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With exposure calculations, there are aperture, shutter speed, and ISO. Correspondingly, flash and light meters gives you the aperture (as an f-number) and shutter speed for a given ISO.

However, light intensity drops drastically with distance.

This may not be a problem for through-the-lens metering since that measures the amount of reflected light from the subject that reaches the camera—but an external flash meter reads the light hitting the subject yet gives you an f-number and shutter speed regardless of where the camera is located.

Why is this so? I've seen a lot of photographers that take a measurement from under the chin and then move around—how is this a proper metering? Shouldn't the exposure settings needed change depending on how far away the meter is from the camera?

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You do get less light as you move further from a object. However, that less light is focused on a smaller area. It so happens that the two effects cancel out, and the focused image of a object will be the same brightness as distance is changed, assuming the f-stop is held the same.

For example, moving twice as far away means the lens intercepts ¼ the light from the same object. However, the size of the focused image shrinks by two in linear dimension, meaning a reduction of area by 4. So ¼ the light is focused on ¼ the area, resulting in the same brightness image.

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  • \$\begingroup\$ Thank you that helped a lot, but yet it raises another question, why the focal length is also absent from reading? I think your answer rely on keeping the focal length the same? \$\endgroup\$
    – user174174
    Jan 5, 2018 at 22:09
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    \$\begingroup\$ Focal length does matter, but its effect is already wrapped into f-stops. The f-stop is the ratio of the effective lens diameter to the focal length. As the focal length gets longer, and the same light is therefore spread over more image area, keeping the f-stop the same means the lens diameter goes up too, which compensates. \$\endgroup\$ Jan 5, 2018 at 22:36
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    \$\begingroup\$ Whoever disagrees with this, what exactly do you think is wrong? \$\endgroup\$ Jan 5, 2018 at 22:38
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    \$\begingroup\$ It is not wrong, it is precisely right. 1/4 the light (due to distance) on 1/4 the area (due to distance) is the same exposure of the surface there. This can soothe those who think ISL should be involved. But this result can be expressed as that the "same" exposure is independent of the camera distance, as of course we easily see is true. For example, the full moon is rather distant, but it is lighted by our same sun, and approximately the same Sunny 16 works well on it. The big difference is that it's a 12% gray body. \$\endgroup\$
    – WayneF
    Jan 6, 2018 at 4:33
  • \$\begingroup\$ Very good answer! \$\endgroup\$
    – Itai
    Jan 8, 2018 at 5:28
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The inverse square law applies to the distance between the light source and the subject. It doesn't apply to the distance between the subject reflecting the light and the camera in the same way.

This is because as the camera distance is increased, the area covered by the same subject in terms of the camera's field of view is decreased by an inverse amount. The two cancel each other out. If you double the distance to the subject you reduce the area the subject covers on the film/sensor by a factor of four. One fourth as much light covering one fourth the area on the film or sensor is the same field density, which is what we measure for exposure: light per unit area.

If we double the distance and also double the focal length to keep the same subject framing, then our entrance pupil must also double in diameter (a four-fold increase in area) to maintain the same f-stop. So we are right back to the same field density of light falling on the sensor or film.

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The exposure is based on the amount of light hitting the subject intertwined with how much light is reflected from the subject. Thus the exposure remains a constant regardless of camera to subject distance. While this might seem to violate the fact that light falls off with distances, it doesn’t because this is a special case.

Light falloff with distance is called the “law of the inverse square”. Suppose a lamp 1 meter from a surface delivers 1000 units of light. If we double the lamp to subject distance by backing off the lamp to 2 meters, the light falloff is 2 squared = 4. Now the light intensity at the subject plane is 1000 ÷ 4 = 250 units. But, you recognized this fact so what’s happening with our photo setup?.

The law of the inverse square only strictly applies only if the lamp is a point source like a tiny bare light bulb. As soon as we place this lamp in a reflector, or impose as diffuser, this law goes out the window. Maybe not completely gone, the degree violation is a variable, depending on the situation.

Suppose the lamp is put in a collimating reflector and the beams become parallel like a spot light? Now the spot does not obey, the falloff is practically nonexistent. Same for a laser beam, they practically never falloff, they can hit the moon with well-nigh no loss.

If the light bulb is in an umbrella and totally diffused, now the light is called a “broad” and this law goes out the window, you can move the subject around quite a bit and the exposure will be highly constant.

So what about a portrait subject illumined for an exposure of f/5.6? The light reflections from the face and clothing consist of highly diffused light beams. They don’t even come close to obeying the law of the inverse square. You move the camera all over the place and the exposure remains constant. However, just pat a bare bulb lamp and change lamp to subject distances and the exposure dances.

By the way, the popularity of the umbrella lighting and their origin, a broad, is due to the diffusion they bring to the table due to the fact they almost completely slay the law of the inverse square.

Added thoughts: Spotlights output parallel beams. It is this parallelism that thwarts ray scattering thus the output of the spotlight is preserved over distance. Now most illuminated objects do not have polished surfaces thus they reflect light rays that scatter in all possible directions. Most of this reflected light from objects will be lost to us and our camera. If we draw trace lines of the light rays reaching our eyes and camera, the trace reveals, these image forming rays are arriving as parallel or nearly so. It is this parallelism that quashes the inverse square law. This explains why commonplace objects do not brighten or dim as distance changes and why we need not change camera setting as subject distance changes, and why spot light meter reading do not change with distances.

