Fourier Transform Property of a Lens

Hold a slide in front of a converging lens, shine a laser through it, and look one focal length behind the glass. What you find there is not a blurred picture of the slide — it is the slide's spatial-frequency spectrum, its two-dimensional Fourier transform, spread out on a plane and computed at the speed of light. Fine detail in the object lands far from the axis; coarse structure crowds near the centre; a plain uniform patch collapses to a single bright dot on axis. A passive piece of glass, doing no arithmetic, has performed a continuous Fourier integral in parallel over the whole beam. The mechanism is a phase accident that is too clean to be coincidence. A converging lens is thick in the middle and thin at the rim, so it retards the wavefront most on axis: to paraxial order it multiplies the field by a quadratic phase, one focal length of curvature's worth. Free-space (Fresnel) propagation also multiplies by a quadratic phase, of the opposite sign. Put the object exactly one focal length in front of the lens and propagate to the back focal plane, and the two quadratic phases annihilate term by term, stripping the diffraction integral down to a bare Fourier kernel. Position in the focal plane then reads off spatial frequency directly, $f_X = x/\lambda f$ . This single result is the hinge of Fourier optics. It says that the Fraunhofer diffraction pattern of any aperture — the Airy ring of a hole, the sinc of a slit — is simply that aperture's Fourier transform brought from infinity to a plane you can reach with a ruler. It says imaging is two Fourier transforms back to back, so a mask in the focal plane edits the picture's frequencies by hand: the 4f processor, phase contrast, matched filtering. The lens is the analog computer that made spatial-frequency thinking physical.

Key takeaways

A thin lens of focal length $f$ imprints a quadratic phase $\exp[-ik(x^2+y^2)/2f]$ that is the exact conjugate of the Fresnel propagator's quadratic phase.
Place the object in the FRONT focal plane; the field in the back focal plane is the exact 2-D Fourier transform of the object, with no leftover phase.
Focal-plane position maps to spatial frequency as $f_X = x/\lambda f$ : a tilted plane wave (one Fourier component) focuses to $x = f\theta$ .
The lens delivers the Fraunhofer pattern of any aperture at finite distance $f$ instead of at infinity — no huge Fresnel-number path needed.
Move the object off the front focal plane and the transform magnitude survives but a residual quadratic phase $\propto(1-d/f)$ appears — invisible to a detector, fatal to phase filtering.
Key idea: with the object one focal length in front, a lens turns the Fresnel diffraction integral into a pure Fourier transform, laying the object's spatial spectrum out on the back focal plane at frequency $f_X = x/\lambda f$ .

The core insight

A thin lens multiplies the field by a quadratic phase that exactly cancels the Fresnel propagator's quadratic phase. With the object in the front focal plane, the back-focal-plane field is the exact 2-D Fourier transform of the object, and focal-plane position reads spatial frequency directly, $f_X = x/\lambda f$ .

Where this lives today

Every optical spatial filter on a laser table is this equation at work: a microscope objective Fourier-transforms a noisy beam, a pinhole at the focus passes only the clean low-frequency core, and a second lens recollimates — the ubiquitous "spatial filter" that turns a speckled beam into a smooth Gaussian. In microscopy the focal plane hosts Zernike's phase-contrast dot and dark-field stops, both Fourier-plane masks. Optical correlators for pattern recognition place a reference spectrum in the focal plane so the output is a correlation peak. And the whole architecture scales to the 4f processor and, conceptually, to every diffraction-limited imaging system, where the pupil is nothing but the aperture of this transform. The lens-as-transformer is also the mental model behind Fourier-transform infrared spectrometers' imaging optics and behind the design of phase masks for extended depth of field in modern camera modules.

The physical picture

Decompose the object into a sum of sinusoidal transmission gratings. A grating of spatial frequency $f_X$ splits an incoming plane wave into diffracted beams tilted by $\theta \approx \lambda f_X$ . A lens focuses every parallel bundle to a single point, and a tilt $\theta$ lands at height $x = f\theta$ in the back focal plane. So the lens is a sorter: it takes each spatial-frequency component of the object — each grating, each tilt — and drops it at its own address $x = \lambda f\, f_X$ . That sorting, done for all frequencies at once, is precisely what the Fourier transform does on paper. Why is the result exact and not merely "far-field-like"? Because the lens supplies the one phase that free-space propagation is missing. Propagation over a distance $z$ carries the paraxial factor $\exp[+ik r^2/2z]$ — a spreading spherical shell. The converging lens carries $\exp[-ik r^2/2f]$ — the same curvature with the opposite sign. Tune $z = f$ and place the object at the front focal point, and the two curvatures cancel exactly, leaving only the linear phase $\exp[-i2\pi f_X x_0]$ that is the Fourier kernel. The transform is diffraction with its quadratic phase surgically removed by the glass.

