Cloudy day

—1. In a thin cloud the light coming from the top is scattered only once, or not scattered at all. If the observer is below the cloud, not looking directly into the Sun, the scattered light can reach the observer's eye. That makes the cloud look white.

In a thick cloud the light is scattered multiple times before reaching the observer. As a result, most of the light is scattered back to the top of the cloud, or sideways, making the bottom of the cloud dark.

To answer the question we need to estimate the thickness of the cloud layer that results in more than one scattering. We will use $r=10^{-5}m$ for the radius of the water droplets and $n=200/cm^3=2\times 10^{8} m^{-3}$ for the density of the droplets. Let us consider a vertical column of base area $A$ and height $h$ . What should be the value of $h$ so that looking from the top down the whole area is obscured by water droplets?

One droplet covers and area of $a=r^2\pi=(10^{-5}m)^2 (3.14) =3.14 \times 10^{-10}m^2$ . We need $N=\frac{a}{A}=\frac{A}{3.14 \times 10^{-10}m^2}=A (3.2 \times 10^9)$ droplets to cover an area of $A$ . The volume of the column under consideration is $V=Ah$ and there are $N=nV= (2\times 10^{8} m^{-3})A h$ particles in that volume. This yields

$A(3.2 \times 10^9)=(2\times 10^{8}m^{-1}) A h$

or $h=16m$ . Accordingly, a 16m thick cloud will scatter the light, in average, once. In the multiple choices the lowest value for the thickness that can result in significant multiple scattering is 100m.

Note: The very same reasoning can be used to estimate the distance of visibility in a dense fog.

—2. Solution with dimensional analysis (corrected after reading the comment by Blake Farrow): We are given a length ( $d=20\mu$ m) and a number density, with units of 1/ $m^3$ . The question is a length. The simplest combination of these quantities that yields a length is $h=1/(d^2/n)=12.5m$ .