Fourier Optics and Spatial Filtering

Fourier optics takes one theorem from linear systems — that a shift-invariant operation is diagonalised by the Fourier transform — and one equation from electromagnetism — the monochromatic scalar Helmholtz equation $(\nabla^2+k^2)U=0$ — and welds them into a single claim: the propagation of light through free space is a linear filter acting on the transverse spatial-frequency content of the field. Decompose a field $U(x,y)$ into plane waves, its angular spectrum, and each component sails off with a definite transverse wavevector $(k_x,k_y)$ and picks up only a phase set by $k_z=\sqrt{k^2-k_x^2-k_y^2}$ . Propagation is diagonal in the spatial-frequency basis, so the whole toolkit of transfer functions, impulse responses and convolutions transfers, word for word, from circuit theory to optics. The practical payoff is a plane you can put your hand into. A thin converging lens turns each transverse spatial frequency into a transverse position in its back focal plane, so the abstract Fourier transform of an object becomes a real light pattern you can mask, stop, or phase-shift. Modify the field there, re-transform with a second lens, and you have realised an arbitrary spatial filter — the 4f processor, and with it coherent image formation itself. This one picture unifies phenomena that look unrelated in ray optics: the resolution limit of a microscope set by numerical aperture, the diffraction rings around a star, dark-field and Zernike phase-contrast imaging, apodization, matched filtering, and holography. It also draws the sharp line between coherent light, where amplitudes interfere and the transfer function is a hard-edged pupil cut off at $\mathrm{NA}/\lambda$ , and incoherent light, where intensities add and the optical transfer function is the pupil autocorrelation, reaching to $2\,\mathrm{NA}/\lambda$ .

Key takeaways

Free-space propagation is a spatial-frequency filter: each plane-wave component is multiplied by the transfer function $H(k_x,k_y;z)=e^{ikz}e^{-i(k_x^2+k_y^2)z/2k}$ in the paraxial (Fresnel) regime.
In the far field (Fraunhofer, Fresnel number $N_F\ll1$ ) the diffracted field is the 2D Fourier transform of the aperture, with spatial frequency mapped to angle, $f_x=x/\lambda z$ .
A converging lens performs an exact Fourier transform: the back-focal-plane field is proportional to $\tilde U_0$ of the front-focal-plane field, evaluated at $(x/\lambda f,\,y/\lambda f)$ .
A mask in that Fourier plane multiplies the object spectrum by a filter; a second lens inverse-transforms, giving the 4f coherent imaging and processing system.
Coherent imaging is linear in complex amplitude with a sharp cutoff at $\mathrm{NA}/\lambda$ ; incoherent imaging is linear in intensity with cutoff $2\,\mathrm{NA}/\lambda$ , its transfer function the autocorrelation of the pupil.
Key idea: free space and a lens are both linear filters acting on the transverse Fourier spectrum of the field; a lens maps spatial frequency to position in its focal plane, so a mask there filters spatial frequencies directly.

The core insight

Free space and a lens are both linear filters acting on the transverse Fourier spectrum of the field; a lens maps spatial frequency to position in its focal plane, so a mask placed there filters spatial frequencies directly.

Where this lives today

The Fourier plane is now a control surface. In modern microscopes a liquid-crystal spatial light modulator sits in the pupil and imposes a programmable complex mask, so phase contrast, dark field, differential-interference contrast and aberration correction all become a matter of loading a different pattern — adaptive optics for biology, driven by the same $H(f_x,f_y)$ of equation (7). Fourier ptychography stitches many low-NA images, each under illumination from a different angle, into a reconstruction whose spatial-frequency support far exceeds the objective's own cutoff, synthesising a large effective NA in software; coherent diffractive imaging pushes further, recovering an object from its far-field intensity alone by iterating between object plane and Fourier transform. The same reasoning runs through technology far from the microscope. Metasurface holograms and diffractive optical elements are designed by prescribing an aperture whose Fraunhofer transform is the desired far-field image, then etching the required phase profile. Grating couplers that feed light from a fibre into an integrated photonic chip are matched filters in disguise, shaping a free-space beam's spatial spectrum to overlap the guided mode's — the bridge from this article to numerical aperture and on-chip waveguides. And optical correlators, the descendants of Vander Lugt's 1963 filter, still perform pattern recognition at the speed of light — a 4f system is, quite literally, an analog computer for convolution.

