Fresnel Zones and Zone Plates

Stand behind an illuminated hole and ask what the light does at one point on the axis. Huygens' recipe says add a secondary wavelet from every patch of the wavefront — but the patches arrive with every possible phase, and the bookkeeping looks hopeless. Fresnel's move was to slice the wavefront into rings drawn so the path to your eye grows by exactly half a wavelength from one ring's edge to the next. Adjacent rings then arrive in antiphase and nearly cancel, and the intractable sum collapses to something readable almost by inspection: the whole open wavefront delivers only half the amplitude of its innermost ring, so the free-space intensity is a mere quarter of what that first zone alone would give. That single accounting trick — the half-period zone — drives a family of surprising facts. Stopping a beam down to just the first zone makes the axis four times brighter, not dimmer. It explains the bright spot Poisson thought would demolish the wave theory, sitting at the centre of a circular shadow. And turned into hardware it becomes the zone plate: a flat patterned screen that focuses light with no curved glass at all, simply by blocking every other zone so the survivors add in step. Because that focusing is pure diffraction, the plate is fiercely chromatic and carries a comb of higher-order foci — drawbacks in a camera, but exactly why it is the workhorse lens of soft-X-ray and extreme-ultraviolet microscopy, where no glass can refract. This article builds the zone construction, derives the focal length and efficiency, and reads the physics off the vibration spiral.

Key takeaways

Fresnel zones are concentric annuli on a wavefront drawn so each successive edge is $\lambda/2$ farther from the observation point $P$ ; adjacent zones arrive in antiphase and cancel in pairs.
The $m$ -th zone edge sits at $r_m=\sqrt{m\lambda f}$ , and every zone has the same area $\Delta S=\pi\lambda f$ , so their amplitudes are comparable but alternate in sign.
The slowly decreasing alternating series sums to $A\approx A_1/2$ for the whole open wavefront ( $I\approx I_1/4$ ), while a single open first zone gives $A_1$ , i.e. $4I_0$ on axis.
A zone plate blocks alternate zones so the survivors add in phase and focus light; its primary focus is $f_1=r_1^2/\lambda$ , with odd higher foci $f_p=f_1/(2p+1)$ .
A binary amplitude zone plate sends $1/\pi^2=10.1\%$ of the light to the first-order focus; a $\pi$ -shifting phase plate reaches $4/\pi^2=40.5\%$ , a blazed kinoform nearly $100\%$ .
Key idea: partition a wavefront into half-wave annuli of equal area $\Delta S=\pi\lambda f$ at radii $r_m=\sqrt{m\lambda f}$ ; adjacent zones cancel so the whole wave gives only $A_1/2$ , and blocking or phase-flipping alternate zones converts that cancellation into a bright diffractive focus at $f_1=r_1^2/\lambda$ .

The core insight

Slice a wavefront into annular zones whose edges differ in path to $P$ by $\lambda/2$ . The zones have equal area and alternating sign, so the full open wave sums to about half the first zone's amplitude ( $I\approx I_1/4$ ). Remove or phase-invert every other zone — a Fresnel zone plate — and the cancellation becomes reinforcement: the odd zones add in phase to focus light at $f_1=r_1^2/\lambda$ , with a discrete set of weaker odd-order foci and strong chromatic dispersion, all fixed purely by diffraction.

Where this lives today

The zone plate reigns at the short-wavelength frontier. At soft-X-ray and extreme-ultraviolet synchrotron and free-electron-laser beamlines, where indices sit within a thousandth of unity and materials absorb ferociously, diffractive zone plates with outer zones of tens of nanometres are the standard focusing elements — scanning transmission X-ray microscopes reach sub- $20$ nm resolution on biological and magnetic samples, imaging the interior of a cell or the domains of a hard disk. The same $\sqrt{m}$ geometry scales everywhere: flat Fresnel zone antennas replace parabolic dishes for microwaves; stepped Fresnel lenses focus ultrasound; and in EUV lithography, zone-plate sensors align and meter the exposure wavelength through their built-in chromaticity. Even in a lecture hall the device proves a patterned opaque screen can concentrate light — focusing by interference, the purest demonstration that diffraction and imaging are one phenomenon.

