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Sandbox Physics

Optics 061 · Diffraction, Fourier optics, and computational imaging

Talbot Carpet

Why does light behind a grating repeat the grating’s pattern at certain distances? The grating's diffraction orders pick up different phases as they travel, and at the Talbot distance those phases line up again, so the pattern reappears. Halfway there, the odd orders are flipped in sign and the copy is shifted by half a period. At simple fractions of the distance, several shifted copies overlap and the pattern looks finer. A finite grating makes every revival imperfect.

Slide an absorbing sensor behind finite amplitude or phase gratings. Find shifted and fractional Talbot images, crossed two-dimensional patterns, the effect of tilting the source and diffraction from the grating edges, and compare each with an ideal periodic reference.

Interactive modelTalbot Carpet
Full unshifted recurrence / actual sensor distance zT,zz_{\rm T},z0.500.50
Mask transmission / finite-window mean intensity T,I‾windowT,\overline I_{\rm window}50%50\%
Finite carrier side / sensor pixel pitch L,ΔxL,\Delta x0.00π0.00\pi
Model regimevalid model regime\text{valid model regime}

Physics tutorial

Audit the physics of Talbot Carpet

BackgroundA field that repeats forever across the beam contains only a discrete set of transverse spatial frequencies. Fresnel propagation shifts their phases at different rates, and they all line up again after zT=2d2/λz_{\rm T}=2d^2/\lambda. At this distance the whole field comes back with no shift. Some texts use half this distance as their definition instead.

Why it mattersWhy does light behind a grating repeat the grating’s pattern at certain distances?

Start with the essentials

Focus question
Why does light behind a grating repeat the grating’s pattern at certain distances?
One-sentence intuition
At half this distance every even harmonic keeps its sign and every odd harmonic reverses it: uzT/2per(x)=tper(x−d/2)u^{\rm per}_{z_{\rm T}/2}(x)=t^{\rm per}(x-d/2). That pattern of signs shifts the whole field sideways by half a period. If the grating changes only the phase, the intensity can still look uniform.

Core mathematical model

One mask, two physical behaviours

A(x)=1∣x∣<Nd/2gD(x)=∑n1∣x−nd∣<Dd/2tamp=AgDtphase=A[1+(eiϕ−1)gD]\begin{aligned}A(x)&=\mathbf1_{|x|<Nd/2}\\g_D(x)&=\sum_n\mathbf1_{|x-nd|<Dd/2}\\t_{\rm amp}&=A g_D\\t_{\rm phase}&=A[1+(e^{i\phi}-1)g_D]\end{aligned}

The opening fraction sets the actual slit width. The amplitude mask blocks light outside the openings. The phase stripes change the phase of the field but not the transmitted intensity. The aperture around the grating is finite. The phase step is set directly for the chosen wavelength; it is not derived from any unspecified material.

The field behind the finite mask

uz(x)=1iλz∫−Nd/2Nd/2t(ξ)eiπ(x−ξ)2λz dξUz(x,y)=uz(x)vz(y)\begin{aligned}u_z(x)&=\frac{1}{\sqrt{i\lambda z}}\int_{-Nd/2}^{Nd/2}t(\xi)e^{\frac{i\pi(x-\xi)^2}{\lambda z}}\,\mathrm d\xi\\U_z(x,y)&=u_z(x)v_z(y)\end{aligned}

The common phase from propagation along the axis is left out. Each slit contributes differences of the Fresnel integrals C and S taken at its two edges. A striped mask is clear over a finite vertical window; a crossed mask is the product of two gratings, one in each direction. This simulated measurement uses no periodic boundary.

Separate infinite-periodic reference

uzper(x)=∑m=−MMcme2πimx/d−iπλzm2/d2cmamp=D sinc⁡(mD)cmphase=δm0+(eiϕ−1)cmampM=96\begin{aligned}u^{\rm per}_z(x)&=\sum_{m=-M}^M c_m e^{2\pi imx/d-i\pi\lambda zm^2/d^2}\\c_m^{\rm amp}&=D\,\operatorname{sinc}(mD)\\c_m^{\rm phase}&=\delta_{m0}+(e^{i\phi}-1)c_m^{\rm amp}\\M&=96\end{aligned}

Here sinc of an argument means the sine of pi times that argument, divided by pi times the argument. The zero order must be included. The periodic transmission has no envelope from the finite carrier. Because the reference keeps only a finite number of harmonics, it rings near edges and misses a small tail of power. It is neither the finite-mask simulation nor a full-wave prediction.

