Michelson's Interferometer — Question 6

Question 6

Two spectral lines produce recurring fringe-visibility packets

A Michelson is illuminated by two mutually incoherent lines of equal intensity at wavelengths $\lambda_1=600\,\mathrm{nm}$ and $\lambda_2=602\,\mathrm{nm}$. Each line interferes with itself. For optical path difference D the summed modulation is proportional to $\cos(2\pi D/\lambda_1)+\cos(2\pi D/\lambda_2)$.

1. Factor the modulation and derive the visibility envelope in terms of the synthetic wavelength $\Lambda=(1/\lambda_1-1/\lambda_2)^{-1}$.

2. Find the first mirror displacement from equal paths giving zero contrast and the mirror displacement between successive full-contrast revivals.

Question 6 - Solution

Two spectral lines produce recurring fringe-visibility packets

(a) Method and calculation

$$\cos A+\cos B=2\cos\frac{A-B}{2}\cos\frac{A+B}{2}.$$$$\boxed{V(D)=|\cos(\pi D/\Lambda)|,\quad\Lambda=\frac{\lambda_1\lambda_2}{\lambda_2-\lambda_1}=180.6\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$V=0\ \text{first at}\ D=\Lambda/2;\quad D=2x\Longrightarrow\boxed{x_{\mathrm{zero}}=\Lambda/4=45.15\,\mu\mathrm m}.$$$$\Delta D_{\mathrm{revival}}=\Lambda\Longrightarrow\boxed{\Delta x_{\mathrm{revival}}=\Lambda/2=90.30\,\mu\mathrm m}.$$

Physical check

The slowly varying envelope is distinct from the rapid optical fringes. Different spectral lines add intensities rather than maintaining a common optical phase.