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A Fresnel number compares squared aperture size with the product of wavelength and propagation distance. For an optical slit of full width $D=2a$, explicitly distinguish the full-width and half-width conventions:The full-width convention is used throughout the Fresnel slit propagation example below. The definition $D^2/(\lambda z)$ also appears in the Delft treatment.^[ref]

For uniform incident wave phase, $N_D\ll 1$ makes the omitted aperture wave phase small and gives the Fraunhofer diffraction regime. When $N_D$ is of order one or larger, retain it and use Fresnel diffraction, provided the paraxial approximation still holds. These are continuous changes in accuracy, rather than a physical boundary at one selected number. For a rectangular aperture, check both transverse dimensions. Incident wavefront curvature contributes its own wave phase through $U_0$; the aperture-to-screen Fresnel number alone does not describe a spherical source's illumination.

The scaling is revealing: doubling the optical slit width at fixed wavelength and distance multiplies the Fresnel number by four. Maintaining the same Fresnel number then requires four times the distance. Two systems can consequently have the same dimensionless pattern despite very different physical lengths.

Fresnel diffraction retains the quadratic transverse propagation phase; Fraunhofer diffraction neglects it and leaves an aperture Fourier transform. Expanding the propagation distance gives $z+y^2/(2z)+\cdots$. The aperture phase variation is of order $\pi {\rho }^2/(\lambda z)$, so the crossover scale is $z\sim {\rho }^2/\lambda$, with far-field propagation requiring a much smaller Fresnel number than1.

Here $\lambda \simeq 0.0100\mathrm{m}$ and the full gap width gives ${\rho }^2/\lambda =25\mathrm{m}$, comparable with the16m range. More decisively, measured from the grazing line of sight to the far gap edge, the quadratic phase reaches $\pi (0.5{)}^2/(0.01\times 16)=4.91\mathrm{rad}$. Fresnel diffraction is required; conventions using a half-width change the quoted Fresnel number but not this phase test.

Define $F(w)=C(w)+iS(w)$ with the normalization in the paper. The grazing-edge gap spans $w=0$ toThe Cornu chord power normalization uses a received field proportional to $F(w_g)-F(0)$, while no fence gives $F(\infty )-F(-\infty )=1+i$. Numerical Fresnel integration gives $C(w_g)=0.3245$, $S(w_g)=0.4818$, and thereforeThe two labelled child answers specify the corresponding vectors.

For the largest contiguous Fresnel aperture, a freely chosen single contiguous opening must maximize the chord between two points of the Cornu spiral. Its longest chord joins $F(-w_{\ast })$ to $F(w_{\ast })$, where the first maximum radius satisfies $C(w_{\ast })\cos (\pi w_{\ast }^2/2)+S(w_{\ast })\sin (\pi w_{\ast }^2/2)=0$. This gives $w_{\ast }=1.2094$ andHere x is signed displacement along the fence from the grazing optical axis. The pointwise Fresnel gain versus integrated transmitted power distinction explains why an aperture can concentrate more power at one receiver than the fully open field through constructive interference; this is not an increase of integrated transmitted power.

The existing-opening constraint in Fresnel aperture optimization matters literally: the original gap already occupies $[0,0.5]\mathrm{m}$. The symmetric optimum would require filling its far end. If only further removal is permitted and the final opening must be connected, maximize the chord subject to $w_{-}\le 0$ and $w_{+}\ge 1.768$. The constrained optimum is approximatelyThese are different optimization assumptions; both are stated so the original gap is not silently replaced. The model treats the fence as an ideal opaque screen and ignores ground reflections or additional propagation paths. ^[ref]

Far-field transform. For transverse aperture scale $b$ and screen distance $z$, the quadratic phase across the aperture is negligible when $k{b}^2/(2z)\ll 1$, equivalently a small Fresnel number up to geometrical factors. Expanding the source-to-screen distance at transverse screen coordinate $X$ givesDropping the final aperture-dependent term and treating the prefactor as constant givesThus intensity is the modulus squared of the aperture Fourier transform. A lens focal plane supplies the same angular transform without a very distant screen. The infinitely extended vertical dimension is treated per unit vertical extent.

half-slit phase-reversal diffraction. Put $h=-1$ on the left half and $h=1$ on the right half. With $q=k\sin \theta$ and $s=qb/4$,ThereforeThe apparent singularity at zero angle is removable: $H\to 0$, so the centre is dark. Normalizing by the uncoated slit's on-axis intensity $I_0$ gives $I/I_0={\sin }^{4}s/s^2$. Noncentral zeros have $s=j\pi$, or $\sin \theta =2j\lambda /b$. The two largest lobes are symmetric, at the first nonzero root of $\tan s=2s$, $s\simeq \pm 1.166$, giving $\sin \theta \simeq \pm 0.742\lambda /b$ when those directions propagate. Smaller lobes lie farther out under a decaying envelope.