The physical picture

Looking back from $P$ at the lit plane, draw a circle wherever the travel distance to you jumps by another half-wavelength. The bull's-eye this paints is the zone system: each ring hands you light, and because consecutive rings are half a wave out of step, they fight each other. They form a tidy alternating series because they have equal area — the $m$ -th zone runs from $r_{m-1}$ to $r_m$ with $r_m^2\propto m$ , so $\pi(r_m^2-r_{m-1}^2)$ is constant — and equal-area rings at nearly equal distance send nearly equal amplitude. A zone plate is then almost embarrassingly simple: it paints over the cancellers.

Figure

Fresnel half-period zones on the wavefront: equal-area annuli at radii rₘ = √(mλf); odd zones (amber) reinforce at P, even zones cancel — a zone plate blocks the even ones

Now picture the running sum of those amplitudes as a chain of phasors laid head to tail. Because the phase advances smoothly outward, the chain curls into a spiral — the vibration curve — winding onto a central point. The first zone alone traces a half-turn whose chord is the diameter $A_1$ ; the fully open wavefront runs all the way to the centre, and the vector from start to centre is only the radius, half that diameter. That is $A\approx A_1/2$ in one picture — the punchline the next figure makes quantitative, along with what a zone plate does to it.

Figure

The vibration spiral: the open wavefront sums to A₁/2 (I = I₁/4 = I₀) while one open zone gives A₁ (I = 4I₀); a zone plate stacks the odd zones in phase — 1st-order efficiency 1/π² = 10.1% (amplitude), 4/π² = 40.5% (phase)

The two panels are the same physics read twice. The open wave wastes its light in the tight inner curl, arriving at $A_1/2$ ; the zone plate deletes the backward-pointing phasors so the survivors stack into a long resultant — a focus built from destruction avoided. The price is on the plot: even perfect, a binary amplitude plate throws only $1/\pi^2=10.1\%$ into the first-order focus, while $\pi$ -phase zones recover the absorbed half and reach $4/\pi^2=40.5\%$ .

The core equations

The construction begins with a single geometric demand: mark the wavefront wherever the path to $P$ has grown by another half-wave. Imposing that on the paraxial path length fixes both the zone radii and their areas at once — and because the radii go as $\sqrt{m}$ , the equal spacing in $r^2$ makes every annulus the same size.

r_m=\sqrt{m\lambda f},\qquad \Delta S_m=\pi\!\left(r_m^2-r_{m-1}^2\right)=\pi\lambda f \qquad (1)

where: $r_m$ — radius of the $m$ -th zone edge; $m$ — zone index ( $1,2,3,\dots$ ); $\lambda$ — wavelength; $f$ — axial distance from the wavefront to $P$ (the focal length when the plate is used as a lens); $\Delta S_m$ — area of the $m$ -th annular zone. The radii bunch together outward ( $r_2/r_1=\sqrt2$ , $r_4/r_1=2$ ) while every zone keeps the same area $\pi\lambda f$ — the reason the amplitudes stay comparable. Turn the transparent-odd-zone screen into a lens and equation $(1)$ hands you its focal length directly: setting $m=1$ pins $f$ to the first zone radius. Because the same rings can also satisfy the reinforcement condition with three, five, seven, … half-waves per zone, a zone plate focuses not once but at a comb of shorter odd-order foci.

f_1=\frac{r_1^2}{\lambda},\qquad f_p=\frac{f_1}{2p+1}\quad(p=1,2,3,\dots) \qquad (2)

where: $f_1$ — primary (first-order) focal length; $f_p$ — the $p$ -th higher-order focus at $f_1/3,\,f_1/5,\,\dots$ ; $p$ — diffraction-order index. Only odd multiples appear, because the even orders correspond to every zone contributing an integer number of whole waves, which the alternating pattern suppresses. With the geometry fixed, the on-axis amplitude of the unobstructed wave is a signed sum of the zones. Each term is set by the constant area, the slowly growing distance, and a fading obliquity factor, so the magnitudes decrease smoothly while the signs alternate — a series that sums to half its first term.

