Lost in a thicket

Electricity and Magnetism Level 4

Folk tales are filled with cautionary tales about getting lost in the middle of a dark forest. This trope is not just a convenience for story tellers, it is a lesson about light scattering. As one goes farther into a forest, photons have had more opportunities to hit a tree and bounce backward out of the forest.

Suppose you find yourself $\text{10 m}$ into a dark forest which is illuminated from a (one) side. The canopy is extremely thick so that no light penetrates from above. Moreover, the forest consists of cylindrical trees with average radius $\bar{r}$ which are randomly distributed with number density $\rho$ (i.e. 2 trees per square meter).

What fraction of the incoming light the can you see, on average?

Assumptions

Ignore second scattering events, if a photon hits a tree once, it exits the forest forever.
$\bar{r}=6\text{ cm}$
$\rho = 2\text{ tree m}^{-2}$

The answer is 0.09071795328941251.

1 solution

Josh Silverman Staff
Jun 1, 2014

The idea here is simple, the further one walks into the forest, the more trees have had a chance to to stand between the edge of the forest and your position. If a tree stands at a depth $l$ into the forest, it will block light for all positions $l^\prime>l$ .

The effect of all the trees compound such that very deep into the forest, no light can penetrate at all, as illustrated in the diagram below.

On average there are $\rho$ trees in every square meter patch of forest, and each tree blocks light entering a $2\bar{r}$ wide strip of forest. We expect each step of length $\Delta l$ into the forest to reduce the illumination, $L$ , by a factor $\left(1-2\bar{r}\rho\Delta l\right)$ , i.e.

$\frac{\Delta L}{\Delta l}=-L2\bar{r}\rho\Delta l$

This has the solution $\boxed{L(l)=L_0e^{-2\bar{r}\rho l}}$ .

Therefore, we expect the illumination in the forest to drop off exponentially with the distance walked into the forest.

Looks interesting, but I'm really lost.

First of all, why will it block light for all positions $l'>l$ ?

Second, why with each step $\Delta l$ the illumination L reduce by a factor of ( $1-2\overline{r} p \Delta l$ )? And how did you acquire the relation for $\frac{\Delta L}{\Delta l}$ ?

At last, how did you solve it for L(l)?

Thank you very much.

Zero Mech - 6 years, 9 months ago

I've been thinking about what makes certain materials opaque, translucent, or transparent – how light interacts in a lattice of atoms. Usually they follow an exponential decay law opacity .

I still don't really know what the primary causes of opacity are. Density is not a good measure, for diamonds are transparent. Its allotrope graphite makes the question more elusive. Perhaps, there are many reasons, I don't know. I was wondering if you would know.

Steven Zheng - 6 years, 9 months ago

Lost in a thicket

The answer is 0.09071795328941251.

1 solution

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