Figure

The lens as a spatial-frequency sorter — each plane-wave tilt θ focuses to its own address x ≈ f·θ in the back focal plane

The scaling knob is $\lambda f$ . Because $x = \lambda f\, f_X$ , a longer focal length or a longer wavelength spreads the spectrum out — more millimetres per line-pair — while a short lens compresses it. This is exactly how you match a spectrum to the pixel pitch of a spatial light modulator or the size of a filter mask: pick $f$ .

Figure

Punchline — the 2f transform geometry with real numbers: at λ = 633 nm, f = 200 mm, a point 10 mm off axis reads a spatial frequency of 79 line-pairs per mm

The core equations

A thin lens acts as a pure phase screen. Its thickness falls off quadratically from the axis, so, dropping an irrelevant constant, it multiplies the field by a quadratic phase whose curvature is set entirely by the focal length. The sign is fixed by demanding that this phase cancel the spreading phase of Fresnel propagation — a converging lens must undo divergence.

t_L(x,y)=\exp\!\Big[-\,i\,\frac{k}{2f}\,(x^2+y^2)\Big] \qquad (1)

where: $t_L$ — complex amplitude transmittance of the ideal thin lens; $k=2\pi/\lambda$ — wavenumber; $f>0$ — focal length (converging); $x,y$ — transverse coordinates in the lens plane. Read it physically: the on-axis field ( $x=y=0$ ) is retarded most; the phase is negative and increasingly so toward the rim, the exact curvature a diverging spherical wave from the focus would carry — which is why the lens collimates a point source at $f$ . Free-space propagation over a distance $z$ , in the same convention, contributes the opposite-sign quadratic phase. This is the object the lens is built to cancel.

h_z(x,y)=\frac{e^{ikz}}{i\lambda z}\,\exp\!\Big[+\,i\frac{k}{2z}(x^2+y^2)\Big] \qquad (2)

where: $h_z$ — paraxial (Fresnel) free-space impulse response over distance $z$ ; the sign of the quadratic phase is $+$ , opposite to the lens. The cancellation $t_L$ against $h_f$ at $z=f$ is the whole story. With the object $U_0$ in the front focal plane, propagating to the lens, multiplying by $t_L$ , and propagating one more focal length yields a bare Fourier integral.

U_f(x,y)=\frac{e^{i2kf}}{i\lambda f}\iint U_0(x_0,y_0)\,\exp\!\Big[-i\frac{2\pi}{\lambda f}(x x_0+y y_0)\Big]\,dx_0\,dy_0 \qquad (3)

where: $U_f$ — back-focal-plane field; $U_0$ — front-focal-plane (object) field; the kernel $\exp[-i2\pi(x x_0)/\lambda f]$ is the Fourier kernel with frequency $f_X = x/\lambda f$ . In compact form the statement is unmistakable:

U_f(x,y)=\frac{e^{i2kf}}{i\lambda f}\;\hat U_0\!\left(\frac{x}{\lambda f},\frac{y}{\lambda f}\right) \qquad (4)

where: $\hat U_0$ — the 2-D Fourier transform of $U_0$ ; $f_X=x/\lambda f$ and $f_Y=y/\lambda f$ — spatial frequencies mapped to focal-plane position. The prefactor $e^{i2kf}/i\lambda f$ is a global phase and scale, irrelevant to the pattern's shape. If the object sits a general distance $d$ in front of the lens rather than at $d=f$ , the transform survives but drags a residual quadratic phase whose coefficient measures the mismatch.

U_f(x,y)=\frac{e^{ikf}}{i\lambda f}\,\exp\!\Big[i\frac{k}{2f}\Big(1-\frac{d}{f}\Big)(x^2+y^2)\Big]\;\hat U_0\!\left(\frac{x}{\lambda f},\frac{y}{\lambda f}\right) \qquad (5)

where: $d$ — object-to-lens distance; the factor $(1-d/f)$ is the residual-phase coefficient, and it vanishes identically at $d=f$ . Intensity $|U_f|^2$ is untouched by this phase; complex-amplitude work is not.

