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Suppose I have a setup with 1 light source, 1 subject to be photographed, and 1 camera.

Keeping all parameters constant, if the light source gets closer to the subject, it will be more exposed, leading to a brighter image, according to the inverse square law (as expected).

But if I keep the light source at constant distance, and change only the distance from the subject to the lens (by moving the camera backwards), the exposure will not change, even though light still has to travel a bigger distance from the subject to the camera.

Why is the exposure affected in the first case but not in the second?

Distance 1 affects exposure, while Distance 2 does not

As seen in the image, Distance 1 affects exposure, while Distance 2 does not.

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8 Answers 8

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My simple diagrams to the rescue, this time stealing yours.

I segmented some zones of the apple into little squares. Each square reflects some amount of light. When the distance is one unit (whatever this is), as a reference, the camera receives the amount of light = 1.

enter image description here

When you double the distance, the light received on the camera by each little square actually is reduced to 1/4, but now you have 4x zones of the apple shooting photons to the same area of the sensor.

enter image description here

You have two inverse square laws compensating each other.

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Another way of looking at this effect is to look at the definition of the aperture number, also known as f-number. If, for example, you have an f-number of f/2.8, that means that the diameter of the light collecting circle, the iris, of the lens is 2.8 times smaller than the focal length.

For example a 50mm lens, shot at f/2.8 has an iris diameter of 50mm/2.8 ≈ 17.9mm, whereas a 100mm lens at f/2.8 has an iris diameter of 50mm/2.8 ≈ 35.7mm. This means that the light-collecting area of the 100mm/2.8 lens is four times that of the 50mm/2.8 lens.

Therefore if you increase the subject-to-camera distance by a factor two, but want to keep the subject the same size in the frame, then you have to also double the focal length of your lens (e.g. going from 50mm to 100mm). To keep the exposure the same, i.e. keep the same amount of light hitting each pixel, you use the same f-number, e.g. f/2.8, but we just saw that doubling the focal length while keeping the same f-number, will increase the light-collecting area by four times. Therefore, even though the light intensity from the subject has decreased by a factor four, when you increase the focal length to compensate the framing of the subject, while keeping the same f-number, you also increase the amount of light collected by a factor four, thus compensating for the drop in intensity due to the increased distance.

We use the f-number when talking about exposure, precisely because this definition ensure that the exposure will be the same for different focal-length lenses.

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    \$\begingroup\$ And since doubling the exposure while keeping the f-number the same means doubling the light collecting area, a lens that can operate at a high focal length with a low f-number is likely to be large and therefore expensive - and cheaper zoom lenses may only be able to operate at low f-number while the focal length is short. \$\endgroup\$
    – bdsl
    Dec 10, 2022 at 19:53
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Subject distance does vary the amount of light in accordance with the inverse square law.

If you increase the subject distance by 2x it becomes 1/2 the size LxW at the image plane... .5L x .5W is .25 the area, and 1/4 the amount of light... or factor 2 square inversed (the ISL formula). I.e. a smaller area at the same brightness IS less light.

So the exposure, light per area, doesn't change; but the total light from the source does... This is the same thing as looking at situation 1 from the perspective of the apple... the brightness of the bulb doesn't change, but the amount (intensity) of light at the apple does.

This is also the same thing that happens when you zoom in/out on a subject with a constant aperture zoom lens. And it is why larger sensors tend to perform better in low light; because they actually receive more light (subject is larger on larger sensor for same composition).

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For this purpose the camera is the same as a human eye. When you look at something from further away it doesn't look dimmer, it looks smaller.

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Light falls off with distance. Most often this falloff follows the law of the inverse square --- but not always. This law pertains to point sources thus large area lamps do not precisely follow.

As a rule of thumb, the subject will not be a light source, it will likely be a reflector of light. As such the amount of light it receives and thus reflects only changes if the amount of light energy playing on the subject changes.

In the case of changing camera-to-subject distance, as the camera advances or retreats, the subject’s image size increases or decreases. As an example, if the camera position is moved 10x further away, the image’s area size as discerned by the camera shrinks. This change will be a shrinkage that results in the image reducing to 1/100 of its former area (10 x 10). The result is the apparent size change and the apparent brilliance change with distance cancels; the exposure setting remains constant.

However, if the camera position or focal length is being adjusted, it is likely that a background or foreground change as to characteristic will occur. Such a change is likely to weigh the exposure and induce a change. This type of exposure change has nothing to do with the law of the inverse square.

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In addition to the other answers: actually, the registered brightness of an object does depend on the distance, if its perceived size is smaller than the resolution of the system. For example, more distant stars seem dimmer. This is because the amount of light reaching the camera indeed decreases with the inverse square of the distance, while the size of the image cannot get any more small.

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  • \$\begingroup\$ Oh, death by analogy! A classic physics question is that because of the number of stars, why isn't the night sky a solid blaze of light? The answer has nothing to do with inverse square law, there are enough stars in the sky that every location you look is a point source of light. The reason the sky is dark is DUST! \$\endgroup\$ Dec 13, 2022 at 2:42
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If the light were large enough (i.e. wall-shaped) then that distance wouldn't matter either!

So for example, if you used a large, flat light box to illuminate a small object and shot it from the side, it would be nearly the same brightness 5 cm away from the box as it would be 2 cm away, and not ~six times dimmer.

Let's refer to @Rafael's answer and exploit the idea of pixels, and first think of why when we walk away from a wall it stays the same brightness and does not get dimmer per the inverse square law.

Each fixed size "pixel" (let's say one micron square) of the wall definitely gets dimmer per inverse square, but the number of "wall pixels" that illuminate a given camera pixel increases as the square of the distance so that the total illumination of the camera pixel (or "eye pixel") remains about constant.

Same thing with the "illumination pixels" of the light box; as our sample moves away from its surface the light from a given 1 micron square drops by inverse square, but the number of illumination pixels that contribute to say say a 60° or 75° cone of illumination on our object pixel increases as the square of the distance, so once again, the illumination remains roughly constant.

Once your light box gets far enough away that it stops filling that large cone of illumination, the object will appear dimmer and dimmer.

Likewise, once your apple gets far enough away that you can't resolve it, it will also of course get dimmer and dimmer per the inverse square law.

Why does the object have to be so far away as to be unresolved before it begins to obey the inverse square law, but the light box only needs to be a few tens of centimeters?

Because the object is (likely) a diffuse scatterer, so that large cone of illumination feeds each camera pixel.

If you apple was polished so much that you could see the directly reflected light from its mirror-smooth surface, then the only thing that would matter for the reflection's brightness would be whether the reflected (virtual) image of the light source as seen by the camera was resolved, or not.

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The light spreads out over the distance, making it dimmer. The subject has an inherent brightness that does not depend on distance, however the subject gets smaller therefore decreasing the total amount of light received by the camera with the same inverse square law if the whole subject fits in the frame.

Think of lighting a soccer ball in a vantablack room. Now position the camera far enough away so that the ball barely fits in frame. If you move to twice the distance, the ball will appear 1/2 the size and 1/4 the light will reach the camera.

When you get far enough away that the camera can no longer resolve the shape of the ball and all the light fits in one pixel (like when photographing stars in astrophotography), moving further away will start decreasing the brightness of that pixel.

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