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A Fresnel half-period zone is a region of an aperture whose optical path to an observation point differs from the axial path by an interval of half a wavelength. For an axial optical point source a distance $s$ before the aperture and an axial observer a distance $r$ beyond it, the paraxial path excess at aperture radius $u$ is $u^2(s^{-1}+r^{-1})/2$. The outer radius of zone $n$ is thereforeSuccessive zones contribute approximately opposite phases to the Huygens-Fresnel principle integral. The finite source distance matters: using the plane-wave formula $u_n^2=n\lambda r$ requires $s\to \infty$.

The Huygens-Fresnel principle describes propagation by coherent addition of secondary-wave contributions from a wavefront or aperture. Their complex amplitudes, including relative wave phases, must be summed before calculating light intensity. The Fresnel propagator follows by retaining quadratic propagation wave phase in the paraxial approximation.

Each point on a wavefront acts as a source of secondary wavelets, whose envelope gives the next wavefront. The geometric construction gives refraction from secondary wavelets. The Huygens-Fresnel principle additionally keeps the amplitudes and phases needed to calculate interference and diffraction.

The Huygens-Fresnel principle constructs a later wavefront as the envelope of secondary wavelets emitted from the earlier one. At a planar interface, take a time interval during which one point on the incoming front travels distance $v_1\Delta t$ before reaching the interface, while the point that reached it first generates a wavelet of radius $v_2\Delta t$. The two right triangles share the interface segment, giving $\sin i/\sin r=v_1/v_2$. Since the refractive index is $n=c/v$, this yields Snell's law:The angles are measured from the interface normal. This geometric construction assumes homogeneous media and uses the propagation speeds at the given frequency.

In the Huygens-Fresnel principle construction, after a time $\Delta t$ the incident disturbance advances $v_1\Delta t$ while the secondary disturbance in the second medium advances $v_2\Delta t$. The two right triangles share the segment along the interface. HenceHere both angles are measured from the normal vector, and $v_j=c/n_j$ are phase velocities. This is Snell's law. For $n_1>n_2$, no real transmitted angle exists whenThis is total internal reflection; equality is the limiting grazing geometry. The statement concerns an ideal transparent interface, not material absorption or losses at a nearby second boundary.

Let $\beta$ be the internal angle to the optical axis in the straight step-index optical fiber. At its flat entrance, Snell's law gives $\sin {\theta }_i=n_1\sin \beta$. At the side wall a meridional ray has angle of incidence $\pi /2-\beta$. The limiting total internal reflection condition is $\cos \beta =n_2/n_1$. Thus the optical fiber acceptance angle satisfiesThe numerical aperture expression assumes the square root is at most one for air; otherwise all forward launch angles are admitted by this ideal ray test. Strictly confined meridional rays lie inside the limiting cone.

For $n_1=1.48$ and $n_2=1.46$, $\mathrm{NA}=0.242487$ and ${\theta }_{i,\max }=14.0334^{\circ }$. A light ray at internal angle $\beta$ travels distance $L/\cos \beta$ along a guide of axial length $L$. In the nondispersive approximation its travel time is $t=n_{1}L/(c\cos \beta )$. ThereforeHere $L=1000\mathrm{m}$ and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$; using $3.00\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ gives $5.001\mu \mathrm{s}$ and $4.950\mu \mathrm{s}$. The halving applies to the external launch angle, not to $\beta$. For actual short pulses the group refractive index $n_{\mathrm{g}}$ replaces the prefactor $n_1$ in $t=n_{\mathrm{g}}L/(c\cos \beta )$; the source supplies no optical dispersion data. The difference calculated here is geometrical intermodal dispersion in a nondispersive geometrical optics model.