The physical picture

A lens is a Fourier machine because it sorts directions into places. Every plane wave leaving the object at a given angle — that is, every single transverse spatial frequency — is a bundle of parallel rays, and a converging lens brings any bundle of parallel rays to one point in its back focal plane. Steeper angle, farther from the axis. So the lens lays the object's whole angular spectrum out across a plane, ready to be touched. Fine detail is rapid variation, high spatial frequency, large angle, and lands far out; the smooth background is the near-axis light. Block the outer region and you erase fine detail (a blur, a low-pass); block the centre and you keep only edges (a high-pass).

Figure

A lens sorts plane-wave directions into positions. A grating of period d = 20 µm under λ = 633 nm light sends its 0 and ±1 diffraction orders to distinct spots in the back focal plane of an f = 200 mm lens, the ±1 orders landing at x = ±λf/d = ±6.33 mm — spatial frequency has become position

Diffraction and imaging are the same physics at two distances. Just past an aperture (the Fresnel, near-field regime) every component has accumulated a quadratic phase, so the field is a rippling version of the aperture. Far away — or at the focal plane of a lens — that phase collapses and you see the bare Fourier transform of the aperture. The referee for which regime you are in is the Fresnel number $N_F=a^2/\lambda z$ : large means shadow, small means transform. The last distinction is coherent versus incoherent, a statement about which quantity the system is linear in. Coherent light superposes amplitudes, so spatial frequencies interfere and edges can ring; its transfer function is the pupil, cut off at $\mathrm{NA}/\lambda$ . Incoherent light superposes intensities, so its transfer function is the pupil autocorrelation, reaching twice as far, to $2\,\mathrm{NA}/\lambda$ , but tapering smoothly and never ringing.

Figure

Coherent versus incoherent transfer for the same objective, NA = 0.65 at λ = 550 nm. The coherent transfer function is the flat pupil, passing amplitude unchanged to a hard cutoff f_c = NA/λ = 1182 cycles/mm; the incoherent optical transfer function is the pupil autocorrelation, a triangle reaching twice as far to 2f_c = 2364 cycles/mm but falling steadily to zero

The core equations

Everything begins by writing the field as a superposition of plane waves — the angular spectrum, which is just the 2D Fourier transform of the transverse field reinterpreted physically. Each Fourier component is a plane wave heading in a direction set by $(k_x,k_y)$ , and because plane waves are the eigenfunctions of free-space propagation, this is the basis in which propagation is simple.

U(x,y,0)=\iint \tilde{U}(k_x,k_y)\,e^{i(k_x x+k_y y)}\,\frac{\mathrm{d}k_x\,\mathrm{d}k_y}{(2\pi)^2} \qquad (1)

where: $U(x,y,0)$ — complex scalar field in the input plane $z=0$ ; $\tilde{U}(k_x,k_y)$ — 2D Fourier transform, the angular spectrum; $k_x,k_y$ — transverse wavevector components, i.e. spatial frequencies. Physically, the field is a fan of plane waves and $|\tilde U|^2$ says how much power heads in each direction. Propagating a distance $z$ multiplies each plane wave by the phase it accumulates along $z$ . Since $k_z$ depends on $(k_x,k_y)$ , this is a frequency-dependent multiplier — a transfer function — that in the paraxial limit reduces to a quadratic phase.

H(k_x,k_y;z)=e^{iz\sqrt{k^2-k_x^2-k_y^2}}\ \xrightarrow{\ \text{paraxial}\ }\ e^{ikz}\,e^{-i\frac{(k_x^2+k_y^2)z}{2k}} \qquad (2)

where: $k=2\pi/\lambda$ — wavenumber in the medium; $z$ — propagation distance; $H$ — the phase-only multiplier on each spatial frequency. Physically, high spatial frequencies accumulate phase faster, which is exactly why a sharp feature spreads as it propagates. The impulse response is the inverse transform of $H$ , so in real space propagation is a convolution of the input with a single spreading kernel — the Fresnel point-spread function, a diverging spherical wave in paraxial form.

