Fog visibility

The total cross-section area of the droplets in $1cm^3$ is: $(10\times10^{-6})^2\times\pi\times200=2\pi\times10^ {-8} (m^2)$

Hence as the light travel for 1cm, the ratio of scattered light is: $\frac{2\pi\times10^{-8}} {10^{-4}}=2\pi\times10^{-4}$

As the light travel for 1cm, the ratio of unscattered light is: $1-2\pi\times10^{-4}=0.99937$ (approximately)

So the distance in cm is: $log_{0.99937} 0.1=3654$ (approximately)

Thus the answer is 35m

First, we will compute the distance a light beam must travel to intersect one droplet, on average. Let us call this distance $L_0$ . Instead of modelling the light ray as a line and computing its intersection with spherical droplets, it will be easier to consider only the centers of all the droplets and determine the expected number of points within a distance $r_{droplet}$ of our light ray. This defines a cylindrical region around our light ray with a volume of $\pi r^2 L_0$ ; if a droplet has its center within this region, it will intersect our ray and scatter it. Since the average density of droplets is $N_{droplet} = 200$ per cubic cm, the expected number of droplets whose center lies within our region is $\pi r^2 L_0 N$ .

Setting this to one and solving for $L_0$ , we find that $L_0 \approx 15.9$ meters (this calculation left as an exercise to the reader). It follows that at any distance $L$ , we expect to find $\frac{L}{L_0}$ droplets in our path. However, just knowing the expected number of droplets does us no good (yet) - we want to know how many beams are scattered, which requires knowing the probability that there are no droplets in our path.

We can model the droplets as being independently and identically distributed. Therefore, the expected number of droplets intersecting our path of length $L$ can be modeled as a Poisson distribution with a rate parameter of $\lambda = \frac{L}{L_0}$ . We want $90\%$ of the light rays to be scattered, meaning that the probability of no droplets must be $0.1$ . The probability that no events occur in our Poisson distribution is $\frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}$ , which we set equal to $0.1$ . Solving yields $\lambda = \ln{10} \approx 2.303$ , and therefore $L = \lambda L_0 \approx \boxed{36.6\,\text{meters}}$ .

Another method, giving slightly different results, but with the same order of magnitude: consider the volume ratio of the droplets

$r=\frac{N_{droplet} \times \frac{4}{3} \times \pi \times r_{droplet}^{3}}{10^{-6}}$

$r = 8,4 \times 10^{-7}$

Let us now discretize the problem, by considering that space is filled with elementary cubes of the same volume as the droplets, and that any given cube may or may not contain a droplet (but there are no overlapping of a droplet on several cubes). The dimension of such elementary cubes is:

$d_{eff} = r_{droplet} \times (\frac{4}{3} \times \pi)^{1/3}$

$d_{eff} = 1,61 \times 10^{-5} m$

If droplets are randomly distributed in space, $r$ is then the probability of finding a droplet in any given elementary cube.

Now if we observe an object from a distance $l$ , the rays of light need to crosse $\frac{l}{d_{eff}}$ elementary cubes without hitting a droplet. The probability of such an event is:

$p(l) = (1-r)^\frac{l}{d_{eff}}$

When this probability is $10 \%$ , $90 \%$ of the rays of light do not get (directly) to us, and so $90 \%$ of the image is scrambled. This happens when:

$\boxed{l \approx 44 m}$

p = 0.000628319 is the ratio of scattered light per centimeter. So the distance is ln(1-0.9)/ln(1-p) = ln(10)/p = 3665 cm.