A recent discussion thread on the Strobist Flickr group did something that Internet discussions frequently do: it diverged rapidly from the original topic, and devolved into an argument over terminology.
The argument centered on the made-up term “light depth of field”. (For the record, I find this term silly; even worse, I find it misleading. I won’t be using it again, and I suggest that you forget it was ever mentioned.)
The context of the discussion, however, is both useful and instructive to examine. The key issue in that context is:
Given the inverse square law, how do you determine the minimum distance away from your subjects to place a light, in order that the exposure variation across those subjects due to falloff stays within acceptable limits?
It sounds complicated, but it really isn’t. There’s a smidgen of math involved, but it’s nothing harder than very basic high school algebra.
For those who’ve forgotten, let’s go over the inverse square law:
The intensity of light falling on a subject at a certain distance from a light source is inversely proportional to the square of the distance from the light source.
In layman’s terms, when you double the distance, you get
the light intensity. Triple the distance, you get
the light intensity. Move n times further out, you get
times the light intensity.
We photographers like to deal in “stops” of light. Each “stop” is a doubling or a halving of light intensity. Put in math terms, the intensity difference in stops is the base 2 logarithm of the ratio of the intensities in foot-candles or lux.
And now, a little bit of high-school math:
Don’t worry — I’ll do the math for you. You can feel free to skip to the end, if you just want the answer….
First, we describe the setup. We have a light source (“L”) that is a certain distance (“x”) away from the closest subject (“C”). A second subject (“F”) is further away. How far? Subject F is “y” distance past subject C.
Something like this:
L -------------------x----------------- C ----y----F
So, the distance from L to C is x, and the distance from L to F is (x+y).
If a light intensity falling on C (“x” distance units away from L) is “i” lux, then that same light falling on F at “x + y” distance units away has an intensity of
, which is a change of
stops.
We’re looking to find out how far away to put our light in order to get the light falling over our subjects to be even within a certain number of stops, when the group is spread out over a certain distance from the closest to the furthest subject. In other words, we’re looking for “x”, given the number of stops and “y”:
![stops = \log_{2}\left [ \frac{x^2}{(x+y)^2} \right ]](http://latex.codecogs.com/gif.latex?stops&space;=&space;%5Clog_%7B2%7D%5Cleft&space;[&space;%5Cfrac%7Bx%5E2%7D%7B%28x+y%29%5E2%7D&space;%5Cright&space;])
If y is a positive number (which it will be, as we’ve laid out the diagram), then the number of stops will be a negative number, indicating that the light is getting dimmer the further away that we go. While mathematically accurate, most photographers aren’t used to be thinking in negative numbers. To make the equation more useful, let’s flip the fraction over, which will reverse the sign on the logarithm. That way, we can specify the number of stops down as a positive number, rather than having to think of it as a negative number of stops up.
![stopsdown = \log_{2}\left [ \frac{(x+y)^2}{x^2} \right ]](http://latex.codecogs.com/gif.latex?stopsdown&space;=&space;%5Clog_%7B2%7D%5Cleft&space;[&space;%5Cfrac%7B%28x+y%29%5E2%7D%7Bx%5E2%7D&space;%5Cright&space;])
Now, we just need to solve for “x”.
![stopsdown = \log_{2}\left [ \frac{x+y}{x} \right ]^2](http://latex.codecogs.com/gif.latex?stopsdown&space;=&space;%5Clog_%7B2%7D%5Cleft&space;[&space;%5Cfrac%7Bx+y%7D%7Bx%7D&space;%5Cright&space;]%5E2)
![2^{stopsdown} = \left [ \frac{x+y}{x} \right ]^2](http://latex.codecogs.com/gif.latex?2%5E%7Bstopsdown%7D&space;=&space;%5Cleft&space;[&space;%5Cfrac%7Bx+y%7D%7Bx%7D&space;%5Cright&space;]%5E2)




And, finally….

There is our answer. It’s a bit ugly, but it’ll tell us what we need to know.
If the closest subject and the furthest subject are y feet apart, and we want the exposure across the group to be consistent within “stopsdown” stops, we can calculate the distance x in front of the group at which to place our light.
As an example, if we want at most a variation of 1 stop from front to back across a group of people, and the distance between the closest and furthest people is 3 feet, then our light has to be approximately:
in front of the group.
If we want to limit ourselves to 1/2 stop of variation across the same group, we need to place our light further away:
in front of the group.
The Bottom Line:
The further away your light source is from your group, the less variation in light intensity you’ll have across the group due to falloff. The overall intensity will, of course, be lower. That’s not in question. However, the important thing to remember is that there will be less variation of intensity across the group. You’ll have more even — although dimmer — light.
If you know how much variation in light intensity you can tolerate (in stops) and you know the size of the group (from front to back, as seen from the perspective of your light source), then you can fairly-easily calculate the minimum distance to your light source.

To make this easy for those of you without a scientific calculator, I’ve pre-calculated a few factors that you can use in the following equation, just by multiplying:

| # of stops |
factor |
| 2 |
1.00 |
| 1.5 |
1.47 |
| 1 |
2.41 |
| 2/3 |
3.85 |
| 1/2 |
5.29 |
| 1/3 |
8.17 |
| 1/4 |
11.05 |