silly inverse square law question.

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gpop
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p.1 #1 · silly inverse square law question.

I understand the inverse square bit about light / exposure (within reason) but I'm confused about something...
if you light a subject with off camera light source, but move the camera away from the light and subject will this effect exposure to the same degree as if you moved the light farther from the subject?

as an example if you have two subjects, each with its own light source at the same fixed distance and power (lets assume each subject and light source are identical) but one subject/light is three times as far from the camera, would the exposure fallow the same inverse square law?

I would assume it does, but I've never read any mention of it.

I don't have a practical situation I'm trying to solve, just trying to better understand the theory.

pathetic illustration:
L2/S2 <-20 feet-> L1/S1 <-10 feet-> camera.

(light source and subject #2 is 30 feet from the camera, light source and subject #1 is 10 feet from camera.)

Oct 31, 2012 at 07:42 PM

PeterBerressem
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p.1 #2 · silly inverse square law question.

Don't worry, you get the same exposure no matter what subject distance.
The inverse law bases on the fact that light from a point source is emitted at an angle.
To illustrate this: draw lines from a point to a subject. The more distant the subject the more lines pass it without touching it. Now look at the 'lines' that your camera (or eye) sees - they are the same, regardless of distance.

Oct 31, 2012 at 07:58 PM

alohadave
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p.1 #3 · silly inverse square law question.

The lighting exposure will remain the same because the light on the subjects hasn't changed.

What will change is the total exposure for the scene (assuming that you don't change your focal length), because your subjects will fill less of the frame than when the camera is closer. Whether you want to adjust for the scene change is a creative decision.

Oct 31, 2012 at 10:44 PM

gpop
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p.1 #4 · silly inverse square law question.

thanks for the replies, it's starting to sink in.

Oct 31, 2012 at 11:14 PM

RustyBug
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p.1 #5 · silly inverse square law question.

Kinda like if we both stand about 10' from a wall (camera and light same distance from subject). I'm shining a flashlight on the wall. Then, if you (camera) walk up closer or farther from the wall ... the amount of light (me) on the wall remains constant whether you (camera) are 2', 10' or 100' away from the wall (as long as I keep holding the flashlight 10' from the wall.

Nov 01, 2012 at 02:47 AM

curious80
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p.1 #6 · silly inverse square law question.

When you move away from the subject, the light reaching from the subject to the camera of course goes down as per the inverse square law. However at the same time the subject is becoming smaller in your frame, i.e. taking up less pixels. Thus the amount of light falling per pixel remains the same. So the exposure remains the same.

Nov 01, 2012 at 02:54 AM

blob loblaw
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p.1 #7 · silly inverse square law question.

if a lonely flash goes off in the woods and there's nobody to see it, how much light did it throw? sorry could not resist!

Nov 01, 2012 at 03:00 AM

BrianO
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p.1 #8 · silly inverse square law question.

curious80 wrote: When you move away from the subject, the light reaching from the subject to the camera of course goes down as per the inverse square law.

Nope.

curious80 wrote: ...Thus the amount of light falling per pixel remains the same. So the exposure remains the same.

If that were true, an 8Mp camera and an 18Mp camera would need different aperture and shutter speed settings to get the same exposure using the same lens, light, and distances.

Re-read PeterBerressem's explaination; he has it correct.

Nov 01, 2012 at 09:23 PM

curious80
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p.1 #9 · silly inverse square law question.

BrianO wrote:
curious80 wrote: When you move away from the subject, the light reaching from the subject to the camera of course goes down as per the inverse square law.

Nope.

curious80 wrote: ...Thus the amount of light falling per pixel remains the same. So the exposure remains the same.

If that were true, an 8Mp camera and an 18Mp camera would need different aperture and shutter speed settings to get the same exposure using the same lens, light, and distances.

Re-read PeterBerressem's explaination; he has it correct.

What I have described is completely correct, except I should have said that the "intensity of light" reaching the camera goes down as per the inverse square law. It follows simply from physics laws, and there is no ambiguity or doubt about it. I didn't completely follow PeterBerressem's explanation but I think he is just saying what I am saying in a different way.