A=A_1-A_2+A_3-\cdots\approx\frac{A_1}{2}\quad\Rightarrow\quad I\approx\frac{I_1}{4} \qquad (3)

where: $A_m$ — amplitude contributed by the $m$ -th zone ( $A_1>A_2>A_3>\cdots\to0$ ); $A$ — resultant on-axis amplitude; $I_1$ — intensity a single open first zone would deliver. The open wave gives $A_1/2$ ; a lone first zone gives $A_1=2A$ , hence $4$ times the free-space intensity on axis. Finally, treating a binary zone plate as a diffraction grating in the squared-radius coordinate $\rho=r^2$ (where the quadratic lens phase becomes linear) gives the power in each order; the first-order coefficients set the two plate efficiencies.

\eta_1^{\text{amp}}=\frac{1}{\pi^2}\approx10.1\%,\qquad \eta_1^{\text{phase}}=\frac{4}{\pi^2}\approx40.5\% \qquad (4)

where: $\eta_1^{\text{amp}}$ — first-order efficiency of an opaque/transparent (amplitude) zone plate; $\eta_1^{\text{phase}}$ — first-order efficiency of a $\pi$ -shifting (phase) zone plate. The amplitude plate loses $25\%$ to absorption and $25\%$ to the undiffracted zeroth order; the phase plate recovers the absorbed quarter, quadrupling the focus.

Derivation AZone radii and the zone-plate focal length

Our goal is to derive the $\sqrt{m}$ radius law and the focal length $f_1=r_1^2/\lambda$ from nothing but the half-wave path condition, and to see why the plate carries a comb of odd-order foci. We place an axial point $P$ a distance $f$ from the wavefront, demand successive zone edges differ in path by half a wavelength, expand paraxially, and read the radii and foci off.

Step 1. Write the exact path from a zone edge at radius $r_m$ to the axial point $P$ and set its excess over the axial path to $m$ half-waves: $\sqrt{f^2+r_m^2}-f=m\,\tfrac{\lambda}{2}$ . Why: this is the defining property of a half-period zone — each edge is one more $\lambda/2$ of travel away from $P$ , so consecutive zones arrive in antiphase.

Step 2. Expand for $r_m\ll f$ : $\sqrt{f^2+r_m^2}=f\sqrt{1+r_m^2/f^2}\approx f+\dfrac{r_m^2}{2f}$ . Why: paraxially the sagitta $r_m^2/2f$ is the leading correction to the axial distance, as in the thin-lens and Fresnel-number bookkeeping; the dropped term is of order $r_m^4/f^3$ .

Step 3. Equate the two: $\dfrac{r_m^2}{2f}=m\,\dfrac{\lambda}{2}$ , hence $r_m^2=m\lambda f$ and $r_m=\sqrt{m\lambda f}$ . For the first zone this is $r_1^2=\lambda f$ , i.e. $f_1=r_1^2/\lambda$ . Why: once the outer radius $r_1$ is chosen, the wavelength and the half-wave rule fix the focus completely — a purely diffractive focal length with no index of refraction anywhere in it.

Step 4. Ask when the same fixed rings refocus. If each zone spans $(2p+1)$ half-waves, $\dfrac{r_1^2}{2f_p}=(2p+1)\dfrac{\lambda}{2}$ gives $f_p=f_1/(2p+1)$ . Why: the pattern reinforces whenever a zone holds an odd number of half-waves, so the plate images at $f_1,f_1/3,f_1/5,\dots$ but not the even orders — the origin of its multiple foci. *The binomial expansion confirms $r_m=\sqrt{m\lambda f}$ and $f_1=r_1^2/\lambda$ symbolically; numerically, $r_1=0.5$ mm and $\lambda=500$ nm give $f_1=0.5$ m and $f_3=0.167$ m.*

Go deeper · Derivation B · 1 minWhy the whole wavefront gives only half the first zone

Derivation BWhy the whole wavefront gives only half the first zone

The most counter-intuitive result in the subject is that the whole open wavefront delivers just half the amplitude of its innermost zone — so stopping a beam down to one zone makes the axis brighter. Our aim is to prove $A\approx A_1/2$ by summing the signed zone series and to draw out the fourfold-intensity consequence, treating each zone as a phasor set by area, distance, and obliquity, with alternating sign.