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    \$\begingroup\$ I think you need to explain this a little more. A large, diffuse light source still follows the inverse square law; it's just that instead of having a single point source, the large source acts like a huge array of point sources. Close to a broad source, the falloff from one little part is made up for by light from other parts. With a point, the light spreads in all directions from one place. With a broad, it spreads in all directions from many points. But if you move farther away, that broad starts acting more and more like a point, and you're back to inverse square behavior. \$\endgroup\$
    – Caleb
    Jan 5, 2018 at 22:12
  • \$\begingroup\$ And when you're talking about the reflection from part of a face in a portrait, you don't have to be far at all before that inverse square behavior kicks in, because a face isn't very large in the first place. \$\endgroup\$
    – Caleb
    Jan 5, 2018 at 22:13
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    \$\begingroup\$ @ Caleb - You are flat wrong! You can shoot a face from 1 meter distance or 1000 meters and the exposure will virtually be unchanged. The thing I need to add - The reflected light meter averages and as you move about its field of view changes and so does the reading. A spot meter does a good job and an incident meter is likely the best bet. \$\endgroup\$ Jan 5, 2018 at 22:31
  • \$\begingroup\$ @ Caleb -- Broad lamps and umbrellas begin to act like point sources after a certain distance. The studio needs to have lots of maneuvering room else they remain reasonably constant. \$\endgroup\$ Jan 5, 2018 at 22:35
  • \$\begingroup\$ As I was and still am confused I did a little experiment. I used a bulb and set my camera to spot metering and fix the iso and aperture. Reading from the bulb, shutter speed was dropping exactly by half for every meter I was going back. Then I pointed the bulb against the wall reading from it and shutter speed literally didn't change by distance. \$\endgroup\$
    – user174174
    Jan 6, 2018 at 0:54
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Exposure simply does not depend on camera distance. However, the Inverse square law does depend on light source distance (but not on camera distance).

So both the photo of your dog in backyard at ten feet, and the photo of the mountain at 30 miles away are both the same Sunny 16 exposure (assuming no clouds). Because both are 93 million miles from the light source, so a few more feet or miles are not significant. Even our astronauts on the Moon were at an insignificant different distance (at most about 1/4 of 1% difference from here on Earth). Mars will be a little different.

Flash is a little different in that it is in the same room with us at close distance, so flash distance very definitely matters. But in a studio portrait situation, still only the flash to subject distance matters (which likely does not move). The camera distance does not matter whether it moves or not.

Or said another way is the way Olin said it. Which is correct of course, but it is the reason that it still boils down to "the camera distance does not affect exposure". However, cameras at different distances can then see radically different scenes to meter, which is a different factor.

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    \$\begingroup\$ As far as the camera is concerned, every object in the frame is a light source. Whether the light is emitted or reflected doesn't matter -- it still follows the inverse square law (more or less). Olin's answer nails it -- the change in size compensates for the reduction in the total amount of light. I don't think this answer addresses the question. \$\endgroup\$
    – Caleb
    Jan 5, 2018 at 20:58
  • \$\begingroup\$ You're imagining that, but simple tests show it's not true. Incident metering of ISL from a reflected umbrella works rather well (past the fabric) if you count distance along the actual light path from the source (flash tube). ISL does not hold up if you try to count from the fabric. Same for a softbox if counted from the source distance, but not if you count from the fabric. The flash tube is the source. You should try it once before you try to explain it. And if these light a subject, the camera exposure of that subject is the same from any camera distance. This is of course clearly obvious. \$\endgroup\$
    – WayneF
    Jan 6, 2018 at 4:20
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Your question is bit hard to understand, A hand held light meter simply measures the light that is hitting it, (ambient or flash) It does not need to know the distance of the light source in order to measure it, nor does it need to know where the camera is. It is simply measuring the amount of light.

The reason for moving the meter around is because there may be ( by design or not ) a difference in the amount of light on one side of the face or the other. The photographer wants to know everything about the light so they can make a good photo or change the light to meet their preconceived artistic vision of what they want the photo to look. They may want 2 stops less light on the opposite of the side of the face that is lit by the KEY light. They may want 1.5 stops more light from the rim light placed behind the subject. Each of these lighting zones need to be measured in order to adjust them and set the camera to correspond to them. You have to tell the light meter what ISO you are going to be setting your camera to so you get the correct measurements.

I am unclear as to what "exposure triangle" has to do with your question about light meters.

IN MY OPINION "Exposure triangle" is a miss leading concept. Exposure is the amount of light you let into the camera by changing the aperture and\or the shutter speed, changing the ISO is changing the the sensitivity of the sensor that is capturing the amount of light you are allowing to enter the camera.

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    \$\begingroup\$ I edited the question for clarity — I don't think you're answering what I'm pretty sure is the key part of the question. \$\endgroup\$
    – mattdm
    Jan 5, 2018 at 20:45
  • \$\begingroup\$ On including ISO in "exposure", see this — I'm afraid you're fighting an uphill battle. Although maybe not as uphill as my battle to stop people from saying "exposure triangle" because triangle makes no sense. :) \$\endgroup\$
    – mattdm
    Jan 5, 2018 at 20:48
  • \$\begingroup\$ @mattdm Clarity? the meter still does not need to know the distance of the light it is measuring. The key part of the question seams to be about metering light. \$\endgroup\$
    – Alaska Man
    Jan 5, 2018 at 21:11
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    \$\begingroup\$ The fact is that language reflects usage. "Exposure" in any of these senses is jargon which does not necessarily map to the original common definition of the word in any case. \$\endgroup\$
    – mattdm
    Jan 5, 2018 at 21:36
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    \$\begingroup\$ @Alaskaman The OP's question is basically: Why is the meter reading you get when you're close to a subject still valid when the camera is far away? This answer doesn't address that at all. \$\endgroup\$
    – Caleb
    Jan 5, 2018 at 21:39

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