Fresnel Diffraction Regime

Fresnel diffraction is what light does in the awkward middle distance — too far from an aperture for the sharp-edged shadow of ray optics, too near for the clean angular pattern of the far field. Push a screen up close behind a lit hole and you get a bright patch with the aperture's shape and a fringe of ripples on its rim. Pull the screen back and those ripples march inward, the patch fills with concentric rings that pulse bright and dark, and the on-axis point flickers between brilliant and black as you walk backward. Only much farther out does the churning settle into the smooth, fixed lobe of Fraunhofer diffraction. Everything in that restless near-field zone is Fresnel diffraction. The restlessness comes from one retained term. Every point of the aperture re-radiates a spherical wavelet — the Huygens–Fresnel picture — and the field at your eye is their coherent sum. Rayleigh–Sommerfeld theory (recalled by name as the rigorous parent) gives that sum exactly; Fresnel diffraction is its paraxial reduction, keeping the quadratic curvature of each wavelet's path but dropping higher orders. That quadratic phase depends on the screen distance $z$ , so the pattern is dynamic: move the screen, re-weight the wavelets, and the interference reshuffles. Drop the phase entirely and you fall into the far field. The controlling number is the Fresnel number $N_F=a^2/\lambda z$ , counting the half-period zones your eye sees — many ( $N_F\gg1$ ) is the fuzzy shadow, one-ish ( $N_F\sim1$ ) the richest structure, a fraction ( $N_F\ll1$ ) the far field.

Key takeaways

Fresnel (near-field) diffraction keeps the quadratic transverse phase in the Huygens sum, so the propagated field is the aperture field convolved with a paraxial spherical-wave kernel.
Unlike Fraunhofer diffraction, a Fresnel pattern changes shape as the screen moves — the retained phase depends on the distance $z$ , so the pattern is dynamic, not merely rescaled.
The regime dial is the Fresnel number $N_F=a^2/\lambda z$ : $N_F\gg1$ is the geometric shadow with edge fringes, $N_F\sim1$ is full Fresnel structure, $N_F\ll1$ collapses to Fraunhofer.
On the axis behind a circular hole the intensity is $I=4I_0\sin^2(\pi N_F/2)$ — bright ( $4I_0$ ) at odd $N_F$ , dark at even — the on-axis cousin of the Poisson–Arago spot.
Free propagation is a pure phase filter on the angular spectrum, $H=e^{ikz}\exp[-i\pi\lambda z(f_x^2+f_y^2)]$ — the engine of numerical beam propagation and digital holography.
Key idea: retain the aperture's own quadratic phase and light propagates by convolution with a paraxial spherical wave, making the near-field pattern depend on distance through $N_F=a^2/\lambda z$ ; drop that phase ( $\pi N_F\ll1$ ) and Fresnel diffraction reduces to the Fraunhofer Fourier transform.

The core insight

Fresnel diffraction is the paraxial near field: expand the exact Rayleigh–Sommerfeld path length to quadratic order, keep that phase in the fast exponential, and the diffracted field becomes the aperture field convolved with the Fresnel propagator $h=\frac{e^{ikz}}{i\lambda z}\exp[ik\rho^2/2z]$ . Because the retained phase carries the distance $z$ , the pattern evolves as the screen moves — from geometric shadow, through zone-structured Fresnel patterns, to the far field. The dial is $N_F=a^2/\lambda z$ ; when $\pi N_F\ll1$ the aperture's curvature is negligible and the integral becomes the Fraunhofer transform.

Where this lives today

Fresnel diffraction is the working regime of computational wave optics, because most real optical distances are near-field. The transfer function $H=e^{ikz}\exp[-i\pi\lambda z(f_x^2+f_y^2)]$ is the kernel of the angular-spectrum and split-step methods that simulate laser cavities, design computer-generated holograms and diffractive optical elements, and reconstruct digital holograms — each an FFT, a phase multiply, an inverse FFT. Lensless in-line holographic microscopes record a Fresnel pattern and numerically back-propagate it to recover the object, trading a lens for a computer. The same physics runs at the short-wavelength frontier. Propagation-based X-ray phase-contrast imaging places the detector in the Fresnel regime so that edge fringes encode the phase a weakly absorbing sample imprints, turning soft-tissue boundaries invisible to absorption into sharp bright-dark rims. In EUV lithography and proximity X-ray masks the mask-to-wafer gap sits at $N_F$ of order tens, so the printed image is a Fresnel pattern, not a geometric shadow.