Fractional copies add as amplitudes

uzT/4per(x)=1−i2tper(x)+1+i2tper(x−d/2)uzT/3per(x)=∑j=02ajtper(x−jd/3)aj=13∑m=02e−2πim2/3+2πimj/3\begin{aligned}u_{z_{\rm T}/4}^{\rm per}(x)&=\frac{1-i}{2}t^{\rm per}(x)+\frac{1+i}{2}t^{\rm per}(x-d/2)\\u_{z_{\rm T}/3}^{\rm per}(x)&=\sum_{j=0}^2a_jt^{\rm per}(x-jd/3)\\a_j&=\frac13\sum_{m=0}^2e^{-2\pi im^2/3+2\pi imj/3}\end{aligned}

These equalities describe the untruncated infinite-periodic limit. They do not describe the finite simulation or the truncated reference. The quarter-plane coefficients come from the phases of the even and odd harmonics. When the amplitude openings are narrower than one third of a period, the copies at the third plane stay separate. Wider copies overlap coherently, so multiplying intensity stripes does not work as a general rule. Finite edges spoil exact repetition.

Tilt, pixels and transmission mean different things

Iθ(x,z)=I0(x−zθx,z)Iab=1Δx2∬pixelab∣Uz∣2 dx dyTamp=D1+b,Tphase=1\begin{aligned}I_{\theta}(x,z)&=I_0(x-z\theta_x,z)\\I_{ab}&=\frac1{\Delta x^2}\iint_{\mathrm{pixel}_{ab}}|U_z|^2\,\mathrm dx\,\mathrm dy\\T_{\rm amp}&=D^{1+b},\quad T_{\rm phase}=1\end{aligned}

Tilt shears the Talbot carpet and moves the pattern across the sensor. The mean intensity inside the six-period window is not the total transmitted power. The camera view and the enlarged data use the same area-averaged pixels. The carpet is a separate scan of point intensities, computed on its own and not rebuilt from one frame. Display gain never changes the raw values.

Why real self-images are not perfect

kz=k2−kx2−ky2kz≃k−kx2+ky22kΔx=d/32\begin{aligned}k_z&=\sqrt{k^2-k_x^2-k_y^2}\\k_z&\simeq k-\frac{k_x^2+k_y^2}{2k}\\\Delta x&=d/32\end{aligned}

The solver uses the quadratic paraxial phase, perfectly coherent light and thin masks with sharp edges. High-angle components from the edges, finite bandwidth, limited coherence and pixel integration all limit fine detail. Of these, only diffraction from the finite carrier and pixel integration are included here. The Lab claims neither an infinitely fine fractal nor any outside scientific certification.

Common difficulties

Uniform intensity does not mean no self-image

Typical misconceptionEvery Talbot plane should show bright copies of the phase grating.

Better mental modelRight behind a lossless phase grating, the intensity is already uniform. An ideal revival brings back that phase pattern, so it shows no visible amplitude stripes either. Intensity contrast can appear at fractional planes. The edges of the finite carrier add extra diffraction.

Run the experiment

  1. 01

    Scene 1: Unshifted finite revival

    Start with the amplitude stripes. Compare the half-distance and full-distance revivals, then look at the finite openings in the mask and the front face of the sensor.

    What to observe: At half the distance the ideal pattern is shifted by half a period; at the full distance the field returns unshifted.
  2. 02

    Scene 2: One third · three copies

    Use the third-plane preset. Widen the openings until the three coherent copies overlap, and compare the result with the periodic reference.

    What to observe: Narrow amplitude openings keep the fractional copies apart. Wide copies interfere with each other; they do not simply multiply the intensity stripes.
  3. 03

    Scene 3: Phase to intensity · crossed grid

    Switch to crossed phase stripes. Compare the contrast at a quarter of the distance with the uniform intensity at the full distance, then reduce the number of periods.

    What to observe: A revived phase pattern can look uniform in intensity. The finite carrier and the area-averaging detector keep the repetition from being perfectly sharp.
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