Please shed some flash on this

[Colin Southern]

The reason you don't notice any change in brightness when approaching a sunlit wall is because the distance you travel in getting closer to the wall is piddling compared to the distance the sunlight has already travelled from the sun to the wall.

The distance the light has traveled from the sun to the wall doesn't enter the equation (it would only do so if we were on board a space craft heading towards the sun). The only thing we're interested in is the light that reflects from the object back towards us. That WILL increase and decrease according to "the law" but so does the apparent size, so the energy density remains constant, which is why we don't perceive a change in brightness.

[Glenn NK]

Colin:

Suppose you set up an extremely large and perfectly flat and perfectly reflecting mirror aimed at the sun so that it reflected towards you. Should it matter if you were up above the earth a few miles, or on the surface of the earth a few miles away?

There is one other factor with sunlight that overshadows pure reflectance - skylight - it diffuses from all around the sky above us, and in many cases it more important than direct sunlight, so in any event, the reflective aspect is a rather moot point.

However none of this sun business answers Alexander's question does it?

Do you have any other ideas what's going on? I'm stuck, but still think there is something missing from the information provided to us.

Glenn

[pnodrog]

He probably forgot to mention he is using flash bounced of the ceiling...

[Colin Southern]

Suppose you set up an extremely large and perfectly flat and perfectly reflecting mirror aimed at the sun so that it reflected towards you. Should it matter if you were up above the earth a few miles, or on the surface of the earth a few miles away?

Direct reflections are a slightly different case, but regardless, once it's been reflected, the inverse square law "starts" from the object doing the reflecting; the light being reflected has no knowledge or memory of how far it's traveled to get to that point.

There is one other factor with sunlight that overshadows pure reflectance - skylight - it diffuses from all around the sky above us, and in many cases it more important than direct sunlight, so in any event, the reflective aspect is a rather moot point.

Sorry, but that doesn't make sense - yes, diffuse skylight is often present, but that's irrelevant; if you're able to see an object in the distance then it's because light is being reflected from it, and that light coming towards you will be subject to the inverse square law. It makes zero difference in terms of relative brightness whether it's a hard light from a relatively small light source like the sun, a relatively large light source like we get on an overcast day, or whether the light that illuminates the object is near or far.

With regards to the original question (post 1), I've already answered it. The exposure remains the same because the relative energy density remains the same (less energy received, but spread over fewer pixels).

[Glenn NK]

1) Light takes a determinant amount of time to go from point A to point B. If it's reflected back from point B to point A, the journey takes twice as long as from A to B. Light has no knowledge or memory of how far it's traveled, but obviously the timing and the ISL starts from the source.

2) I didn't mean to suggest that skylight had anything to do with the problem posed by Alexander, it was the fact that in the open in sunlight, diffuse skylight is more important than the sun shining directly on an object - otherwise we couldn't take images in the shade. Diffuse light is the primary (not the only) reason we can do photography after the sun is below the horizon.

3) If energy density doesn't diminish with distance, then the ISL doesn't work, because that's what the ISL is all about. If Earth was one half the distance from the Sun than it is, we'd have been fried long ago.

[Colin Southern]

1) Light takes a determinant amount of time to go from point A to point B. If it's reflected back from point B to point A, the journey takes twice as long as from A to B. Light has no knowledge or memory of how far it's traveled, but obviously the timing and the ISL starts from the source.

In terms of the light we're observing being reflected from an object and in terms of the inverse square law, the "source" of the light is the object. It makes no difference whether the light illuminating that object is 1 meter away or millions of miles away. so statments like:

The reason you don't notice any change in brightness when approaching a sunlit wall is because the distance you travel in getting closer to the wall is piddling compared to the distance the sunlight has already travelled from the sun to the wall.

... are just plain wrong I'm afraid Glenn.