Step 1. Write each zone's amplitude as $A_m\propto \dfrac{\Delta S_m}{\ell_m}\,K(\chi_m)$ , with $\Delta S_m=\pi\lambda f$ constant, $\ell_m$ growing slowly, and the obliquity $K(\chi)=\tfrac12(1+\cos\chi)$ decreasing monotonically. Hence $A_1>A_2>\cdots\to0$ , with alternating sign. Why: equal-area zones at nearly equal distance radiate nearly equal amplitude, so only the obliquity falloff distinguishes them — the precondition for the alternating-series trick.

Step 2. Regroup the signed sum symmetrically: $A=\dfrac{A_1}{2}+\Big(\dfrac{A_1}{2}-A_2+\dfrac{A_3}{2}\Big)+\Big(\dfrac{A_3}{2}-A_4+\dfrac{A_5}{2}\Big)+\cdots$ . Why: splitting each interior term across two brackets isolates the first half-term and leaves brackets that measure the sequence's curvature.

Step 3. Each bracket nearly vanishes because a smooth sequence satisfies $A_m\approx\tfrac12(A_{m-1}+A_{m+1})$ , leaving $A\approx\dfrac{A_1}{2}+\dfrac{A_{\text{last}}}{2}$ ; the obliquity factor drives $A_{\text{last}}\to0$ , so $A\approx\dfrac{A_1}{2}$ and $I\approx\dfrac{I_1}{4}$ . Why: the interior zones cancel in pairs and only the first survives — the resultant is the radius, not the diameter, of the vibration spiral.

Step 4. Read off the consequence: a single open first zone gives the full $A_1=2A$ , so its on-axis intensity is $(2A)^2=4$ times the free-space $I_0=A^2$ . Why: this is the $N_F=1$ bright peak $I=4I_0$ , and blocking the central zones instead leaves the outer zones summing to the bright Poisson–Arago spot. *The alternating-series identity gives $A=A_1/2\Rightarrow I=I_1/4$ , and a single open zone yields $I=4I_0$ — consistent with the general expression $4I_0\sin^2(\pi N_F/2)$ evaluated at $N_F=1$ .*

Go deeper · Derivation C · 1 minDiffraction efficiency of a binary amplitude zone plate

Derivation CDiffraction efficiency of a binary amplitude zone plate

A perfect opaque-and-clear zone plate still sends only about a tenth of the light to its focus, and knowing why — and how to do better — separates a toy from an X-ray optic. Our aim is to compute $\eta_1=1/\pi^2$ by recognising the plate as a diffraction grating in the squared-radius coordinate, then show a $\pi$ -phase plate quadruples it to $4/\pi^2$ , Fourier-decomposing the binary transmittance and tallying the power order by order.

Step 1. Change variables to $\rho=r^2$ . In $\rho$ the zone edges $r_m^2=m\lambda f$ are equally spaced, so the transmittance $t(\rho)$ is a $50\%$ -duty square wave of period $\rho_1=r_1^2$ , and the quadratic lens phase $\propto r^2$ becomes linear. Why: a grating periodic in $\rho$ diffracts into orders that are precisely the foci $f_1/p$ — the zone plate is a circular grating in disguise.

Step 2. Expand the square wave: $t(\rho)=\dfrac12+\dfrac{2}{\pi}\displaystyle\sum_{p\,\text{odd}}\dfrac{(-1)^{(p-1)/2}}{p}\cos\!\dfrac{p\pi\rho}{\rho_1}$ . Each cosine splits into a converging and a diverging lens of focal length $f_1/p$ . Why: the Fourier coefficients of a square wave are $2/(p\pi)$ for odd $p$ and zero for even $p$ — the analytic origin of the odd-only foci.

Step 3. Tally the power. The DC term $c_0=\tfrac12$ carries $|c_0|^2=\tfrac14$ (undiffracted zeroth order); the opaque half absorbs another $\tfrac14$ ; and the first-order amplitude coefficient is $|c_1|=1/\pi$ , so $\eta_1=|c_1|^2=1/\pi^2\approx0.101$ . Why: half the incident energy is simply lost (absorbed or undeflected) before any focusing, capping a binary amplitude plate near $10\%$ .