Bending lowers the outer-wall angle of incidence, allowing a previously guided light ray to escape into the cladding. To calculate a definite meridional ray bend threshold, continue the on-axis launch assumption and take a meridional ray in the plane of the circular bend. Its initial internal angle to the centreline tangent obeys $\sin \beta =\sin 10^{\circ }/n_1$. Its straight path has distance $b_{\mathrm{ray}}=R\cos \beta$ from the bend centre. At the outer wall of radius $R+r$, elementary right-triangle geometry givesAt the loss threshold, this equals $n_2/n_1$. Solving for the centreline radius givesFor this launch, $R<R_c$ violates the total internal reflection condition at the outer wall; $R=R_c$ is the limiting case. A real optical fiber has gradual bend loss governed by its electromagnetic wave modes, rather than a universal abrupt cutoff.

Launch dependence. The final sentence does not separately specify the entry position or orientation of the bend relative to the incident light ray. For a planar launch displaced by $u$ from the centreline toward the outer wall, replace the numerator of $\sin i_{\mathrm{out}}$ by $(R+u)\cos \beta$. For example, a launch at the inner edge, $u=-r$, instead givesThus $14.93\mathrm{mm}$ is the result for the stated on-axis continuation; it is not an entry-independent threshold. No missing figure is needed for these explicitly specified ray models.

The Huygens-Fresnel principle treats every point of an advancing wavefront as a source of secondary spherical waves; their forward envelope constructs the next wavefront. Consider two points where an oblique plane wavefront meets a plane interface a distance $AB$ apart. After its first point reaches $A$, let time $t$ elapse until the second reaches $B$. In medium 1, the incident travel distance is $v_{1}t=AB\sin i$. Meanwhile, a secondary wave from $A$ has travelled distance $v_{2}t$ in medium 2. The tangent from $B$ to that secondary wavefront gives $v_{2}t=AB\sin r$. ThereforeHere $i,r$ are the angle of incidence and angle of refraction, measured from the interface normal vector, and $n_j=c/v_j$ are the refractive indices at the common frequency. This is Snell's law.

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.

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.

Let the aperture occupy the $z=0$ plane, with coordinates $(x,y)$ and opening $\Sigma$. Write ${\psi }_{\Sigma }(x,y)$ for the incident scalar complex amplitude and $h(x,y)$ for its dimensionless amplitude transmission, including any imposed phase. The observation point has transverse coordinates $(X,Y)$, corresponding to the PDF's $(x_0,y_0)$, and axial distance $R$. Also $k=2\pi /\lambda$ is the wavenumber and $i^2=-1$.

The Huygens-Fresnel principle expresses the propagated field, up to common prefactors and an approximately constant obliquity factor, asIn the paraxial approximation, expand the path asThe last term is negligible in the Fraunhofer diffraction limit, while the first two terms and $1/r\simeq 1/R$ are common across the aperture. Factoring them out leavesIt is the two-dimensional Fourier transform of the transmitted aperture field, evaluated at transverse wavevector $(kX/R,kY/R)$. For appreciable observation angles the corresponding exact directional components are used rather than $X/R,Y/R$.

The expression is a scalar diffraction theory integral for a monochromatic wave. The complex quantity ${\psi }_P$ represents a field complex amplitude, with physical field obtained by taking the real part after multiplication by $e^{-i\omega t}$. The light intensity is proportional to $|{\psi }_P{|}^2$. The coordinates $x,y$ label transverse positions in the aperture plane, $\Sigma$ is its transmitting region, and $h(x,y)$ is the incident field multiplied by the aperture's complex transmission. The wavenumber is $k=2\pi /\lambda$ in the propagation medium, and $R$ is the axial propagation distance if $h$ already includes the incident field.

To derive the Fresnel diffraction, start from Huygens-Fresnel principle contributions with propagation factor $e^{iks}/s$. With the coordinate origin at the projection of $P$ into the aperture plane,The paraxial approximation requires $\rho /R\ll 1$, where ${\rho }^2=x^2+y^2$. Replace the slowly varying $1/s$ and the obliquity factor by their axial values, but retain the quadratic variation of wave phase. Neglect of the next wave phase term requires $k{\rho }^4/(8R^3)\ll 1$ over the relevant aperture. Thus, apart from a common constant,The quadratic wave phase may be of order one or much larger; setting it to zero would require the stronger conditions for Fraunhofer diffraction. The scalar treatment also neglects the detailed vector field at aperture edges and assumes dimensions large enough compared with wavelength for this approximation to be useful.