2) I didn't mean to suggest that skylight had anything to do with the problem posed by Alexander, it was the fact that in the open in sunlight, diffuse skylight is more important than the sun shining directly on an object - otherwise we couldn't take images in the shade. Diffuse light is the primary (not the only) reason we can do photography after the sun is below the horizon.

I agree, but I'm talking about the inverse square law -- and that applies to both sorts.

3) If energy density doesn't diminish with distance, then the ISL doesn't work, because that's what the ISL is all about. If Earth was one half the distance from the Sun than it is, we'd have been fried long ago.

My apologies - I meant to write the RELATIVE energy density (since corrected). Yes - of course the energy received at the camera falls off as the camera to subject distance increases, but SO DOES THE AREA OF THE SENSOR THAT THE THOSE PHOTONS ARE RECEIVED AT. Say you had 4 pixels being used to capture an image at 1 meter -- move the object twice as far away and now it's only 1/2 as high and 1/2 as wide, and only 1/4 as bright - BUT - because it's further away it's only going to received by ONE pixel. So what was X brightness covering 4 pixels is now the same object 1/4 the original brightness when it's distance is doubled, but that 1/4 power is now concentrated into a single pixel (not spread out over 4 pixels) so the RELATIVE (got it right this time!) energy remains the same, hence the reason the exposure doesn't change.

Or to put that another way for clarity ... when the object was at one meter and taking up 4 pixels, lets assume enough energy reached the sensor to take all 4 pixels to "100%". Move the object away (double the distance) and the "100% worth" of energy is now only 25% - but what was 4 pixels worth at 100% is now only 1 pixel, but getting 4 lots of 25% energy from the area of the object that previously took up 4 pixels (25% + 25% + 25% + 25% = 100%) (so that 1 single pixel from the object at 2 meters is now exposed EXACTLY the same as the 4 pixels where when the object was only 1/2 the distance). Relative energy density is unchanged thus the exposure is unchanged.

[xpatUSA]

I guess this is one of those days when my brain has failed. After finding the hard re-boot button on my head I realize that most of what I said was incorrect. As long as the light from a source is diverging then the inverse square law does apply. It is only in the case where the light does not diverge (laser) that the rule falls apart

John

No, most of what you said was right, I reckon. But we can not now say "As long as the light from a source is diverging then the inverse square law does apply" for the following reasoning:

First, imagine a perfectly collimated beam for example a point source of light with a collimating lens focused so that all the rays out of the lens are parallel. (found in spectroscopes but not in car headlights which must obviously have some spread). Now, if a camera frames the lens across the sensor diagonal some distance away then the sensor would be illuminated by so many lux (sorry to introduce actual optical units into this thread ;-). Move the camera towards the collimating lens and sensor is still illuminated by the same number of lux. This is because, in theory and on a clear day, when the rays are parallel the lumens/square meter, i.e. lux, within the beam are the same at any distance. Thus, just like in the OP, no adjustment in exposure is required.

Now imagine the collimating lens being slightly out of focus such that rays are slightly divergent. Now, as a camera is moved back and forth, the lux varies slightly - but not by the inverse square law. Indeed, as the divergence approaches zero, so does the variation in luminance or illuminance if you will. That is to say, for small angles of divergence so created, doubling a distance does not necessarily reduce the lux by one-fourth. We could also say that the divergence of the sun's rays is not particularly large here at the surface of the planet, so Glenn is right too.

And so we must conclude, as most have already, that the law only applies to a point source of light with no other obfuscatory items influencing its rightful emissive path through the aether ;-)

Angles: the key to all understanding

OT (opposite to the topic) but is there an opposite scenario, where illuminance squares as it approaches a sensor?

