Many have heard of the Inverse Square Law, some even understand it, but everybody feels its effect on a daily basis. It’s what prevents gravity from pulling us all into the ground and it’s also what allows the sun to illuminate the planet without blinding us.
Essentially, the Inverse Square Law (let’s just call it the ISL from here, or this could get annoying) states that any point source which spreads its influence equally in all directions will see the strength of that influence reduced in a manner inversely proportional to the square of the difference in distance (or to put it more simply, when you double the distance from the source, you quarter the strength (1/2²) of that point source’s influence, when you triple the distance you get 1/9th (1/3²) the power, quadruple the distance you get 1/16th (1/4²) the power, and so on).
This is simply geometrical fact. Point sources such as gravitational force, light, sound, radiation and electricity all conform to the ISL. If the earth were twice the diameter it currently is, yet had the same mass, we’d weigh a quarter of what we currently do (much to the delight of some). If we cut the distance between the Earth and the Sun in half, daytime would be 4x as bright (although, as a nasty side effect to the improvement in British weather, we’d cook from the heat!).
So, how does that apply to us, as photographers?
Well, you’ll notice one of the things I mentioned above that’s affected by the ISL is light, and given that photography depends entirely upon light (or a lack thereof) that’s a bit of a biggy, especially if you use flash (although the principle still applies to ambient light shooters too).
Let’s have a look at why this happens, with the help of the image below.
Remember, the point source spreads its influence equally in all directions. So, given that the point source is at the centre of the sphere, at a set distance you’ll get the exact same readings from any point on the surface of that sphere. You can see above that as the sphere grows it has to spread that influence out to cover the new larger surface area, which results in a reduction of power.
When you double the radius or diameter of the sphere, the surface area is multipled by 4. The inverse of this is 1/4, so when you double the distance, you get 1/4 the power.
Oddly enough, while making the above image, I discovered that radial gradients (or “ramps” as AE calls them) can also obey the ISL.
Looking at our two spheres in the image above, if the sphere on the left has a diameter of 100cm (50cm radius) and the sphere on the right has a diameter of 200cm (100cm radius) that pans out as follows.
- Left Sphere : 4 x π x 50² = 31,416cm²
- Right Sphere : 4 x π x 100² = 125,664cm²
Our point light source is in the dead centre of the sphere, spreading out evenly in all directions, we’ve doubled the distance, but have a look at the surface area.
Do you know what you get when you divide 125664 by 31416? You get exactly 4.
The same amount of energy is now spread out over 4 times as much surface area, therefore it is now at 1/4 of its former strength at any given point on the surface of the larger sphere.
I think I’m starting to understand, so how does that work in practical terms?
For our purposes, there are two basic things to keep in mind. Light falls off from the source (it gets weaker the further away it is) and diminishes at a predictable rate, and the greater the distance between the light source and your subject, the more even spread of light your flash will provide over a larger area (at the expense of power).
Woah, wait, let’s take this slowly
Ok, let’s have a look at the first bit. Light falling off and diminishing at a predictable rate.
Let’s assume that we have a bare flash which meters f/16 at a distance of 10ft from our subject. We lose approximately a stop of light every time we multiply the distance by 1.4, and we lose two stops whenever we double the distance. With our light metering f/16 at 10ft, it would meter f/11 at 14ft (a loss of 1 stop, or 1/2 as much light), and f/8 at around 20ft (a loss of 2 stops, or 1/4 of the light.
It stands to reason, then, that this also works the other way, if we divide the distance by 1.4, we gain a stop of light, and if we divide the distance by 2, we gain two stops of light. So, at 7ft, we meter f/22 (a gain of 1 stop, or double the light), and at 5ft we meter f/32 (a gain of 2 stops, or 4x as much light).
Aperture follows a formula very similar to the inverse square law, however it is based on the area of the circle rather than the surface area of a sphere.
To get the aperture of a circle based on its radius we use the equation πr². To get the surface area of a sphere with the same radius, we simply multiply that number by 4 (the spherical surface area equation being 4πr²).
If you had a lens that opened up all the way to f/1, and stopped it down to f/1.4, you’d lose one stop of light. If you stopped it down to f/2 you’d lose two stops, f/2.8, three stops, f/4 four stops.
If you have your light 1 ft away from your subject, and moved it 1.4ft away from your subject, you’d lose a stop of light. If you move it 2ft away you’d lose two stops, 2.8ft away you’d lose three stops, and 4ft away you’d lose 4 stops.
You have just learned how to increase or decrease the amount of light hitting your subject from your light without adjusting any settings on your camera, and without adjusting the flash’s power setting. This can have other side effects which may or may not be desired. One is the angle and convergence of shadows casting onto and by your subject, another brings us on to point number two.
