In the Michelson interferometer setup shown below, first I obtain the interferogram of the source (without sample). Then I put in the sample. If the sample is not dispersive (its refractive index is constant over all frequencies), then it would simply just translates the interferogram by $x_0=2d(n-1)$.

But there's a general formula used in papers, which gives the refractive index in general using the interferogram:

$$n(\sigma) = 1+\frac{x_0}{2d}+\frac{1}{4\pi \sigma d}\tan^{-1}\frac{Im\{F\{I(x-x_0)\}\}}{Re\{F\{ I(x-x_0)\}\}}$$ where $F\{\}$ denotes Fourier transform, $I$ is the interferogram and $\sigma$ is $1/\lambda$.

How is this relation actually obtained?

Note that the relation between the spectrum $B(\sigma)$ and the interferogram $I(\Delta)$ (function of $\Delta$, the displacement of the moving mirror) is the Cosine transform:

$$B(\sigma) = \int_0^\infty I(\Delta) \cos (2\pi \sigma \Delta) d\Delta$$ $$I(\Delta) = \int_0^\infty B(\sigma) \cos(2\pi \sigma \Delta) d\sigma $$

The refractive sample modifies the optical path length that each frequency in $B(\sigma)$, shomehow affecting the interferogram.

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