Intensity in Single-Slit Diffraction, Qualitatively — Question 9

Question 9

A common random phase does not wash out a diffraction pattern

An entire coherent aperture acquires a spatially uniform random phase $\phi(t)$. Its instantaneous far-field amplitude is $F(q,t)=e^{i\phi(t)}F_0(q)$. Compare this with two equal aperture patches whose relative phase is uniformly random during an exposure; each patch alone gives intensity $I_p$.

1. Calculate the time-averaged intensity for the common-phase fluctuation.

2. Calculate the average intensity of the two independent-phase patches and compare with their fully in-phase value.

Question 9 - Solution

A common random phase does not wash out a diffraction pattern

(a) Method and calculation

$$I(q,t)\propto|e^{i\phi(t)}F_0(q)|^2=|F_0(q)|^2.$$$$\boxed{\langle I(q)\rangle=I_0(q)}\quad\text{for arbitrary common phase noise}.$$

(b) Interpretation and further calculation

$$I=2I_p(1+\cos\delta),\quad\langle\cos\delta\rangle=0\Longrightarrow\boxed{\langle I\rangle=2I_p}.$$$$\delta=0:\quad\boxed{I_{\mathrm{coherent}}=4I_p,\quad\langle I\rangle/I_{\mathrm{coherent}}=1/2}.$$

Physical check

Diffraction depends on relative phase across the opening, so common phase noise and differential phase noise have different consequences.