Solution
= Solution
For source and screen distances $s,r$, the paraxial optical path excess at <aperture> radius $u$ is $u^2(1/s+1/r)/2$. The outer boundary of <Fresnel half-period zone> $n$ has excess $n\lambda/2$, hence
$$
u_n^2=\frac{n\lambda sr}{s+r},\qquad \boxed{u_2=\sqrt{\frac{2(500\times10^{-9})(0.5)^2}{1.0}}=0.500\,\mathrm{mm}.}
$$
This is the outer radius of the second zone, measured from the axis.