Diffraction — Question 7

Question 7

An aperture integral reveals why rescaling the opening rescales the pattern

The far-field amplitude of a uniformly illuminated slit is proportional to $A_a(q)=\int_{-a/2}^{a/2}e^{iqu}\,du$, where $q=k\sin\theta$. Define the normalized amplitude $F_a(q)=A_a(q)/A_a(0)$. Consider a new slit of width $sa$, with positive scale factor s.

1. Use a change of integration variable to derive the scaling law for the normalized amplitude.

2. For $s=3$, determine the new angle corresponding to a fixed feature originally at $\theta=6.0^\circ$ and give its small-angle limit.

Question 7 - Solution

An aperture integral reveals why rescaling the opening rescales the pattern

(a) Method and calculation

$$A_{sa}(q)=\int_{-sa/2}^{sa/2}e^{iqu}du=s\int_{-a/2}^{a/2}e^{i(sq)v}dv=sA_a(sq).$$$$A_{sa}(0)=sA_a(0)\Longrightarrow\boxed{F_{sa}(q)=F_a(sq)}.$$

(b) Interpretation and further calculation

$$sk\sin\theta^\prime=k\sin\theta\Longrightarrow\boxed{\theta^\prime=\sin^{-1}(\sin6^\circ/3)=1.99675^\circ}.$$$$|\theta|\ll1:\quad\boxed{\theta^\prime\simeq\theta/s=2.0^\circ}.$$

Physical check

Normalization removes the increase in on-axis amplitude; only the angular contraction remains in this comparison.