[Colin Southern]

No, most of what you said was right, I reckon. But we can not now say "As long as the light from a source is diverging then the inverse square law does apply" for the following reasoning:

First, imagine a perfectly collimated beam for example a point source of light with a collimating lens focused so that all the rays out of the lens are parallel. (found in spectroscopes but not in car headlights which must obviously have some spread). Now, if a camera frames the lens across the sensor diagonal some distance away then the sensor would be illuminated by so many lux (sorry to introduce actual optical units into this thread ;-). Move the camera towards the collimating lens and sensor is still illuminated by the same number of lux. This is because, in theory and on a clear day, when the rays are parallel the lumens/square meter, i.e. lux, within the beam are the same at any distance. Thus, just like in the OP, no adjustment in exposure is required.

Now imagine the collimating lens being slightly out of focus such that rays are slightly divergent. Now, as a camera is moved back and forth, the lux varies slightly - but not by the inverse square law. Indeed, as the divergence approaches zero, so does the variation in luminance or illuminance if you will. That is to say, for small angles of divergence so created, doubling a distance does not necessarily reduce the lux by one-fourth. We could also say that the divergence of the sun's rays is not particularly large here at the surface of the planet, so Glenn is right too.

And so we must conclude, as most have already, that the law only applies to a point source of light with no other obfuscatory items influencing its rightful emissive path through the aether ;-)

Angles: the key to all understanding

OT (opposite to the topic) but is there an opposite scenario, where illuminance squares as it approaches a sensor?

Some great theory there Ted, unfortunately, most of it is wrong.

Highly collimated light (eg laser light) is a different set of rules, but ...

(a) sunlight isn't like laser light, and the inverse square law very much applies to sunlight and starlight (it's one way astronomers calculate how far a star is away)

(b) In terms of using the same exposure with changing camera-to-subject distances (assuming the same illumination), it has ZERO to do with how far away the light source is or how collimated it is.

(c) In terms of exposure - as I've explained a couple of times now - it's the same because as the energy decreases due to the distance, the size of the image on the sensor also decreases, so a doubling of the distance quarters the energy from a given object but also quarters the size of the object on the sensor - which quadruples the energy those remaining pixels now receive because they're getting light from 4 times the area (that's 1/4 strength) - so the RELATIVE energy density remains that same.

That's all there is to it. It's dead easy when folks stop and think about it.

[Glenn NK]

(c) In terms of exposure - as I've explained a couple of times now - it's the same because as the energy decreases due to the distance, the size of the image on the sensor also decreases, so a doubling of the distance quarters the energy from a given object but also quarters the size of the object on the sensor - which quadruples the energy those remaining pixels now receive because they're getting light from 4 times the area (that's 1/4 strength) - so the RELATIVE energy density remains that same.

That's all there is to it. It's dead easy when folks stop and think about it.

So it doesn't matter what size of flash I use, or how far away the subject is, I'll always have enough flash power to expose the subject?

I can set the ISO, f/stop, shutter speed for a close-up, and no matter how far I move from the subject the exposure will be the same.

This is what Alexander seems to have said too - no matter what distance he was from the subject, the exposure was the same.

[Colin Southern]

So it doesn't matter what size of flash I use, or how far away the subject is, I'll always have enough flash power to expose the subject?

No. Not sure how you got that. The energy from the flash simply needs to be appropriate for the exposure; doesn't matter if you do that "inefficiently" (eg a thermonuclear reactor called the sun millions of miles away that wastes 99.99999% of it's energy, or via an efficient flash that's zoomed and placed appropriately close. It only needs to be the right amount of light regardless of how you get it.

I can set the ISO, f/stop, shutter speed for a close-up, and no matter how far I move from the subject the exposure will be the same.

For the portion that you were originally exposing for, yes (keep in mind that the field of view will change as you pull back -- so I have to qualify that in terms of only the original scene because there's no way of knowing what lighting is on the "new" parts of the composition that are introduced by pulling back the camera. There will also be a point where the camera won't be able to resolve the original scene, and in theory, things like atmospheric contamination may also absorb some of the light, but provided we're not talking extremes, then the answer is yes).