Step 4. Replace each opaque zone with one that transmits but adds a $\pi$ phase shift. Absorption vanishes and the blocked zones now contribute with reversed sign — in phase with the focus — doubling the first-order amplitude to $2/\pi$ , so $\eta_1=4/\pi^2\approx0.405$ . Why: recovering the absorbed quarter and reclaiming the antiphase zones is a fourfold gain; a blazed (kinoform) profile pushes it toward $100\%$ . *The Fourier square-wave coefficients yield $\eta_1^{\text{amp}}=1/\pi^2=0.1013$ and $\eta_1^{\text{phase}}=4/\pi^2=0.4053$ .*

Limiting cases

Case	Result	Meaning
Single open first zone	$A\approx A_1=2A_\infty\Rightarrow I\approx 4I_0$	stopping a beam down to one zone makes the axis four times brighter
Full open wavefront	$A=\tfrac12 A_1$	the alternating series sums to half the first zone — the ordinary free field $I_0$
Zone plate (even zones opaque)	$A_{\text{ZP}}=A_1+A_3+A_5+\cdots$	odd zones add in phase → a bright diffractive focus, up to $\sim10\%$ (amplitude) of the light
Central disc obstacle	$I_{\text{axis}}\neq0$	block the inner zones and the outer ones sum to the Poisson–Arago bright spot
Many zones, $N_F=a^2/\lambda b\gg1$	geometric shadow	the aperture spans many zones; $N_F$ counts them, linking to the Fresnel number

Common trapThe half-period zones are equal-width rings and a zone plate is a clean single-focus lens

Both halves are wrong, and both bite. The zones are equal in area, not width: their radii grow as $r_m=\sqrt{m\lambda f}$ , so the annuli get thinner outward — the outermost width $\Delta r_N=\lambda f/2r_N$ is the finest feature and limits resolution and manufacturing. And a zone plate is not a tidy lens: it carries real and virtual foci at $f_1/(2p+1)$ , a strong undiffracted zeroth-order beam, and — for a binary amplitude plate — only $\sim10\%$ in the primary focus. Design a zone-plate system as a single-focus, full-throughput lens and you will be ambushed by higher-order ghosts and a bright zeroth-order background, with the chromatic shift $f\propto1/\lambda$ compounding it.

Worked examples

1 · Focal length and foci of a visible-light zone plate (confidence builder)

A zone plate has first-zone radius $r_1=0.5$ mm and is used with green light, $\lambda=500$ nm. Find its primary focal length, the radius of its tenth zone, and its third-order focus.

Step 1. Primary focus from equation $(2)$ : $f_1=\dfrac{r_1^2}{\lambda}=\dfrac{(5\times10^{-4}\,\text{m})^2}{5\times10^{-7}\,\text{m}}=\dfrac{2.5\times10^{-7}}{5\times10^{-7}}=\mathbf{0.5\ m}$ . A patterned flat screen focuses at half a metre with no curved surface anywhere.

Step 2. Tenth-zone radius from $(1)$ : $r_{10}=\sqrt{10}\,r_1=3.162\times0.5\ \text{mm}=\mathbf{1.58\ mm}$ . The zones from the first to the tenth are packed into an extra $1.08$ mm of radius — they crowd together outward, exactly as the $\sqrt{m}$ law demands.

Step 3. Higher-order foci from $(2)$ : $f_3=f_1/3=\mathbf{0.167\ m}$ , $f_5=f_1/5=0.10$ m. The plate simultaneously forms fainter, shorter-focus images at each odd order — a feature to baffle out in any real instrument.

Step 4. Chromaticity check: since $f=r_1^2/\lambda$ , blue light ( $\lambda=450$ nm) focuses at $0.556$ m, $56$ mm farther than green — dispersion opposite in sign to glass and far stronger. $f_1=0.5$ m, $r_{10}=1.58$ mm, $f_3=0.167$ m.

2 · Resolution of a soft-X-ray zone plate (exam level)

A soft-X-ray microscope uses a zone plate whose outermost (finest) zone has width $\Delta r_N=50$ nm, operating in the water window at $\lambda=2.4$ nm. Estimate its diffraction-limited resolution and its numerical aperture.

Step 1. The Rayleigh-limited spot size is set by the finest zone: $\delta\approx1.22\,\Delta r_N$ , borrowing the $1.22$ Airy constant carried across from the Rayleigh Resolution Criterion. Here $\delta=1.22\times50\ \text{nm}=\mathbf{61\ nm}$ — a resolution no glass lens reaches, on radiation no glass lens can even focus.

