= Solution
Take the <aperture> in $z=0$ and the observation point $P=(0,0,b)$, with $b>0$. Under <scalar diffraction theory>, the <Huygens-Fresnel principle> sums <complex amplitudes> from area elements, rather than summing their <intensities>. Apart from a common normalization,
$$
\psi(P)\propto\iint h(x,y)K(\theta)\frac{e^{ikR}}R\,dx\,dy,
\qquad R=\sqrt{b^2+x^2+y^2},\quad k=2\pi/\lambda.
$$
Here $h$ is the complex <aperture> transmission, $b$ is the axial distance, and $\theta$ is the angle of the secondary ray to the <aperture> normal. The <obliquity factor> $K$ describes the direction dependence of the secondary contribution; one Kirchhoff convention is $K=(1+\cos\theta)/2$ for normal illumination. Other scalar boundary approximations have slightly different angular weights, all tending to unity near the axis.
Put $r^2=x^2+y^2$. The <paraxial approximation> gives $R\simeq b+r^2/(2b)$ in the phase and $1/R\simeq1/b$ in the amplitude. Removing common $e^{ikb}/b$ yields
$$
\boxed{\psi(P)\propto\iint h(x,y)K(\theta)
\exp\left[\frac{i\pi(x^2+y^2)}{\lambda b}\right]dx\,dy.}
$$
Assumptions are uniform normally incident plane-wave illumination, coherent monochromatic propagation, <aperture> dimensions large enough for the scalar description, $r/b\ll1$, and phase accuracy $kr^4/(8b^3)\ll1$ over the relevant <aperture>. The last condition justifies dropping the quartic distance term; geometric smallness alone does not guarantee phase accuracy.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-2-fresnel-zones.png]
{title=Axial Fresnel geometry and phasor addition with alternate half-period zones blocked}
For a circular <aperture>, $h=1$ for $r\leq a$ and zero elsewhere. Rotational symmetry and $s=r^2$ give $dx\,dy=r\,dr\,d\varphi$, $r\,dr=ds/2$, and
$$
\psi\propto2\pi\int_0^a K(\theta)e^{i\pi r^2/(\lambda b)}r\,dr
=\pi\int_0^{a^2}K(\theta)e^{i\pi s/(\lambda b)}ds.
$$
The constant $\pi$ can be absorbed into the proportionality factor, producing the required one-dimensional <integral>.
At a focus a distance $f$ away, half-period-zone boundaries obey
$$
\sqrt{f^2+r_n^2}-f=\frac{n\lambda}{2},\qquad
r_n^2=n\lambda f+\frac{n^2\lambda^2}{4}\simeq n\lambda f.
$$
With slowly varying obliquity and illumination, each zone has almost the same area and its integrated <phasor> has the opposite sign to its neighbour. A <zone plate> blocks alternate zones, so the remaining <phasors> add instead of nearly cancelling. Choosing an open central zone, the first three open regions have outer boundaries $n=1,3,5$:
$$
\boxed{r_1=0.100\,\mathrm{mm},\qquad r_3=0.173\,\mathrm{mm},\qquad r_5=0.224\,\mathrm{mm}.}
$$
These are outer radii, not zone widths; the second open region is the annulus between $r_2$ and $r_3$.
For the specified outer radius, $R^2/(\lambda f)=49$: there are approximately 49 half-period zones, of which 25 odd zones are open. Normalize to the field $\psi_0$ of the unobstructed <plane wave>. The normalized Fresnel <aperture> <integral> is
$$
\frac{\psi}{\psi_0}=\frac{1}{i\lambda f}\int h\,e^{i\pi r^2/(\lambda f)}dA.
$$
For any odd open annulus this gives $2$, so the whole plate gives $\psi/\psi_0\simeq50$ and
$$
\boxed{I_{\rm focus}/I_{\rm unobstructed}\simeq4(25)^2=2.50\times10^3.}
$$
The <axial intensity gain of an amplitude zone plate> grows as the square of the number of open zones. Treating the number of zones continuously gives $49^2\simeq2.40\times10^3$, the same accuracy as an estimate. This large local <intensity> gain is consistent with <conservation of energy> because light is concentrated into a small focal spot. It is not a <power> efficiency; the ideal first-order <power> efficiency of a binary amplitude <zone plate> is only $1/\pi^2$.<Zone Plates by Janos Kirz and David Attwood>[X-Ray Data Booklet, §4.4, first-order efficiency paragraph]{ref}
“Without the <zone plate>” has here meant removing the finite plate and its outer stop, giving the unobstructed incident <wave>. If instead a clear circular <aperture> of radius $R$ is retained, its axial field is $\psi_{\rm clear}/\psi_0=1-e^{i49\pi}=2$, so \b[the ratio to that finite-aperture reference is $2500/4=625$]. The two reference experiments differ; neither ratio should be confused with total transmitted <power>.
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