Intensity in Double-Slit Interference — Question 6

Question 6

A finite detector pixel averages over part of a fringe

The local pattern is $I(y)=A[1+V_0\cos(2\pi y/p)]$, where p is the fringe spacing. A uniform rectangular pixel of width w centered at $y_0$ reports the spatial average across its width.

1. Integrate the signal and derive the recorded visibility.

2. For $V_0=1$ and $w=p/2$, find the recorded visibility; determine the first positive w that removes all modulation.

Question 6 - Solution

A finite detector pixel averages over part of a fringe

(a) Method and calculation

$$\overline I(y_0)=\frac1w\int_{y_0-w/2}^{y_0+w/2}I(y)dy.$$$$\boxed{\overline I=A\left[1+V_0\frac{\sin(\pi w/p)}{\pi w/p}\cos(2\pi y_0/p)\right]}.$$

(b) Interpretation and further calculation

$$\boxed{V_{\mathrm{pixel}}=V_0\left|\frac{\sin(\pi w/p)}{\pi w/p}\right|};\quad w=p/2\Rightarrow\boxed{V_{\mathrm{pixel}}=2/\pi}.$$$$\pi w/p=\pi\Longrightarrow\boxed{w=p\ \text{at the first zero}}.$$

Physical check

A detector can report poor contrast even for a perfectly coherent optical field if its pixels integrate over a significant fraction of a fringe.