed amplify the light at the film plane, relative to the subject.

No one seems to believe me on this point, but the reality is that physics proves it. The 2nd point you make about amplifying light is a little harder to wrap my head around, but I believe there's an equation out there somewhere that can show this as well. (I gotta be careful touting the "math", because if you asked me to do it, I'd have trouble...)

Here, the first two equations on the wikipedia exposure value page (http://en.wikipedia.org/wiki/Exposure_value) should be able to prove it if you feel like plugging in some numbers.

When you said take the meter reading at the scene like normal, how do you do that? Using incident or reflected light meter? Same thing when you said meter at the film plane, incident or reflected. And if measuring reflected light at the film plane how would you do it?
• 09-20-2012, 10:00 AM
holmburgers
Johnielvis, now I'm really confused! But, it sounds like the boat's still holding water, right? Can you extrapolate on this a bit, it sounds interesting but I'm not sure I've wrapped my noggin around it yet.

Chan, the easiest way to do this would be to get a gray card and measure that with a reflective meter outside of the camera. However, an incident reading in the same position as the card should give the same reading. Then, measure the image of that gray card being projected at the film plane of your camera with an incident meter, or if you're really small (or have an enormous camera) you could take a reflective reading.
• 09-20-2012, 10:42 AM
Chan Tran
Quote:

Originally Posted by holmburgers
Johnielvis, now I'm really confused! But, it sounds like the boat's still holding water, right? Can you extrapolate on this a bit, it sounds interesting but I'm not sure I've wrapped my noggin around it yet.

Chan, the easiest way to do this would be to get a gray card and measure that with a reflective meter outside of the camera. However, an incident reading in the same position as the card should give the same reading. Then, measure the image of that gray card being projected at the film plane of your camera with an incident meter, or if you're really small (or have an enormous camera) you could take a reflective reading.

If you use a reflected light meter and measuring the subject (gray card) and then with an f/1.0 lens and assuming there is no light loss due to flare, fall off, whatsoever and then you measure the illumination at the film plane with an incident meter the meter would indicdate 4.5 stops less light.
• 09-20-2012, 08:13 PM
johnielvis
ok--that was early in the AM..I just noticed that .2 factor....where the devil did THAT come from? so I started up again with a diagram of object lens and image...then I thought...there's gotta be MORE out there...and there is...turns out that it's along thelines I was thinking--it DOES depend on object distance or magnification...however it simplifies for infinity focus to a parameter called "lens throughput" which is T=1/(4*F^2) where F is the fstop number of the lens. So that .2 factor is also wrong..it should be .25...however it can be lower for objects which are closer than infinity, so that's probably why they chose the .2 factor. the total energy passed through the lens non infinity is the throughput times 1/(1+M)^2 where M is the magnification--this is only good for a circular objects (corresponding circular images) and a circular aperture of course...

anyways...the 1/F^2 factor is dealt with using the f/1 setting trick...the rest you get by setting the iso by a factor of .25 (or .2 or whatever)....and these are of course gross approximations....but thats for an object omitting some energy...we can MEASURE what is hitting the lens with a light meter, in fact, that's what we're interested in...so I'm gonna continue looking...but short answer is use f/1 but compensate by moving 2 to 2.5 stops for the film speed......I'll see if I can come up with those other formulae myself...or I'll just keep looking for the derivation...it's much easier than doing it...and less prone to mistakes...and more fun.
• 09-20-2012, 09:40 PM
Chan Tran
The 0.2 or 0.25 (assuming the lens has no loss due to focusing closer than infinity, object not at center, lens flare etc...) with the condition that the subject reflects 100% the light incident on it. If the subject is a gray card (or some kind of average scene) then it is 0.18 x 0.25 = 0.045 and that is -4.47 stops less.
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