Quantifying Exposure

Stephen Benskin · May 9, 2013

Occasionally I like to review what I think I know about an area of the photography. It's lead to new insights and challenged a few long held beliefs. With this in mind, I find it interesting that there is often some uncertainty among photographers regarding the fundamental question of what the exposure meter reads and its connection to film exposure. The discussion usually centers broadly around percentages, Zones, or other nonspecific terms, but does this really explain it? This is a subject that Ive covered before, although its never been the topic of a thread.

benjiboy · May 9, 2013

There comes a point in photography when one can so embroiled in the technicalitys, testing equipment and materials that we lose sight that it's about taking good pictures .

Stephen Benskin · May 9, 2013

Michael R 1974 said:
Is it not reading the illuminance (intensity) of reflected light and then using Exposure = Illuminance x Time?

Edit: I feel like I'm missing something here...

Your basically right, but I'm talking more about image illuminance, Eg. Although the subject illuminance is part of it. So the equation is more like Hg = Eg x t. So what then is Eg, and what is Hg (g = mean)?

Stephen Benskin · May 9, 2013

Michael R 1974 said:
I'm still confused. When you say Hg are you talking about the image plane? What is image illuminance?

Yes, H is the illuminance at the film plane. E is the camera illuminace. You used the classic simplified exposure equation E = I x t. Are you able to put it to use? What I mean is that that equation is in most photo books, but none of them actually seem to define it or show how it works with examples. So, we think we know what it is about but only in the abstract and not in practical terms.

Chan Tran · May 9, 2013

A specific subject luminance will result in a specific illuminance at the film plane. It would depend on aperture, focal length, focus distance, angle from lens axis, lens flare, light loss due to glass absorption etc.. I think that's what Steve was talking about. This illuminance x the exposure time will result in a specific density depending on the film characteristic and processing.

Stephen Benskin · May 9, 2013

Let's not forget exposure meters. People talk about them seeing 12%, or 18%, or Zone V. These terms don't really define how the meter works or how it relates to camera exposure.

Stephen Benskin · May 9, 2013

Michael R 1974 said:
Chan Tran: That's all fairly obvious so I'm still suspicious. Underlying all the variables is the usual question of whether or not the various image exposures at the film plane based on the metering and camera setting will generate the expected film densities (ie the first transition in the tone reproduction diagrams), but I don't think that's what Stephen is getting at.

That is what I'm getting at. It should be obvious, but how many people are comfortable with it. In camera testing, such as the Zone System method, use the exposure meter for calibrating the film speed, yet few can define what the meter is actually doing or what the exposure at the film plane should be.

Bill Burk · May 9, 2013

I have a couple simple questions.

Is it just that the meter wants to put the same amount of light on the film as sensitometric testing would put on the film?

Then the meter has the extra work to try to back out the optics of the meter and factor in the optics of the camera and estimate the camera flare. (while an in-camera meter is excused from some of that work)?

Stephen Benskin · May 9, 2013

Michael R 1974 said:
Well, I'll admit, for example, the section in Henry's book on meters is fairly tough for me to get through every time.

Henry was somewhat non-specific in that section.

Stephen Benskin · May 9, 2013

Bill Burk said:
I have a couple simple questions.

Is it just that the meter wants to put the same amount of light on the film as sensitometric testing would put on the film?

That's bascially how the two are related.

Then the meter has the extra work to try to back out the optics of the meter and factor in the optics of the camera and estimate the camera flare. (while an in-camera meter is excused from some of that work)?

Yes, hand held exposure meters have to estimate the value of q, while in camera meters measure the actual luminance passing through the optical system.

Stephen Benskin · May 9, 2013

Michael R 1974 said:
Why do you say that? It gets into quite a bit of detail on calibration, flare etc.

Not very specific for my taste, and while we're talking about it, not very accurate either.

Stephen Benskin · May 9, 2013

Michael R 1974 said:
In any case it doesn't matter much to me since there is not much I can do about it anyway. I always get a laugh out of the meter modifications. I'll get right on that as soon as Leica builds me a precision 4x5.

Take a look at his flare graph. He starts it at 0.10 over Fb+f and has the amount of flare based from that point, when it is actually based off a stop below. A flare factor of 2 will not double the exposure at 0.10. His exposure testing also uses Zones and stopping down 4 stops.

What about the technique of shooting a step tablet with a camera, whether it is inside like with Shaffer's method or outside like WBM? The camera is the exposing device and the exposure meter determines the exposure. Wouldn't it be beneficial to understand what the film plane exposure should be (Hg)?

