Intensity in Single-Slit Diffraction, Qualitatively — Question 10

Question 10

Reversing a symmetric defocus phase leaves the intensity unchanged

A real even aperture amplitude A is multiplied by $e^{icu^2}$, where real c represents defocus. Define $F_c(q)=\int A(u)e^{icu^2}e^{-iqu}du$. Integrals extend over a symmetric aperture.

1. Use complex conjugation and parity to relate $F_{-c}(q)$ to $F_c(q)$.

2. Determine whether one intensity image can distinguish positive and negative c, and state a phase-sensitive way to remove the ambiguity.

Question 10 - Solution

Reversing a symmetric defocus phase leaves the intensity unchanged

(a) Method and calculation

$$F_c(q)^*=\int A(u)e^{-icu^2}e^{iqu}du=F_{-c}(-q).$$$$A(-u)=A(u)\Longrightarrow F_{-c}(-q)=F_{-c}(q)\Longrightarrow\boxed{F_{-c}(q)=F_c(q)^*}.$$

(b) Interpretation and further calculation

$$\boxed{|F_{-c}(q)|^2=|F_c(q)|^2}:\quad\text{defocus sign is ambiguous}.$$$$I_{\mathrm{ref}}=|F_c+E_r|^2=|F_c|^2+|E_r|^2+2\Re(F_cE_r^*).$$

Physical check

A coherent reference measured at controlled phase steps can recover the complex field and distinguish conjugate solutions.