U(x,y,z)=U(x,y,0)\ast h(x,y;z),\qquad h(x,y;z)=\frac{e^{ikz}}{i\lambda z}\,e^{\frac{ik}{2z}(x^2+y^2)} \qquad (3)

where: $\ast$ — 2D convolution over the input plane; $h$ — Fresnel free-space impulse response; $\lambda$ — wavelength in the medium. Physically, each object point launches a quadratic-phase wavelet and the field downstream is their coherent sum — Huygens' principle made quantitative. When the observation distance is large enough that the source-plane quadratic phase can be dropped ( $N_F\ll1$ ), the convolution collapses to a pure Fourier transform. This is Fraunhofer diffraction: the far field is the spectrum of the aperture, position mapped to spatial frequency through $f_x=x/\lambda z$ .

U(x,y,z)=\frac{e^{ikz}\,e^{\frac{ik}{2z}(x^2+y^2)}}{i\lambda z}\,\tilde{U}_0\!\left(\frac{kx}{z},\frac{ky}{z}\right) \qquad (4)

where: $\tilde{U}_0$ — Fourier transform of the input aperture field; $x/\lambda z$ — spatial frequency $f_x$ read off at position $x$ , distance $z$ . Physically, a small aperture gives a broad pattern and a large one a narrow pattern — the reciprocal spreading that signals a Fourier transform. A thin lens acts, to paraxial order, as a pure phase plate: thicker at the centre, it retards the on-axis field and imprints a converging quadratic phase. That single factor is the whole optical action of an ideal lens.

t_{\text{lens}}(x,y)=e^{-i\frac{k}{2f}(x^2+y^2)} \qquad (5)

where: $f$ — focal length; $t_{\text{lens}}$ — multiplicative transmittance (paraxial, aperture ignored). Physically, the lens phase is the exact negative of the free-space Fresnel phase over a distance $f$ , which is why the two can be made to cancel. Cascading a Fresnel propagation, this lens phase, and a second Fresnel propagation — object at the front focal plane, screen at the back — cancels every stray quadratic phase and leaves a clean transform. This is the equation that turns Fourier optics into an instrument.

U_f(x,y)=\frac{e^{i2kf}}{i\lambda f}\,\tilde{U}_0\!\left(\frac{x}{\lambda f},\frac{y}{\lambda f}\right) \qquad (6)

where: $U_f$ — field in the back focal plane; $U_0$ — field in the front focal plane, one focal length ahead. Physically, the Fourier plane sits a finite distance $f$ away, not at infinity, so the transform is something you can build on a bench and reach into. Placing a mask $H(f_x,f_y)$ in that Fourier plane and re-transforming with a second lens is the 4f processor. The output is the object convolved with the mask's impulse response, so choosing the mask chooses the filter.

U_{\text{img}}(x,y)=\mathcal{F}^{-1}\{\,H(f_x,f_y)\,\tilde{U}_0(f_x,f_y)\,\} \qquad (7)

where: $H(f_x,f_y)$ — amplitude and/or phase transmittance of the Fourier-plane mask; $\tilde U_0$ — object spectrum from the first lens; $U_{\text{img}}$ — filtered image. Physically, this is the optical twin of applying a digital filter to an FFT, executed at the speed of light in glass. Finally, imaging itself has a transfer function that depends on coherence: under coherent light the system is linear in amplitude and its transfer function is the pupil; under incoherent light it is linear in intensity and its transfer function is the pupil autocorrelation, reaching twice as far.

H_{\text{coh}}=P(\lambda f_x,\lambda f_y),\qquad \mathrm{OTF}=P\star P,\qquad f_{c,\text{coh}}=\frac{\mathrm{NA}}{\lambda},\qquad f_{c,\text{incoh}}=\frac{2\,\mathrm{NA}}{\lambda} \qquad (8)

where: $P$ — pupil (aperture) function; $\star$ — autocorrelation; $\mathrm{NA}=n\sin\theta_{\max}$ ; $f_c$ — highest transmitted spatial frequency. Physically, the same objective resolves finer detail in incoherent light because the autocorrelation of a disc is wider than the disc itself.

Derivation AA lens performs a Fourier transform

The claim that a single lens computes a two-dimensional Fourier transform deserves a proof, not a slogan. We assemble it from three operations we already trust — Fresnel propagation over a distance, multiplication by the lens quadratic-phase plate, and a second Fresnel propagation — then choose the one geometry, object at the front focal plane and screen at the back, in which every stray quadratic phase cancels and only the bare kernel $e^{-i(k/f)(x\xi+y\eta)}$ survives. All the bookkeeping lives in the coefficients of the quadratic terms in the exponent.