This is what Alexander seems to have said too - no matter what distance he was from the subject, the exposure was the same.

Yes - this is perfectly normal and expected behaviour. The exposure would only change if the lighting were changed.

You DO need to make sure that you're comparing apples with apples though; if you focus on a models face at 1m and setup for a correct exposure then if you move back to 2m or 3m then the face will be smaller (obviously), but it'll still be exposed just fine (if it wasn't then when I have a model in the studio - the lights are setup - and I get my light meter out, I'd also have to enter/factor camera to subject distance into the meter to get a valid reading - and clearly that's not the case). The histogram on the other hand may or may not change; that depends on what the rest of the scene comprises of after the new scenery comes into the expanded FoV.

In terms in the inverse square law, just think of it in terms of lighting to subject distance (ie if you vary the position of the subject relative to the LIGHTING (if the lights are relatively close) then "the law" comes into play. If the light source is already a long long way away then it makes little to no difference). Forget about camera to subject distance in terms of exposure; only the image size changes, not the exposure (because the relative energy emitted/reflected by the subject remains constant).

In contrast, and this might be what's confusing some people; having lights and model at a set distance and then moving the camera back is NOT NOT NOT NOT the same as having the camera and lights at a set distance and then moving the model back; in the 2nd example the exposure changes because the light on the model falls off due to the increased distance, whereas in the first example the the exposure remains the same; the image just gets smaller. HUGE difference.

[xpatUSA]

In contrast, and this might be what's confusing some people; having lights and model at a set distance and then moving the camera back is NOT NOT NOT NOT the same as having the camera and lights at a set distance and then moving the model back; in the 2nd example the exposure changes because the light on the model falls off due to the increased distance, whereas in the first example the the exposure remains the same; the image just gets smaller. HUGE difference.

Yes, it is easy to see how that happens in the world of models and strategically placed lights and there would indeed be a HUGE difference. But I'm not sure that the statement is universally applicable to all forms of lighting and all models. With some forms, the difference could be quite TINY. I can see that, if the distance between a model and a flash bulb or a halogen lamp were changed, then the luminance from the model would change. We all know that and the data for lamps, is normally published in the form of polar plots and lumens and stuff. However, the only forms of lighting that obey the inverse square law are point sources which, in the Real World, simply do not exist. So, there's more to luminance at a distance than simply quoting the theoretical inverse square law, I reckon.

I'm imagining a hypothetical studio in Japan with a paper wall lit evenly by our friend the sun. No other lighting is in use (we're after that soft, mysterious, occidental look). There, some 6 ft away, sits a Geisha. We expose for the face but want more of the torso in. "Could you move back twice as far to 12ft?", we ask.

Does the light on the model fall off due to the increased distance?

Does the exposure for the face change?

I'm imagining a model lit from the front by a couple of those umbrella thingies which are illuminated by a couple of flashes which face away from the model. Or a diffuser sheet placed between a light source and the subject (I do that a lot). Same questions.

So, clearly, there are factors involved other than just distance and changes thereto.

[Colin Southern]

Yes, it is easy to see how that happens in the world of models and strategically placed lights and there would indeed be a HUGE difference. But I'm not sure that the statement is universally applicable to all forms of lighting and all models.

It is.

With some forms, the difference could be quite TINY. I can see that if the distance between a model and a flash bulb or a halogen lamp were changed, then the luminance from the model would change. We all know that and the data is normally published in the form of polar plots and lumens and stuff. However, the only forms of lighting that obey the inverse square law are point sources which, in the Real World, simply do not exist.

They exist everywhere. If you don't believe me then just get out a light meter and a tape measure - make sure you don't contaminate the reading with additional reflections - and measure it for yourself.

So, there's more to luminance at a distance than simply quoting the theoretical inverse square law, I reckon.

No, there's not.