The physical picture in one identity: angle equals frequency

x = f\theta,\qquad f_X=\frac{\theta}{\lambda}=\frac{x}{\lambda f} \qquad (6)

where: $\theta$ — tilt of an incident plane-wave component; $x$ — its focus in the back focal plane. A plane wave $\exp[i k \theta x_0]$ is a single Fourier component at frequency $\theta/\lambda$ , and the lens sends it to $x=f\theta$ . The Fourier transform and the ray picture are the same statement read twice.

Derivation AThe Fourier-transform property from two Fresnel integrals and one lens

The goal is to show, without hand-waving, that the back-focal-plane field is the exact Fourier transform of a front-focal-plane object — no residual phase. The strategy is a paraxial sandwich: Fresnel-propagate the object one focal length to the lens, multiply by the lens quadratic phase, Fresnel-propagate one more focal length to the back plane, and watch every quadratic term in the lens-plane variable cancel. Work in one dimension for clarity; the 2-D result is a product of identical factors. The single load-bearing move is the sign match between lens and propagator.

Step 1. Propagate $U_0$ from the front focal plane to the lens: $U_L^-(x_1)=\frac{e^{ikf}}{i\lambda f}\int U_0(x_0)\,e^{\,i\frac{k}{2f}(x_1-x_0)^2}\,dx_0$ . Why this form: the Fresnel integral over distance $f$ carries the $+$ quadratic kernel of equation (2), and expanding $(x_1-x_0)^2$ will separate a pure lens-plane phase from the cross term that becomes the transform.

Step 2. Expand $(x_1-x_0)^2 = x_1^2 - 2x_1 x_0 + x_0^2$ . Why it matters: the $x_1^2$ piece is a phase living entirely in the lens plane — exactly what the lens is positioned to cancel — while $-2x_1x_0$ is the seed of the Fourier kernel and $x_0^2$ is an object-plane phase to be dispatched later.

Step 3. Apply the lens: $U_L^+(x_1)=U_L^-(x_1)\,e^{-i\frac{k}{2f}x_1^2}$ . The lens factor is exactly $-x_1^2$ curvature, so it annihilates the $+x_1^2$ generated by propagation. Why this is the crux: with the $x_1^2$ phase gone, the remaining $x_1$ dependence is purely linear, and a linear phase integrates to a delta-like Fourier kernel rather than a Fresnel spread.

Step 4. Propagate the lens field one focal length to the back plane, $U_f(x)=\frac{e^{ikf}}{i\lambda f}\int U_L^+(x_1)\,e^{\,i\frac{k}{2f}(x-x_1)^2}\,dx_1$ , and collect the coefficient of $x_1^2$ : it is $\tfrac{k}{2f}-\tfrac{k}{2f}=0$ (lens-cancelled term plus the new propagation term). Why the object had to sit at exactly $z=f$ : only then do the surviving $x^2$ and $x_0^2$ phases pair up and cancel, leaving no chirp on the transform; any other object distance leaves the $(1-d/f)$ residue of equation (5).

Step 5. The $x_1$ integral is now $\int \exp[-i\frac{k}{f}(x+x_0)x_1]\,dx_1$ , a linear phase whose integration ties $x$ to $x_0$ through the kernel $\exp[-i\frac{k}{f}x x_0]=\exp[-i\frac{2\pi}{\lambda f}x x_0]$ . Assembling the surviving factors gives $U_f(x)=\frac{e^{i2kf}}{i\lambda f}\int U_0(x_0)\,e^{-i\frac{2\pi}{\lambda f}x x_0}\,dx_0=\frac{e^{i2kf}}{i\lambda f}\hat U_0(x/\lambda f)$ — the claimed exact transform. Verified numerically and symbolically in verify_fourier-transform-lens.py: a discrete Fresnel–lens–Fresnel cascade on a test object reproduces its FFT (phase-free) to machine precision only when the object is at $z=f$ ; the residual-phase coefficient matches $(1-d/f)$ for $d\ne f$ .

Go deeper · Derivation B · 1 minWhere the object must sit — the residual quadratic phase for general distance d

Derivation BWhere the object must sit — the residual quadratic phase for general distance d

The goal is to quantify the price of misplacing the object: how the back-focal-plane field departs from a clean transform when the object is a distance $d\ne f$ in front of the lens. The strategy repeats the sandwich of the previous card but keeps the first propagation distance as a free parameter $d$ , tracking the one quadratic phase that no longer cancels. The payoff is a single coefficient, $(1-d/f)$ , that tells you when phase-sensitive filtering is safe.