The physical picture

Slice the wavefront at the aperture into a mosaic of point emitters. Each launches a spherical wavelet, and at your eye they arrive with path lengths Fresnel's approximation computes to quadratic order in the transverse offset. Because those path differences depend on where the screen sits, the sum is a moving target. Near the aperture the wavelets are still nearly in step and the field is a faithful ray-optics projection with a fringe where the balance first breaks; farther out, more of the aperture's own curvature enters the phase, the interior fills with rings, and the whole pattern breathes with distance.

Figure

The same circular aperture at three distances: near (N_F ≈ 10) a sharp shadow with edge fringes, mid (N_F ≈ 1) full Fresnel rings, far (N_F ≈ 0.2) a single smooth lobe — the pattern reshapes as z grows

The Fresnel number counts the half-period zones the aperture subtends: many zones is a busy, shadow-like near field; a fraction of a zone is the smooth far field. Nowhere is the dynamic character sharper than on the axis behind a circular hole, where the exposed zones add as alternating-sign phasors and the intensity swings between four times the unobstructed value and total darkness as the screen moves. The next figure is that oscillation, with the real numbers on it.

Figure

The punchline: on-axis intensity behind a circular aperture, I/I₀ = 4 sin²(πN_F/2) — bright 4I₀ at odd N_F, dark at even N_F, and the worked case N_F = 0.25 giving 0.586

Read the curve as zones. Each unit of $N_F$ exposes one more half-period zone, contributing a field roughly equal in size but opposite in phase to the last. An odd number of open zones leaves a net single zone and the axis blazes at $4I_0$ ; an even number cancels in pairs and it goes dark. Between the integers, at $N_F=0.25$ , only a quarter-zone is open and $I/I_0=0.586$ — the far field has barely begun to assemble.

The core equations

The starting point is the first Rayleigh–Sommerfeld solution — the exact Huygens sum with obliquity $\cos\theta$ — recalled by name from Rayleigh–Sommerfeld Diffraction. Fresnel diffraction is what that integral becomes many wavelengths downstream at modest angle: expand the path length to quadratic order, keep that term in the fast phase, freeze the slow amplitude. The result is a convolution of the aperture field with a paraxial spherical-wave kernel.

U(x,y,z)=\frac{e^{ikz}}{i\lambda z}\iint U_0(\xi,\eta)\,\exp\!\left[\frac{ik}{2z}\big((x-\xi)^2+(y-\eta)^2\big)\right]\mathrm{d}\xi\,\mathrm{d}\eta \qquad (1)

where: $U(x,y,z)$ — field on the screen a distance $z$ downstream; $U_0(\xi,\eta)$ — aperture-plane field; $k=2\pi/\lambda$ — wavenumber; the kernel $\frac{e^{ikz}}{i\lambda z}\exp[ik\rho^2/2z]$ is the Fresnel propagator $h$ , so $(1)$ reads $U=U_0*h$ . The distance $z$ sits inside the exponent — exactly why the pattern is not self-similar and reshapes as the screen moves. Because $(1)$ is a convolution, it becomes a multiplication in the frequency domain. Fourier-transforming the propagator — equivalently, paraxially expanding the exact angular-spectrum kernel $k_z=\sqrt{k^2-4\pi^2(f_x^2+f_y^2)}$ — turns free propagation into a pure phase filter on the plane-wave components. This is the transfer-function view a computer actually uses.

H(f_x,f_y)=e^{ikz}\exp\!\left[-i\pi\lambda z\,(f_x^2+f_y^2)\right] \qquad (2)

where: $f_x,f_y$ — spatial frequencies (cycles per length) of the angular spectrum; $H$ — the multiplicative filter applied to $\tilde U_0(f_x,f_y)$ ; the quadratic phase dephases high spatial frequencies (fine detail) fastest, which is why sharp edges smear into fringes as $z$ grows. Propagate by $\tilde U(z)=H\,\tilde U_0$ : one FFT, one multiply, one inverse FFT. The regime is fixed by one dimensionless group. Comparing the aperture half-width $a$ to the transverse scale $\sqrt{\lambda z}$ — over which the spherical-wave phase accrues about a radian — and squaring gives the Fresnel number, which also counts the half-period zones the aperture subtends and sets their radii.