I'm imagining a hypothetical studio in Japan with a paper wall lit evenly by our friend the sun. No other lighting is in use (we're after that soft, mysterious, occidental look). There, some 6 ft away, sits a Geisha. We expose for the face but want more of the torso in. "Could you move back twice as far to 12ft?", we ask.

Yes you can. If you don't believe me then tell me where on my lightmeter do I enter the camera to subject distance since according to your theory this would be needed to calculate the "correct exposure".

Does the light on the model fall off due to the increased distance?

Yes, it does - but the number of pixels used to record the previous area drops off in the EXACT same proportion so the relative energy density (and thus the exposure) remain the same.

Does the exposure for the face change?

No, it doesn't change. Only the relative size changes.

I'm imagining a model lit from the front by a couple of those umbrella thingies which are illuminated by a couple of flashes which face away from the model. Or a diffuser sheet placed between a light source and the subject (I do that a lot). Same questions, possibly different answers, though.

Same answer. You get the same exposure whether you're 1m or 10m away. Only the apparent size changes.

So, clearly, there are other factors involved other that just distance and changes thereto.

No Ted - there aren't "clearly other factors". You've presented invalid hypothesis and then used them to reach an incorrect conclusion. Think about it;

(a) Lightmeters don't have distance scales to say how far away the camera is - why? Because distance is irrelevant. The ONLY thing it's interested in is how much light is falling on the model; it doesn't give a rats if it's sunlight, moonlight, flash light, a mixture all it need to know is how much.

(b) Photographers exploit this hundreds of thousands of times a day. Don't you think that if the exposure changed when we pulled back we'd have noticed by now?

(c) If you still don't believe me, just measure it for yourself -- it's easy enough to prove.

About the only time you'd change lighting if you increased the camera to subject distance would be if the increased distance then revealed parts of the subject that were never lit correctly, eg if I've been doing 3/4 shots with appropriate lighting and I backup to make it a full length shot, I'll then have to consider additional lighting for the feet and lower legs.

[pnodrog]

Yes, it is easy to see how that happens in the world of models and strategically placed lights and there would indeed be a HUGE difference. But I'm not sure that the statement is universally applicable to all forms of lighting and all models. With some forms, the difference could be quite TINY. I can see that, if the distance between a model and a flash bulb or a halogen lamp were changed, then the luminance from the model would change. We all know that and the data for lamps, is normally published in the form of polar plots and lumens and stuff. However, the only forms of lighting that obey the inverse square law are point sources which, in the Real World, simply do not exist. So, there's more to luminance at a distance than simply quoting the theoretical inverse square law, I reckon.

I'm imagining a hypothetical studio in Japan with a paper wall lit evenly by our friend the sun. No other lighting is in use (we're after that soft, mysterious, occidental look). There, some 6 ft away, sits a Geisha. We expose for the face but want more of the torso in. "Could you move back twice as far to 12ft?", we ask.

Does the light on the model fall off due to the increased distance?

Does the exposure for the face change?

I'm imagining a model lit from the front by a couple of those umbrella thingies which are illuminated by a couple of flashes which face away from the model. Or a diffuser sheet placed between a light source and the subject (I do that a lot). Same questions.

So, clearly, there are factors involved other than just distance and changes thereto.

Correct ....

If you take a macro shot 10cm away from your large reflective wall and then move it to 20cm the exposure change will be insignificant. The wall is to all intents and purposes an infinite source of light not a point source so the inverse square law does not apply. Move back to 50 meters take an exposure then move back to 100m and the difference in exposure will almost be 4X as per the inverse square law because the light source is now behaving more like a point source.

The exposure will be dependent on a function that approaches the inverse square law as we get further away.

[Colin Southern]

Correct ....

If you take a macro shot 10cm away from your large reflective wall and then move it to 20cm the exposure change will be insignificant. The wall is to all intents and purposes an infinite source of light not a point source so the inverse square law does not apply. Move back to 50 meters take an exposure then move back to 100m and the difference in exposure will almost be 4X as per the inverse square law because the light source is now behaving more like a point source.

