Inverse square law, sort of...

jcc · Apr 9, 2014

Let me preface this by saying that it's past 4 am (late to bed, not early riser), I can't quite think of the answer, and it's unimportant enough that I will forget it the next day. But, it's a curious thought...

(1) Inverse square law states that light will fall off exponentially (the equation doesn't matter for the purposes of this discussion--or does it?!).
(2) The camera records light that is reflected off of the subject.
(3) Light rays travel from source to subject, then bounce off the subject, then into the camera. The light travels the distance of D_total = D_source-to-subject + D_subject-to-camera.
(4) Assume the source and subject remain at a constant distance from each other.
(Q) Shouldn't changing the distance between subject and camera change light intensity (via inverse square law), since the light has to travel that much more/less? Let's say we have a properly exposured photo at X distance from subject. Why does changing subject-to-camera distances, by X/2 or 2X, require no stop-compensation?

Steve Smith · Apr 9, 2014

Yes, but as you increase the camera to subject distance, the area the subject occupies on the film reduces by the same ratio so the light density remains the same.

Steve.

jcc · Apr 9, 2014

Steve Smith said:
Yes, but as you increase the camera to subject distance, the area the subject occupies on the film reduces by the same ratio so the light density remains the same.

Steve.

Using a constant-aperture zoom lens to keep composition the same, while changing distance. You'd still use the same shutter speed.

Dr Croubie · Apr 9, 2014

Think of it more in terms of glass and apertures than film.
Say you're somewhere where you need 400mm to fill the frame with the subject, the aperture according to the light is f/4, 100mm diameter.
Then move a bit closer, you need 50mm to fill the frame. You still need f/4 for the light, but now that means 12.5mm diameter.
Move closer, smaller 'light gathering area' ie the aperture. Square law. Yeah.

cowanw · Apr 9, 2014

The inverse square law works for light falling on the subject; not reflected light from the subject, which is not a point source.

markbarendt · Apr 9, 2014

At practical distances individual light "rays" can be thought of as not changing in intensity.

As light moves toward a subject from the light source, the "rays" spread out.

Lets say that at a distance of 1X the measured luminance might be 1 and a given set of rays might cover an area of 1.

At a distance of 2X the light has spread out and the same number of rays now cover an area 4 times larger.

So a light meter's sensor has a fixed size, lets call it 1. So at a distance of 1X the meter will measure a certain amount of light rays.

At a distance of 2X the light has spread out, but the meter hasn't. The meter area still equals 1, but now the area the light covers equals 4. So the meter will only see 1/4 of the light rays it saw before. Just like the meters sensor, there are only 1/4 of the light rays available to be bounced back from a given size subject, at that distance, toward the camera.

What the meter sees at that distance will give the right info needed to decide on the right exposure for that subject. Remember that individual light rays don't diminish.

The same thing does though happen going back the other way but it is easier for me to think about this as the size and importance of the subject matter in the composition changing (the subjects size on the film).

chrisaisenbrey · Apr 9, 2014

I think the at different object camera distance the inverse square law is simply compensated by the size of the object on the film:

If we double the distance between camera and object 1/4 of the light arrives at the film plane but the object has now half of the size on the film plane. Hence the object has 1/4 of the area on the film plane and the light energy per film area is constant.

cramej · Apr 9, 2014

cowanw said:
The inverse square law works for light falling on the subject; not reflected light from the subject, which is not a point source.

Exactly. If the inverse square law worked for reflected light, the whole world would be dark.

andrew.roos · Apr 9, 2014

jcc said:
Using a constant-aperture zoom lens to keep composition the same, while changing distance. You'd still use the same shutter speed.

Correct, but the light collecting area of your lens would increase as you moved away from the subject, making up for the effect of the inverse square law fall-off of the light.

For example, suppose you stand 4m from the subject with a 40mm f/4 lens. The diameter of the lens aperture is 40mm/4 = 10mm.

Now you move 12m away and zoom to 120mm, still f/4, to keep the same framing. Now the clear aperture diameter is 120mm/4 = 30mm.

Although the f-stop remains the same, the actual diameter of the aperture opening has increased by a factor of 3 and the light gathering area, which is proportional to the square of the diameter, increases by a factor of 9.

At the same time since you are now 3 times further away from the subject, the light falling per unit area has reduced by a factor of 9 due to the inverse square law.

However since the effective light gathering area of the lens has increased by the same amount, these effects cancel out and the amount of light reaching the sensor (and hence the required exposure) remains the same.

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andrew.roos · Apr 9, 2014

cowanw said:
The inverse square law works for light falling on the subject; not reflected light from the subject, which is not a point source.

Actually it is also applicable to reflected light. If the light is coherently reflected, for example by a mirror, then the fall-off follows the inverse square law for the total path length (source to reflector + reflector to observer).

If it is scattered, for example by illuminating a non reflective subject, then the subject acts like a whole lot of point sources and the law applies twice - once to the distance from source to subject and once for the distance from the subject to the observer.

(This is similar to the so-called radar range equation, which states that the reflected power received from a target decreases with the fourth power of distance since the inverse square law applies separately to the transmitted and reflected wave paths.)

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mopar_guy · Apr 9, 2014

If you are outside, the light source is the sun. To get twice as far from it, you would need a space ship.

John Koehrer · Apr 9, 2014

I like mopar_guy's answer.

I think most of the answers provided are assuming a constant light source, not flash. Some solutions are using the same f number on different FL lenses 10 vs 30 etc. not the distance the light is traveling from your light source.

