The unaswerable photography question...

RDKirk wrote in post #11409217
However, you can also stay in the same spot and mount wider lenses on the camera to create the same size image on the sensor. And if you do move back, you can put a telephoto lens on the camera to maintain the same size image on the sensor.

In all three cases, whether you move back and change lenses or move back and not change lenses or stay in position and change lenses, the exposure necessary to create the same image density of the monitor screen does not change. That's because in all cases, the image on the sensor is still being created by the same bundle of rays.

The exposure does not change but the physical size of the aperture does change as the focal length changes to compensate for the amount of light (for instance, f4 on a 50mm lens is a physically smaller opening than f4 on a 100mm lens). This is why the size of the aperture is directly related to the focal length of the lens....so that you'll maintain the same exposure no matter what focal length you use.

RDKirk wrote in post #11409217
However, you can also stay in the same spot and mount wider lenses on the camera to create the same size image on the sensor. And if you do move back, you can put a telephoto lens on the camera to maintain the same size image on the sensor.

In all three cases, whether you move back and change lenses or move back and not change lenses or stay in position and change lenses, the exposure necessary to create the same image density of the monitor screen does not change. That's because in all cases, the image on the sensor is still being created by the same bundle of rays.

It's CAN'T be created by the same bundle of rays. They HAVE to spread out or when you backed up there would be places missing the image altogether.

Here's a good discussion of how this works:
http://photo.net …t-techniques-forum/00Fzjx

I think of it like this: as you double your distance, the intensity is quartered but the area shown in the viewfinder is 4x the original area = 4x the light I.e. it cancels out!

Conversely stay still and zoom in and you will see the exposure required increases.

Macro and astro are moving to the limits of slr phoography - the main problem is same in both cases - a lack of light - and it's the same thing - you are looking at very small sections of the "Sphere" of light reaching you. Wide angle landscapes are conversely the easiest shots technically.

There are no unanswerable photography questions, just ones with unknown answers.

This is not one of them.

The question basically is, when using a flash or other (relatively) small light source, the inverse-square law is used for source-to-subject distances, but not for subject-to-camera distances. Why?

The answer is simple. Subject-to-camera distances are already compensated for in the nature of the optics of the camera (or the eye). There is no need to compensate twice, i.e, both ways.

Look at it this way. Presume the subject is lit with natural light at an acceptable level and the camera exposure is appropriately set. In this case, the camera will record the subject properly.

Now, the subject has insufficient light, and a flash must be used. The amount of flash, with the inverse-square law in play, is computed to bring the overall light upon the subject up to an acceptable level. Again, the camera will record the subject properly.

Notice that the light reaching the camera is acceptable, i.e., approximately the same, in both cases. Nothing need be done about the inverse-square law for the subject-to-camera distance. That's done "automatically" in the camera.

It is the light falling upon the subject that matters. If it is acceptable, it is acceptable, regardless of its source. If the source is a flash, or other artificial lighting, then we must worry about the inverse-square law and other factors, but only for the light falling upon the subject.

The light reaching the sensor/film will take care of itself.

Neither works well in falling rain, snow, or fog. So you're going to need a Inverse Fog Law.

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PhotosGuy wrote in post #11409887
Neither works well in falling rain, snow, or fog. So you're going to need a Inverse Fog Law.

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But then, popping a bright flash in a heavy fog does get rid of the point-source problem.

Lots of good tries, but not even the physicist's answer was logical.

Point source of light doesn't matter. The source-to-subject distance exposure works for point sources, bounced sources (umbrellas) or any kind of modified light source.

What makes light diminish is whether it's coherent or non-coherent light ("regular" light vs. laser). Not even using a lens over a coherent light removes the diminishing effect.

The paradox still remains: if non-coherent light is falling on the subject from the source, it's still non-coherent light being reflected to the observer. It should diminish at the same rate.

I have yet to find an answer....

BestVisuals wrote in post #11414317
Lots of good tries, but not even the physicist's answer was logical.