The exposure will be dependent on a function that approaches the inverse square law as we get further away.

Grrrr. No!

The exposure DOESN'T CHANGE.

*** Bashes head against the wall in question ***

Think about it everyone; if the exposure changed with camera to subject distance changes then that would have to be factored into exposure calculations. Light meters tell you what settings to use to get a correct exposure - you put them next to the subject and take a reading. Put those settings in your camera and your exposure will be correct. NOWHERE IS THE CAMERA TO SUBJECT DISTANCE ENTERED because IT DOESN'T MATTER.

People really need to start thinking this through a bit more.

[pnodrog]

I specifically said macro so the subject to camera distance is a constant. The only light source in the scenario is the white wall. I totally agree people need to think about it. ....

[Colin Southern]

I specifically said macro so the subject to camera distance is a constant. The only light source in the scenario is the white wall. I totally agree people need to think about it. ....

No - what you said was ...

Move back to 50 meters take an exposure then move back to 100m and the difference in exposure will almost be 4X as per the inverse square law because the light source is now behaving more like a point source.

The exposure will be dependent on a function that approaches the inverse square law as we get further away.

And this is blatantly incorrect. I'm really finding it almost surreal that I'm having to keep going over this.

For goodness sake - if you think a significant distance alters the exposure, take a shot at 50m - open your aperture 2 stops - take another shot at 100m - open another couple of stops and repeat at 200m

By your (incorrect) theory you should have 3 perfect exposures - but you won't have - it'll be over-exposed on the 100m shot and most likely blown on the 200m shot because the energy density remains the same and opening the aperture will upset the balance.

Folks need to remember that there's two parts to the equation, and they balance each other out EXACTLY. Doubling the the camera to subject quarters the reflected energy but it also quarters the sensor area that it's received on - so what was exposing 1 pixel of the sensor at "full strength" now has FOUR TIMES that area exposing it at quarter strength.

PS: Up close you don't have an infinite light source instead of a point light source; in essence what you have is an infinite number of point sources - and they ALL follow the "the law".

[xpatUSA]

Hello again, y'all . . .

Anticipating some resistance from a certain quarter to my posts, I retired to the shed to try a quick and dirty test.



Although the lighting was fairly even, the diffusers were a little small. Undaunted, I dug out the cheap lux-meter and found that, when it was close up to the paper, I couldn't read the stupid display. So, I had to settle for measurement at 10cm and 20cm - a doubling of distance which, according to some, should have 'quartered' the lux and which, according to others, should not have changed the lux at all. In practice the reading was about halved, which I attribute to the lux meter having a largish angle of view. Sadly inconclusive - must move to Japan . .

So, took a 'model' (Kodak white card) and tried it at the same distances:



Pointed a camera at it and set the exposure to 0 EV in each position. There was a difference. The light fell off by about 2/3 stops. But it did NOT fall off by two stops.

I am still talking about illuminance of the entire sensor from a large, evenly-lit area - so as make the point that illuminance at the sensor is NOT changed by distance under that particular circumstance and therefore that the inverse square law is not quite as universal as is claimed. I am NOT talking about some smaller item whose framing changes with distance - and that may be why my point is apparently not understood.

PS: A point source has no diameter. In my book, infinity times 0 = 0

PPS: can we stop using caps? My little finger aches from pushing the shift key

[pnodrog]

Sorry Colin, I think we are talking at cross purposes partly because I did not make my self clear. I was referring to the light falling on a macro subject that was being progressively moved away from a large light source. It seemed easier to visualize it that way than moving a wall. So the exposure varies according to the distance of the light source but not in accordance with the inverse square law and I see Ted has been kind enough and energetic enough to prove it.