Step 1. Propagate the object a distance $d$ (not $f$ ) to the lens, apply $t_L=e^{-ik x_1^2/2f}$ , then propagate $f$ to the back plane. Why keep $d$ general: the lens still cancels the $x_1^2$ term that the SECOND propagation over $f$ produces, but the FIRST propagation over $d$ now deposits an $x_0^2$ -and- $x^2$ bookkeeping that depends on $d$ .

Step 2. Carry the algebra of the two Fresnel integrals with the lens between them; the transform kernel $\exp[-i2\pi x x_0/\lambda f]$ emerges unchanged, so the magnitude $|\hat U_0(x/\lambda f)|$ is independent of $d$ . Why the scale never moves: the frequency map $f_X=x/\lambda f$ is fixed by the final propagation distance $f$ alone, which is why every text quotes $x/\lambda f$ regardless of object placement.

Step 3. The uncancelled phase collects into $\exp[i\frac{k}{2f}(1-\frac{d}{f})(x^2+y^2)]$ , equation (5). At $d=f$ it is unity; at $d=0$ (object against the lens) it is the full $\exp[+ikx^2/2f]$ . Why detectors do not care but holograms do: $|U_f|^2$ discards the phase, so intensity spectra and Fraunhofer patterns are pristine at any $d$ , whereas a phase-contrast dot or a holographic matched filter reads the complex field and is corrupted by the chirp. *Verified in verify_fourier-transform-lens.py: the simulated focal-plane phase across the field fits $\tfrac{k}{2f}(1-d/f)r^2$ for a scan of $d$ , and $|U_f|^2$ is $d$ -independent.*

Limiting cases

Case	Result	Meaning
Object in front focal plane, $d=f$	$U_f\propto \hat U_0(x/\lambda f)$ , no residual phase	The ideal configuration — the true complex spectrum, ready for amplitude or phase filtering.
Object against the lens, $d=0$	$U_f\propto e^{ikr^2/2f}\,\hat U_0(x/\lambda f)$	Correct transform magnitude, full residual chirp; intensity is unaffected, complex filtering is not.
Plane-wave illumination of an aperture $A$	$I_f\propto\|\hat A(x/\lambda f)\|^2$	The Fraunhofer pattern of $A$ (Airy ring, sinc) appears at finite distance $f$ .
Double the focal length	$x=\lambda(2f)f_X$ , intensity $\propto 1/f^2$	Spectrum spreads out twofold (more mm per line-pair), at the cost of $4\times$ dimmer.

Common trapThe clean transform lives ONLY in the front focal plane — anywhere else you keep the magnitude and lose the phase

It is tempting to think "the lens Fourier-transforms whatever is in front of it." It transforms the magnitude from any distance, but the phase-free transform — the true complex spectrum $\hat U_0$ with no chirp — exists only when the object sits exactly one focal length in front, $d=f$ . Put the slide against the lens, or at an arbitrary distance, and a quadratic phase $\exp[i\frac{k}{2f}(1-d/f)r^2]$ multiplies the spectrum. Your power meter and your photograph of the Fraunhofer pattern will never notice. But a phase-contrast plate, a holographic matched filter, or any downstream coherent stage reads the complex amplitude and will be spoiled by the hidden curvature. When the physics is phase-sensitive, the front-focal-plane placement is not a nicety — it is the whole game.

Worked examples

1 · A grating's diffraction orders, measured on the bench (confidence builder)

A transmission grating of period $\Lambda = 10\ \mu\text{m}$ sits in the front focal plane of a lens with $f = 200\ \text{mm}$ , illuminated by a He–Ne laser, $\lambda = 633\ \text{nm}$ . Where do its diffraction orders focus?

Step 1. A grating of period $\Lambda$ has spatial frequency $f_X = 1/\Lambda = 100\ \text{mm}^{-1}$ in its fundamental, with harmonics at $m/\Lambda$ . The lens maps each to $x_m = \lambda f\, (m/\Lambda)$ . Why the orders are evenly spaced: the grating's spectrum is a comb of delta functions at $m/\Lambda$ , and the linear map $x=\lambda f f_X$ preserves even spacing.

Step 2. Compute the order spacing $\Delta x = \lambda f/\Lambda = (633\times10^{-9}\,\text{m})(0.200\,\text{m})/(10\times10^{-6}\,\text{m})$ . Evaluating gives $\Delta x = \mathbf{12.66\ mm}$ between successive orders in the focal plane.

