 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,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 $1/R\simeq 1/b$ in the amplitude. Removing common $e^{ikb}/b$ yieldsAssumptions are uniform normally incident plane-wave illumination, coherent monochromatic propagation, aperture dimensions large enough for the scalar description, $r/b\ll 1$, and phase accuracy $k{r}^4/(8b^3)\ll 1$ over the relevant aperture. The last condition justifies dropping the quartic distance term; geometric smallness alone does not guarantee phase accuracy.

For a circular aperture, $h=1$ for $r\le a$ and zero elsewhere. Rotational symmetry and $s=r^2$ give $dxdy=rdrd\phi$, $rdr=ds/2$, andThe 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 obeyWith 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 $n=1,3,5$: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 isFor any odd open annulus this gives $2$, so the whole plate gives $\psi /{\psi }_0\simeq 50$ andThe 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 $49^2\simeq 2.40\times 10^3$, 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 $1/{\pi }^2$.^[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 ${\psi }_{\mathrm{clear}}/{\psi }_0=1-e^{i49\pi }=2$, so 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.