Introduction to Fourier Optics

Place a fine wire mesh in a laser beam, set a simple lens one focal length behind it, and look one focal length beyond the lens. You will not see the mesh. Instead you find a sharp constellation of bright dots hanging in the air — the mesh's spatial-frequency spectrum, sorted by angle and laid flat. Slide a pin into that plane to cover one dot and the mesh reappears in the image with that one periodicity erased. The lens has, at the speed of light, computed a two-dimensional Fourier transform of the field that crossed the mesh — and you have edited a picture by editing its frequencies. That single fact — a lens turns an object's spatial-frequency content into a physical pattern you can reach in and change — is the seed of Fourier optics. The deeper statement is that free-space propagation is linear and shift-invariant, hence diagonalised by complex exponentials: the natural basis is tilted plane waves, each labelled by a spatial frequency, and in that basis propagation is nothing but multiplication by a phase — the transfer function of free space. This viewpoint unifies the diffraction results treated separately elsewhere: the Fraunhofer pattern is the transform of the aperture, the Fresnel regime a quadratic-phase convolution, imaging transform-filter-inverse-transform. It also fixes the diffraction limit, since frequencies above $1/\lambda$ are evanescent and never reach the lens. Here we build the transforms, derive the lens transform from phase cancellation, and read off the $\lambda/2$ limit.

Key takeaways

Free-space propagation is linear and shift-invariant, hence diagonal in the plane-wave basis: each component $e^{i2\pi(f_x x+f_y y)}$ acquires a phase $H=e^{i2\pi z\sqrt{1/\lambda^2-f^2}}$ .
A lens performs a physical 2-D Fourier transform: an object one focal length in front appears in the back focal plane as its exact transform, with spatial frequency mapped to position by $x=\lambda f\,f_x$ .
Fraunhofer diffraction is that transform evaluated at infinity; the lens brings the same pattern to a finite distance $z=f$ .
Imaging is a cascade of two transforms — build the spectrum, low-pass it with the pupil, transform back — so blur is the loss of high spatial frequencies.
Spatial frequencies above the cutoff $f_{\max}=1/\lambda$ are evanescent and decay before the far field, capping the finest detail at $\Delta x_{\min}\approx\lambda/2$ .
Key idea: propagation is diagonal in spatial frequency; a lens (or the far field) converts that frequency-domain picture into a physical Fourier transform of the field, and the evanescent cutoff $f_{\max}=1/\lambda$ sets the diffraction limit $\Delta x_{\min}\approx\lambda/2$ .

The core insight

Free-space propagation is diagonal in the plane-wave basis: each tilted plane wave $e^{i2\pi(f_x x+f_y y)}$ picks up a phase $H=e^{i2\pi z\sqrt{1/\lambda^2-f^2}}$ . A lens (or the far field) converts this into a physical Fourier transform, mapping spatial frequency to position by $x=\lambda f\,f_x$ . Frequencies above $f_{\max}=1/\lambda$ are evanescent — exactly the diffraction limit $\Delta x_{\min}\approx\lambda/2$ .

Where this lives today

Fourier optics is the design language of coherent imaging and optical information processing. In photolithography the projection lens is analysed through its pupil and transfer function, and the resolution-enhancement tricks — off-axis illumination, phase-shift masks, source-mask optimisation — are Fourier-plane manipulations that reshape which spatial frequencies reach the wafer. In microscopy, phase contrast, dark field, and structured-illumination super-resolution are Fourier-plane edits, and Fourier ptychography stitches many low-NA transforms into one high-resolution image. The $4f$ correlator performs optical convolution at the speed of light, and spatial light modulators synthesise holograms, shape flat-top beams, and generate optical-tweezer arrays for single atoms. The same relations set the ceiling on every instrument. Because a finite pupil is a hard low-pass filter with cutoff $\mathrm{NA}/\lambda$ , the Optical Transfer Function fixes contrast versus spatial frequency, and the resolution limit that astronomers fight with larger mirrors and lithographers with shorter wavelengths is just the evanescent cutoff seen through an aperture. Reading a lens as a transform and an aperture as a filter is what makes all of this one design problem instead of many.

The physical picture

Think of any transverse field as a chord of tilted plane waves. A wave tilted by $\theta_x$ carries spatial frequency $f_x=\sin\theta_x/\lambda$ : the steeper the tilt, the faster its phase oscillates across a screen. Because space is homogeneous, propagation cannot mix these components — each simply advances in phase. That is "diagonal in the spatial-frequency basis," and it is why the far field, sorting waves by direction, displays the input's Fourier spectrum directly. A lens stamps on a quadratic phase $e^{-ik(x^2+y^2)/2f}$ , cancelling the phase Fresnel propagation over one focal length accumulates and leaving a clean transform in the back focal plane. It converts direction into position: every component leaving at angle $\theta$ reaches height $x\approx\lambda f f_x$ , so the focal plane maps the object's spatial-frequency content — the Fourier plane you reach into.