Point source of light doesn't matter. The source-to-subject distance exposure works for point sources, bounced sources (umbrellas) or any kind of modified light source.

What makes light diminish is whether it's coherent or non-coherent light ("regular" light vs. laser). Not even using a lens over a coherent light removes the diminishing effect.

The paradox still remains: if non-coherent light is falling on the subject from the source, it's still non-coherent light being reflected to the observer. It should diminish at the same rate.

I have yet to find an answer....

Coherency has little to nothing to do with it. None of the light from a conventional flash unit is coherent. Neither is it monochromatic.

Collimation, however, does affect the inverse-square law..

I gave the correct answer to the OP's question. If you choose not to believe it, that's not my problem. I recommend an Optics 101 text for further clarification.

And it also remains true that there is no paradox. This is simple optics, after all. No flux capacitors are involved.

PhotosGuy wrote in post #11409887
Neither works well in falling rain, snow, or fog. So you're going to need a Inverse Fog Law.

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Heh! With me, it's usually the Inverse Late Afternoon Law!

BestVisuals wrote in post #11414317
Point source of light doesn't matter. The source-to-subject distance exposure works for point sources, bounced sources (umbrellas) or any kind of modified light source.

What makes light diminish is whether it's coherent or non-coherent light ("regular" light vs. laser). Not even using a lens over a coherent light removes the diminishing effect.

Au contraire. The point nature of the source does matter. A true point source radiates outward in a sphere if increasing diameter, the number of photons emitted from the source is fixed, so as the diameter increases the number of photos per unit area are decreasing, resulting in the Inverse Square relationship.

For an infinitely long linear source (think fluoresent tube infinitely long), the relationship is Inverse Linear (rather than Inverse Square_.

For an infinitely large 2D planar source, the falloff is much less than inverse square, linear in its relationship.

For real world planes (i.e. softboxes) this is true when the subject is within 2x longest dimension of the softbox. At greater distances the softbox begins to behave like a virtual point because its size is diminshing with distance.

From Wikipedia

Uncle.

BestVisuals wrote in post #11407166
I love proposing this technical dilema, and it's fascinating to see how many how many people swear they know the answer. To everyone's defence, no pro photographer I have ever asked has the solution.

Q: Why does the inverse square law of lighting intensity apply only to source-to-subject distance, and not apply to subject-to-observer distance?

The question is unanswerable, in a sense, because your base assumption is incorrect. It is not true that subject-observer distance doesn't matter.

Consider a non-photographic example: the planet Venus.

The subject (Venus) is lit quite brightly by the sun, agreed?

Take a look at Venus in the night sky. Not all that bright, is it?

Because of the inverse square law. The surface of Venus is reflecting a certain amount of light, which disperses in accordance with the inverse square law. Only a small portion of that light hits the Earth, so we see Venus as dim.

-js

RDKirk wrote in post #11408002
I saw this explained by a physicist in a newsgroup back in the late 80s. It's essentially the same reason the sunny/16 rule works when taking pictures of the moon.

Imagine for a moment firing a shotgun at a wall at five feet and the pellet at the very center of the buckshot spread happens to hit an unlucky fly on that wall. Now, what if you'd fired the shotgun at the wall from 10 feet? That spread of buckshot would be wider, but that same center pellet would still hit the fly. Now, what if you had fired from 20 feet from the wall? That same center pellet would still hit the fly, even though the spread of buckshot would be even greater.

The light reflected from an object does spread out, but the particular rays of light that strike our eyes (or a camera lens) to form our visual image is an essentially parallel bundle (or "pencil") out of that spread (like that single buckshot). They aren't absolutely parallel, but the bundle is close enough to parallel that it would take parsecs to make a difference in exposure. Moving back a few feet or even a quarter of a million miles makes little difference--pretty much all of that same bundle of rays will still strike our eyes.

That's a great explanation, and makes perfect sense.

The answer is faster then the speed of light.