Diffraction and the Wave Theory of Light — Question 9

Question 9

Opposite phases across two aperture halves force an axial null

A slit spans $-a/2<u<a/2$. Its field is $+A_0$ for $u>0$ and $-A_0$ for $u<0$. Define $F(q)=\int A(u)e^{-iqu}du$. Both halves transmit equal power.

1. Calculate the on-axis field and evaluate F for nonzero q.

2. Expand the result for small q and obtain the leading intensity dependence near the axis.

Question 9 - Solution

Opposite phases across two aperture halves force an axial null

(a) Method and calculation

$$\boxed{F(0)=A_0(a/2)-A_0(a/2)=0}.$$$$F(q)=A_0\int_0^{a/2}(e^{-iqu}-e^{iqu})du=\boxed{-\frac{2iA_0}{q}[1-\cos(qa/2)]}.$$

(b) Interpretation and further calculation

$$1-\cos(qa/2)\simeq q^2a^2/8\Longrightarrow\boxed{F(q)\simeq-iA_0qa^2/4}.$$$$\boxed{|F(q)|^2\simeq A_0^2a^4q^2/16}.$$

Physical check

An on-axis maximum is a property of uniform phase illumination, not an unavoidable property of every single opening.