Coherence — Question 3

Question 3

A finite exposure can average a slowly moving fringe to zero contrast

Two equal beams have a frequency difference $\Delta\nu$. After averaging over the optical carrier, their intensity is $I(t)=2I_0[1+\cos(\delta+2\pi\Delta\nu t)]$. A rectangular exposure extends from $-T/2$ to $T/2$. Define $\operatorname{sinc}x=\sin x/x$.

1. Integrate the recorded intensity and derive the exposure-limited visibility.

2. For $\Delta\nu=100\,\mathrm{Hz}$, find the first positive exposure duration with zero visibility and the visibility for $T=5.0\,\mathrm{ms}$.

Question 3 - Solution

A finite exposure can average a slowly moving fringe to zero contrast

(a) Method and calculation

$$\overline I=\frac1T\int_{-T/2}^{T/2}I(t)dt=2I_0\left[1+\frac{\sin(\pi\Delta\nu T)}{\pi\Delta\nu T}\cos\delta\right].$$$$\boxed{V=|\operatorname{sinc}(\pi\Delta\nu T)|}.$$

(b) Interpretation and further calculation

$$\pi\Delta\nu T=\pi\Longrightarrow\boxed{T_{\mathrm{zero}}=1/\Delta\nu=10.0\,\mathrm{ms}}.$$$$T=5.0\,\mathrm{ms}:\quad\boxed{V=\sin(\pi/2)/(\pi/2)=2/\pi=0.63662}.$$

Physical check

Instantaneous interference can be strong even when an integrating detector records no stationary fringe contrast.