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I have come across many papers but still couldn't find the relationship between index of refraction or atomic scattering factors, and reflectivity.

My flow of thought goes as follows:

  1. Get the tabulated scattering factors with the respective energy($f'$ and $f"$)
  2. Calculate the absorption coefficients at their respective energy (or simply get it from the tabulated scattering factors too)
  3. Obtain the extinction coefficients, and perform Kramers-Kronig to get the refractive indices.
  4. From here onwards, could anyone advise me on how to calculate the reflectivity for p-polarized light, and plot it against angles of incidences to get a graph of Kiessig interference fringes?

Thank you so much! Any help is much appreciated! Please do correct any of the mistakes that I may have in my flow of thought =)

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One problem with Kramers-Kronig is it needs to integrate over all values of $\omega$, so you can do an estimation using it, but most materials don't have robust measurements over wide ranges of $\omega$ in order to accurately obtain the complete complex indices of refraction. –  daaxix Dec 31 '12 at 8:10

1 Answer 1

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The Wikipedia article should provide some help, http://en.wikipedia.org/wiki/Transfer-matrix_method_(optics)#Abeles_matrix_formalism.

The last section relates how the reflectivity of a thin film system can be calculated by considering the interface as a set of slabs of uniform scattering length density. For X-rays the scattering length density, $\rho$, of a material is related to the electron density: $$ \rho = r_e/V_m \times\Sigma_{i} f_i $$ where $f_i$ is the scattering factor of an element $i$ in the material (at a given energy), $r_e$ is the Compton radius ($A$), and $V_m$ is the molecular volume ($A^3$). $f$ and $\rho$ are complex.

Once you have the scattering length densities it is easy to calculate the reflectivity. There are several programs available to do so (e.g. http://www.sourceforge.net/projects/motofit).

The refractive index can be linked to $\rho$ as follows: $$n\approx1-\frac{\rho\lambda^2}{2\pi}$$

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