For a finite-distance axial point source at distance $p$ and an observer at distance $q$, the incident field contributes its own quadratic wave phase. If $h$ instead denotes only the aperture transmission, the combined curvature is described by an effective distanceThese are two equivalent conventions for distributing the incident wave phase between $h$ and the exponential. Both distances must be included in the later calculation.

Snell's law is $n_1\sin {\theta }_1=n_2\sin {\theta }_2$, with angles measured from the interface normal. The Huygens-Fresnel principle explains it by constructing the transmitted wavefront from secondary wavelets: during the same time interval, distances travelled in the two media are proportional to $c/n_1$ and $c/n_2$. The geometry of the advancing plane wavefront then gives $\sin {\theta }_1/\sin {\theta }_2=(c/n_1)/(c/n_2)=n_2/n_1$.

For incidence from index $n$ into vacuum, the transmitted angle is $\alpha =\arcsin (n\sin \theta )$. The ray bends away from the normal, so its positive deviation isFor $0<\theta <{\theta }_c=\arcsin (1/n)$,Thus the graph rises monotonically from zero to a finite endpoint with a vertical tangent:At the endpoint the transmitted ray grazes the interface. Beyond the critical angle for total internal reflection, no propagating transmitted ray exists and total internal reflection occurs.

Inside a prism, the two ray angles with the face normals sum to its apex angle. For the equilateral cross-section, they are $\theta$ and $\pi /3-\theta$. The entry deviation equals the corresponding exit deviation under ray reversal. Adding the two changes of direction therefore givesThis is symmetric about $\theta =\pi /6$. Symmetry alone gives a stationary point; to establish that it is a minimum, differentiate once more:within the transmitted-ray interval. The sum is therefore strictly convex, andThese expressions require $1<n<2$ for a non-grazing path through both faces; $n=2$ is a grazing limit, and $n>2$ admits no such two-face transmitted path. More explicitly, both internal angles must be at most ${\theta }_c$, which is possible only if $2{\theta }_c\ge \pi /3$.

Atmospheric hexagonal ice crystals can present pairs of faces with a $60^{\circ }$ angle, acting as small prisms. Their orientations turn the minimum-deviation cone around the Moon into an optical halo. No transmitted ray of this family is deflected less than the minimum deviation of a prism. Moreover, near the minimum, $\Delta -{\Delta }_{\min }$ is quadratic in the departure from the symmetric ray, so a range of incident directions is compressed into a small outgoing angular range. This caustic makes the inner rim sharp and bright, with the ray density decreasing farther outside. The common 22-degree halo has an angular radius near $22^{\circ }$ about either the Sun or Moon, hence a diameter near $44^{\circ }$.^[ref]

The Huygens-Fresnel principle constructs a new wavefront as the envelope of secondary wavelets emitted by points on an earlier wavefront. Let the phase velocities in the two media be $v_1=c/n_1$ and $v_2=c/n_2$, with refractive indices evaluated at the light's fixed frequency.

For refraction from secondary wavelets, let point $A$ of the incident wavefront reach the interface first. A second point $B$ reaches the interface at $C$ after time $\Delta t$, so $BC=v_1\Delta t$. During the same interval, the secondary wavelet centered on $A$ advances radius $AD=v_2\Delta t$ into the second medium. The refracted wavefront $CD$ is tangent to this wavelet, hence $AD\perp CD$; also $BC\perp AB$ because a ray is normal to its wavefront in these isotropic media.

Let ${\theta }_1,{\theta }_2$ be the incident and refracted ray angles from the interface normal. They also equal the corresponding acute angles between the wavefronts and the interface. The two right triangles have common hypotenuse $AC$, givingTheir ratio eliminates both $AC$ and $\Delta t$:This is Snell's law. In particular, entering a medium of larger refractive index reduces the ray's angle to the normal. The construction assumes a stationary interface and homogeneous isotropic media with a propagating refracted wave.