Step 3. So the $\pm1$ orders sit $12.66$ mm either side of the central (DC) spot, the $\pm2$ orders at $25.3$ mm, and so on until an order's focus runs off the lens's clear aperture and is lost — the physical origin of the resolution cutoff. Why this is a spectrometer: read off $x_m$ , invert $\Lambda = \lambda f/\Delta x$ , and you have measured the grating period optically.

2 · Reading a spatial frequency off the focal plane (exam level)

In the same system, a bright feature appears $x = 10\ \text{mm}$ off axis in the back focal plane. What spatial frequency in the object produced it?

Step 1. Invert the frequency map: $f_X = x/\lambda f = (10\times10^{-3}\,\text{m})/[(633\times10^{-9}\,\text{m})(0.200\,\text{m})]$ . Why no other geometry enters: the map depends only on $\lambda$ and $f$ , not on the object.

Step 2. The denominator is the area $\lambda f = (633\times10^{-9}\,\text{m})(0.200\,\text{m}) = 1.266\times10^{-7}\ \text{m}^2 = 0.1266\ \text{mm}^2$ , so $f_X = 10\,\text{mm}/0.1266\,\text{mm}^2 = \mathbf{79\ line\text{-}pairs/mm}$ . A structure repeating every $1/79\ \text{mm} = 12.7\ \mu\text{m}$ in the object lands there — consistent with the $10\ \mu\text{m}$ grating's slightly-lower-frequency fundamental at $12.66$ mm.

Physics of the portrait

The transform that the lens performs in glass was written down in heat, not light. In 1807 Joseph Fourier, prefect of Isère and veteran of Napoleon's Egyptian expedition, claimed that any function on an interval could be built from sines and cosines — a claim so audacious that Lagrange resisted it and the Academy withheld its prize for years. Fourier needed the series to solve the diffusion of heat in a solid; he could not have imagined that a century and a half later a converging lens would evaluate his integral optically, in parallel, at the speed of light, turning a slide into its spectrum on a plane you can touch. The lens does not know it is doing analysis. It only knows curvature — and curvature, it turns out, is Fourier's transform wearing a different coat.

Origins · 1 minjoseph fourier and how this was discovered

joseph fourier

1768–1830 · French mathematician and physicist; his analytic theory of heat introduced the series and transform that now bear his name

Jean-Baptiste Joseph Fourier rose from an orphaned tailor's son to the scientific elite of post-Revolutionary France, surviving the Terror, governing Egypt's antiquities under Bonaparte, and administering the department of Isère — where, legend has it, his conviction that heat was the fundamental agent led him to overheat his rooms and swaddle himself against draughts. His masterwork, Théorie analytique de la chaleur (1822), solved heat conduction by expanding temperature profiles in trigonometric series, and in doing so gave physics its most reusable tool: the decomposition of any signal into frequencies. Contemporaries balked — Lagrange doubted that a sum of smooth sines could reproduce a corner — but Fourier was right in every way that mattered for physics. Optics inherited the idea whole. A diffraction pattern is a Fourier transform of an aperture; a lens computes that transform; coherent imaging is Fourier analysis followed by synthesis. When we say the focal plane "is the spectrum," we are speaking Fourier's language literally.

History

1807

— Fourier submits his memoir on heat conduction, asserting arbitrary functions expand in trigonometric series

1822

— Publication of Théorie analytique de la chaleur, the analytic theory of heat

1873

— Ernst Abbe explains microscope image formation as diffraction plus reconstruction, making the spectrum physical

1935

— Frits Zernike invents phase contrast, a Fourier-plane phase filter, later a Nobel-winning microscopy method

1964

— Anthony VanderLugt builds the holographic matched filter, realising optical correlation in a lens's focal plane

Going further

References. Goodman, Introduction to Fourier Optics, 3rd ed., Ch. 5 (Wave-Optics Analysis of Coherent Optical Systems); Hecht, Optics, 5th ed., §11.3 and §13.2 (Fourier methods); Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 4 (Fourier Optics); Born & Wolf, Principles of Optics, Ch. 8.

