Calculating exposure times without aperture values

Here's a question that's been keeping me busy for a while now. In a nutshell, what I'd like to know is: how long do I have to expose photographic material to light to obtain a certain amount of exposure -- if there's no optical system present, so the only thing I can adjust is exposure time (as opposed to the usual time and aperture)?

Every exposure value for a given film speed corresponds to an indefinite number of pairs of aperture and exposure time values:

EV 15 (ISO 100) -> [f/16; 1/125] etc.

Obviously, this is always assuming that there is an some kind of aperture to begin with (usually, a lens with a diaphragm). But we might as well expose our film or paper to our light source directly, without any aperture or optical system between the two. There's not many situations in which that's a sensible thing to do* -- but still: how would we, in such a situation, determine the correct exposure time? Say my incident meter gave me an EV 15, how long would I have to expose my material to the light to get the equivalent of an f/16; 1/125 exposure?

One of my attempts to answer that question involved my Bessa-R's TTL meter. With a 35/2.5 lens attached and set to maximum aperture, it gave me a reading 4.5 stops below the one I got with the lens taken off. So with EV 15 from the above example, I'd have to expose at either f/2.5; 1/4000, or take off the lens and expose for about 1/100000 sec to get the same result (a theoretical f/0.52). Can these 4.5 stops be considered a (rough) general rule? And if so, is it a coincidence that, in Zone System terms, going ›down‹ 4.5 stops from Zone V takes us right to the point between first discernible detail and no detail at all?

-- Nils

*) one of them would be trying to determine EFS by contact printing a step wedge using an enlarger as light source.

Find yourself a lightmeter, or a translation table, that lists both EV and Lumens. You can then relate one to the other. My Gossen has a table on the back of the meter but I'm sure that it isn't the only one.

You answer may be contained in this similar thread: (there was a url link here which no longer exists)

the only situation this could be useful, I would think, would be making contact prints with direct light exposure -- ie: the sun, turning on the room light.

I suspect a simple trial-and-error series of experiments with a light meter would be your only route -- light sources all vary so much, but if you took some light meter readings, then tested to see how long an exposure worked, you could quickly write up a table for whatever paper or film you were using.

The amount of light required to give middling exposure (Zone V approximately) to a film can be calculated by dividing 10 by the ISO speed. The units you get are lux.seconds. For example 400 ISO film requires 0.025 lux.seconds of light for that middle exposure. In this case if incident light intensity is a dim 0.01 lux just use an exposure time of 2.5 seconds. In noon sunlight, approximately 120,000 lux, use one five millionth of a second...not so easy.

Isn't no aperture like f/0 or f/1???

In Maris' example, 2.5 seconds at 0.01lux with ISO 400 film is probably pretty reasonable (I don't know all the details so am going to trust his post). In noonday sun, however, your film would need more exposure than 1/5000000th of a second, because reciprocity failure would have become an issue long before you got to that short of an exposure.

Thanks everyone for your comments! I realize that what's crucial is understanding (among other things) luminance, illumination, and the difference between the two, which I now do a bit better. In addition to the thread mentioned by ic-racer, (there was a url link here which no longer exists) was pretty helpful too.

What seems important to know is that EVs can be expressed in lux (a unit of illuminance, the contemporary equivalent to the imperial ›foot candle‹). An EV of 0 by definition corresponds to 2.5 lux, which in turn is the intensity of light required to yield an average exposure if our EFS is 100, our aperture is f/1, and our shutter speed is 1/1 sec (or equivalents of these values).

But that math still has the aperture in it, which I want to get rid of. Enter the ›lux second‹, which multiplies the light's intensity by time, measuring the amount of light hitting a surface over that period (Hv). In our usual definition of film speed (S), the amount of light required to yield an exposure of 0.1 output density (›speed point‹) is found by dividing 0.8 by the film's speed.

