advanced30 min studyLesson 213 in Physics

Fresnel diffraction and near fields

Retain quadratic propagation phase, estimate Fresnel-zone scales, and distinguish paraxial near fields from subwavelength evanescence.

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What you’ll understand

  • Classify a propagation distance using a declared aperture half-width.
  • Explain cancellation between neighbouring Fresnel zones.
  • Calculate how a periodic field can reproduce its transverse structure.

Near an aperture, a screen samples a changing interference pattern rather than a rescaled copy of one fixed angular distribution. The curvature of the propagation phase across the opening matters. Fresnel diffraction retains that curvature while keeping the paraxial simplification of the exact distance. Here near field means a propagation regime relative to aperture size and wavelength; it does not automatically mean distances smaller than one wavelength or the presence of evanescent components.

Two approximation tests answer different questions

Define the Fresnel number using an explicitly chosen transverse aperture scale aa, taken here as a radius or half-width. The aperture-dependent quadratic phase reaches approximately πNF\pi N_F at that radius on axis. Large or order-one Fresnel number warns against discarding that phase. A small Fresnel number supports a Fraunhofer approximation, but the required smallness depends on the permissible field or intensity error, rather than a universal boundary exactly at one.

Fresnel propagation itself still neglects higher terms of the distance expansion and assumes small propagation angles. It can remain accurate for a large Fresnel number when those higher phase terms are small. Conversely, a label such as far away in metres cannot establish a far-field regime without the aperture scale and wavelength. Doubling an aperture dimension requires four times the distance to keep the same Fresnel number, which often makes an apparently minor experimental change consequential.

NF=a2λz,Δϕquad(a)≃πNF,∣Δϕquartic∣≃πρ44λz3N_F=\frac{a^2}{\lambda z},\qquad \Delta\phi_{\mathrm{quad}}(a)\simeq\pi N_F,\qquad |\Delta\phi_{\mathrm{quartic}}|\simeq\frac{\pi\rho^4}{4\lambda z^3}
a is the declared aperture radius or half-width, rho the largest contributing transverse separation, and all lengths are in metres. NF is dimensionless. The phase estimates are paraxial, with the quartic estimate testing the Fresnel truncation separately.↔ Scroll wide equations horizontally.

Zones organize the phase cancellation

On axis, the excess path from radius rr is approximately r2/(2z)r^2/(2z). Boundaries where this excess increases by half a wavelength define half-period Fresnel zones. Neighbouring zones then contribute roughly opposite phases. Their areas are approximately equal because squared radius, rather than radius, increases by a constant increment. Slowly varying amplitudes allow strong cancellation between neighbouring annuli; stopping the aperture at a different zone can therefore increase or decrease the central brightness.

This reasoning explains why more open area need not increase irradiance at a particular screen point. Energy conservation constrains power integrated over an appropriate whole surface, not monotonic brightness at every pixel. Zone arguments also require care at edges: the contributions form a curved path in the complex plane, and treating each zone as one exactly equal arrow is a qualitative approximation. The actual Fresnel integral remains the quantitative statement.

rm≃mλz,rm22z=mλ2,m=1,2,3,…r_m\simeq\sqrt{m\lambda z},\qquad \frac{r_m^2}{2z}=\frac{m\lambda}{2},\qquad m=1,2,3,\ldots
rm is the radius in metres of the mth half-period zone for a normally incident plane wave observed on axis a distance z away. The formula assumes rm is small relative to z and does not include finite-source curvature.↔ Scroll wide equations horizontally.

Periodic near fields can self-image

A periodic aperture field can be expanded into transverse Fourier orders of pitch dd. Under paraxial propagation, order mm acquires a relative phase −πλzm2/d2-\pi\lambda z m^2/d^2, after removing the common carrier phase. Unlike a far-field picture that spatially separates orders, a near-field plane lets them overlap. Their changing relative phases can reconstruct, shift, or reshape the original periodic field as distance changes.

At the paraxial Talbot distance 2d2/λ2d^2/\lambda, every integer order gains an integer multiple of 2π2\pi, so an ideal infinite periodic field reproduces itself up to a common phase. At half that distance, the factor is (−1)m(-1)^m, equivalent to a half-period lateral translation. Real finite gratings have edge diffraction and limited useful repetitions. High orders also violate the paraxial approximation, so the simple distance is a model prediction with explicit angular limits.

Choose a propagation regime

A circular aperture of radius 0.50 mm0.50\,\mathrm{mm} is illuminated at 500 nm500\,\mathrm{nm}. A screen is 0.50 m0.50\,\mathrm m downstream. Evaluate its Fresnel number, its aperture quadratic phase, and the estimated on-axis quartic error at the rim. Decide which propagation approximation to start with.

  1. Use the radius specified in the problem: NF=(5.0×10−4)2/[(5.0×10−7)(0.50)]=1N_F=(5.0\times10^{-4})^2/[(5.0\times10^{-7})(0.50)]=1. Substituting the diameter would change this convention by a factor of four.

  2. The rim quadratic phase is πNF=π\pi N_F=\pi rad. It is not negligible, so a Fraunhofer transform without the aperture chirp is not justified for the requested field.

  3. The omitted quartic path is a4/(8z3)=6.25×10−14 ma^4/(8z^3)=6.25\times10^{-14}\,\mathrm m, giving phase 7.85×10−7 rad7.85\times10^{-7}\,\mathrm{rad}. This is tiny, and a/z=0.001a/z=0.001, so Fresnel propagation is a sensible approximation on axis.

The result

Use the Fresnel integral: the retained quadratic phase is π rad while the estimated omitted phase is only 7.85×10⁻⁷ rad.

Explore the model

Interactive lab

Interactive lab · waves

Cancellation depends on phase

For separate runs, Reset and set BOTH Phase difference and Amplitude ratio. Compare 0° and 180° with ratio 0.50, predicting summed amplitudes 1.50 and 0.50, then Play or Step. These fixed-frequency copropagating waves illustrate cancellation between contributions; they are not Fresnel zones, chirped aperture fields, or a near-field solver.

Cancellation depends on phaseTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPartial cancellation can reduce a local field even when both contributions are nonzero; a two-wave lab cannot reproduce a full Fresnel pattern.

Check your understanding

For a fixed aperture radius and wavelength, what happens when the screen distance is multiplied by four?

Try it yourself

An ideal paraxial grating field has period d=20 μmd=20\,\mathrm{\mu m} at wavelength 500 nm500\,\mathrm{nm}. Find its Talbot distance and explain the field at half that distance. Assume only paraxial diffraction orders contribute appreciably.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Require the propagation phase of every integer Fourier order to be a multiple of 2π2\pi. At half the answer, inspect odd and even orders separately.

Compare with the solution

The full self-imaging distance is 2d2/λ=2(20×10−6)2/(500×10−9)=1.6×10−3 m2d^2/\lambda=2(20\times10^{-6})^2/(500\times10^{-9})=1.6\times10^{-3}\,\mathrm m, or 1.6 mm. At 0.8 mm, each order receives e−iπm2=(−1)me^{-i\pi m^2}=(-1)^m, since an integer and its square have the same parity. That factor equals the Fourier phase of a translation by d/2=10 μmd/2=10\,\mathrm{\mu m}. The ideal field is a shifted self-image, rather than necessarily the identical complex field at the same coordinates. Finite edges and nonparaxial orders limit this conclusion.

Take it with you

  • Fresnel number and paraxial truncation test different approximations.
  • Opening additional Fresnel zones can decrease irradiance at one point.
  • Self-imaging is interference between overlapping periodic diffraction orders.

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