And 18Mp camera indeed does get less light per pixel. The smaller pixels require more gain in hardware to bring the berightness to the desired level. This is why the smaller pixels are noisy.

Edited on Nov 02, 2012 at 12:41 AM · View previous versions

Nov 02, 2012 at 12:37 AM

BrianO
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p.1 #10 · silly inverse square law question.

gpop wrote: ...if you have two subjects, each with its own light source at the same fixed distance and power (lets assume each subject and light source are identical) but one subject/light is three times as far from the camera, would the exposure fallow the same inverse square law?

This is something that you can easily test for yourself, and you don't need two subjecst and two light sources; just use one subject and one light source, and take two (or more) shots from different distances without changing your exposure settings. No fair cheating with autoexposure; this is a Manual mode experiment.

We can also extrapolate from our everyday experience shooting outdoors under sunlight; we don't have to change our exposures every time we move closer or further from the subject.

Think about the "Sunny 16 Rule" -- "On a sunny day set the aperture to f/16 and the shutter speed to the reciprocal of the ISO setting, for a subject in direct sunlight." Note that there's no mention of camera-to-subject distance there, because it doesn't change the exposure.

Next, it's not directly applicable to your question, but I thought you and others might find this 12-minute video by Marl Wallace of interest:

http://www.youtube.com/watch?v=nk9cTa3UthM&feature=player_embedded#!

Nov 02, 2012 at 12:39 AM

RustyBug
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p.1 #11 · silly inverse square law question.

curious80 wrote:
When you move away from the subject, the light reaching from the subject to the camera of course goes down as per the inverse square law.

As Brian mentions, this isn't correct.

Also, if that were true, its inverse would be true too ... i.e. that as you moved your camera CLOSER to the subject the light reaching from the subject to the camera would go UP per the inverse square law.

OBVIOUSLY, this does NOT occur ...

Point being ... it is NOT "camera to subject" distance changes that are determinant of exposure, rather "light to subject" distance changes that matter regarding illumination / exposure.

When your flash is on camera, it might seem like the camera to subject distance is making the difference when you move farther away (as both camera & flash are moving simultaneously), but it is the flash to subject distance, not the camera to subject distance that is responsible.

Nov 02, 2012 at 01:14 AM

BrianO
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p.1 #12 · silly inverse square law question.

curious80 wrote: What I have described is completely correct, except I should have said that the "intensity of light" reaching the camera goes down as per the inverse square law. It follows simply from physics laws, and there is no ambiguity or doubt about it.

Your understanding of physics is, in this case, flawed, unless I am misunderstanding what you are saying.

curious80 wrote: ..And 18Mp camera indeed does get less light per pixel. The smaller pixels require more gain in hardware to bring the berightness to the desired level. This is why the smaller pixels are noisy.

Right, but the reduction in light isn't dependent on distance; the smaller sensels receive less light than the larger sensels at 5 feet and at 50 feet, and that amount is unchanged.

The physics are thus: the reason the inverse square law applies to light-source-to-subject-distance is that the light from the source spreads out over a greater area as the distance increases: the same amount of light on more surface. But the light bouncing off the subject and entering the lens of the camera doesn't cover more area; the size of the lens remains unchanged.

Again I say, this is easy to test; do the experiment I described above and you'll see for yourself: camera-to-subject distance changes don't change exposure...light-to-subject distance changes do.

[Edit: In re-reading your post I see that you did stress that the actual exposure remains unchanged with camera-to-subject distance changes, so maybe we're on the same page after all.]

Nov 02, 2012 at 02:40 AM

RustyBug
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p.1 #13 · silly inverse square law question.

curious80 wrote:
However at the same time the subject is becoming smaller in your frame, i.e. taking up less pixels. Thus the amount of light falling per pixel remains the same. So the exposure remains the same.

According to this, you're implying an offsetting change predicated upon camera to subject and image size. A subsequent corollary would be that if I move further away and change lenses to retain (or enlarge) the subject size in the frame (no longer relatively offsetting as you've proposed), it would REQUIRE a change in exposure.