Doubled thickness. Neglecting absorption and interface reflection, the phase delay doubles from $\pi$ to $2\pi$. Both halves then transmit with the same phase, recovering the ordinary single-slit patternIt has a bright centre and first zeros at $\sin \theta =\pm \lambda /b$. The phase layer redistributes transmitted power rather than absorbing it in this idealization.

For a circular structure, cylindrical symmetry reduces the scalar Fresnel-Kirchhoff diffraction integral to concentric annuli. With plane-wave illumination, radius $a$ and distance $z$, the relevant Fresnel number is $N_F=a^2/(\lambda z)$. Successive Fresnel half-period zones have radii $r_j^2\simeq j\lambda z$: their areas are approximately equal while their phases alternate by $\pi$, causing neighbouring-zone cancellation.

For Fresnel diffraction by a circular aperture, the on-axis field follows directly from the radial integral,Odd zone counts reinforce on axis; even counts cancel in this ideal scalar model. An annular opening gives the difference of the corresponding two edge exponentials. A zone plate alternately blocks or phase-shifts zones so that otherwise cancelling contributions reinforce at a focus, with $r_j^2\simeq j\lambda f$ for its principal focal distance.

Babinet's principle gives $U_{\mathrm{disk}}=U_{\mathrm{free}}-U_{\mathrm{ap}}=U_{\mathrm{free}}{e}^{i\pi N_F}$ for Fresnel diffraction by an opaque disk. Thus its ideal on-axis intensity equals the unobstructed intensity, the Arago spot, even within the geometric shadow. This concerns a narrow axial feature, not the whole shadow. Finite source size, imperfect circularity, bandwidth and alignment reduce its visibility. When $N_F\ll 1$, the circular-aperture far-field pattern becomes the Airy disk intensity pattern $[2J_1(u)/u{]}^2$, with $u=ka\sin \theta$ and first minimum near $\theta =1.22\lambda /(2a)$.

Fresnel diffraction is interference during propagation when curvature of the optical path across an aperture matters. It provides a continuous account of fringes near geometric edges and their evolution into the far-field pattern. A screen that is many aperture widths away is not automatically in the Fraunhofer diffraction limit; wavelength also enters the comparison.

For plane-wave illumination of a transmission $h(\xi ,\eta )$ at $z=0$, the Huygens-Fresnel principle in the scalar paraxial approximation givesIt follows by expanding the propagation distance as $z+[(X-\xi {)}^2+(Y-\eta {)}^2]/(2z)$ while treating the amplitude prefactor as slowly varying. Small propagation angles justify dropping higher path terms. Finite coherent illumination, vector edge fields and very small apertures can require a more complete theory. The Fresnel number $N_F=a^2/(\lambda z)$, with a clearly declared aperture scale $a$, measures whether the aperture's quadratic phase is significant. When it is negligible, the remaining linear phase produces the aperture Fourier transform.^[ref]

Define the optical Fresnel integralsTheir limiting values are $C(\infty )=S(\infty )=1/2$, with odd continuation to negative arguments. The Cornu spiral plots $(C(u),S(u))$: the field contributed by a finite interval is proportional to the chord between its endpoints. Intensities require squared chord lengths; summing zone intensities would lose the cancellation mechanism.

For an opaque straight edge transmitting $\xi >0$ and all $\eta$, let $v=X\sqrt{2/(\lambda z)}$, so $X>0$ is the geometrically illuminated side. Integrating in the unrestricted transverse direction and using the half-line in the other giveshenceFar into the shadow, $I\to 0$; far into the illuminated region, $I\to I_{\mathrm{in}}$. At the geometric shadow boundary, $I(0)=I_{\mathrm{in}}/4$, not half the incident intensity: the field is halved there. On the illuminated side, alternating constructive and destructive contributions give bright and dark fringes whose contrast declines away from the edge. Their positions scale with $\sqrt{\lambda z}$ and are not equally spaced. A finite detector averages fine fringes and tends to recover the geometric-optics step. In a controlled experiment, translate a straight edge through a broad collimated beam, scan a small photodetector across the screen and normalize against an unobstructed reference; vary $z$ to test the square-root scaling. Source size, spectral bandwidth, edge roughness and detector aperture wash out contrast and must be distinguished from a failure of the propagation law.