Let's say we are testing a 125 speed film. What should the exposure be at the speed point? What should the exposure be at the metered exposure point? What is the difference between the speed point constant and the exposure constant?

Chan Tran · May 9, 2013

Michael R 1974 said:
Is it not reading the illuminance (intensity) of reflected light and then using Exposure = Illuminance x Time?

Edit: I feel like I'm missing something here...

As Ben said, some of us can get unnecessary too technical but if you feel like you're missing something then yes you do. While it's simple Exposure=Illumninance x Time but the meter reads the luminance( and not Illuminance) of reflected light and thus there is a relatively complex process to arrive at what will be the Illuminance at the film plane.

Stephen Benskin · May 10, 2013

Michael R 1974 said:
Stephen, for the speed point we know the exposure should be Hm in S=0.8/Hm when the ISO conditions are satisfied. Determing Hm is what I find tricky. Simplifying without ISO requirements, even when just targetting an arbitrary fixed density speed point with a step tablet test (in camera, out of camera, whatever), I find it somewhat less than straight forward.

I'm speaking more theoretically. Under the ISO conditions, what should Hm be if the film speed was 125? What would the metered exposure point be? And what is the difference between the two?

This isn't as hard or overly technical as some are suggesting, and by understanding a few basic rules of exposure, it's possible to evaluate the validity of a test method like Schaffer's or WBM. To start with, what are the exposure instructions for the WBM method? It's a simple job of comparing the expected results with the two known exposure points in the above question.

Stephen Benskin · May 10, 2013

Books don't seem to cover this. The more technical books assume their readers are familiar with the values of the variables and don't bother to show examples. More general photography books usually don't attempt to cover it, so the opportunity of working with actual numbers associated with exposure falls through the cracks. I think this deprives people of a very useful tool for analysis or simply for a better understanding of the process. How can someone think to properly analyze something like the ISO speed standard when they don't have the necessary tools.

Bill Burk · May 10, 2013

Isn't it just 10 times? So instead of .8 it's 8...

Stephen Benskin · May 10, 2013

Bill Burk said:
Isn't it just 10 times? So instead of .8 it's 8...

Yup. So what would Hm and Hg be for a 125 speed then?

Bill Burk · May 10, 2013

So, the speed point would be..

125 = 0.8 / Hm
multiply both sides of equation by Hm... 125 Hm = 0.8
divide both sides of equation by 125... Hm = 0.8 / 125
Hm = 0.0064

Likewise, the metered point would be...
Hg = 0.064

Ten times... Seems extremely arbitrary or lucky to be such a round easy to remember number.

Chan Tran · May 10, 2013

Bill Burk said:
So, the speed point would be..

125 = 0.8 / Hm
multiply both sides of equation by Hm... 125 Hm = 0.8
divide both sides of equation by 125... Hm = 0.8 / 125
Hm = 0.0064

Likewise, the metered point would be...
Hg = 0.064

Ten times... Seems extremely arbitrary or lucky to be such a round easy to remember number.

So it takes 0.064 Lux.Sec to produce a density of .10 plus fog on the film?

Bill Burk · May 10, 2013

When the rest of the conditions are met, but yes, that is the idea.

I should have labeled the exposure units...

metre-candela-seconds

Stephen Benskin · May 10, 2013

Chan Tran said:
So it takes 0.064 Lux.Sec to produce a density of .10 plus fog on the film?

No, it takes 0.0064 lxs for a 125 speed film. Got to keep an eye out with those zeros.

Bill Burk · May 10, 2013

Stephen Benskin said:
No, it takes 0.0064 lxs for a 125 speed film. You need to keep an eye on the number of zeros.

Right. are lxs same as mcs?

Stephen Benskin · May 10, 2013

Meter-candle-second is a unit of light-exposure equivalent to one lux for one second. Personally, my preference is mcs.

Hm for film speed
100 = 0.0080 lxs
125 = 0.0064 lxs
200 = 0.0040 lxs
400 = 0.0020 lxs

Bill Burk · May 10, 2013

I'm still very curious how we got exactly 10x from Hm to Hg

It seems very capricious.

Stephen Benskin · May 10, 2013

Bill Burk said:
I'm still very curious how we got exactly 10x from Hm to Hg

It seems very capricious.

Without the 1960 change in the light source temperature the ratio would be 8X. The ratio for transparencies is 0.8X. It's about exposure placement based on the desired results and with black and white negative films that would be the shadows. If you want something to be curious about, consider the convenience of a density 0.10 over Fb+f for a speed point. Wouldn't that be a lot easier to find than something based on some fraction of the gradient?

Quantifying Exposure

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