N_F=\frac{a^2}{\lambda z},\qquad r_m=\sqrt{m\,\lambda z}\quad(m=1,2,3,\dots) \qquad (3)

where: $a$ — aperture radius (or half-width); $N_F$ — Fresnel number, the number of zones seen from the axial point; $r_m$ — outer radius of the $m$ -th half-period zone, over which the path grows by $\lambda/2$ . Setting $r_m=a$ gives $m=N_F$ : the zone count is the Fresnel number. Evaluating $(1)$ on the axis behind a uniformly lit circular hole gives the most quotable near-field result — the oscillation plotted above — from the alternating-phasor sum of the exposed zones, derived in full below.

I_{\text{axis}}(z)=4\,I_0\sin^2\!\left(\frac{\pi N_F}{2}\right) \qquad (4)

where: $I_0$ — unobstructed incident intensity; $I_{\text{axis}}$ — on-axis intensity at distance $z$ . Odd $N_F$ (a net single open zone) gives the maximum $4I_0$ ; even $N_F$ (zones cancelling in pairs) gives darkness — the on-axis relative of the Poisson–Arago spot behind an opaque disc.

Derivation AFrom Rayleigh–Sommerfeld to the Fresnel propagator

The aim is to show that the Fresnel integral is not a new physical law but the paraxial shadow of the exact Rayleigh–Sommerfeld result: keeping just the quadratic term of the path length converts the rigorous Huygens sum into a convolution with a spherical-wave kernel. The point is to expose precisely which terms are kept in the fast phase and which are frozen in the slow amplitude — the single asymmetric choice that defines the regime.

Step 1. Start from RS-I, $U(P)=\frac{1}{i\lambda}\iint_\Sigma U_0\,\frac{e^{ikr}}{r}\cos\theta\,\mathrm{d}\Sigma$ , with $r=\sqrt{z^2+(x-\xi)^2+(y-\eta)^2}$ . Why: RS-I is the exact scalar propagator recalled from the parent article, so anchoring here keeps the approximation honest — we know exactly what we drop.

Step 2. Expand the path length for transverse separations small against $z$ : $r\approx z+\frac{(x-\xi)^2+(y-\eta)^2}{2z}$ . Why: the binomial expansion keeps the leading quadratic term — the one that bends light — with the neglected quartic term negligible when $z^3\gg\frac{\pi}{4\lambda}[(x-\xi)^2+(y-\eta)^2]^2_{\max}$ .

Step 3. Treat phase and amplitude asymmetrically: put the expanded $r$ into the fast exponential $e^{ikr}$ (where the quadratic term multiplies the large $k$ ), but set $r\approx z$ in the slow amplitude $1/r$ and $\cos\theta\approx1$ . Why: over a one-wavelength screen shift the phase turns a full cycle while $1/r$ barely changes, so only the phase needs the correction — the defining Fresnel approximation.

Step 4. Collect: $U(x,y,z)=\frac{e^{ikz}}{i\lambda z}\iint U_0\,e^{\frac{ik}{2z}[(x-\xi)^2+(y-\eta)^2]}\mathrm{d}\xi\,\mathrm{d}\eta$ , a convolution $U_0*h$ with $h=\frac{e^{ikz}}{i\lambda z}e^{ik\rho^2/2z}$ ; its Fourier transform is $H=e^{ikz}\exp[-i\pi\lambda z(f_x^2+f_y^2)]$ . Why: recognising a convolution makes propagation a shift-invariant filter — and hands the calculation to the FFT. *The paraxial expansion of $r$ and the propagator–transfer-function Fourier pair follow from the $k_z\approx k-\pi\lambda(f_x^2+f_y^2)$ expansion of the exact angular-spectrum kernel.*

Go deeper · Derivation B · 1 minFresnel to Fraunhofer: when the aperture's own curvature disappears

Derivation BFresnel to Fraunhofer: when the aperture's own curvature disappears

This is the article's owned structural result: an exact criterion for when Fresnel diffraction collapses into the far field. The aim is to isolate the one term carrying the aperture's own curvature and show its largest value across the opening is precisely $\pi N_F$ — so that "far field" is not a vague distance but the sharp statement $\pi N_F\ll1$ , at which the integral becomes a clean Fourier transform of the aperture.