For a rectangular aperture of full widths $a,b$, defineThe two transverse integrals separate:andA long slit leaves only one varying Fresnel factor. For a square centred on axis, $a=b$ and $w=(a/2)\sqrt{2/(\lambda z)}$ giveOpposite edge contributions overlap; changing aperture size or propagation distance changes their relative phase. Consequently a geometrically illuminated centre can brighten or darken, rather than remaining a fixed top-hat plateau. The original rectangular-aperture map shows fringes parallel to both edges and their crossing structure. Outside the geometric rectangle the wave also extends into the shadow.

At large Fresnel number an interior region approaches geometric illumination while edge fringes remain on a scale $\sqrt{\lambda z}$. At intermediate values the opposite-edge fringes interact across the whole aperture image. In the far field, the formula tends to the product of sinc-squared factors,The far-field lobe scale $\lambda z/a$ grows linearly with $z$, unlike the local near-edge scale. The transition is controlled by the neglected phase, not an abrupt physical boundary.

To test the main effect, record normalized profiles at several distances, calculate the Fresnel integral with the measured aperture dimensions and compare fringe positions before fitting a free intensity scale. Use the same transverse coordinate origin and check that the incident beam is sufficiently uniform. Babinet's principle relates complementary aperture and obstruction fields through $U_a+U_o=U_{\mathrm{free}}$; their near-field intensities are not generally equal, because the unobstructed field remains present and interferes with the diffracted field. This is why a far-field obstruction argument cannot simply be carried over as equality of Fresnel intensities.

Circular apertures and circular obstructions illustrate both the usefulness and the limits of geometric shadow arguments. In scalar diffraction theory, each point of a transmitted wavefront contributes a complex secondary field. Their coherent sum, described by the Huygens-Fresnel principle, determines the intensity. Circular symmetry simplifies the integral, but the answer depends on whether propagation is in the Fraunhofer diffraction or Fresnel diffraction regime.

For aperture radius $a$ and axial propagation distance $z$, the Fresnel number $N_F=a^2/(\lambda z)$ measures the importance of quadratic phase across the aperture. The far-field condition is $N_F\ll 1$; a lens can produce the same angular spectrum in its back focal plane. Fresnel propagation retains that quadratic phase. For an axial point source a distance $s$ before the object and an observer a distance $r$ after it, the phase uses the effective distance $z_{\mathrm{eff}}=sr/(s+r)$, because both incident and outgoing spherical paths contribute curvature. Scalar, paraxial propagation and an opening large compared with wavelength are assumed throughout.

A circular hole in the far field gives an Airy pattern. For uniform plane illumination the field is the two-dimensional Fourier transform of the aperture. In polar coordinates, let $Q=k\sin \theta$; the angular integral gives the Bessel function $J_0$:ThusThe limiting central value is one because $J_1(u)\simeq u/2$ near zero. A bright central Airy disk is surrounded by progressively weaker concentric rings. The first dark ring has $ka\sin \theta =3.8317$, or $\sin \theta \simeq 1.22\lambda /D$ with diameter $D=2a$. This scale gives the Rayleigh criterion for a circular optical instrument. Larger apertures narrow the angular pattern; a lens of focal length $f$ maps the first dark ring to radius $1.22\lambda f/D$ in its focal plane.

In the Fresnel regime, the axial intensity oscillates as aperture size or distance changes. Fresnel half-period zones have boundaries at path excesses $m\lambda /2$ and approximately opposite phases in successive zones. For a plane wave, their radii obey $r_m^2=m\lambda z$. A circular opening can therefore expose alternating constructive and cancelling zone contributions. More precisely, Fresnel diffraction by a circular aperture gives, normalized by the unobstructed field,An odd integer number of zones gives a bright axial point with ideal intensity $4I_0$; an even number gives an axial zero. Intensity can locally exceed the unobstructed value because interference redistributes energy; total transmitted power still obeys conservation of energy. The relative normalisation here differs from the Airy expression, which normalises each far-field pattern by its own central intensity.

Off axis, the circular Fresnel field isIts concentric structure evolves with $N_F$ and includes fringes near the geometrical edge. The axial zone argument does not determine the complete off-axis pattern.

Complementary holes and opaque disks are related at the field level. Babinet's principle states $U_a+U_d=U_0$ for complementary aperture and disk transmissions under the same illumination. In the Fraunhofer region away from the direct transmitted beam, the background field is absent and the diffracted fields are opposite in sign. Their scattered intensity profiles therefore have the same Airy angular dependence. Exactly in the forward direction the unobstructed field must be retained; subtracting intensities is incorrect. Finite incident beams also give a finite-width direct beam rather than an ideal angular delta function.

