= Half-slit phase-step diffraction factorization
{title2=$I_\pi/I_{0,\rm peak}=\operatorname{sinc}^2s\sin^2s$}
Two adjacent half-slits of width h with opposite field transmission have centres separated by h. The aperture transform is $h\operatorname{sinc}(qh/2)[e^{-iqh/2}-e^{iqh/2}]$. Thus the phase-reversed pair factor multiplies the finite-half-slit envelope, giving the displayed normalized intensity with $s=qh/2$ and $\operatorname{sinc}s=\sin s/s$. The on-axis field cancels. Positive side maxima satisfy $\tan s=2s$, and total integrated transmission is unchanged by a lossless phase plate even though the angular distribution changes.
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