Step 1. Expand the bracket into three groups: $(x-\xi)^2+(y-\eta)^2=(x^2+y^2)-2(x\xi+y\eta)+(\xi^2+\eta^2)$ — a pure observation term, a cross (Fourier) term, and a pure aperture term. Why: only the last involves the aperture coordinates quadratically, and it is the sole term distinguishing Fresnel from Fraunhofer — the other two are already a Fourier kernel times an outer phase.

Step 2. Take the largest value of the aperture phase $\phi_{\text{ap}}=\frac{k}{2z}(\xi^2+\eta^2)$ across the opening, where $\xi^2+\eta^2=a^2$ : $\phi_{\text{ap}}^{\max}=\frac{ka^2}{2z}=\frac{\pi a^2}{\lambda z}=\pi N_F$ . Why: this reduces a messy phase to one clean number — the maximum wavefront-curvature phase in units of $\pi$ — which is exactly the Fresnel number scaled by $\pi$ .

Step 3. Impose $\pi N_F\ll1$ . Then $e^{i\phi_{\text{ap}}}\approx1$ and drops from the integrand, leaving $U\approx\frac{e^{ikz}e^{ik(x^2+y^2)/2z}}{i\lambda z}\iint U_0\,e^{-\frac{ik}{z}(x\xi+y\eta)}\mathrm{d}\xi\,\mathrm{d}\eta$ — a 2-D Fourier transform of the aperture at $f_x=x/\lambda z$ , $f_y=y/\lambda z$ . Why: with the self-curvature gone the aperture looks flat, and diffraction reduces to reading off its spatial-frequency content — the Fraunhofer regime handed to the next article.

Step 4. The residual observation-plane phase $e^{ik(x^2+y^2)/2z}$ has unit modulus, so it never touches the intensity — which is why the Fraunhofer pattern only rescales with $z$ (through $f_x=x/\lambda z$ ) while the Fresnel pattern reshapes. Why: a phase with $|e^{i\phi}|=1$ is invisible to a detector measuring $|U|^2$ — the reason the far-field shape is stationary. *The edge-phase identity $\frac{ka^2}{2z}=\pi N_F$ holds symbolically; at $N_F=0.25$ it gives $\pi N_F=0.785$ rad.*

Go deeper · Derivation C · 1 minOn-axis intensity behind a circular aperture

Derivation COn-axis intensity behind a circular aperture

The oscillating on-axis curve — bright at odd $N_F$ , dark at even — is the near field's signature, and it falls out of the Fresnel integral by an elementary integration on the axis. The aim is to evaluate $(1)$ at $x=y=0$ for a uniformly lit disc of radius $a$ , reduce the radial integral to a single complex exponential, and read off the $4I_0$ ceiling and the integer- $N_F$ bright/dark rule. The zone-counting behind the name is developed in Fresnel Number.

Step 1. On the axis there is no linear (cross) phase, only the quadratic term, so $U_{\text{axis}}\propto\int_0^{a}e^{ik\rho^2/2z}\,2\pi\rho\,\mathrm{d}\rho$ with $\rho^2=\xi^2+\eta^2$ . Why: setting $x=y=0$ kills the Fourier kernel — the axial point is the one place the 2-D integral becomes one-dimensional.

Step 2. Substitute $u=\rho^2$ : $U_{\text{axis}}\propto\pi\int_0^{a^2}e^{iku/2z}\mathrm{d}u=\pi\cdot\frac{2z}{ik}\big(e^{ika^2/2z}-1\big)$ . Why: the substitution turns the radial integral into a trivial exponential one — why the on-axis case is clean while off-axis needs the Cornu spiral.

Step 3. Insert $ka^2/2z=\pi N_F$ and factor $e^{i\pi N_F}-1=2i\,e^{i\pi N_F/2}\sin(\pi N_F/2)$ , so $|U_{\text{axis}}|\propto2\,|\sin(\pi N_F/2)|$ . Normalising to the unobstructed $I_0$ (amplitude equal to half a single zone) gives $I_{\text{axis}}=4I_0\sin^2(\pi N_F/2)$ . Why: that identity produces the $\sin^2$ oscillation, and the factor of two between one open zone and the full wave lifts the ceiling to $4I_0$ .

