The unaswerable photography question...

Martin Dixon wrote in post #11409495
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!

This is the only thing that has made sense to me so far in this thread. Can anyone direct me to a book or some other resource that would explain this in relatively simple terms?

Google is your friend...to get you started on the topic

Illustration of the physics concept
Another illustration of the principle

Illustration expressed in photographic terms

This is the kind of stuff that makes my head hurt...

If you understand that the flash-to-subject distance is what controls the light on your subject, the reason why it does it is unimportant, at least to the average photographer.

There has been a couple of correct answers, but I'll try to give a simple explanation (I wont use correct terms, they exist but they could be more confusing):

1) OP:s assumption is wrong: the inverse square law applies to all light, sources and reflected.

2) Lets picture a white wall, illuminated with light (doesnt matter how). Wall is 2x2 meters with total area of 4 square meters.

3) Lets take a photo of the wall with a 100mm f/4 lens so that the wall fills the whole picture. There is certain amount of total light hitting the sensor, lets assume that equal to 1. (imaginary measure)

4) Ok, lets double our distance from the wall. Now, according to inverse square law, the amount of light should be 1/4 of what it used to be. This is correct, and at the same time the apparent dimensions of the wall have been cut by 1/2. Now the apparent area of the wall is also 1/4 (1/2*1/2) of what it used to be.

5) Humen (and cameras) see brightness of an object as (amount of light) per (apparent area). In this case, after backing down with the camera, the brightness of the wall is still the same (1/4 : 1/4 = 1). However, the size of the wall in a photo is 2x smaller (1/4 area)

6) Now, as we doubled our distance to the wall, we need to use a 200/4 lens to achieve same framing and exposure (first we had 100/4). Since the F-number (f/4 in this case) is only a ratio of focal length and actual light-gathering aperture opening, we have to take that into account.

Lets see what is the ratio of the light-collecting area of 100/4 and 200/4 lenses. Area of a circle is pi*radius^2 and as we double focal lenght and apertures radius, the area of the aperture grows 4 times larger.

In our example, this compensates the 1/4 diminishing of the light by having a 4-times larger hole to collect the light.

Case closed?

herzi wrote in post #11424674
There has been a couple of correct answers, but I'll try to give a simple explanation (I wont use correct terms, they exist but they could be more confusing):

1) OP:s assumption is wrong: the inverse square law applies to all light, sources and reflected.

2) Lets picture a white wall, illuminated with light (doesnt matter how). Wall is 2x2 meters with total area of 4 square meters.

3) Lets take a photo of the wall with a 100mm f/4 lens so that the wall fills the whole picture. There is certain amount of total light hitting the sensor, lets assume that equal to 1. (imaginary measure)

4) Ok, lets double our distance from the wall. Now, according to inverse square law, the amount of light should be 1/4 of what it used to be. This is correct, and at the same time the apparent dimensions of the wall have been cut by 1/2. Now the apparent area of the wall is also 1/4 (1/2*1/2) of what it used to be.

5) Humen (and cameras) see brightness of an object as (amount of light) per (apparent area). In this case, after backing down with the camera, the brightness of the wall is still the same (1/4 : 1/4 = 1). However, the size of the wall in a photo is 2x smaller (1/4 area)

6) Now, as we doubled our distance to the wall, we need to use a 200/4 lens to achieve same framing and exposure (first we had 100/4). Since the F-number (f/4 in this case) is only a ratio of focal length and actual light-gathering aperture opening, we have to take that into account.

Lets see what is the ratio of the light-collecting area of 100/4 and 200/4 lenses. Area of a circle is pi*radius^2 and as we double focal lenght and apertures radius, the area of the aperture grows 4 times larger.

In our example, this compensates the 1/4 diminishing of the light by having a 4-times larger hole to collect the light.

Case closed?

Sounds good to me...That's basically the point I was attempting to explain although you did it much more elegantly

herzi wrote in post #11424674
There has been a couple of correct answers, but I'll try to give a simple explanation (I wont use correct terms, they exist but they could be more confusing):

1) OP:s assumption is wrong: the inverse square law applies to all light, sources and reflected.

2) Lets picture a white wall, illuminated with light (doesnt matter how). Wall is 2x2 meters with total area of 4 square meters.

3) Lets take a photo of the wall with a 100mm f/4 lens so that the wall fills the whole picture. There is certain amount of total light hitting the sensor, lets assume that equal to 1. (imaginary measure)

4) Ok, lets double our distance from the wall. Now, according to inverse square law, the amount of light should be 1/4 of what it used to be. This is correct, and at the same time the apparent dimensions of the wall have been cut by 1/2. Now the apparent area of the wall is also 1/4 (1/2*1/2) of what it used to be.

5) Humen (and cameras) see brightness of an object as (amount of light) per (apparent area). In this case, after backing down with the camera, the brightness of the wall is still the same (1/4 : 1/4 = 1). However, the size of the wall in a photo is 2x smaller (1/4 area)

6) Now, as we doubled our distance to the wall, we need to use a 200/4 lens to achieve same framing and exposure (first we had 100/4). Since the F-number (f/4 in this case) is only a ratio of focal length and actual light-gathering aperture opening, we have to take that into account.

Lets see what is the ratio of the light-collecting area of 100/4 and 200/4 lenses. Area of a circle is pi*radius^2 and as we double focal lenght and apertures radius, the area of the aperture grows 4 times larger.

In our example, this compensates the 1/4 diminishing of the light by having a 4-times larger hole to collect the light.

Case closed?

What if the entire wall still filled the viewfinder when you doubled your distance?

itzcryptic wrote in post #11435802
What if the entire wall still filled the viewfinder when you doubled your distance?

The portion that filled it when you were closer would still only occupy 1/4 the area, therefore delivering 1/4 of the light. Assuming the wall was evenly lit, you would now be capturing additional light from the section of the wall that wasn't visible in the frame when you were standing closer (therefore you're capturing light that you weren't capturing when you stood closer).

jra wrote in post #11435864
The portion that filled it when you were closer would still only occupy 1/4 the area, therefore delivering 1/4 of the light. Assuming the wall was evenly lit, you would now be capturing additional light from the section of the wall that wasn't visible in the frame when you were standing closer (therefore you're capturing light that you weren't capturing when you stood closer).

So if you used the same lens, you would set it to the same f-stop?

itzcryptic wrote in post #11438643
So if you used the same lens, you would set it to the same f-stop?

Absolutely, the exposure would not change.

BestVisuals wrote in post #11407166
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?

Because the image on the sensor is getting smaller at the same time with the intensity, leaving the number incident photons per measured area to be the same regardless of distance.

Rad

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.

Radtech1 wrote in post #11439573
Because the image on the sensor is getting smaller at the same time with the intensity, leaving the number incident photons per measured area to be the same regardless of distance.

Rad

this sounds right to me. the lens is only a very small hole. so regardless of the spread, as long as that small hole is filled with reflected light, photography happens

btw, the question is not unanswerable...