Please shed some flash on this

[thatguyfromvienna]

I'm very obviously too stupid to wrap my mind around this - or I'm just acting on a false assumption.

We all know that light falls off according to inverse-square law which means that when the distance x between light source and subject is three times bigger than before, the power of light getting to the subject is not 1/3x but 1/x³.

Why doesn't the same principle apply to my camera then?
Because no matter whether the distance to my subject is x, 2x or 3x, I still take pictures with the same ISO, aperture and shutter speed.

I'm pretty sure I'm currently just too stubborn to understand this so I'd be very grateful if someone could give me a hand here.

[FrankMi]

Think of it this way, Alexander, the amount of light the camera receives is also proportionately greater with the distance so at three times the distance the camera can take in nine times the surface area and nine time the amount of light.

I have the same kind of problems trying to understand the treadmill. Perhaps you know?

Why is it harder to walk on a treadmill at 15° than one at level? I'm not lifting my body any higher so why is it harder?

[Glenn NK]

The main source of light for photographers is the sun. Sunlight has already travelled 93 million miles and the inverse square law has been taken into account on its way here.

However, but the remaining distance it travels between your subject and your camera is so small in comparison to the 93 million miles, that the effect is negligible.

For fun, try the math: see what the difference in illumination is between: the sun and the subject vs the sun and your camera that is a mile from the subject.

Take the square root of 93,000,000 and 93,000,001 and compare.

sqrt (93,000,000) = 9643.65076
sqrt (93,000,001) = 9643.65081

Not significant is it?

I'm familiar with metric, but didn't feel like converting miles to km (the 93,000,000 was drummed into our heads when we were children and non-metric).

[thatguyfromvienna]

FrankMi
Why would the surface be bigger? It's a constant size.
Regarding the treadmill: You are actually lifting your body. The treadmill is just moving you back again to where you started. Which reminds me of my job.

Glenn
This applies to sunlight without a doubt.
But when I shoot strobes only in an otherwise totally dark room?

[FrankMi]

FrankMi
Why would the surface be bigger? It's a constant size.
Regarding the treadmill: You are actually lifting your body. The treadmill is just moving you back again to where you started. Which reminds me of my job.

I may have wrong but if the camera is able to take in 1 square meter of say, a blank wall it will receive the amount of light reflected off that size surface. If the camera is moved back so that it takes in the light reflected from 9 square meters of wall, wouldn't be receiving 9 times the amount of light?

Obviously something is making the treadmill harder to walk on so you are probably right. It just seems that if I haven't lifted my body any higher...

[Glenn NK]

FrankMi
Why would the surface be bigger? It's a constant size.
Regarding the treadmill: You are actually lifting your body. The treadmill is just moving you back again to where you started. Which reminds me of my job.

Glenn
This applies to sunlight without a doubt.
But when I shoot strobes only in an otherwise totally dark room?

Alexander:

The inverse square law applies to flash/strobe, particularly in a totally dark room. And the fall-off is dramatic in this case. If the ambient light is strong, and you're using fill-in flash, then the fall-off effect could be much less.

I have an image entitled Red Lily at Night (link below) which was taken well after sunset using the onboard flash. The background was a mix of other plants and foliage, but the BG was about three feet (900 mm) away. The light fall-off resulted in the BG getting very much less light. The distance from the lens to the subject was maybe six inches (150 mm).

There is one thing I will stand by - the inverse law applies and it's not likely ever broken.

The light travels from the sun to the subject and to the camera - the inverse square law applies.

[thatguyfromvienna]

FrankMi
That's true for a wall, no doubt.
But the person or object I'm taking a picture of won't grow any bigger when I'm moving away.

Regarding the treadmill: Switch it off. Walk. When you're on top, stand still and have somebody switch it on again and you'll experience in two steps what's actually happening continuously.

Glenn
Absolutely right. You're describing what's happening to the light emitted by your flash and I have to admit that even after using strobes for quite a while, the quick falloff still fascinates me.
Nonetheless, this doesn't answer why the same doesn't apply for the flash light that's being reflected by my object and recorded by my camera.

[Glenn NK]

Nonetheless, this doesn't answer why the same doesn't apply for the flash light that's being reflected by my object and recorded by my camera.

I don't think I quite follow you. Please elaborate.

[thatguyfromvienna]

Glenn
When the light traveling from flash to subject is affected by reverse-square law, why isn't the light traveling from subject to camera affected?

[Glenn NK]

Glenn
When the light traveling from flash to subject is affected by reverse-square law, why isn't the light traveling from subject to camera affected?

It is - your camera/flash system has adjusted for it - it meters the light coming into the lens that is reflected off your subject.