Yes, you can think of it as an infinite array of point light sources but moving away when are very close you can think of it that as you reduce the illumination from one point you move into some illumination from another. I am not suggesting there is no reduction in light intensity but only that at this point the (effective) lighting does not follow the inverse square law.

[Colin Southern]

Anticipating some resistance from a certain quarter to my posts

That'll be me -- the guy with the large wound on his forehead from the banging against the wall.

So, took a 'model' (Kodak white card) and tried it at the same distances:

Pointed a camera at it and set the exposure to 0 EV in each position. There was a difference. The light fell off by about 2/3 stops. But it did NOT fall off by two stops.

Your testing methodology is fatally flawed. When you moved the camera to the furthest position you were metering an area that was 4 times the area of what was being metered the first time. Four times the area multiplied by 1/4 the lux arriving at the sensor equals exactly the same exposure. The difference is probably due to unevenness of the lighting.

I am still talking about illuminance of the entire sensor from a large, evenly-lit area - so as make the point that illuminance at the sensor is NOT changed by distance under that particular circumstance and therefore that the inverse square law is not quite as universal as is claimed.

Yes - it IS universal - you're just not correctly accounting for the change in the size of the light source. Think of the original area being metered as being like 100 soldiers at full strength and at the doubled distance, soldiers at only 1/4 strength but 4 times as many of them.

I am NOT talking about some smaller item whose framing changes with distance - and that may be why my point is apparently not understood.

It doesn't matter if you are or aren't - it doesn't change anything. Whenever you double a camera-to-something distance you ALWAYS quadruple the field of view. That quadrupled filed of view contributes reflected photons that the sensor captures (ie "energy"). The only difference is whether you consider a portion of the scene (if that's what's lit) (eg a model in a studio) or you consider the whole scene (if that's what's lit).

Some additional points ...

1. If it's not highly collimated light from a laser then it follows the inverse square rule. If that's not the result someone gets then they need to re-examine their setup because it's a fundamental law of physics. And yes I know that the law refers to a single point of light, but it still applies here because for the purposes of the rule a diffuse light source (ie everyday object reflecting light) is simply an infinite number of point light sources. And no, putting a reflector behind a bulb and shooting it out through a grid doesn't collimate it (and make it "almost laser like") (see how far a 1/2w laser travels -v- a 1/2w bulb with a reflector behind it and something to focus the beam in front of it -- not even close).

Please - "absorb this information" once and for all; if varying the camera to subject distance changed the exposure then light meters wouldn't be able to calculate the correct settings without knowing the distance. Do they calculate correct exposures without knowing distance? YES - they DO. So please, have a think about that -- when you acknowledge that fact then you'll be forced to acknowledge the fact that the distance is irrelevant because it has no effect on exposure.

Next - logically, take that a step further ... if the distance is doubled and the size of an image is quartered - and the exposure doesn't change then the light MUST have also reduced to 1/4 of it's original value because if it didn't then the exposure would have changed, which it doesn't - therefore it MUST be reducing according to the square law.

[xpatUSA]

Sorry Colin, I think we are talking at cross purposes partly because I did not make my self clear. I was referring to the light falling on a macro subject that was being progressively moved away from a large light source. It seemed easier to visualize it that way than moving a wall. So the exposure varies according to the distance of the light source but not in accordance with the inverse square law and I see Ted has been kind enough and energetic enough to prove it.

I too agree that we are talking at cross-purposes. In fact, I needn't have bothered with a trip to the shed. I just put a white screen up on the monitor, thereby simulating the hypothetical Japanese studio wall. Put the camera hard up against the screen and then retired a good few paces with no change in exposure whatever. Not what one would expect with a universal inverse square law in effect. Kept walking back and at 8 or so feet the exposure required by the camera started to rise but not by much - 1/20 sec fell to 1/10 sec at 12 feet (aperture priority, spot metering). Had I walked out thru the kitchen into the front yard (too bloody cold for that) I'm sure the exposure would have tended toward "the law" but would never have approached it.