Step 2. Numerical aperture from $\text{NA}=\lambda/(2\Delta r_N)$ : $\text{NA}=2.4\ \text{nm}/(2\times50\ \text{nm})=\mathbf{0.024}$ . Tiny by visible-microscope standards, yet with a $2.4$ nm wavelength it still resolves tens of nanometres, because resolution scales as $\lambda/\text{NA}$ .

Step 3. Such a plate might carry $N\sim500$ zones, so it images sharply only over $\Delta\lambda/\lambda\lesssim1/N\approx0.002$ — doubling as a monochromator at a broadband synchrotron. At X-ray wavelengths every material has index within $10^{-3}$ of unity and absorbs fiercely, so diffraction is the only way to focus, and the zone plate — pushed to sub- $20$ nm outer zones — is the defining optic of soft-X-ray microscopy. $\delta\approx61$ nm, NA $=0.024$ .

Physics of the portrait

The portrait is not a person but an idea: the half-period zone, the decomposition that sums an infinity of Huygens wavelets by pairing them off. Its hardware embodiment, the zone plate, has outlived nearly everything else in classical diffraction because it goes where lenses cannot.

Origins · 1 minfresnel zones and how this was discovered

fresnel zones

half-period annuli that turn a wavefront into a phasor sum · physical optics — Fresnel diffraction

The zone construction was Augustin-Jean Fresnel's device, introduced in his 1818 prize memoir to the French Academy: partition the wavefront into rings differing by $\lambda/2$ in path, and the Huygens–Fresnel sum becomes a manageable alternating series. Its vindication came at once — Poisson, hostile to the wave theory, deduced from Fresnel's own equations that a bright spot must sit at the centre of a circular shadow, an "absurdity" Arago promptly observed, turning a would-be refutation into a triumph. Half a century on the idea became a device: in 1875 the Swiss physicist Jacques-Louis Soret built the first zone plate, a screen of alternate opaque rings that focuses purely by diffraction, and Lord Rayleigh soon proposed the far more efficient phase-reversal plate. In the 1970s the Göttingen group of Günter Schmahl and Dietmar Rudolph fabricated zone plates fine enough to focus soft X-rays, founding modern X-ray microscopy. Today's plates, with outer zones a few nanometres wide, remain the sharpest lenses in the extreme ultraviolet — a two-century-old trick still setting the resolution record where refraction fails.

History

1818

— Fresnel's prize memoir introduces the half-period zone construction of the wavefront

1818

— Poisson deduces, and Arago observes, the bright spot at the centre of a circular shadow

1875

— Soret builds the first zone plate, focusing light by blocking alternate zones

1888

— Rayleigh proposes the phase-reversal (phase) zone plate, quadrupling the efficiency

1974

— Schmahl and Rudolph fabricate high-resolution X-ray zone plates for microscopy

Going further

Texts. Hecht, Optics, 5th ed., §10.3 (Fresnel diffraction; zones and zone plates) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 4 (Fresnel diffraction) and Ch. 24 (diffractive optics) · Born & Wolf, Principles of Optics, 7th ed., §8.2 (Fresnel zones) · Attwood, Soft X-Rays and Extreme Ultraviolet Radiation, Ch. 9 (zone-plate optics) · Demtröder, Laser Spectroscopy, Vol. 1 (diffractive focusing elements). The key results are: (1) $r_m=\sqrt{m\lambda f}$ from the binomial expansion of the half-wave path condition; (2) equal zone areas $\Delta S_m=\pi\lambda f$ ; (3) the alternating-series result $A\approx A_1/2$ giving $I\approx I_1/4$ ; (4) higher-order foci $f_p=f_1/(2p+1)$ ; (5) the Fourier-decomposition efficiencies $\eta_1=1/\pi^2\approx10.1\%$ (amplitude), $4/\pi^2\approx40.5\%$ (phase); (6) resolution $\delta\approx1.22\,\Delta r_N$ and bandwidth $\Delta\lambda/\lambda\lesssim1/N$ . The worked example $f_1=0.5$ m follows directly from the given $r_1$ and $\lambda$ .