Step 1. Propagate the input $U_0(\xi,\eta)$ a distance $d_1$ to the lens with the Fresnel integral, $U_-(u,v)=\frac{e^{ikd_1}}{i\lambda d_1}\iint U_0\,e^{\frac{ik}{2d_1}[(u-\xi)^2+(v-\eta)^2]}\,\mathrm{d}\xi\,\mathrm{d}\eta$ . Why: this is equation (3) as an integral, the only propagation law needed — everything after is algebra on the exponent.

Step 2. Multiply by the lens phase, $U_+=U_-\,e^{-\frac{ik}{2f}(u^2+v^2)}$ . Why: the lens subtracts a converging quadratic phase from the diverging one propagation added, and this subtraction is what makes the $u^2,v^2$ terms cancel in the right geometry.

Step 3. Propagate a further distance $d_2$ to the screen, $U_f=\frac{e^{ikd_2}}{i\lambda d_2}\iint U_+\,e^{\frac{ik}{2d_2}[(x-u)^2+(y-v)^2]}\,\mathrm{d}u\,\mathrm{d}v$ . Why: composing the three maps object plane to output plane, and the whole thing is now one nested Gaussian integral over $\xi,\eta,u,v$ .

Step 4. Set $d_1=d_2=f$ and collect the coefficient of $u^2$ : $\frac{1}{d_1}-\frac{1}{f}+\frac{1}{d_2}$ evaluated at $d_1=d_2=f$ leaves only the cross terms pairing $u$ with $x$ and $\xi$ . Why: with the object exactly at the front focal plane the pure- $u^2$ curvature vanishes, so the $u,v$ integrals are unmodulated Gaussians linking output position to input position.

Step 5. Do the integrals; the residual $(\xi,\eta)$ phase cancels too, leaving $U_f=\frac{e^{i2kf}}{i\lambda f}\iint U_0\,e^{-i\frac{k}{f}(x\xi+y\eta)}\,\mathrm{d}\xi\,\mathrm{d}\eta\propto\tilde U_0(\tfrac{x}{\lambda f},\tfrac{y}{\lambda f})$ . Why: only the kernel $e^{-i(k/f)(x\xi+y\eta)}$ remains — the definition of the 2D Fourier transform with $f_x=x/\lambda f$ , phase-free at the finite distance $f$ . *Verified: removing the lens ( $f\to\infty$ , replaced by a distance $z$ ) reproduces the Fraunhofer result (4); matches Goodman Ch. 5 and Saleh and Teich Ch. 4.*

Go deeper · Derivation B · 1 minThe Fresnel number and the near-to-far crossover

Derivation BThe Fresnel number and the near-to-far crossover

Equations (3) and (4) differ by a single factor — the quadratic phase in the source coordinate — so the whole near-versus-far question reduces to when that factor may be set to one. This makes the crossover quantitative for a circular aperture of radius $a$ at distance $z$ , produces the dimensionless Fresnel number that controls it, and reconciles the two textbook far-field criteria. The estimate is a phase budget: how many radians of extra phase the rim of the aperture accumulates relative to its centre.

Step 1. Isolate the source-coordinate quadratic phase $e^{\frac{ik}{2z}(\xi^2+\eta^2)}$ inside the Fresnel integral. Why: this is the only term standing between a genuine convolution and a pure Fourier transform, so its size is the whole story.

Step 2. Bound its phase at the rim $\xi^2+\eta^2=a^2$ : $\Delta\phi_{\max}=\frac{ka^2}{2z}=\frac{\pi a^2}{\lambda z}\equiv\pi N_F$ , defining $N_F=a^2/\lambda z$ . Why: writing the worst-case phase as $\pi N_F$ turns an integral condition into one back-of-envelope number.

Step 3. Demand $\pi N_F\ll1$ , i.e. $z\gg a^2/\lambda$ : the source phase is then near unity and the integral is a Fourier transform — the Fraunhofer regime. Why: small edge phase means the aperture radiates as one coherent piece rather than as a shadow, which is what "far field" means.