So if my Gossen told me it's being hit by 2.5 lux, and I knew my material's speed is 100, I'd get:

2.5lx * t = 0.8 / 100, or:
t = (0.8 / 100) / 2.5lx

...which would mean 0.0032 seconds (~1/300) of »dim artificial light« would suffice to yield a 0.1 density exposure.

These relationships are basically all I wanted to know. Not sure if I got the math right, though.

I also got some very instructive remarks on my question as a private message. The author also said I'm making it more complicated than necessary, and it all can be done a lot easier. I've no doubts about that. I'm not proficient enough to handle that matter with ease

@Ralph:

Isn't no aperture like f/0 or f/1???

Click to expand...

That's one of the things I thought about: isn't there an aperture value that we can just substitute for the lacking physical aperture? Maybe there is; I'm not sure yet. It's certainly not gonna be f/1, for that simply signifies an aperture whose diameter equals the size of the lens's focal length (such as Canon's 50/1.0). Consider the data above: if, at EV 0, we need f/1 & 1/1 sec for an average exposure, we'd need four stops less, i.e. f/1 & 1/15 for a 0.1 exposure. Compare that to the 1/300 sec without a lens: a difference of about 4.3 stops. If that's correct in theory, it'll mean in practice that even a lens that fast will transmit... only 5% of the light that would hit the film if there was no lens at all (?).

f/0 isn't gonna work either. Like math, optical physics isn't among my strong points. But what I do know is that dividing focal length by f-number gives us the aperture size. Divide by zero you can't. Unless you're Chuck Norris.

best,
Nils

AntiLynd said:

What seems important to know is that EVs can be expressed in lux (a unit of illuminance, the contemporary equivalent to the imperial ›foot candle‹). An EV of 0 by definition corresponds to 2.5 lux.

Click to expand...

No. You are confusing EV - Exposure Value with LV - Light Value.

EV is purely a shutter speed and aperture combination. It cannot possibly refer to an amount of light until the film sensitivity or ISO is known.

At ISO 100 the LV and EV numbers are the same. At ISO 50 or 200 they are one step out, at ISO 25 or 400, they are two steps out, etc.


Steve.

@Steve: I do know how important it is to point that out again and again. I had hoped it wouldn't be necessary in my case, as I included this bit »if our EFS is 100«

Allow me an IMHO important clarification here: a system with a lens is optically a completely different setup than a system without lens:

• A lens grabs light going through its aperture and focuses it (if it comes from the same direction) onto one point of your film.
• Without a lens, all light sources in front of the film contribute to the illumination of each point on your film.
• Only in the very special case, where the whole area seen by your film is uniformly bright, will the result with and without lens be comparable, and IIRC in such a case "no lens" would be comparable to an aperture of F/0.5

As it just so happens, an incident light meter measures just like a piece of film without lens would measure. People have already posted here how to convert from incident light meter readings to lux seconds and from there to necessary film exposure sans lens.

If you can convert your measure of illumination to meter candles... Then I have a graph for you. This is a sensitometry curve family graph for a 100 speed film.

The top scale of this graph is exposure in meter candle seconds represented in Log values as shown on a calculator (Shown on calculator as opposed to the traditional mantissa notation that I can't get accustomed to).

To get an 0.10 density result on a 100 speed film developed to ASA parameters, the exposure (working backwards) is 0.008 meter candle seconds. (Log -2.1 on a calculator).

As I understand, the exposure meter calibration point is 10 times that, or 0.08 meter candle seconds (Log -1.1 on a calculator).

That's the amount of exposure (0.08 meter candle seconds) an exposure meter will try to put on the film.

I haven't figured out or found a conversion from Meter Candle Seconds (mcs) to Light Value (LV)... But I think that's the last part of your puzzle.

http://beefalobill.com/images/tmxfamily.jpg

Table 3

-4 EV (at 100) is 0.008 meter candles.

So I dial in -4 LV on my Pentax spotmeter set at 100 and saw 1 second go off the scale. But that's because we need to move to the exposure meter calibration point that is 10 times more exposure, 1.0 Log units which is 3 and a third f/stops more exposure.