Provenance. The lens phase (eq. 1) and its lensmaker origin follow Goodman §5.1 and Hecht §5.7; the sign is $t_L=\exp[-ik r^2/2f]$ for a positive lens (Goodman Eq. 5-10), opposite to the Fresnel kernel's $+$ so the two cancel. The main derivation reproduces Goodman Eqs. (5-19)–(5-24); verify_fourier-transform-lens.py recomputes the cascaded Fresnel–lens–Fresnel integrals and confirms the $x_1^2$ cancellation, the mapping $f_X=x/\lambda f$ , and the $(1-d/f)$ residual coefficient. Base-error repair: the dry base wrote the lens phase with a $+$ sign in its equation card and one derivation step (inconsistent with its own tldr, key-idea, and cancellation step, and with Goodman); corrected here to $-$ so it genuinely cancels the Fresnel $+$ phase, with the sign match made the load-bearing point of the derivation. Numerical anchors $\Delta x=\lambda f/\Lambda=12.66$ mm and $f_X=x/\lambda f=79$ line-pairs/mm are recomputed in the verifier.

Builds on: Fraunhofer Diffraction Regime · Fresnel-Kirchhoff Diffraction Integral · Fresnel Number Leads to: Spatial Filtering and the 4f System · Circular Aperture Diffraction and the Airy Pattern · Fourier Optics

Is the lens really computing a Fourier transform, or just something that looks like one?

Literally computing it. Equation (4) is an exact equality (in the paraxial, scalar regime) between the back-focal-plane complex field and the mathematical 2-D Fourier transform of the object, evaluated at $f_X=x/\lambda f$ . It is not an analogy or an approximation once the object is in the front focal plane: verify_fourier-transform-lens.py reproduces a numerical FFT of a test object, phase and all, from a simulated propagate–lens–propagate cascade. The only caveats are the usual paraxial and scalar-field assumptions and finite aperture; within those, the lens is an analog Fourier processor.

What exactly maps to position in the focal plane — angle, frequency, or size?

All three, because they are the same thing seen differently. A plane wave tilted by angle $\theta$ is a single Fourier component of spatial frequency $f_X=\theta/\lambda$ , and it focuses to $x=f\theta=\lambda f f_X$ . So a large focal-plane radius $x$ means a large tilt, a high spatial frequency, and therefore fine object detail (small period $1/f_X$ ). Coarse features cluster near the axis; the finest detail lives at the rim, which is exactly why a finite lens aperture sets the resolution limit.

Why is the sign of the lens phase negative when the base and some textbooks show positive?

The sign is convention-relative but not free: it must be opposite to whatever sign the Fresnel propagator carries, so the two quadratic phases cancel. With the propagator written $\exp[+ik r^2/2z]$ (as in eq. 2 and Goodman's convention), a positive lens must be $\exp[-ik r^2/2f]$ (Goodman Eq. 5-10). A text using the opposite time convention flips both signs together and the cancellation still works. The error to avoid — which the dry base contained — is writing the lens with the same sign as the propagator, in which case nothing cancels and there is no clean transform. Always check that lens and propagator disagree in sign.

If intensity is unaffected by object placement, why insist on the front focal plane?

Because more and more optics is phase-sensitive. If all you do is photograph the Fraunhofer pattern or read a power spectrum, any object distance works — $|U_f|^2$ ignores the residual chirp. But the moment you place a phase mask (Zernike phase contrast), record a hologram, build a matched filter, or feed the field into a second coherent stage, the complex amplitude matters, and the hidden $\exp[i\frac{k}{2f}(1-d/f)r^2]$ curvature distorts the result. The front-focal-plane rule is the guarantee that the complex spectrum is clean.

How does this connect to ordinary imaging, where I don't place things at focal planes?

Ordinary imaging is two of these transforms back to back. A single lens forming a real image can be analysed as: object spectrum formed at the pupil (a Fourier transform), then a second transform from pupil to image. The pupil plane of any imaging system is a Fourier plane — it holds the object's spectrum, band-limited by the aperture stop. That is why the aperture controls resolution and why the optical transfer function (a sibling article) is defined there. The 4f system just makes the pupil accessible so you can reach in and filter it.

Does a longer wavelength change where frequencies land?

Yes, through the same scale factor $\lambda f$ . Since $x=\lambda f f_X$ , red light ( $633$ nm) spreads a given spatial frequency farther from the axis than blue ( $450$ nm) in the same lens — the spectrum is wavelength-scaled. In white-light Fourier optics this smears each frequency into a short radial spectrum, which is why coherent, monochromatic illumination (a laser) is used whenever the focal plane must be a clean, single-wavelength map of spatial frequency.