Ok, hit me!
In the second point I mentioned that the greater the distance between your light source and your subject, the more even the spread of light it provides. This is because the further away you get from your light source, the slower the light it emits falls off.
Let’s look at our example again. At 5ft, we meter f/32, at 7ft we meter f/22, at 10ft we meter f/16, at 14ft we meter f/11 and at 20ft we meter f/8.
At 5ft, we lose a whole stop of light 2ft further away from our subject. At 14ft, we have 6ft before we lose another stop. This is handy to know if you’re shooting several people at once. The further away your lights are from the subject, the more evenly lit each of your subjects will be as they stand closer to or further away from the light relative to each other.
Look at the images below for an example. If you’re doing a group portrait of 3 people stood side by side, your light is on the left, and your closest subject is 5ft away from the light, metering at f/32, your subject furthest from the light may be 10-11ft away.
In reality, you probably wouldn’t light a group of people this way, but just for the sake of demonstration, let’s say you would.
If your lens is set to f/32, the person on the left will meter beautifully and be accurately exposed, the person in the middle will be about a stop and a half underexposed, and the person on the right will be over 2 stops underexposed. If you went the other way and set your lens to about f/16 to expose the subject on the far right, the person closest to the light will probably be pure white with no detail at all.
We have over 2 stops of difference between a distance of 5ft and 11ft from the light source, or over 4 times as much light on our closest subject.
If we move all our subjects about 15ft to the right (our right, their left), you can see that they’re all lit fairly similarly.
In fact, the variance here between the closest and furthest subject is only about a 20-30% difference in the amount of light received (we don’t lose another whole stop of light until we’re 28ft away from our light). This means that all three subjects are going to receive a fairly similar amount of exposure (much less than a stop of difference between the nearest and farthest subject). You could bump up the amount of light on the subject furthest from the light with a little bit of fill, or you could position your subjects so that the one with the darkest skin tone is closer to the light, and the one with the lightest skin tone is furthest away, then all will appear more consistent in your shot.
If we set our aperture somewhere between f/5.6 and f/8 we can get a fairly even exposure on all three subjects.
Aha, I get it now. So, how might this affect the background behind our subject?
Because the falloff of light from its source is predictable, you can also position your lights, subject and background in positions relative to each other so that you can turn a white background into a grey or even black background. If your subject meters f/8 at a distance of 4ft from your lights, your background 12ft behind your subject (16ft away from the light) would only be metering at about f/4 and become very dark grey (even if it’s actually white in reality).
If your subject stepped forward by 2ft, and you knocked your flash power down by 2 stops, so that your subject still metered f/8, your background would now be metering f/2 and virtually black.
Or, you can position them in such a way that your background receives about the same amount of light as your subject. If your subject is 20ft away from your light source, and the background is only 2ft behind your subject, both will receive a similar amount of light (of course, your subject will also now cast a shadow on the background as they’re so close – which may or may not be desirable – this is why you sometimes need to light your background separately from your subject).
So, how does this work for ambient light?
It works in exactly the same way. Let’s say you’re shooting by a large window or inside a doorway on a cloudy, but still rather bright, day. If your subject is 2ft away from the window, and is metering f/8, you would see a reading of f/4 if you moved your subject 4ft away from the window, and you would see a reading of f/2 if you moved them 8ft away from the window. This happens because the window, rather than the sky, has become your light source, so what matters is the distance between your subject and the window (or doorway).
So what about shooting outside in ambient light then? If I’m outside on a sunny day and my subject is 6ft away from me, why are the people 50ft behind them just as bright?
When the sun is your light source, the inverse square law still applies, however it involves such huge distances that you don’t notice any difference. The Sun is approximately 93 million miles away from the Earth.
In order to lose just one stop of light, the Earth would need to be about another 37 million miles further away from the Sun (at which point your camera would probably seize up anyway with it being so cold n’ all, although that probably wouldn’t be your biggest concern). A difference of only 50ft is so small it can’t even be measured when compared to 93 million miles.
Well, that makes sense, I think. I’m going to read through all that again a couple of times, and get back to you!
Don’t worry if it’s not fully sinking in just yet. While the Inverse Square Law is a fairly basic concept, its implications in the photographic world can be a bit difficult thing to initially grasp. Just experiment. The more you play around and practise, the more you’ll understand it, and one day it’ll just click, and the whole thing will make complete sense.
Happy shooting, and if anything above confused you even more than it helped you and you want to ask further questions, or you notice that I made any big fat obvious mistakes in this article please post a comment below.