Inverse Square Law (Camera-to-Subject Distance)

Hi,

I understand that as you move a flash unit (or light source) away from a subject, the amount of light the subject receives is reduced. For example, if the flash-to-subject distance is doubled, the subject receives 2 stops less light.

OK, so far so good... but what happens if the flash-to-subject distance remains the same but instead the camera-to-subject distance is doubled? Does the film/sensor receive 2 stops less light? I don't think this is the case because for example, a spot meter is used from a distance, and in the studio a flash meter can be used to measure incident light on the subject which determines exposure regardless of camera-to-subject distance.

In all my years of photography the thought never occurred to me, until now, and now it is bugging me. Can someone explain (perhaps even the math) as to why the camera-to-subject distance doesn't affect exposure... or does it?

Thanks,
~Dom

Just to confirm - if the light falling on a subject stays the same then, as the camera is moved to various distances from the subject, the exposure settings need to be kept constant to keep the exposure of the subject in the images the same.

As you move the camera further from the subject, it receives less light. But the subject will be a smaller portion of the total FOV, so that light covers a smaller area of the sensor. The result is you get the same number of photons per photocell, so the exposure remains the same.

Example: "Correct" exposure for a full moon on a clear night is 1/100 f/16 ISO 100, the same as "correct" exposure for a sunny scene on earth, even though camera-to-subject distance is a bit different.

Thanks for the explanation... I think I will be able to sleep tonight!
~Dom

Curtis N wrote in post #7905156
As you move the camera further from the subject, it receives less light. But the subject will be a smaller portion of the total FOV, so that light covers a smaller area of the sensor. The result is you get the same number of photons per photocell, so the exposure remains the same.

No, because if you use a telephoto lens to keep the subject the same size on the sensor, the exposure will still remain the same.

The reason it remains the same is because it's the same bundle of rays forming the image at 10 feet as it is at 100 feet or 100 miles (if you've got a telephoto big enough).

The inverse square law predicts the effects of the spread of light from a point source.

But when you view the subject, the image is formed by a quite parallel bundle of rays striking upon your sensor. Even when you move back a great deal, that parallel bundle of rays has spread only imperceptibly.

Yes, light may be spreading from the object in all directions, but from whatever direction you happen to be looking, the image is being formed on your retina (or sensor) by a parallel bundle, which stays parallel even as you move away from it.

An old thread, but this thought has bothered me for long. This forum is great

Curtis N wrote in post #7905156
As you move the camera further from the subject, it receives less light. But the subject will be a smaller portion of the total FOV, so that light covers a smaller area of the sensor. The result is you get the same number of photons per photocell, so the exposure remains the same.

Example: "Correct" exposure for a full moon on a clear night is 1/100 f/16 ISO 100, the same as "correct" exposure for a sunny scene on earth, even though camera-to-subject distance is a bit different.

I thought I had this figured out and now I am confused.

http://photo.tutsplus.​com …g-the-inverse-square-law/

What are you confused about?
The inverse square law applies to distance from light source to subject. To expand on the moon example, the correct exposure for the sunny side of the moon is the same as for a subject on earth in direct sunlight. The exposure doesn't change because the distance from light source to subject hasn't changed (the earth and the moon are about the same distance from the sun).

I was under the impression that if I move my subject back I need to adjust either my flash output or aperture as the link I provided is stating.

Curtis N wrote in post #7905156
As you move the camera further from the subject, it receives less light.

The "it" in Curtis' statement above should be translated as "the camera".

digital paradise wrote in post #13244610
I was under the impression that if I move my subject back I need to adjust either my flash output or aperture as the link I provided is stating.

If you are changing the distance between the light source and the subject, then yes you will need to change the exposure setting on the camera or change the brightness level of the light source to maintain the same exposure level of that subject in the recorded image.

However, if you are photographing a subject that is lit by a light source that stays at the same distance from the subject, you can position the camera anywhere you like and the exposure settings would not need to change to record that subject at the same brightness in the image.

As an example of the above, assume someone is sitting in their garage at night with a light in the garage illuminating him. The light is bright enough so that you could use ISO 400, 1/30 shutter speed and f/4 aperture if you were standing just outside the garage and get a nice image of the person. Now, back up 100 yards and re-shoot with the same exposure settings. The result will be precisely the same exposure of the person.

That makes perfect sense. I just misunderstood that. Curtis was talking about the light source not the camera. Thanks.

Ok. So if you're taking a picture of Jupiter, you'll need to crank up the ISO because it's further away from the sun. Conversely, Venus is very bright because it's so close to the sun, and has been referred to as "the morning star" because of its brightness.

It's all about distance from light to subject. Make sense?

Yes it does. Thanks