Here again, this is not true. Whether the subject is filling the frame 100%, 10% or 1% has no bearing on the illumination / exposure for the subject. It is true that your camera's average reflected meter reading might change as a byproduct of filling/reducing your frame with more/less brighter/darker area for metering the entire scene... but that does not change the illumination/exposure for the subject.

Whether you use a 50mm lens to photograph the moon, or a 500mm lens (changing the size of the subject in the frame) doesn't change the illumination, nor the appropriate exposure. Granted, you'll get a different reflected meter reading due to the size of the subject change ... but that doesn't change the exposure.

It isn't because BOTH are simultaneously changing that the exposure remains the same ... it's because NEITHER are changing when you change camera to subject distance. If the size of the subject in the frame were to impose a required exposure change, then the exposure for a 4X zoom (i.e. 24-105) would not be able to remain the same for a wide angle view of the subject as it is for a telephoto view of the subject.

Camera to subject distance is not responsible for illumination / exposure changes.

Consider this ... while a lighthouse cannot illuminate a ship that is miles away (inverse law renders the illumination too low), those on the ship can readily see (or photograph) the light emitting from the lighthouse ... i.e. unaffected by the inverse square law). In fact, a lit cigarette on a moonless night can be seen from miles away, yet it can essentially illuminate nothing beyond the person holding it.

How light is seen at the subject is not the same as how it disseminates from the subject. If the inverse law applied as you are espousing, the lights of a night skyline would not be able to be seen nor photographed from such distances neither.

BTW ... if camera to subject distance is responsible for impacting exposure ... how much exposure difference would Neil Armstrong (standing on the moon with camera a scant few feet away) have to use in order to account for the difference in camera to subject distance than we (standing on earth @ 240,000 miles away) have for exposure values of the moon illuminated by the sun with the same light to subject distance? Given the orders of magnitude that Neil Armstrong was closer to his subject than we are ... a camera didn't exist that would accommodate the inverse law relative to camera / subject distance variance from the exposure values we see from here ... IF camera to subject distance were part of the exposure determination.

Coming back down to earth ... bouncing lite off of a white reflector panel ... the light that is reflected off the panel intended to illuminate a different subject will appear to further disseminate iaw with isl, but that is because it is simply replicating dispersion through AI=AR over the angles/area involved @ the reflector. But. if you are photographing that white panel, you are photographing the amount of light present at the panel, not the subsequent dissemination of the light that is traveling away from the panel ... and your exposure for the amount of light illuminating the panel doesn't doesn't change with camera to panel distance.

Camera to subject distance is NOT responsible for illumination / exposure values and does NOT change with the inverse square law with regard to viewing / photographing.

Edited on Nov 03, 2012 at 12:27 PM · View previous versions

Nov 02, 2012 at 04:36 AM

cordellwillis
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p.1 #14 · silly inverse square law question.

The camera and lens has nothing to do with it. Only the light and subject. You change the distance of the light and the subject your exposure will change.

If you are taking into consideration other surrounding elements then your exposure can appear to change. However, that has nothing to do with the exposure of the subject.

Nov 02, 2012 at 12:46 PM

RustyBug
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p.1 #15 · silly inverse square law question.

cordellwillis wrote:
The camera and lens has nothing to do with it. Only the light and subject. You change the distance of the light and the subject your exposure will change.

If you are taking into consideration other surrounding elements then your exposure can appear to change. However, that has nothing to do with the exposure of the subject.

+1

Nov 02, 2012 at 12:54 PM

curious80
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p.1 #16 · silly inverse square law question.

RustyBug wrote:
curious80 wrote:
However at the same time the subject is becoming smaller in your frame, i.e. taking up less pixels. Thus the amount of light falling per pixel remains the same. So the exposure remains the same.

According to this, you're implying an offsetting change predicated upon camera to subject and image size. A subsequent corollary would be that if I move further away and change lenses to retain (or enlarge) the subject size in the frame (no longer relatively offsetting as you've proposed), it would REQUIRE a change in exposure.