Step 4. In the opposite limit $N_F\gg1$ the edge phase runs through many cycles, contributions localise, and the field is the geometric shadow with Fresnel fringes at its rim. Why: many phase cycles across the aperture is the signature of ray optics with diffraction confined to edges. *Verified: with $D=2a$ this gives $z\gg D^2/4\lambda$ , of the same order as the engineering rule $z\gtrsim 2D^2/\lambda$ (the prefactor depends on the tolerated phase error); consistent with Hecht Ch. 10.*

Go deeper · Derivation C · 1 minZernike phase contrast as a Fourier-plane filter

Derivation CZernike phase contrast as a Fourier-plane filter

A transparent object — a cell, a clean phase grating — changes only the phase of the light, so to first order it is invisible: intensity is amplitude squared, and a pure phase factor has unit modulus. Zernike's insight was that the object is not truly invisible, only mis-tuned: its scattered light sits in phase quadrature with the unscattered background, and a single $\pi/2$ touch on the zero spatial frequency rotates them into alignment so phase becomes brightness. Here a weak phase object is carried through the 4f system and the linear-in-phase image emerges.

Step 1. Write the object as $U_0=e^{i\phi}\approx1+i\phi(x,y)$ with $|\phi|\ll1$ : a strong background $1$ plus a small scattered part $i\phi$ . Why: the factor $i$ is the point — the scattered light is 90° out of phase with the background and so cannot beat against it to make contrast.

Step 2. Form the unfiltered intensity $I=|1+i\phi|^2=1+\phi^2\approx1$ . Why: to first order the intensity is flat — the precise statement that a phase object is invisible in bright field.

Step 3. In the Fourier plane $\tilde U_0(f)=\delta(f)+i\tilde\phi(f)$ : the background collapses to a bright point at the origin, the scattered light spreads around it. Why: the unscattered beam is one plane wave and focuses to the zero-frequency point — exactly the spot a small dot can act on alone.

Step 4. Insert a $\pi/2$ dot on the origin, $\delta(f)\to i\,\delta(f)$ , and inverse-transform: $U_{\text{img}}=i+i\phi=i(1+\phi)$ . Why: advancing only the background by $\pi/2$ multiplies it by $i$ , putting it back in phase with the scattered $i\phi$ so the two now interfere.

Step 5. Take the intensity, $I=|i(1+\phi)|^2\approx1+2\phi$ . Why: phase now maps linearly onto brightness — Zernike phase contrast; a $-\pi/2$ dot flips the sign to negative contrast. *Verified: standard Zernike result in Born and Wolf and Goodman; the map $I\approx1+2\phi$ and the sign reversal for a $-\pi/2$ plate both hold.*

Limiting cases

The single propagation integral reduces to recognisable regimes according to the Fresnel number $N_F=a^2/\lambda z$ (aperture radius $a$ , distance $z$ ), and a lens supplies a fourth, finite-distance realisation of the far-field transform.

Case	Result	Meaning
Near field, $N_F\gg1$	$U(x,y,z)\approx U(x,y,0)\,e^{ikz}$	Geometric shadow: the field reproduces the aperture with little spreading; ray optics suffices, fringes only at edges.
Fresnel, $N_F\sim1$	$U(\mathbf{r})\propto\!\int U_0(\mathbf{r}')\,e^{\frac{ik}{2z}\|\mathbf{r}-\mathbf{r}'\|^2}\,\mathrm{d}^2r'$	Quadratic-phase diffraction: near-field fringes, self-focusing, the Talbot effect.
Fraunhofer, $N_F\ll1$	$U(x,y,z)\propto\tilde U_0\!\left(\tfrac{kx}{z},\tfrac{ky}{z}\right)$	Far field is the Fourier transform of the aperture; angle maps to spatial frequency.
Back focal plane of a lens	$U_f(x,y)\propto\tilde U_0\!\left(\tfrac{x}{\lambda f},\tfrac{y}{\lambda f}\right)$	Exact Fourier transform at the finite distance $f$ : the practical realisation of spatial filtering.

Common trapA coherent laser microscope must out-resolve an incoherent one because the light is cleaner

It does not — the incoherent system reaches finer periodic detail, by a factor of two, because the two systems are linear in different quantities with different transfer support. A coherent system is linear in complex amplitude, its transfer function is the pupil itself, so it passes spatial frequencies only up to $f_c=\mathrm{NA}/\lambda$ . An incoherent system is linear in intensity, its transfer function is the pupil autocorrelation; the autocorrelation of a disc of radius $R$ has radius $2R$ , so the cutoff doubles to $2\,\mathrm{NA}/\lambda$ . "Cleaner" laser light buys coherence, not bandwidth — and coherence brings its own baggage, since coherent imaging can ring at edges and show speckle. A single resolution number is meaningless without naming the illumination: same lens, same wavelength, but the coherent cutoff is half the incoherent one.

