The inverse law is irrelevant in incidental light readings (after all, if the light comes from the sun -92 million miles- does it matter which side of the Grand Canyon your'e standing on in relation to your subject?)
However, what does matter is that both your subject and the meter are in the same light (not subject in the shade, you in the sun etc).
Point the dome on the meter towards the camera (away from subject) along a line parallel to that between the camera and subject.
Can't be any simpler!
People, it just works, leave it alone. Don't go all quantum-mechanic on me
It's amazing that something as simple as incident light measurement is so confusing to so many people. There are similar threads currently on Rangefinderforum and l-camera-forum. Galah's previous post is 100% accurate. I've been doing it that way for about 50 years and it works just fine.
But you don't show that you understand, and certainly do not answer the question.
Originally Posted by Galah
Incident light metering meters the light incident on the subject.
So far so good. It indeed does not matter, if the light falling on the subject comes from very far away, where exactly you hold the meter, an inch (or any other unit that's relatively small, compared to the distance to the source of light) closer to the subject or closer to the source.
But after falling on the subject, that light has to travel from the subject to the camera to be captured on film.
The subject acts as the source of the light you will allow to fall on the film in your camera.
The question is why, while traveling that distance, it would not spread out and behave according to the inverse square law.
And if it does, why the reading would not change, depending on how far your camera is from the subject. Why do those mountains lit by a distant sun, appear equally bright when you move a considerable distance away?
And the answer to that is that it certainly does behave according to the inverse square law.
With increasing distance, less light reaches the light gathering area of your lens.
The same 'mechanism' (distance), the same geometry however is also responsible for the apparent size of the subject: it diminishes at the same pace as the amount of energy available to create the image of the subject gets less.
The thing that varies with distance is the size of the solid angle, the size of the area that both determines how much light is captured and how big the thing will appear.
The result is that the same exposure will create an equally bright image, no matter how close or far away you are.
Perhaps it's easier to see if you think of light as paint: You do get less of it if you put a bucket of a given diameter further away from a paint spray. But that lesser amount will suffice to cover a lesser sized area in an equally thick layer of paint.
The location of the camera is only relevant in incident metering as it relates to composition. There is no technical reason that the location of the camera should even be a factor in how you take an incident light reading, except as it relates to what parts of the subject are shown in the frame. (E.g. one generally would measure the light falling on a part of the subject that is visible in the frame.) Think of using an incident meter only in terms of where the light is in relation to the subject, not where the camera is in relation to the subject. They are great tools because they take the camera's location and the composition out of the equation. No need to compromise their usefulness by reintroducing the factors that you have deliberately nixed by choosing an incident meter in the first place.
Do that if you want to average the light that is falling on both sides of the subject (i.e., from the area covered by the dome). But when using light that is uneven, why would you want to average it as a matter of course? Sometimes you do, yes. But not always; not even often, I would say, for my own pix. Do this in deliberate ratio lighting, e.g. a side-lit portrait, and you overexpose. The more contrasty the lighting ratio, the more overexposed you will be.
Originally Posted by michaelbsc
Point it at the light for which you want to correctly expose - the "main light." You have gone to all this trouble to craft light, or choose a location and time of day, that will sculpt the subject the way you want it, why would you then average that light with the dark side, which you have intentionally made dark? You crafted, or chose to use, the light that way because you want the dark side to be dark. If you don't want it to be dark, then change the fill ratio. Don't average the exposures for the light side and the dark side as a matter of course; it does not make sense. For best results, one meters the main light source, unless in very flat, even light, in which case one could point the dome practically anywhere and get the same reading.
Overexposed negs are not a problem to print down, so it is easy to get by without realizing that this method is a problem. But try the "point at the camera every time" method with positive film, and you are screwing yourself in anything but even light.
Last edited by 2F/2F; 02-09-2011 at 05:29 AM. Click to view previous post history.
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Just what I was going to write, the explanation is simple the light source the Sun is a fixed distance away, and since you are measuring the light falling on the subject not reflected of the inverse square law is irrelevant.
Originally Posted by Galah
So we have a floodlamp in the midst of the night, in the Grand Canyon, there is no moon, and it is bloody dark. The floodlight illuminates a piano, that we placed near the floodlight. The camera is 300 m away, it has got a 1000 mm lens, and we suppose for the sake of the argument that it frames the piano quite exactly.
The floodlamp light falling on the piano follows the inverse square law. We measure the light coming from the lamp with an incident light meter, without dome, pointing at the light source. We calculate the exposure based on that reading.
Then we walk 300 m to our camera, and set that exposure on it. Will it work?
Or, even, we don't have a 5000 mm but an ordinary 50 mm and the piano looks quite tiny in our frame. Nonetheless, is the tiny spot that represents the piano going to be correctly exposed?
"Experience" seems to suggest that the piano is going to be well exposed.
My (shaky) understanding of physics seems to suggest to me that the light reflected by the piano should indeed obey to the inverse square law, and the reading I take with the incident light meter should produce underexposure. If we had placed the lamp at twice the distance, the light we would have measured would have been 2 stops less. Now we walk 300 m away from the piano reflecting our light in all directions, we should observe a fall of light reflected from the piano with distance.
PS In this example, the piano fills the image. We observe the light coming under that smaller angle that is given by our tele lens. We are not gathering on film more light than is necessary to describe the piano. A reflected light meter should give indicate a less bright object.
Last edited by Diapositivo; 02-09-2011 at 07:10 AM. Click to view previous post history.
You're still not getting the question.
Originally Posted by benjiboy
And besides, incident metering works the same, whether the light source is millions of miles away, or two feet from your subject.
The question is why, if you measure at your subject (which you certainly should with a light two feet away from the subject), will the reading stil be correct if you set the cmaera up two miles from your subject.
The light seen by the camera has to travel those two miles after being reflected off the subject.
So before anyone else gives the same non-answer you gave, tell us why you can use the same setting you should use with the camera not two miles, but six feet from your subject.
It's easy enough to do that, since the answer has been given a couple of times now.
And it's definitely not "the light source the Sun is a fixed distance away, and since you are measuring the light falling on the subject not reflected of the inverse square law is irrelevant".
Last edited by Q.G.; 02-09-2011 at 07:33 AM. Click to view previous post history.
Reason: Spelling. One day, i will learn to type. I promise.
Originally Posted by Diapositivo
There is less light, reflected off the piano, reaching the camera than at 150 m, or 10 m.
But don't forget that the image of that piano will be proportionally smaller.
As mentioned before: both image size and light intensity follow the same geometry. So though there's less light, that light has to fill in a smaller spot on film. The intensity per area unit will be the same at 300 m as it would be at 150 m, or 10 m.
So the setting to use is the same too.
Light at double the distance is reduced to a quarter of its previous level. However at twice the distance it also covers four times the area.
Not sure what that has to do with the actual question though!
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