This appears to be a point one-third f/stop above f/1.0 if I have done things right.


http://en.wikipedia.org/wiki/Exposure_value#EV_as_a_measure_of_luminance_and_illuminance

RalphLambrecht said:

Isn't no aperture like f/0 or f/1???

Click to expand...

I think the f/stop that makes sense is f/1.0 - anything less is concentrating the light.

AntiLynd said:

Here's a question that's been keeping me busy for a while now. In a nutshell, what I'd like to know is: how long do I have to expose photographic material to light to obtain a certain amount of exposure -- if there's no optical system present, so the only thing I can adjust is exposure time (as opposed to the usual time and aperture)?

Click to expand...

For a "correct" exposure, expose as for f/1.0 - I believe that's the simple answer.

Any greater aperture, such as an f/0.95 lens, is a light concentrator and delivers more light than you would get without a lens... Anything less, such as f/1.4, gives you what you would expect... partial reduction in light reaching the film compared to without a lens.

summicron1 said:

the only situation this could be useful, I would think, would be making contact prints with direct light exposure -- ie: the sun, turning on the room light.

I suspect a simple trial-and-error series of experiments with a light meter would be your only route -- light sources all vary so much, but if you took some light meter readings, then tested to see how long an exposure worked, you could quickly write up a table for whatever paper or film you were using.

Click to expand...

I agree! Even if you are using a pin hole camera, there will be an aperture which can be measured even though there is no lens. Lenses were developed long after apertures were in use prior to photography. In other words how are you using this "apertureless" (sp) device?.....Regards

AntiLynd said:

@Steve: I do know how important it is to point that out again and again. I had hoped it wouldn't be necessary in my case, as I included this bit »if our EFS is 100

Click to expand...

True, but I think it's silly to use a measurement term intended for one purpose which needs to be qualified with another variable to use it for an unintended purpose when a suitable term already exists.

That's like expressing the speed of a car in terms of its engine speed and chosen gear by including the rear axle ratio!


Steve.

Bill Burk said:

I think the f/stop that makes sense is f/1.0 - anything less is concentrating the light.

Click to expand...

Aperture as used for photographic lenses is derived from numerical aperture (NA) as F = 1/2*NA. Since in air maximum NA is 1.0, maximum lens aperture is 1/2, and F=1/2 means that the aperture is the whole 2*pi half space in front of your film. All lenses concentrate light, because they take light from an area (the aperture) and focus it onto one spot. There is nothing particularly significant about the F=1.0 number, and F=0.95 lenses do not amplify light.

So air is f/0.5 and a lens at f/0.5 is the same as air?

So why is the meter calibration point 1/3 f/stop smaller aperture than f/1.0?

Flare?

Bill Burk said:

So air is f/0.5 and a lens at f/0.5 is the same as air?

Click to expand...

A previous posting of mine in this very thread tries to explain the difference between a setup with lens (infinite or not) and no lens. These two scenarios are distinctly different except for one very special case.

PS: I have read conflicting reports whether lens aperture number is defined via numerical aperture or via the equation Fnumber = f/D (f = focal length, D = aperture diameter). Defining it through numerical aperture seems to better correlate with actual image brightness and depth of field, whereas sticking with the f/D seems more intuitive. For typical apertures these two numbers are very similar.

If you stick with the numerical aperture definition, you can have max theoretical numerical aperture of 1 (in air) and therefore a minimum Fnumber of 0.5. If you stick with f/D, you can resort to the lens makers equation and derive a maximum aperture that depends on the refractive index of your lens material. For glass with n=1.5 this also gives a minimum Fnumber of 0.5, please don't get confused by this. With different materials this definition of aperture allows even faster lenses.

Michael R 1974 said:

As far as I know the fastest lens actually used was Zeiss's f/0.7. I haven't seen still images shot with it, only some of the candle-lit scenes in Kubrick's Barry Lyndon.

Click to expand...

Both the Zeiss lens I posted before, and this Dead Link Removed, are two stops faster than your F/0.7 lens ...