Michelson's Interferometer — Question 2

Question 2

Evacuating a cell measures the gas refractivity

A rigid gas cell of length $\ell=0.100\,\mathrm m$ is placed in one Michelson arm and traversed twice. Removing the gas shifts the phase through 84 complete cycles at vacuum wavelength $632.8\,\mathrm{nm}$. The windows, geometry, and the other arm remain unchanged; the final cell index is 1.

1. Infer the initial gas index.

2. Find the signed optical-path change on evacuation and the mirror motion needed in the cell arm to restore the initial optical path.

Question 2 - Solution

Evacuating a cell measures the gas refractivity

(a) Method and calculation

$$N\lambda_0=2\ell(n-1)\Longrightarrow\boxed{n=1+\frac{84(632.8\times10^{-9})}{0.200}=1.000265776}.$$$$\boxed{\Delta\mathcal L_{\mathrm{evacuation}}=-2\ell(n-1)=-53.1552\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$\Delta\mathcal L_{\mathrm{evacuation}}+2\Delta x=0.$$$$\boxed{\Delta x=+26.5776\,\mu\mathrm m\ \text{away from the beam splitter}}.$$

Physical check

The fixed windows cancel from the before-and-after difference, while the gas contribution doubles because of the return pass.