As light goes a certain distance through some medium, three things can happen to any photon:
- Nothing: It can just pass through, proceeding, after that distance, in exactly the same direction in was moving at the beginning of that distance.
- It can be absorbed by the medium. That is, the photon disappears.
- It can be scattered by the medium. That is, the photon leaves a region of the medium traveling in a different direction than the one it had before the scattering event took place.
Think of a thin slice of material (it may be a solid, a liquid, or a
gas), with a beam of light incident on the left hand "surface." (There
doesn't need to be a real surface, it can be an imaginary plane we picture
in our minds.) The question is how much light comes out of the surface
on the right hand side? That depends, of course, on what the medium is
made of, and the energy of the photons. Any photons that are absorbed inside
the material do not come out. And any that are scattered may come out the
right hand side, but they will be going in some other direction from the
incident beam.
Now, any photon that is traveling through this medium has a certain probability of either being absorbed or scattered. For now, let's combine the absorption and scattering into a single word: extinction. Let's call the probability of extinction in this slice of the medium P. And let's call the flux of photons incident on the left-hand side of the slice Io. Then the flux of photons coming out of the right hand side, and in the same direction they were going on entry I is related to Io and P by:
I = Io (1-P)
Now, if the amount of absorbing or scattering material in the slice is made smaller, then the probability of extinction P is decreased. We can cut P about in half if we either cut the width of the slice in half, or we reduce the concentration of absorbing or scattering material by half. In order for this to be true, P has to be pretty small, which may mean that the slice has to be pretty thin.
How to we handle the situation in the atmosphere, where the distance
a photon travels from top to bottom is very large? The answer is, in our
imagination, we chop the atmosphere into thin slices. Here's how it works.
Imagine we have 1 cm of material, and we want to know, given a certain
flux in a beam coming in from the left, how much light leaves, going in
the same direction toward the right. If we divide this material into N
slices, each having a width 1/N cm, we can ask how much light comes
out to the right from each slice, and take that as the amount of light
that goes into the next slice from its left (See figure). Let's use k
for the probability of extinction per cm. Then the probability of extinction
in one of these slices is k/N. And we will call
In
the flux that has gone through n slices.
So we start with Io, the flux that hasn't gone through any slices. Then the flux that comes out of the first slice is
This flux enters the second slice. The flux that comes out of the second slice is
I2 = I1 (1-k/N) = Io (1-k/N)(1-k/N) = Io(1-k/N)2.
The flux that comes out of the third slice is
I3 = I2 (1-k/N) = Io (1-k/N)3 .
We proceed from layer to layer in this manner, until we come out of the other side of the 1 cm sample, where the exiting flux is
IN = Io (1-k/N)N
Why is it necessary to do this? It is because, if the probability of extinction is very high, then there will be a lot fewer photons going through the last 1 mm than the first 1 mm of the 1 cm sample. Thus, the number of photons available to be absorbed in the last 1 mm is much less than the number available to be absorbed in the first 1 mm. We must treat the material as a large number of very thin slices, so that there is essentially no difference in the probability of absorption in the left and right sides of the slices. In fact, though it would be impossible to actually do this in the laboratory, we can take the number of slices be greater and greater, and eventually infinite.
Here is a result you can find in any calculus book.
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The symbol e stands for a certain number, called the base of the natural logarithms; this number is approximately 2.7818 . Just substituting k for x in this, and comparing to our expression just above for IN , we have
I = Io e-k
Finally, we can relate the extinction probability to the concentration of the absorbing or scattering material, and to the overall length of the light's path through the material. That is, if we double the concentration, we expect to double the extinction probability. That means if the concentration is c then k must be directly proportional to c, that is k = ac . What happens if we double the path length? Well, we can do something similar to what we just did. If we have a 2 cm path, then we can take it apart into two 1 cm paths. The flux leaving the first half is
I1 = Io e-ac
and the flux leaving the second half is
I2 = I1 e-ac = Io e-2 a c
Thus, we see that k is proportional to the length. Now we have
k = sl c
where l is the length of the path, and s is a constant of proportionality. In fact, s is the only quantity we have not yet taken care of: It is proportional to the extinction cross section of the material.
Putting these ideas all together, we come up with the usual expression for the Beer-Bouger-Lambert law:
I = Io e-s.lc