Feynman lectures and apparent area of a nucleus In paragraph 5.7 of this lecture, Feynman explains how to calculate the apparent area of the nucleus, in a sheet of unspecified material.
In the note Feynman says:

"This equation is right only if the area covered by the nuclei is a small fraction of the total, i.e., if $\frac{n1-n2}{n1}$ is much less than 1. Otherwise we must make a correction for the fact that some nuclei will be partly obscured by the nuclei in front of them"

Do you have any idea how to apply this correction factor to the previous formula? 
 A: That "apparent area" is called the cross section, usually denoted with a sigma, $\sigma$.
Suppose (as Feynman et al. do) that you're interested in the probability that scattering from a nucleus removes a particle from the beam.  If the thickness $\ell$ of your target is small enough that the overlap between nuclei is negligible, and the number density of the target nuclei is $n$ nuclei per unit volume, then the probability that a particle from your beam makes it through undeflected is
$$
p_\text{thin} = 1 - n\sigma\ell.
$$
If your target has large thickness $L$ so that this approximation doesn't apply, you can divide it up into many thin targets; the probability of transmitting through all the layers is the product of the probabilities of making it through each layer.  That is,
\begin{align}
p_\text{thick} &= \prod_\text{all layers} p_\text{layer}
= \left( 1 - \frac{n\sigma L}N \right)^N,
\end{align}
if you divide the target into $N$ thin layers.
The continuum result is
\begin{align}
\lim_{\text{smooth}} p_\text{thick} &= \lim_{N\to\infty} (p_\text{thin})^N
= e^{-n\sigma L}
\end{align}
The transmission through a thick target is exponential in the length of the target.
