Exposure vs Light Intensity vs Distance - something I have wondered about forever

[rpcrowe]

There is probably a pretty simple explanation to this question. I have discussed it over the years and have not been able to solve this problem to my satisfaction. Maybe someone with a more scientific mind can help...

I have added numbers to the illustration in the very excellent tutorial on exposure and metering to illustrate what I am talking about...



If the distance between the light source and the subject (#1) is increased, the light intensity decreases to the square of the distance and the exposure must be increased to compensate for that decrease.

However, the distance between the subject and the camera (#2) is irrelevant for exposure. As an example: if distance #2 is one meter and the exposure for that subject is f/16 @ 1/100 second but, if distance #2 is increased to, say, 10 meters, the exposure will remain as f/16 @ 1/100 second.

Why doesn't the light intensity of the reflected light decrease as the #2 distance increases and why does the exposure remain the same.

[ktuli]

Richard,

I'm not sure I have an answer for you. However, I do know that in underwater photography, you have to be aware of the total distance the light has to travel (ie: #1 + #2) and increasing or decreasing one or both of them has an affect.

I don't understand why that would not be the case above water... I would think the total path of light is the total path of light, regardless of the medium it passes through.

- Bill

[benm]

A very interesting question. I think it depends on what the light source is. For a flash unit the exposure will change if distance #2 changes because the total distance (#1 + #2) changes appreciably. If the light source is the sun then the exposure does not change because the total distance remains the same (distance #1 is so large and distance #2 is negligibly small).

[Colin Southern]

Hi Richard,

I'm not strong on these kinds of things, but as I see it, it's not so much the light "intensity" as it is the light "energy" (number of phototons hitting the sensor). As one backs away from the camera - for a given focal length - the light from a given source is spread over fewer photosite on the sensor (with the reduction in sensor area used also being a square-law function).

[PhotomanJohn]

Colin has it right. It is true that a light source that is not focused radiates light off in all directions and the light intensity falls off by the square of the distance. This is because as the distance increases, the light from the source is spread out onto a bigger area (proportional to the square of the distance) so the intensity is less at any one point.

Now for the camera part. Remember that the brightness of the object is a function of the amount of light energy per unit area. As we move back from a lighted object, the light energy from every point on that object (say a square inch of the objects surface) drops off by the square of the distance but at the same time as we increase the distance these points get smaller and get closer together (by the square of the distance) as seen by the sensor. So the brightness of the object as seen by the sensor remains the same because even though there is less light energy coming from the object, the energy is concentrated in a smaller area so the brightness stays the same.

So did that help or totally confuse everyone?

John

[FrankMi]

Hi Richard, let me try to make sense of it. The amount of light falling on the subject and being reflected is pretty much constant for a given distance between the light source and the subject so metering in close will produce a given exposure value.

But what happens when you move the light meter (in camera, you say?) further away from the subject? The further back you move the meter, the proportionally smaller the subject is in frame and the more the meter is reading objects other than the subject. So, assuming that only the subject is providing reflected light, the further away the meter from the, the less light, as proportion of the total image, is available to be metered. Did the amount of light on the subject change? No, that is fixed by the distance from the light source to the subject. The proper exposure for the subject then hasn't changed but the exposure for the entire scene has. I think this is what Colin is saying.

This is the reason the camera's meter can be fooled by a dark scene with a bright subject, such as the full moon at night. In both cases to get the proper exposure of the 'subject', you need to set the exposure value for the light being reflected from the subject, not for the scene as a whole.

I suspect that in underwater shots, the water itself acts like a filter that reduces the light reflecting from the subject proportional to the distance from the camera.

WE also don't want to forget that we can easily double or triple the distance from a flash to the subject so we need to consider the square of the distance to determine the amount of light from a flash to the subject, but if we are using light from further away this becomes increasingly inconsequential. For example, how would you double the distance from the sun to the subject in order to get the square of the distance to have any effect? I hope this makes sense.

[darekk]

Someone has answered this partially in the middletime, but anyway ... In my opinion because the light from given single point is getting weaker reversibly proportionally to the distance raised to the square power, but number of such points / area unit of a field of view (their density) is increasing proportionally to the distance raised to the square power.
#1: ... * 1/r^2
#2: ... * 1/r^2 * ... * r^2 = always ... * 1
(* r^2 because of lens focusing PICTURE)
In this case - after doubling the distance #2
4 times weaker light from one point, but 4 times more pictures of points / area unit of the sensor:

[darekk]

Area is proportional to any linear dimension raised to the square power. Then at n times larger distance the light energy is shed on n^2 larger area (what makes it n^2 weaker / area unit), but area catched as picture by given pixel is also n^2 larger (n^2 times more points in the field of view).

[Mick]

In a studio set up, Reflected light is NOT an issue. So why worry about it? You are only interested in Incident light.

[arith]

Cor; it is complicated. So, what Colin said.

It is irritating how often he is just plain right only jokin, but I will add if you move further away you tend to zoom in as well.

[Colin Southern]

In a studio set up, Reflected light is NOT an issue. So why worry about it? You are only interested in Incident light.

Actually, we use both in the studio

[Mick]

Zoom has nothing to do with it.

If you have lit your subject and determined the exposure to, let's say, f8 @ 1/250.
You can move your camera any where with in the room, with what ever medium at the film plane, CCD or Film, any thing that is lit in that exposure range will be recorded.

[arith]

Zoom has nothing to do with it.

If you have lit your subject and determined the exposure to, let's say, f8 @ 1/250.
You can move your camera any where with in the room, with what ever medium at the film plane, CCD or Film, any thing that is lit in that exposure range will be recorded.

I'm merely pointing out if you zoom in your have smaller fov, a smaller area.

Take a photo of a candle in the dark with both wide and telephoto; the wide is darker.

Also I was taught studio photography at school, and never used incident light. How does that work, photo of a blond the same exposure as a brunette. How do you know the EV range?

I just will do it the way I know, if it is f4 in a dark room at 50mm subject filling the frame, then you receive the same number of photons at f4 with 100mm but twice as far away with the subject the same size and same DOF of course everything else fixed.