Intensity in Single-Slit Diffraction, Qualitatively — Question 3

Question 3

A real aperture profile produces an even far-field intensity

A real aperture amplitude $A(u)$ may be asymmetric in u. Its Fourier amplitude is $F(q)=\int A(u)e^{-iqu}du$. There is no imposed complex phase profile.

1. Relate $F(-q)$ to $F(q)$ and prove the symmetry of the intensity.

2. Explain why unequal illumination of the two slit edges alone cannot shift the centroid of a fully collected, integrable far-field intensity.

Question 3 - Solution

A real aperture profile produces an even far-field intensity

(a) Method and calculation

$$F(-q)=\int A(u)e^{iqu}du=\left[\int A(u)e^{-iqu}du\right]^*.$$$$\boxed{F(-q)=F(q)^*,\quad I(-q)=I(q)}.$$

(b) Interpretation and further calculation

$$\int_{-\infty}^{\infty}qI(q)dq=0\quad\text{when the integral exists}.$$$$\boxed{\langle q\rangle=\frac{\int qI(q)dq}{\int I(q)dq}=0}.$$

Physical check

Amplitude asymmetry can fill minima and reshape lobes while preserving reflection symmetry in the far field.