Intensity in Single-Slit Diffraction, Qualitatively — Question 8

Question 8

An illumination gradient fills a null while leaving the central field unchanged

Within $-a/2<u<a/2$, let $A(u)=A_0(1+2\epsilon u/a)$ with real $|\epsilon|<1$. Use $F(q)=\int A(u)e^{-iqu}du$. At the first uniform-slit zero $q_1=2\pi/a$, the needed integral is $\int_{-a/2}^{a/2}u e^{-iq_1u}du=-ia/q_1$.

1. Find the axial field and the residual field at $q_1$.

2. For $\epsilon=0.20$, calculate the residual intensity relative to the axial intensity.

Question 8 - Solution

An illumination gradient fills a null while leaving the central field unchanged

(a) Method and calculation

$$F(0)=A_0\int_{-a/2}^{a/2}(1+2\epsilon u/a)du=\boxed{A_0a}.$$$$F(q_1)=\frac{2\epsilon A_0}{a}\left(-\frac{ia}{q_1}\right)=\boxed{-i\epsilon A_0a/\pi}.$$

(b) Interpretation and further calculation

$$\boxed{I(q_1)/I(0)=\epsilon^2/\pi^2}.$$$$\epsilon=0.20\Longrightarrow\boxed{I(q_1)/I(0)=0.00405285}.$$

Physical check

The odd amplitude gradient integrates to zero on axis but spoils exact cancellation at the original minimum.