Let us say we send light with wavenumber $\bar \nu \pm \frac{\Delta \bar \nu}{2}$ through a Michelson Interferometer. Using the intensity at the center of the interference pattern $I(x)$ (where $x$ is the separation of the mirrors), how could we measure experimentally the value of the ratio: $$\frac{\Delta \bar \nu}{\bar \nu}$$

I know you could just measure $\Delta \nu$ and $\bar \nu$ separately (using the modulating envelope and high frequency oscillations) but I think there must be an easier way. Any ideas?


It is probably worth mentioning that the intensity distribution takes the form: $$I(x)=\frac{I_0}{2}(1+\cos(\bar \nu \pi x)\cos(\Delta \bar \nu \pi x))$$


I am not sure that this is the best method but it is the only one I can think of.

  1. Find a location of minium ($0$) visibility (which is easy to detect) [1].
  2. Looking at the center of the intensity distribution, move one the mirrors to until you reach the next location of minium ($0$) visibility.
  3. When moving the mirrors count the number $n$ of minium is achieved at the center.
  4. You then have the following:$$\frac{\Delta \bar \nu}{\bar \nu}=\frac{1}{n}$$
  5. You could do the normal things, like repeating at different minium visibilities etc. to further improve accuracy.


[1] Shepherd, G.G. 2002. Spectral Imaging of the Atmosphere. Volume 82. London: Academic Press (p135)


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