Step 4. Read the integers: odd $N_F$ gives $\sin^2(\pi N_F/2)=1$ (net one zone, bright $4I_0$ ); even $N_F$ gives $0$ (pairs cancel, dark). Why: each unit of $N_F$ flips the newest zone's phasor, so parity of the zone count decides constructive versus destructive interference. *As a check: $4\sin^2(\pi N_F/2)=4$ at $N_F=1$ , $=0$ at $N_F=2$ , and $=0.586$ at $N_F=0.25$ .*

Limiting cases

Case	Result	Meaning
Geometric-shadow limit	$N_F\gg1$	aperture spans many zones; the pattern is the sharp geometric shadow decorated with edge fringes — ray optics is the leading behaviour
Fresnel regime	$N_F\sim1$	one to a few zones exposed; the full quadratic-phase integral is needed — Cornu-spiral edge fringes, on-axis oscillations, zone-plate focusing
Fraunhofer limit	$N_F\ll1\;(z\gg a^2/\lambda)$	aperture self-curvature phase $\to1$ ; the field becomes the Fourier transform of the aperture and the pattern shape stops changing (only rescales)
On-axis, odd $N_F$	$I_{\text{axis}}=4I_0$	a net single open zone — four times the unobstructed intensity (the on-axis Poisson–Arago maximum)
On-axis, even $N_F$	$I_{\text{axis}}=0$	exposed zones cancel in pairs — a dark point on the axis
Paraxial validity	$z^3\gg\frac{\pi}{4\lambda}[(x-\xi)^2+(y-\eta)^2]^2_{\max}$	the quartic path term is negligible, so the parabolic expansion of $r$ is accurate

Common trapThe "near field" of Fresnel diffraction is the same as the sub-wavelength evanescent near field

These are two unrelated meanings of "near field" that share a name. Fresnel diffraction lives in the paraxial near field — the region $N_F=a^2/\lambda z\gtrsim1$ where wavefront curvature across the aperture matters — which for visible light through a millimetre aperture extends from centimetres to a couple of metres. The evanescent near field of sub-wavelength optics is something else: it sits within roughly a wavelength of a structure smaller than $\lambda$ , built from non-propagating components with imaginary $k_z$ that decay exponentially and never reach any screen. Fresnel diffraction is a fully propagating, paraxial phenomenon and knows nothing of evanescent waves. Reaching for evanescent intuition to explain Fresnel fringes, or vice versa, mixes physics separated by orders of magnitude in length scale.

Worked examples

1 · A millimetre aperture and where its far field begins (confidence builder)

A circular aperture of radius $a=1$ mm is illuminated by $\lambda=500$ nm light. Find the natural far-field scale distance $a^2/\lambda$ , evaluate the Fresnel number at a screen $z=8$ m away, and compute the on-axis intensity there relative to the unobstructed beam.

Step 1. The far-field scale distance is $a^2/\lambda=(10^{-3})^2/(5\times10^{-7})=\mathbf{2.0\ m}$ . Genuine Fraunhofer diffraction demands $z\gg2$ m, which is why a lab brings the far field into a lens's focal plane rather than walking metres downstream.

Step 2. At $z=8$ m, $N_F=a^2/\lambda z=2.0/8=\mathbf{0.25}$ . The maximum wavefront-curvature phase is $\pi N_F=0.785$ rad — not yet $\ll1$ , so you are on the threshold of Fraunhofer, not deep inside it.

Step 3. The on-axis intensity is $I_{\text{axis}}/I_0=4\sin^2(\pi N_F/2)=4\sin^2(0.3927)=4\,(0.3827)^2=\mathbf{0.586}$ — dimmer than the unobstructed beam because only a quarter-zone is open, so the diffracted field has not yet reassembled the full wave.

Step 4. At $8$ m a millimetre hole is still faintly Fresnel; only for $z\gtrsim10\,a^2/\lambda\approx20$ m does the pattern become a clean Airy transform, which is why far-field measurements on small apertures almost always use a focusing lens. $a^2/\lambda=2.0$ m; $N_F=0.25$ and $I_{\text{axis}}/I_0=0.586$ at $8$ m.

2 · The axial point flickering bright and dark (exam level)

The same $a=1$ mm aperture, $\lambda=500$ nm. Starting far away and pulling the screen inward, at what distances does the on-axis point first go bright, then dark, then bright again? Confirm the zone radii are consistent.