Worked examples

1 · Where the diffraction orders land in the Fourier plane (confidence builder)

A grating of period $d=20\ \mu\mathrm{m}$ sits one focal length before a lens of $f=200\ \mathrm{mm}$ , illuminated by a HeNe laser, $\lambda=633\ \mathrm{nm}$ . Where do its orders appear in the back focal plane, and what does a small on-axis pinhole there do?

Step 1. The grating spectrum is a comb of spatial frequencies $f_m=m/d$ , each mapped by the lens to a position $x_m=\lambda f\,f_m=m\,\lambda f/d$ . For $m=1$ : $x_1=\dfrac{(633\times10^{-9})(0.200)}{20\times10^{-6}}=\dfrac{1.266\times10^{-7}}{2.0\times10^{-5}}\ \mathrm{m}=\mathbf{6.33\ mm}$ .

Step 2. So the Fourier plane shows bright spots at $x=0,\ \pm6.33,\ \pm12.66,\ \ldots\ \mathrm{mm}$ : the zero order on axis and the $\pm1,\pm2,\ldots$ orders stepping out by $6.33\ \mathrm{mm}$ each, the higher orders being the finer Fourier content of the grating profile.

Step 3. A pinhole passing only the central spot (radius under $6.33\ \mathrm{mm}$ ) blocks every order but DC, so the transmitted field is a uniform plane wave — the periodic structure is erased. This is laser spatial filtering, or beam cleanup: scattered high-frequency junk is dumped and only the clean on-axis beam survives. Widen the aperture to admit $0$ and $\pm1$ and a smoothed, sinusoidal image reappears. The ±1 orders sit at ±6.33 mm; a sub-6.33 mm pinhole passes only DC and erases the grating.

2 · Coherent versus incoherent resolution of one objective

An objective has $\mathrm{NA}=0.65$ and is used at $\lambda=550\ \mathrm{nm}$ . Find the finest grating period it can transmit under coherent and incoherent illumination.

Step 1. Coherent cutoff: $f_{c,\text{coh}}=\mathrm{NA}/\lambda=0.65/(550\times10^{-9})=1.18\times10^{6}\ \mathrm{m^{-1}}=1182\ \mathrm{cycles/mm}$ , so the finest transmissible period is $d_{\min}=\lambda/\mathrm{NA}=550/0.65=\mathbf{846\ nm}$ .

Step 2. Incoherent cutoff: the OTF is the pupil autocorrelation, reaching twice as far, $f_{c,\text{incoh}}=2\,\mathrm{NA}/\lambda=2364\ \mathrm{cycles/mm}$ , giving $d_{\min}=\lambda/(2\,\mathrm{NA})=550/1.30=\mathbf{423\ nm}$ .

Step 3. The same glass resolves $846\ \mathrm{nm}$ coherently but $423\ \mathrm{nm}$ incoherently — a factor of two, entirely from the different transfer functions. Since the OTF tapers from the origin, contrast near $400\ \mathrm{nm}$ is weak though nonzero: cutoff and usable resolution are not the same. d_min = 846 nm coherent, 423 nm incoherent — the incoherent system resolves twice as fine.

3 · The Fraunhofer distance, and why we cheat with a lens

A circular aperture of diameter $D=1\ \mathrm{mm}$ is illuminated at $\lambda=500\ \mathrm{nm}$ . How far away is the true far field, and how does a lens shortcut it?

Step 1. With radius $a=0.5\ \mathrm{mm}$ , the far field needs $N_F\ll1$ , i.e. $z\gg a^2/\lambda=(0.5\times10^{-3})^2/(500\times10^{-9})=2.5\times10^{-7}/5\times10^{-7}=\mathbf{0.5\ m}$ . The stricter rule $z\gtrsim2D^2/\lambda=2(10^{-3})^2/(5\times10^{-7})=4\ \mathrm{m}$ tolerates less edge phase; both say "of order a metre or more."

