Number of fog droplets in the air

Question

I have an issue with the official solution to this problem from BelPhO:

Visibility on the road is 100 m. Assuming that the diameter of a fog droplet is 1 micron, estimate the concentration of fog droplets in the air.

The official solution is:

If a drop of water gets in the way of a photon, the photon will be scattered or absorbed. To estimate the average "flight" length of a photon, the following reasoning can be carried out: in cylinder of diameter equal to the diameter equal to the diameter of the droplet and length equal to the free span on average there should be one droplet, i.e.

$$l \frac{\pi d^2}{4} n \approx 1 \Rightarrow n \approx \frac{4}{\pi d^2 l} \approx \frac{4}{\pi(10^{-6})^2 \cdot 100} \approx 10^{10} \mathrm{m}^{-3}$$

I cannot see how this physical situation described by the solution can best fit the model of the initial problem. The only information we are given is the visible radius $l$ and the size of a droplet. Do you not think (as I think) that the solution admits the untrue assumption that the visibility distance is equal to the mean free path? Doesn't it seem more reasonable to you that the fraction of the surface of a sphere of diameter $d$ which is obscured depends directly on the number of droplets within the visible radius, which is something proportional to the concentration $n$ of the fog droplet times a volume?

Any advice or detailed explanation is appreciated. Please let me know what you think. Thank you.

Answer 1

Well the easiest solution is to say $L$ is inversely proportional to $n$ and $d^2$, so:

$$ \frac 1 L \propto nd^2$$

or:

$$ n \approx \frac 1 {d^2L} = 10^{10}\,{\rm m^{-3}}$$

and be done.

A more serious attempt is to consider the extinction of out going photons with a number density of $N(r)$:

$$ \frac{dN}{dr} = -\frac {da} A $$

where (using capitals for lengths for clarity):

$$ da = n dV \pi D^2$$

is the area covered at radius $r$ by the number of particles in the shell:

$$ dV = n(4\pi r^2 dr) $$

and

$$ A = 4\pi r^2 $$

is the total area of a shell at radius $r$.

So:

$$ \frac{dN}{dr} = -\frac{\pi D^2} 4 $$

which has a solution:

$$ N(r) = N_0e^{-r/L} $$

with

$$ L = \frac 4 {\pi D^2 n} $$

which is exactly with the cylinder argument gives.

The cylinder argument works because the $4\pi$'s in $dV$ and $A$ cancel.