Step 1. Integer Fresnel numbers mark the extrema: $N_F=a^2/\lambda z=n$ gives $z=a^2/(n\lambda)=2.0/n$ m. So $N_F=1$ at $z=\mathbf{2.0\ m}$ , $N_F=2$ at $z=\mathbf{1.0\ m}$ , $N_F=3$ at $z\approx\mathbf{0.667\ m}$ .

Step 2. Apply the parity rule $I=4I_0\sin^2(\pi N_F/2)$ : at $z=2.0$ m ( $N_F=1$ , odd) the axis is bright, $4I_0$ ; at $z=1.0$ m ( $N_F=2$ , even) it is dark, $0$ ; at $z=0.667$ m ( $N_F=3$ , odd) it is bright, $4I_0$ again. Pulling the screen in cycles the axis bright–dark–bright.

Step 3. Cross-check with the zone radii $r_m=\sqrt{m\lambda z}$ . At $z=2.0$ m the first zone has $r_1=\sqrt{500\times10^{-9}\cdot2.0}=\sqrt{10^{-6}}=\mathbf{1\ mm}=a$ : exactly one zone fills the aperture, so $N_F=1$ . At $z=1.0$ m, $r_1=0.707$ mm, so the $1$ mm aperture holds two zones ( $N_F=2$ ) — consistent with the dark axial point.

Step 4. This flicker with distance is the cleanest laboratory signature of the near field — exactly what a Fraunhofer pattern cannot do, since its axis brightness is fixed and merely rescales. Watching the axis breathe as you move the screen is watching $N_F$ tick through the integers. Bright at $2.0$ m and $0.667$ m ( $N_F=1,3$ ); dark at $1.0$ m ( $N_F=2$ ).

Physics of the portrait

The portrait here is not a person but a regime — the near-field zone of wave propagation — carrying the name of Augustin-Jean Fresnel, the engineer whose zone construction first made it calculable.

Origins · 1 minfresnel diffraction and how this was discovered

fresnel diffraction

the near-field regime where wavefront curvature across the aperture cannot be dropped · physical optics

Fresnel diffraction names the propagation regime in which the quadratic curvature of the Huygens wavelets — the $\frac{k}{2z}(\xi^2+\eta^2)$ phase across the aperture — must be retained, so the pattern depends on the observation distance and evolves as the screen moves. It is bracketed by geometrical optics ( $N_F\gg1$ , the shadow) and Fraunhofer diffraction ( $N_F\ll1$ , the far-field transform), and it is the home of edge fringes, the Cornu spiral, zone plates, and the Poisson–Arago spot. The name honours Augustin-Jean Fresnel (1788–1827), a French civil engineer who, working in his spare time and in poor health, married Huygens' wavelets to Young's interference to build the first quantitative wave theory of diffraction. His 1818 prize memoir divided the wavefront into the half-period Fresnel zones that still bear his name — the very construction behind the on-axis $4I_0\sin^2(\pi N_F/2)$ oscillation. When the committee's Siméon Poisson used Fresnel's integrals to deduce the "absurd" bright spot at the centre of a disc's shadow, François Arago observed it, and the theory was vindicated on the spot. Fresnel also invented the stepped lighthouse lens before dying of tuberculosis at thirty-nine.

History

1818

— Fresnel submits his prize memoir on diffraction to the French Academy, building the wavefront from half-period zones (prize awarded 1819)

1818

— Poisson deduces the bright axial spot behind a disc from Fresnel's integrals; Arago observes it, confirming the wave theory

1882

— Kirchhoff places the Huygens–Fresnel construction on a rigorous integral footing, obliquity factor and all

1968

— Goodman's Introduction to Fourier Optics systematizes the transfer-function view and numerical Fresnel propagation

Going further

Texts. Hecht, Optics, 5th ed., ch. 10 (Fresnel diffraction, Cornu spiral, zones) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 4 (Fresnel approximation, transfer function) · Goodman, Introduction to Fourier Optics, 4th ed., ch. 4 (Fresnel and Fraunhofer diffraction) · Born & Wolf, Principles of Optics, 7th ed., ch. 8. The key results are: the paraxial expansion of $r$ and the resulting convolution kernel; the transfer function $H=e^{ikz}\exp[-i\pi\lambda z(f_x^2+f_y^2)]$ as the Fourier transform of $h$ ; the Fresnel-number criterion $N_F=a^2/(\lambda z)$ and the Fraunhofer onset $N_F\ll1$ ; the on-axis result $I=4I_0\sin^2(\pi N_F/2)$ and zone radii $r_m=\sqrt{m\lambda z}$ ; and the numerical estimate $a^2/\lambda=2$ m for $a=1$ mm, $\lambda=500$ nm.