Here again, this is not true. Whether the subject is filling the frame 100%, 10% or 1% has no bearing on the illumination / exposure for the subject. It is true that your camera's average reflected meter reading might change as a byproduct of filling/reducing your frame with more/less brighter/darker area for metering the entire scene... but that does not change the illumination/exposure for the subject.

Whether you use a 50mm lens to photograph the moon, or a 500mm lens (changing the size of the subject in the frame) doesn't change the illumination, nor the appropriate exposure. Granted, you'll get a different reflected meter reading due to the size of the subject change ... but that doesn't change the exposure.

It isn't because BOTH are simultaneously changing that the exposure remains the same ... it's because NEITHER are changing when you change camera to subject distance. If the size of the subject in the frame were to impose a required exposure change, then the exposure for a 4X zoom (i.e. 24-105) would not be able to remain the same for a wide angle view of the subject as it is for a telephoto view of the subject.

Camera to subject distance is not responsible for illumination / exposure changes.

Consider this ... while a lighthouse cannot illuminate a ship that is miles away (inverse law renders the illumination too low), those on the ship can readily see (or photograph) the light emitting from the lighthouse ... i.e. unaffected by the inverse square law). In fact, a lit cigarette on a moonless night can be seen from miles away, yet it can essentially illuminate nothing beyond the person holding it.

How light is seen at the subject is not the same as how it disseminates from the subject. If the inverse law applied as you are espousing, the lights of a night skyline would not be able to be seen nor photographed from such distances neither.

BTW ... if camera to subject distance is responsible for impacting exposure ... how much exposure difference would Neil Armstrong (standing on the moon with camera a scant few feet away) have to use in order to account for the difference in camera to subject distance than we (standing on earth @ 240,000 miles away) have for exposure values of the moon illuminated by the sun with the same light to subject distance? Given the orders of magnitude that Neil Armstrong was closer to his subject than we are ... a camera didn't exist that would accommodate the inverse law relative to camera / subject distance variance from the exposure values we see from here ... IF camera to subject distance were part of the exposure determination.

Coming back down to earth ... bouncing lite off of a white reflector panel ... the light that is reflected off the panel intended to illuminate a different subject WILL disseminate iaw with isl. But. if you are photographing that white panel, you are photographing the amount of light present at the panel, not the subsequent dissemination of the light that is traveling away from the panel ... and your exposure for the amount of light illuminating the panel doesn't doesn't change with camera to panel distance.

Camera to subject distance is NOT responsible for illumination / exposure values and does NOT change with the inverse square law with regard to viewing / photographing.

I think you are misunderstanding what I am saying. I am in agreement that the camera to subject distance does not impact the exposure. The question is only about the exact mechanics of WHY that happens. Light always follows the inverse square law, it doesn't know whether it is coming from a source to a subject or from a subject to another subject. Its the combined effect of light and lens operation which makes the inverse square law irrelevant in determining the exposure of a subject.

Just as someone else said, when the light leaves a point light source, it spreads and as you move the light away or move the subject away, the portion of the light that falls on the subject goes down as per the inverse square law. I believe we are all in agreement on that. The situation is somewhat similar with real life sources which are not point sources. If you have a rectangular light source, you can just think of it as lots of point sources arranged in a rectangle.

Now lets switch over to the other side. When light hits the subject, it gets reflected. That reflected light then gets to our eye or the camera and thats how we see the subject. We are all in agreement on that as well. Now the light that gets reflected from the subject is in no way different in its behavior then the light that gets emitted from the original light source. You can roughly think of the subject as lots of small point sources sending light in all directions. And again the light spreads as it moves away from the subject. A portion of that light falls on the lens. Now since the light is still spreading as it moves forward so the inverse square law still applies. If you move the camera away, the lens will see less of the light. There is just no way to deny this fact. However what happens is that since the camera has been moved away so the lens will now focus this smaller amount of total light on a similarly smaller portion of the sensor. The result is that the light intensity at the sensor does not change, even though the light intensity seen by the lens goes down. A similar effect occurs in the eye as well

I hope that clears things up.

Nov 02, 2012 at 08:38 PM

curious80
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p.1 #17 · silly inverse square law question.