Glenn

[thatguyfromvienna]

When my settings are ISO 100, f/2.8, 1/150s then those settings will apply no matter whether I'm 1' or 10' away.
When my flash is set to 1/8 power at 1' and I move it away to 10' my subject will be underexposed.

[shreds]

When you are metering, it depends on whether you are spot metering/CW Metering/Matrix metering. (if your camera / meter can do that) If your camera only can produce one set of readings, then I can understand your confusion.

The readings can be different. e.g. the area where the light is being measured from, does vary, so your camera settings, if on manual, may need to be adjusted. So the distance you are standing away can be significant. Try it and see on the different meter settings. (Of course outside on a bright day, the differences will be minimal).

(If on auto, the camera will make the adjustments for you).

[thatguyfromvienna]

I guess I'm expressing things in a bad way, so let me try again.

This is the situation:
We're in a room with no sunlight and very little artificial light.
I want to shoot an object.
My equipment: 1 strobe, my camera, a light meter.
I set up the strobe using my flash light meter so that the camera settings are ISO 100, f/2.8, 1/125s.
Whether I'm so close to the object that I shoot at 24mm or so far away that I have to zoom in to 200mm, my camera settings don't change even though the reflected light is traveling a much longer way from the object to the camera.
Now if I move the strobes just a tad further away from the person, I'll need to set the strobe to emit more light.

So again - why is the light that's traveling from my strobes to the object affected by the inverse-square law while seemingly the light traveling from the object to my camera isn't?

[pnodrog]

The inverse square law applies to light or radiation traveling in all directions emitted by a point source. It is not applicable to any sort of directional control by reflectors, diffracting elements, wave guides etc or any attenuation. e.g. Torch beams, lasers, car headlights or attenuation by fog do not follow the law.

The image being collected by your lenses is only dependent on the angular direction and amount of the light in relation to the lens and it's aperture. Provided there is no attenuation in the light path the distance is irrelevant. You could try thinking that each beam of light being received by the lens has been sent by lasers at different intensities and distances but all aimed at your lens.

[thatguyfromvienna]

Now I'll be able to sleep tonight.
Thanks a bunch.

Edit: Yay! My 100th posting here.

[RustBeltRaw]

The inverse square law applies to light or radiation traveling in all directions emitted by a point source.

Quoted for truth. A photon does not lose power as it moves away from the source (that would make astronomy tricky ). Imagine a point source of light with two transparent spheres wrapped around it. One has twice the other's radius. The same number of photons will pass through each sphere. But the number of photons per unit area will be less (in this case, 1/4 as much) on the larger sphere.

A laser is the exact opposite. Its strength basically stays the same regardless of distance.

So, yet again, we find a photographic "rule" that's more of a guideline. Most light mods and sources do not perfectly follow the inverse square law. The softer they are, the closer they'll be. But light from a highly collimated beam, like a fully-zoomed hot shoe flash, will not fall off as quickly as a strobe in a 7ft octabank.

[bambleweeney]

That's a really good question and one that's been giving me cause to ponder for some time. The only difference is that one emits light (the flash) and the other reflects. My theory ... light from a flash, the sun or whatever is directional if you will, the photons come from a point source and the spread of the light from source to target diminishes the light intensity (hence the inverse square law) Depending upon what is reflecting the light (person, landscape etc) the light is reflected or scattered randomly in all directions to start with and so there is not the same spread as there would be from an emitter. I guess it depends on the reflector and whether photons are absorbed and re-emitted or scattered (specular highlights on water or from a mirror are I believe probably different from the manner in which my skin reflects light). I hasten to add that this is speculation and regardless of the illumination I'm really not that bright from any distance
There's never a physicist around when you need one ...

[Colin Southern]

Man there are some suspect answers here!

In reality it's very simple -- if an object is moved further away then the light from that object falls off according to the square-law - but - the surface area of the sensor that the light from that object needs to hit ALSO FALLS OFF ACCORDING TO THE SAME SQUARE LAW.

The net result is that the energy ratio remains the same (ie further away = less energy but it's "spread" over a proportionally smaller area).

[pnodrog]

In this diagram S is the light source but as Colin has pointed out if S is a sensor the inverse of the law is also true. So the fact that it is further away is irrelevant as you are gathering light from a proportionally bigger surface.

[Glenn NK]

So again - why is the light that's traveling from my strobes to the object affected by the inverse-square law while seemingly the light traveling from the object to my camera isn't?

Could you be forgetting that most flash units have variable power, and the amount of light put out is controlled by the camera's metering system?

You set the f/stop, the ISO, and the shutter speed (this part can be tricky).

First you take a shot, then move farther away from the subject for a second shot - the flash unit will put out more power (power = light), and the exposure appears similar.

The flash unit adjusted for the distance - but there will reach a point where the flash cannot put out enough power and the image will be under-exposed.

The inverse square law hasn't been violated - it can't.

Glenn