Checking the math in RAW's mas 88 paper

Treating each electron individually and neglecting absorption effects we can estimate the thickness from eq 8.

We must also be assuming an element is only contained in 1 layer:

But we don’t know the phi(rho z) until we iterate so we use a triangular approximation:

The area of the shaded region = $q *\rho z_i^{f1}-[(\rho z_i^f1)^2*q/\rho z_{r,i}]/2$

The area of the big triangle = $[q*\rho z_{r,i}]/2$

For simplicity lets call:

So we have:

So Waldo’s final answer was correct… for a surface film.

But What about for a buried layer?

The actual (\phi(\rho z)) curve whould be shifted but the idea is the same:

So:

For simplicity let’s call:

The area of the blue box (second region) is:

So we have:

Or:

$\delta_{0}^{fn}=\sum_iR_i\left(1.0-\sqrt{1-k_{i,exp}- \frac{2x}{R}+\frac{\left(x\right)^2}{R^2}}\right)-x \tag{6}$ So GMRFilm is correct but if you look the paper is wrong :(

Finding starting compositions

For surface film and burried layer we can use eqs. \ref{eq2} and \ref{eq4} respectivly.

For the substrate we substitute R for y in eq. \ref{eq4}.