For Fresnel diffraction by an opaque disk, the axial complement is particularly striking:The ideal circular disk has an Arago spot at the centre of its geometric shadow. Circular edge contributions reach the axial point with the same phase. This does not mean the whole shadow is illuminated: off-axis cancellation produces the surrounding shadow and rings, and the axial spot becomes narrower for larger obstruction or shorter propagation distance. An aperture's axial intensity oscillates, while its complementary disk's axial intensity stays constant in this ideal paraxial model.

A practical observation uses a compact coherent source, an accurately circular hole or disk, and an aligned screen or camera. Varying the propagation distance or aperture radius reveals the transition between regimes. A point source, or spatial filtering, prevents different source positions from washing out the fine structure. Finite source size, incomplete coherence, edge roughness, misalignment, vector effects for wavelength-sized objects and lens aberrations reduce the ideal contrast. These experimental limitations explain why a geometrically unexpected axial bright point requires a suitable setup even though its scalar derivation is simple.

In Fraunhofer diffraction, the relative propagation phase across the aperture is linear in aperture coordinates. If $b$ is its largest transverse size, the neglected quadratic path term must be small: for an observer at distance $R$,For a finite-distance point source a distance $s$ away, also require $b^2/(\lambda s)\ll 1$ if the illumination is to be treated as a plane wave. Numerical factors depend on whether $b$ is a radius or full width and on the permitted phase error; the condition expresses a small Fresnel number.

One can place both source and screen sufficiently far away, or collimate a point source with a lens and observe the aperture in the back focal plane of a second lens. In the latter arrangement rays of each outgoing direction focus at one point, with $X\simeq f{\theta }_x$; the lens removes the need for a physically distant screen. A laser beam approximately uniform across the aperture can supply the incident plane wave. The sketches distinguish an actual far field from its focal-plane equivalent.

Let $X$ be horizontal position on the screen, $L=0.50\mathrm{m}$, and use $\sin \theta =X/\sqrt{L^2+X^2}$. The plotted interval is $-50\mathrm{mm}\le X\le 50\mathrm{mm}$. With $a=10\mu \mathrm{m}$, $D=20\mu \mathrm{m}$ and $\lambda =0.500\mu \mathrm{m}$, the four-slit Fraunhofer diffraction intensity is the previous normalized expression evaluated at this angle. The total aperture width is $3D+a=70\mu \mathrm{m}$, giving a Fresnel number of order $0.020$, consistent with the far-field approximation.

Nominal principal array directions satisfy $D\sin {\theta }_m=m\lambda$. Their screen positions areThus the visible nonzero orders are near $\boxed{X=\pm 12.50\mathrm{mm}(m=\pm 1),X=\pm 37.61\mathrm{mm}(m=\pm 3)}$, in addition to the central maximum. The even nonzero orders are missing diffraction orders: at $m=2$ the first envelope zero occurs at $X=\pm 25.03\mathrm{mm}$. The next envelope zero and $m=4$ direction lie at about $\pm 50.25\mathrm{mm}$, just beyond the screen edges.

At the nominal first and third array directions, $I/I(0)=4/{\pi }^2\simeq 0.405$ and $4/(9{\pi }^2)\simeq 0.0450$, respectively. The closest central minima satisfy $D\sin \theta =\pm \lambda /4$, hence $\boxed{X\simeq \pm 3.125\mathrm{mm}}$. Two small subsidiary array maxima occur between adjacent principal directions. The single-slit envelope slightly shifts the true off-axis maxima relative to the nominal principal array angles.

Use Fresnel diffraction, since the detector moves along the axis through successive near-field interference extrema. Write $z$ for its distance from the opening and $\rho$ for the integration coordinate within the opening. For a uniform incident plane wave, the on-axis complex amplitude, in the paraxial approximation, is proportional toThus the light intensity relative to the incident value isThe Fresnel number $F=r^2/(\lambda z)$ is an odd integer at a maximum and an even integer at a minimum. It decreases as the detector moves away. Adjacent extrema therefore differ by one:ConsequentlyAs a check, $F=5$ at $z=1\mathrm{m}$ and $F=4$ at $z=1.25\mathrm{m}$, so these are indeed a maximum followed by the next minimum. Also $r/z\simeq 1.5\times 10^{-3}$, consistent with the paraxial approximation.