= Solution
Both the incoming and outgoing spherical <wavefronts> contribute to the path excess. At <aperture> radius $u$, with axial distances $s$ before and $r$ after the <aperture>,
$$
\Delta\ell=\sqrt{s^2+u^2}+\sqrt{r^2+u^2}-(s+r)
\simeq\frac{u^2}{2}\left(\frac1s+\frac1r\right).
$$
The outer edge of the $n$th <Fresnel half-period zone> has $\Delta\ell=n\lambda/2$. Hence
$$
u_n^2=\frac{n\lambda sr}{s+r},\qquad
u_2=\sqrt{\frac{2(400\times10^{-9})(0.5)(0.5)}{1.0}}.
$$
Therefore \b[the outer radius is] $\boxed{u_2=4.47\times10^{-4}\,\mathrm m=0.447\,\mathrm{mm}}$. The <paraxial approximation> is excellent because $u_2\ll s,r$. Treating the illumination as a <plane wave> would incorrectly omit the source-side path curvature.
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