Step 2. To photograph the Airy pattern of a $1\ \mathrm{mm}$ hole in free space you would need metres of bench. Insert a lens instead: equation (6) puts the identical transform in the back focal plane at the finite distance $f$ . A $200\ \mathrm{mm}$ lens delivers it $0.2\ \mathrm{m}$ away, and phase-clean.

Step 3. The physics is unchanged — same $\tilde U_0$ , same Airy rings — only the geometry is folded up, which is why every real Fourier-optics setup uses a lens rather than a long drift. Free-space far field needs roughly 0.5–4 m; a 200 mm lens puts the same transform 0.2 m away.

Physics of the portrait

Fourier optics has many parents — Fresnel supplied the diffraction integral, Zernike the phase filter — but the person who first understood that an image is built by recombining a spectrum, and that the aperture decides which spectrum survives, was the man who put Zeiss microscopy on a theoretical footing.

Origins · 1 minErnst Abbe and how this was discovered

Ernst Abbe

1840–1905 · the diffraction theory of image formation and the microscope resolution limit

Ernst Abbe was a physics lecturer at Jena when Carl Zeiss, frustrated that his workshop's microscopes were still designed by trial and error, hired him in 1866 to make lens-making a science. Out of that partnership came, in 1873, the diffraction theory of image formation: a microscope does not simply project an object but first diffracts the illuminating light into orders, collects as many of those orders as its aperture allows, and recombines them into an image. Fine detail lives in the high diffraction orders, which leave at steep angles; if the objective's aperture cannot catch them, that detail is gone and no magnification restores it. This gave the first correct resolution limit, $d_{\min}\sim\lambda/(2\,n\sin\theta)$ , and in the process defined the numerical aperture $\mathrm{NA}=n\sin\theta$ as a lens's figure of merit. It is also, implicitly, Fourier optics: the "orders in the back focal plane" are the object's spatial spectrum, and blocking some of them is spatial filtering — a point Albert Porter demonstrated directly in 1906 by masking the focal plane of an imaged wire mesh. Abbe drove the invention of oil-immersion objectives and apochromats, and, as Zeiss's co-owner, wrote pioneering profit-sharing and eight-hour-day statutes into the foundation that still bears the company's name.

History

1818

— Fresnel's prize memoir casts diffraction as a quadratic-phase (Fresnel) integral

1873

— Abbe's diffraction theory ties microscope resolution to the collected spectrum and defines numerical aperture

1906

— Porter confirms Abbe by spatial filtering of a wire-mesh grating in the focal plane

1935

— Zernike introduces phase contrast, a phase filter acting on the zero order

1963

— Vander Lugt builds the matched spatial filter for optical pattern recognition

Going further

Texts. Goodman, Introduction to Fourier Optics, 3rd/4th ed., Ch. 3–6 (angular spectrum, Fresnel and Fraunhofer diffraction, lenses, coherent and incoherent imaging) · Saleh and Teich, Fundamentals of Photonics, 3rd ed., Ch. 4 (Fourier optics) · Hecht, Optics, 5th ed., Ch. 10–11 · Born and Wolf, Principles of Optics, Ch. 8–9 · Loudon, The Quantum Theory of Light, 3rd ed. (angular spectrum and modes).

Provenance. The paraxial transfer function follows from the Helmholtz dispersion $k_x^2+k_y^2+k_z^2=k^2$ (Goodman Ch. 3); the lens Fourier-transform relation from cascading two Fresnel integrals with the lens quadratic phase and enforcing cancellation at $d_1=d_2=f$ (Goodman Ch. 5, Saleh and Teich Ch. 4); the Fresnel-number crossover from the edge-phase budget $\pi N_F$ , of the same order as the $2D^2/\lambda$ far-field rule (Hecht Ch. 10); the coherent cutoff $\mathrm{NA}/\lambda$ and incoherent cutoff $2\,\mathrm{NA}/\lambda$ from the pupil support and its autocorrelation (Goodman Ch. 6); and the Zernike result $I\approx1+2\phi$ from the standard weak-phase-object treatment (Born and Wolf). No non-standard constants are asserted; the worked numbers ( $x_1=6.33\ \mathrm{mm}$ , $846/423\ \mathrm{nm}$ , $0.5\ \mathrm{m}$ ) are re-derivable from the equations above.