Prerequisites. Huygens–Fresnel Principle · Fresnel Diffraction Regime · Fresnel–Kirchhoff Diffraction Integral · Cornu Spiral and Fresnel Integrals. Leads to. Circular Aperture Diffraction and the Airy Pattern · Fresnel Number · Rayleigh Resolution Criterion.

Why does the whole open wavefront give only half the light of its first zone?

Because the zones alternate in sign and fade slowly, so they cancel in pairs. Group the series as $A_1/2+(A_1/2-A_2+A_3/2)+\cdots$ ; each bracket nearly vanishes for a smooth sequence, leaving the leading half-term $A_1/2$ . In the vibration spiral the running phasor winds into a centre, and the resultant is the radius to that centre — half the diameter, which is the first zone alone. The corollary is real and measurable: mask a beam to just the first zone and the on-axis intensity rises to $4I_0$ , because you have removed the zones that were cancelling the first.

Why is a zone plate so much more chromatic than a glass lens?

Because its focusing is entirely diffractive. The zone radii are fixed once the plate is etched, and the focus obeys $f=r_1^2/\lambda$ , so $f\propto1/\lambda$ : blue focuses farther than red, opposite in sense to glass and several times stronger. A glass singlet disperses only through the gentle variation of its index; a zone plate disperses through the wavelength sitting in the focal-length formula itself. The consequence is the bandwidth rule $\Delta\lambda/\lambda\lesssim1/N$ : a plate with $N$ zones images sharply over only about $1/N$ fractional bandwidth, which is why zone plates need quasi-monochromatic light and double as spectral filters.

If only ~10% of the light reaches the focus, where does the other 90% go?

For a binary amplitude plate the budget is stark: half the power is absorbed by the opaque zones, one quarter passes straight through undiffracted as a zeroth-order beam, and the first-order focus takes $1/\pi^2\approx10\%$ . The rest scatters into the weaker higher orders ( $\pm3,\pm5,\dots$ ) and the diverging virtual foci. This is why real instruments add an order-sorting aperture and a central beam stop to reject the zeroth order and unwanted foci. A phase plate erases the absorbed half and reclaims the antiphase zones, lifting the focus to $4/\pi^2\approx40\%$ ; a blazed kinoform approaches full throughput.

Why does blocking alternate zones make a bright focus instead of just a darker screen?

Intuition says covering half the aperture can only remove light — but the zones you cover were the ones subtracting from the sum. In the open wave the even zones arrive in antiphase and cancel most of the odd zones' contribution, leaving only $A_1/2$ . Delete the even zones and the survivors $A_1+A_3+A_5+\cdots$ all point the same way, adding to a resultant far larger than $A_1/2$ — an amplitude that grows with the number of open zones. You lose half the area but convert destructive interference into constructive, and the net on-axis field jumps. It is focusing by removing the cancellers, the same logic as the Poisson spot run in reverse.

What is the Poisson–Arago spot, and how do zones explain it?

Put an opaque disc in a beam and, at the centre of its geometric shadow, there is a bright point. Zones make it obvious: the disc blocks the innermost zones seen from the axial point, but the outer zones are untouched and sum — by the same alternating-series argument — to a finite on-axis amplitude, roughly that of the first unblocked zone. Poisson derived this in 1818 from Fresnel's equations expecting an absurdity; Arago observed it, and it became decisive evidence for the wave theory. The spot is the exact complement of a zone-plate focus: a disc keeps the outer zones, a zone plate the alternate ones.

Why are zone plates used for X-rays but never for ordinary cameras?

Two reasons, both from the physics above. At visible wavelengths glass lenses are cheap, achromatic, and nearly $100\%$ efficient, whereas a zone plate is $\sim10\%$ efficient, wildly chromatic, and riddled with higher-order ghosts — no contest. But in the soft-X-ray and extreme-ultraviolet, every material has an index within $\sim10^{-3}$ of unity, so refraction is far too feeble to bend rays and absorption is severe: a conventional lens is impossible. Diffraction still works, and a zone plate with nanometre outer zones focuses X-rays where nothing else can — its poor efficiency and narrow bandwidth accepted as the price of having any lens at all. The tool is chosen by where refraction fails, not where it shines.