Diffraction and the Wave Theory of Light — Question 6

Question 6

Translating an aperture changes far-field phase but not intensity

The far-field amplitude is $F(q)=\int t(u)e^{-iqu}du$. Translate the aperture by $u_0$, so its transmission becomes $t(u-u_0)$, without changing the incident plane wave. Ignore finite detector acceptance.

1. Derive the transformed field and its intensity.

2. For $q=2.0\times10^4\,\mathrm{m^{-1}}$ and $u_0=50\,\mu\mathrm m$, find the added phase and explain how it could be detected.

Question 6 - Solution

Translating an aperture changes far-field phase but not intensity

(a) Method and calculation

$$v=u-u_0:\quad F_{\mathrm{shift}}(q)=\int t(v)e^{-iq(v+u_0)}dv.$$$$\boxed{F_{\mathrm{shift}}=e^{-iqu_0}F,\quad|F_{\mathrm{shift}}|^2=|F|^2}.$$

(b) Interpretation and further calculation

$$\boxed{\Delta\phi=-qu_0=-1.00\,\mathrm{rad}}.$$$$|F_{\mathrm{shift}}+E_r|^2=|F|^2+|E_r|^2+2\Re[e^{-iqu_0}FE_r^*].$$

Physical check

A fixed coherent reference converts the otherwise invisible phase factor into an observable intensity change.