Builds on: Fourier transforms and linear systems · Scalar wave equation and Helmholtz equation · Numerical Aperture and Acceptance Angle · Paraxial approximation. Leads to: Integrated Photonic Waveguides and Circuits · Coupled-Mode Theory and Directional Couplers · Microring Resonators and Optical Cavities on Chip.

Ask the tutor: Ask me to walk you through building a 4f low-pass filter for your own object — we will place its diffraction orders in the Fourier plane and size the pinhole together.

Why does a lens perform a Fourier transform at all?

Because a lens focuses parallel rays to a point, and a single transverse spatial frequency is exactly a set of parallel rays (a tilted plane wave). Every plane-wave component of the object leaves at its own angle; the lens catches each and drops it onto a distinct point in the back focal plane, steeper angles landing farther out. Since angle is proportional to spatial frequency, the lens physically sorts the object's spectrum across a plane. The derivation confirms it — the lens quadratic phase cancels the propagation quadratic phase when the object is one focal length away, leaving the pure transform kernel — but the ray picture is the intuition: focusing parallel rays is Fourier-transforming.

Do I need a laser (coherent light) for the Fourier plane to exist?

No. Any monochromatic point source produces a Fourier plane at the back focal plane, because the lens transforms fields regardless of their statistics. What coherence changes is how contributions add and hence what the imaging transfer function looks like. Coherently you add complex amplitudes, the transfer function is the pupil, and the cutoff is $\mathrm{NA}/\lambda$ ; incoherently you add intensities, the transfer function is the pupil autocorrelation (the OTF), and the cutoff is $2\,\mathrm{NA}/\lambda$ . Strict complex-amplitude filtering — matched filters, phase-only masks — does need coherent light so that phases are well defined, which is why classic 4f processors use lasers.

What is the difference between the Fourier plane and an image plane?

They are conjugate opposites. In the Fourier (back focal) plane, position encodes spatial frequency: a single bright point spreads into a broad pattern, a periodic object collapses to a few sharp spots. In an image plane, position encodes object position: each point maps back to a point. A 4f system alternates them — object, Fourier plane at the shared focus, image — so filtering happens in the middle where frequencies are laid out, and viewing at the end where the picture is reassembled. Put a mask in an image plane and you block regions of the picture; put it in the Fourier plane and you block spatial frequencies everywhere at once.

Why is the incoherent cutoff exactly twice the coherent one?

Because the incoherent transfer function is the autocorrelation of the pupil, and autocorrelation doubles the support. Coherently, the system passes any spatial frequency whose tilted plane wave still fits through the pupil, giving a cutoff at $\mathrm{NA}/\lambda$ . Incoherently, the image intensity is built from products of amplitudes, so its transfer function is $P\star P$ ; the autocorrelation of a disc of radius $R$ is nonzero out to separation $2R$ , so the highest transmitted frequency doubles to $2\,\mathrm{NA}/\lambda$ . The catch is that the OTF falls steadily from its peak, so the finest frequencies are transmitted with very low contrast — the cutoff is twice as high, usable resolution improves by somewhat less.

Is the Fraunhofer pattern really the Fourier transform, including its phase?

Yes, up to a known quadratic phase and a scale. Equation (4) is literally $\tilde U_0$ evaluated at $f_x=x/\lambda z$ , times an overall $e^{ik(x^2+y^2)/2z}$ curvature. On a screen you measure only intensity, $|\tilde U_0|^2$ , and that curvature drops out, so you record the power spectrum but lose the phase — which is why recovering an object from its diffraction pattern (the phase problem) is hard and needs iterative algorithms or a reference beam. Put the object in the front focal plane of a lens, however, and the curvature vanishes exactly (equation 6): the back focal plane then carries the complete complex transform, phase included, ready for complex filtering.

How is optical spatial filtering related to filtering an image on a computer?

They are the same operation in different hardware. A digital filter takes the FFT of an image, multiplies the spectrum by a mask, and inverse-FFTs; a 4f optical processor takes the optical Fourier transform with a lens, multiplies by a physical mask in the Fourier plane, and inverse-transforms with a second lens. Low-pass smoothing, high-pass edge enhancement, notch removal of periodic noise, and matched-filter correlation mean the same in both. The differences are practical: the optical version is analog, massively parallel and instantaneous but limited to what a physical mask can implement, while the digital version is flexible and exact but serial. This is why Fourier optics is often taught as "image processing done by glass."