In practice, it becomes more important with flash.
To bounce light it's distance from source to bounced surface to subject. I'm assuming a single light for this example.

cowanw · Apr 9, 2014

I am pretty sure brightness of the sunlight reflected off the moon's surface is the same here on earth as it is standing on the moon.

cliveh · Apr 9, 2014

cowanw said:
I am pretty sure brightness of the sunlight reflected off the moon's surface is the same here on earth as it is standing on the moon.

Don't you think it has a bit further to travel?

ic-racer · Apr 9, 2014

It is obvious, but sometimes people forget that the film only collects light beams directed right at the lens; the "fallen off" beams are irrelevant.

markbarendt · Apr 9, 2014

cowanw said:
I am pretty sure brightness of the sunlight reflected off the moon's surface is the same here on earth as it is standing on the moon.

Yes, the big difference is typically just the size it appears to be.

While one of the cameras that went to the moon might be able to farm the LEM tightly, it would take a very, very, long lens to do the same from earth.

markbarendt · Apr 9, 2014

cliveh said:
Don't you think it has a bit further to travel?

Depends on the phase.

jcc · Apr 9, 2014

cowanw said:
The inverse square law works for light falling on the subject; not reflected light from the subject, which is not a point source.

Huygens' principle: the light reflecting off of a point acts as a source itself. Since the source (let's say a studio strobe to be working in the same order of distances, instead of AU vs. m) to subject distance is constant, the intensity of the reflected light (i.e., "reflective source") should be the same.

The farther we move away from the reflective source, the intensity of light falls off via inverse square. You can see this effect when you point the light sphere of an incident meter on the subject (assuming perfect reflection, no diffraction, etc.), but why does a reflective light meter not change the metered value when changing distances?

After some much needed sleep, I think I agree with the lens answer(s). The inverse square law assumes no diffraction. When you add a lens (e.g., in the case of a camera or a spot meter) the light path is no longer radiating at a constant spherical propagation. The lens focuses the light, so new rules must come into playoptics.

mopar_guy · Apr 9, 2014

What reflects the light cannot be considered to be the "source"

Huygens' principle: the light reflecting off of a point acts as a source itself. Since the source (let's say a studio strobe to be working in the same order of distances, instead of AU vs. m) to subject distance is constant, the intensity of the reflected light (i.e., "reflective source") should be the same.

In this case the "source" would be the studio light, and the distance is the distance from the light to the subject. The intensity at the subject will be 1/4 when the distance of light to subject is doubled.

The farther we move away from the reflective source, the intensity of light falls off via inverse square. You can see this effect when you point the light sphere of an incident meter on the subject (assuming perfect reflection, no diffraction, etc.), but why does a reflective light meter not change the metered value when changing distances?

The correct incident metering technique is to meter from the subject location and point the meter at the light.

jcc · Apr 9, 2014

mopar_guy said:
In this case the "source" would be the studio light, and the distance is the distance from the light to the subject. The intensity at the subject will be 1/4 when the distance of light to subject is doubled.

The correct incident metering technique is to meter from the subject location and point the meter at the light.

Let me simplify it: suppose the subject is a perfect mirror.

markbarendt · Apr 9, 2014

jcc said:
The farther we move away from the reflective source, the intensity of light falls off via inverse square.

Not exactly.

The proper exposure to show details nicely, of the moons lit surface, is roughly sunny 16. Doesn't matter if you are here or there.

What changes between shooting here and there is the size of the moon on the film. A sufficiently narrow spot meter and ZS ideas will get this result. An incident meter needs to be in the same light as the subject, so full sun, which will also return the same settings.

This concept is true at short distances too, say at 4 vs 8 vs 16 feet/meters/miles of camera to subject distance. As long as the subject lighting remains constant, the camera exposure setting for that subject remains constant.

As the moon or any subject gets smaller on the film though, that subject's importance in the composition diminishes and in the exposure calculation for the whole frame gets skewed toward the larger subjects.

mopar_guy · Apr 9, 2014

Let me clarify, neither of those metering tchniques that you describe is incident metering. Both are reflective metering.

markbarendt · Apr 9, 2014

mopar_guy said:
Let me clarify, neither of those metering tchniques that you describe is incident metering. Both are reflective metering.

When I say incident meter in full sun I mean exactly that during daytime pointing the dome at the sun.

jcc · Apr 9, 2014

markbarendt said:
Not exactly.

The proper exposure to show details nicely, of the moons lit surface, is roughly sunny 16. Doesn't matter if you are here or there.

What changes between shooting here and there is the size of the moon on the film. A sufficiently narrow spot meter and ZS ideas will get this result. An incident meter needs to be in the same light as the subject, so full sun, which will also return the same settings.

This concept is true at short distances too, say at 4 vs 8 vs 16 feet/meters/miles of camera to subject distance. As long as the subject lighting remains constant, the camera exposure setting for that subject remains constant.

As the moon or any subject gets smaller on the film though, that subject's importance in the composition diminishes and in the exposure calculation for the whole frame gets skewed toward the larger subjects.

"Sunny 16" vs. "roughly sunny 16" difference is roughly the percentile difference between Sun-to-Earth distance vs. Sun-to-Moon-to-Earth distance that the light has traveled.

mopar_guy · Apr 9, 2014

markbarendt said:
When I say incident meter in full sun I mean exactly that during daytime pointing the dome at the sun.

Yes.

Inverse square law, sort of...

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