BrianO wrote:
curious80 wrote: What I have described is completely correct, except I should have said that the "intensity of light" reaching the camera goes down as per the inverse square law. It follows simply from physics laws, and there is no ambiguity or doubt about it.

Your understanding of physics is, in this case, flawed, unless I am misunderstanding what you are saying.

curious80 wrote: ..And 18Mp camera indeed does get less light per pixel. The smaller pixels require more gain in hardware to bring the berightness to the desired level. This is why the smaller pixels are noisy.

Right, but the reduction in light isn't dependent on distance; the smaller sensels receive less light than the larger sensels at 5 feet and at 50 feet, and that amount is unchanged.

The physics are thus: the reason the inverse square law applies to light-source-to-subject-distance is that the light from the source spreads out over a greater area as the distance increases: the same amount of light on more surface. But the light bouncing off the subject and entering the lens of the camera doesn't cover more area; the size of the lens remains unchanged.

Again I say, this is easy to test; do the experiment I described above and you'll see for yourself: camera-to-subject distance changes don't change exposure...light-to-subject distance changes do.

[Edit: In re-reading your post I see that you did stress that the actual exposure remains unchanged with camera-to-subject distance changes, so maybe we're on the same page after all.]

Well my understanding of physics can certainly be flawed

though I don't believe this is the case here. Maybe my slightly more detailed description that I just wrote above might convince you?

Nov 02, 2012 at 08:52 PM

PeterBerressem
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p.1 #18 · silly inverse square law question.

curious80 wrote:

Just as someone else said, when the light leaves a point light source, it spreads and as you move the light away or move the subject away, the portion of the light that falls on the subject goes down as per the inverse square law.

Yep, that was me.

And again the light spreads as it moves away from the subject. Now since the light is still spreading as it moves forward so the inverse square law still applies. If you move the camera away, the lens will see less of the light.

The parts of the reflected light spreading out in different directions don't matter all all. You / the camera / sensor only receive the rays which are directed towards you and this doesn't change with distance.

Nov 02, 2012 at 10:31 PM

curious80
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p.1 #19 · silly inverse square law question.

PeterBerressem wrote:
The parts of the reflected light spreading out in different directions don't matter all all. You / the camera / sensor only receive the rays which are directed towards you and this doesn't change with distance.

The rays that come in your "direction" spread out just like the rays that go from the light source to the subject. As you move your camera away the number of rays that fall on the lens surface decrease. This is just like what happens when you move a subject away from the light source. Light is light, it travels and acts the same way in both cases.

However, for the sake of argument, lets assume that what you are saying is true. Assume we are taking the picture of a white square 10 feet away which is printed on a black wall. And lets say that from 10 feet the white square is covering the center half of the frame/sensor with all black around it. Now lets move the camera back to 20 feet. You would agree that the square would now cover only 1/4th of the frame / sensor. Now you are claiming that the amount of light reaching the lens from the subject i.e. the white square has not changes at all. However that amount of light is now being concentrated on a smaller portion of the sensor. So if what you are claiming is true then the white square should become brighter, because we have the same total amount of light now covering a smaller area. In reality however, that does not happen because the total light hitting the lens is now less and that is why the brightness and exposure remains identical as before.

If you are still not convinced then I put my hands up in the air and give up

What I have described is a simple and accurate physics of what is happening.

Edited on Nov 02, 2012 at 10:52 PM · View previous versions

Nov 02, 2012 at 10:35 PM

Guari
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p.1 #20 · silly inverse square law question.

curious80 wrote:
When you move away from the subject, the light reaching from the subject to the camera of course goes down as per the inverse square law. However at the same time the subject is becoming smaller in your frame, i.e. taking up less pixels. Thus the amount of light falling per pixel remains the same. So the exposure remains the same.

Everyday one learns something new

I wonder how it was even possible to make light meters prior to the introduction of the concept of "light falling per pixel".

With all due respect, your reasoning is flawed. It does not respect the laws of physics. And physics do not change due to the introduction of new technologies.

Nov 02, 2012 at 10:48 PM

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