Prerequisites. Huygens–Fresnel Principle · Fresnel–Kirchhoff Diffraction Integral · Rayleigh–Sommerfeld Diffraction. Leads to. Fraunhofer Diffraction Regime · Fresnel Number · Cornu Spiral and Fresnel Integrals · Fresnel Zones and Zone Plates.

Why keep the quadratic term in the phase but not in the amplitude 1/r?

Because the two factors respond to the small correction on wildly different scales. In the fast phase $e^{ikr}$ the correction $\frac{k}{2z}\rho^2$ is multiplied by the large wavenumber $k=2\pi/\lambda$ , so a transverse shift of one wavelength turns the phasor a full cycle — the term controls the interference. In the slow amplitude $1/r$ the same correction changes the magnitude by a part in millions and is invisible. Keeping it only in the phase is the defining, justified Fresnel approximation; mixing the two gives an inconsistent, wrong result.

What is the difference between the Fresnel near field and the evanescent near field?

They are unrelated regimes that share the word "near." The Fresnel near field is a paraxial, propagating zone: $N_F\gtrsim1$ , distances of centimetres to metres for visible light and millimetre apertures, where wavefront curvature across the aperture reshapes the pattern with distance. The evanescent near field is sub-wavelength: within about a wavelength of a structure smaller than $\lambda$ , built from non-propagating components (imaginary $k_z$ ) that decay exponentially and carry detail no screen ever sees. Fresnel diffraction is fully propagating and contains no evanescent waves. Same word, physics separated by orders of magnitude.

Why is the on-axis point bright behind a disc but oscillating behind a hole?

Both are the same zone bookkeeping read from opposite sides. Behind an opaque disc the central zones are blocked and the surviving outer zones — all of nearly equal phase on the axis — sum to a bright point: the Poisson–Arago spot, nearly independent of distance. Behind an open hole the zones are all present up to the edge, and as you move the screen the number exposed, $N_F$ , ticks through the integers, so the alternating-phasor sum swings between one net zone (bright $4I_0$ ) and cancelling pairs (dark). The disc fixes which zones survive; the hole lets their count vary with $z$ — which is why one is steady and the other flickers.

Is "Fresnel diffraction" related to the "Fresnel equations" for reflection?

No — they share only the name of the same person. The Fresnel equations give reflection and transmission amplitudes at a plane interface — an electromagnetism boundary-value problem. Fresnel diffraction is about how a wave spreads after passing an aperture in free space, with no interface at all. If a problem mentions two media and an angle of incidence, it wants the equations; if it mentions an aperture and a screen distance, it wants diffraction.

How is the frequency-domain (transfer-function) view the same as the convolution?

They are Fourier transforms of each other, by the convolution theorem. In real space, propagation smears the aperture field with the kernel $h$ : $U=U_0*h$ . Transform to frequency space and the convolution becomes a product $\tilde U=\tilde U_0\cdot H$ , where $H$ is the Fourier transform of $h$ . Both give identical fields within the paraxial approximation; you pick whichever is easier. Edge and slit problems are natural in real space (Cornu-spiral chords); anything you run on a computer is natural in frequency space, where propagation is one multiply between two FFTs.

When do I actually need Fresnel rather than the simpler Fraunhofer formula?

Compute $N_F=a^2/\lambda z$ first. If $N_F\ll1$ ( $z\gg a^2/\lambda$ ), the self-curvature phase is negligible and the far-field transform — sinc for a slit, Airy for a hole — is accurate and much easier. If $N_F\gtrsim1$ , that phase matters: you need the full Fresnel integral, the Cornu spiral for edges, or numerical propagation. Note that a lens brings the Fraunhofer pattern to its focal plane regardless of distance, which is why lab "far-field" measurements are usually done at a focus even when the raw geometry is deep in the Fresnel regime.