The unaswerable photography question...

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?

To explain: we all (hopefully) know that if you double the distance of a light source to its subject, the amount of light falling on that subject is 1/4 the intensity, not half. Quadruple the distance and you've got 1/16th the intensity.

Let's assume you're 5 feet from the subject in your studio as the observer/photographer and the proper exposure is f/8. The light source (flash) is on a stand and will not move. Now you, the photographer, move to 20 feet away...the exposure does not change, it's still f/8. Move to 50 feet...it's still f/8. Don't believe me? Try it yourself.

Go across the street with a telephoto capable of shooting f/8 and the exposure is still the same. Go 2 miles away with a long enough lens and proper exposure is still f/8. Shoot a rock concert using stage lighting and you'll find that the exposure is the same, regardless of whether you're touching the stage or in the very back of the stadium.

The upshot of my quandary: distance is meaningless, subject-to-observer, as far as light intensity is concerned.

So why doesn't the light intensity diminish as you back away from the subject? Light is light, isn't it? The laws of physics that causes light to diminish between the source and the subject (i.e., non-cohesive light) should apply for subject-to-observer intensity, correct?

It only applies to point sources.

But aren't you taking a photo of the light that is reflected off the subject, and your lens glass is bringing that light in through the shutter to record on the sensor. The intensity of that light is still the same as long as the light hitting it is the same. The light intensity hitting the subject doesn't change just because YOU moved. yes? This sort of question makes my brain bleed, but at first glance, that is what I'm thinking.

suecassidy wrote in post #11407659
But aren't you taking a photo of the light that is reflected off the subject, and your lens glass is bringing that light in through the shutter to record on the sensor. The intensity of that light is still the same as long as the light hitting it is the same. The light intensity hitting the subject doesn't change just because YOU moved. yes? This sort of question makes my brain bleed, but at first glance, that is what I'm thinking.

You read it the same as I did. Took me a second to figure out the question.

What we (and our cameras) see is light that has been reflected off of an object.

Light leaves flash, light hits subject, light reflects off of subject and hits the lens. Why doesn't the light that is reflecting off the subject diminish at the same rate?

I tried to think of something logical, but I can't. On that same token, knowing the answer won't make me suck any less as a photographer. I'll let someone else with more time and a higher IQ write it out, then I'll come look.

Best thing I could come up with is that the light DOES diminish at the same rate. If it didn't then you could light someone with a hot shoe flash from 100 yards away as long as it was bounced into an umbrella. You aren't capturing the reflected light per se, but capturing the SOURCE (in this case, the model) of the light on your sensor. The model becomes a "light source" when the light hits them.

Or something like that.

Good question. I dont have a clue for the answer.

But you posed this question for your first new thread in In almost 18 months?

With the original post about distance being meaningless you'll need a longer lens with the same aperture to capture the same amount of light.

Here's something else to throw into the mix... when working with flashes power diminishes at the square of the distance. This makes sense, as the light is spreading out as you go further away. I'm not saying it's relevant... i'm just throwing it into the mix.

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.

Quad wrote in post #11407252
It only applies to point sources.

I believe this means that when the source of the light is a flash or some other point source the light emitted is spreading out in all directions and therefore loses intensity as the distance from the source increases. But the reflected light that you see is not spreading out in all directions, and therefore doesn't lose intensity over distance.

You can.

Let's make sure the scenario is correct: You place someone five feet from an umbrella. It doesn't matter how far away from the subject you place the camera, the exposure will be the same.

Just as the sunny/16 works when taking a picture of the moon because the moon is also lit by the sun, and moon and the earth are essentially the same distance from the sun, if you used a long enough telephoto lens to give you an ample image size, the exposure will be the same from 100 yards as it is from 10 yards.

The light and the subject do not move so the exposure on the subject will not change but as you move away from the subject the light that your light meter sees will diminish IF you do not change the lens. However if can change the lens so that the framing of the subject will be the same as when you were 5' away then your light meter will see the exposure as f/8. You could stand on the surface of the moon and if you had a long enough lens to frame the subject the same as when you were 5' away the exposure would still be f/8.

If you look at the heavens there are stars out there that you can't see with the naked eye but if you use a large enough telescope you can clearly see those stars. The brightness of the star didn't change but by using a telescope you changed your relative distance to the star making it visible.

Viewing light fall off vs Viewing light source

Try this. Light radiates out from a source in a sphere. As the light radiates out, the photons spread out evenly, but at an inverse rate. When we measure the light that has FALLEN onto an object, we are measuring the the portion of photons that radiated out and fell onto that object. When you look back at a light source, you are view ALL of the original photons, not the photons that have spread out.

RDKirk wrote in post #11408037
You can.

Let's make sure the scenario is correct: You place someone five feet from an umbrella. It doesn't matter how far away from the subject you place the camera, the exposure will be the same.

Just as the sunny/16 works when taking a picture of the moon because the moon is also lit by the sun, and moon and the earth are essentially the same distance from the sun, if you used a long enough telephoto lens to give you an ample image size, the exposure will be the same from 100 yards as it is from 10 yards.

I meant that placing the umbrella/flash at 100yds from the subject. If it was the nature of reflected light to NOT diminish over distance (as referenced in the OP) then the ISL would go out the window as long as the light was reflected of something. I was actually being facetious a bit, but the principle is there.

My thought is that light is light, and it all diminishes the same way. The question stands on the erred theory that the sensor is capturing the dispersed light instead of the source (the source being the model).

I think we're thinking on the same plane...you're just MUCH better at explaining this stuff. Now I'm eating and going to bed before I come across another Q&A thread that makes me want to rock back and forth in my chair while in the fetal position.

shaftmaster wrote in post #11408028
I believe this means that when the source of the light is a flash or some other point source the light emitted is spreading out in all directions and therefore loses intensity as the distance from the source increases. But the reflected light that you see is not spreading out in all directions, and therefore doesn't lose intensity over distance.

More or less but the reflected light is still spreading out. Take a photo of say a computer monitor (a dim light source so you can see the the object ) from a foot away now move back 20 feet and use the same settings for a second photo, the monitor will not look dimmer but less light will be getting to the sensor (the monitor will occupy less area in the photo and the histogram will reflect this as the spike will be in the same place horizontally but will have less height [If you used a darkened room and a nice middle gray on the monitor it will show better on the histogram]). Each time you double the distance the monitor will occupy a quarter the area on the image.

The same law still applies.....if you move back and you're using the same focal length lens, you're capturing less and less light from the subject and this shows up as your subject becomes smaller in the frame as you move back. To capture the same amount of light as you move back (in other words, keeping the subject the same size in the frame), you'll need to find a longer lens which will require a larger opening to create the same aperture as the wider lens to capture (or gather) the same amount of light. This is the reason that the physical size of the aperture is directly related to focal length....as the lens gets longer, the opening needs to be larger to maintain the same aperture to gather the same amount of light from a more distant position.

Hopefully that makes sense. I've seen this question posed before and explained much better than what I just attempted to do. but hopefully I got the basic point correct

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.