Index of refraction

The frequency of microwaves is very low compared to visible light, and for most materials the index of refraction $n = c/v$ is determined by the dielectric constant $\epsilon$ for the response of the material to a fixed electric field

$$n = \sqrt{\epsilon}$$

Typically, dielectric constants of non-polar materials are somewhere between $2$ and $10$, but some materials have very large dielectric constants, most notably water with $\epsilon \approx 80$.

One way to measure the index is to use the material to change the path length on one arm of the Michelson interferometer. If the thickness of the material is $t$ and its index is $n$, when it is inserted into one side of the interferometer the path $t$ is replaced by the optical path $nt$. The increase in optical path for one pass is $nt - t = (n-1)t$. When material is present, Mirror A would be closer to the beam splitter by just this amount to give the path difference it would have had without the material. If you can determine this shift when the material is inserted, then you can find the index. When the displacement of the mirror to restore the original condition is $\Delta y$, the index of refraction $n$ is given by

$$n = 1 + (\Delta y / t)$$

Locate a minimum in the signal without paraffin present. Insert the paraffin block in the path to Mirror A between the beam splitter and the mirror. Now move the mirror toward the beamsplitter until you find a new minimum. How much do you have to move the mirror? It would be best to repeat this several times and average the results. This is $\Delta y$. Calculate the the index of refraction $n$, and the dielectric constant $\epsilon$ of paraffin for microwaves.

There is an ambiguity about this measurement, since it is possible that the paraffin shifted the pattern more than one fringe. How would you modify the experiment to be sure that the shift was less than fringe, or to find the true shift if it were more?