Figure

The single-lens (and 4f) Fourier engine: an object one focal length before the lens has its exact 2-D transform Ũ₀(x⁄λf) in the back focal plane, where a mask edits frequencies; a second lens inverse-transforms to the image

Fine spatial detail means high spatial frequency means a steeply tilted plane wave. Push the tilt until it would require $f_x^2+f_y^2>1/\lambda^2$ and the axial wavenumber turns imaginary: the wave stops propagating and clings to the aperture as an evanescent field, decaying within a fraction of a wavelength. A lens in the far field therefore never receives features finer than about $\lambda/2$ . The diffraction limit is not a defect of the glass — it is a statement about which spatial frequencies survive the trip.

Figure

The propagating band and the diffraction limit: the transfer function passes every frequency below the cutoff 1⁄λ = 2000 cyc⁄mm as pure phase (λ = 500 nm); beyond it waves are evanescent, so the finest detail is Δx_min ≈ λ⁄2 = 250 nm

Imaging follows the same script: the first lens builds the spectrum, the pupil truncates it, the second lens transforms back. Blur is the loss of high frequencies, and editing the Fourier plane — blocking, phase-shifting, apodising — edits the image, the mechanism behind spatial filtering, phase contrast, and dark field.

The core equations

The whole framework rests on the two-dimensional Fourier transform, which analyses a transverse field into its plane-wave components; the inverse integral rebuilds the field from its spectrum. Each spatial frequency $(f_x,f_y)$ is the label of a plane wave whose direction cosines are $(\lambda f_x,\lambda f_y)$ .

\tilde U(f_x,f_y)=\iint U(x,y)\,e^{-i2\pi(f_x x+f_y y)}\,\mathrm{d}x\,\mathrm{d}y \qquad (1)

where: $U(x,y)$ — complex scalar field amplitude in the transverse plane; $f_x,f_y$ — spatial frequencies in cycles per unit length, with $f_x=k_x/2\pi$ ; $\tilde U$ — angular spectrum, the amplitude and phase of each plane-wave component. Every step below is a rule for how these weights evolve. Because propagation cannot mix plane waves, it acts on each as a single multiplicative phase — the free-space transfer function, obtained by forcing every component $e^{i(k_x x+k_y y+k_z z)}$ to satisfy the Helmholtz equation, which fixes the axial wavenumber and the phase over $z$ .

H(f_x,f_y;z)=\exp\!\left(i\,2\pi z\sqrt{\tfrac{1}{\lambda^2}-f_x^2-f_y^2}\,\right) \qquad (2)

where: $z$ — propagation distance; $\lambda$ — wavelength in the medium; $f^2\equiv f_x^2+f_y^2$ ; $H$ — the phase (or, past cutoff, decay) on each component. For $f<1/\lambda$ the root is real and $H$ a pure phase; for $f>1/\lambda$ it is imaginary and $H=e^{-|k_z|z}$ decays. It is exact within scalar theory — the propagator of Spatial Frequency and the Angular Spectrum. For small angles the root expands and $H$ collapses to the Fresnel (paraxial) transfer function — a common axial phase times a quadratic defocus chirp, the frequency-domain face of the Fresnel Diffraction Regime.

H(f_x,f_y;z)\approx e^{ikz}\,e^{-i\pi\lambda z(f_x^2+f_y^2)},\qquad k=\frac{2\pi}{\lambda} \qquad (3)

where: $e^{ikz}$ — on-axis carrier phase; $e^{-i\pi\lambda z f^2}$ — quadratic phase of off-axis components, coefficient of $f^2$ exactly $-\pi\lambda z$ . Valid when $\lambda^2 f^2\ll1$ . The centrepiece is the lens. An ideal thin lens multiplies the field by a quadratic phase; with an object one focal length in front, propagating to the back focal plane leaves the exact transform of the object, no residual curvature — the operation the opening mesh experiment performed.

U_f(x,y)=\frac{1}{i\lambda f}\iint U_0(\xi,\eta)\,e^{-i\frac{2\pi}{\lambda f}(x\xi+y\eta)}\,\mathrm{d}\xi\,\mathrm{d}\eta \qquad (4)

where: $U_0(\xi,\eta)$ — field in the front focal plane (object); $U_f(x,y)$ — field in the back focal (Fourier) plane; $f$ — focal length; $1/i\lambda f$ — prefactor. Comparing with $(1)$ , the focal-plane coordinate maps to $f_\xi=x/\lambda f$ — position is spatial frequency, scaled by $\lambda f$ . Finally, a shift-invariant imaging system reproduces each object point as a fixed blur, so the image is the object convolved with the point-spread function; in Fourier space that convolution is a product with the transfer function. The finite pupil low-pass-filters the spectrum, which is what limits resolution.

g=h*f,\qquad H_{\text{coh}}(f_x,f_y)=P(\lambda d_i f_x,\lambda d_i f_y),\qquad \text{OTF}=\frac{P\star P}{\iint|P|^2} \qquad (5)

where: $f$ — ideal geometric image; $h$ — coherent point-spread function; $g$ — actual image; $P$ — pupil function; $d_i$ — image distance; $\star$ — autocorrelation; $H_{\text{coh}}$ — coherent (amplitude) transfer function, a hard low-pass filter; OTF — the incoherent transfer function, the pupil autocorrelation, treated by the Optical Transfer Function.

Derivation AThe free-space transfer function and its Fresnel form

This card shows that "propagation is multiplication by a phase" is not an assumption but a consequence of the Helmholtz equation acting on a plane-wave basis. The goal is to derive $H$ in equation $(2)$ from first principles and reduce it to the quadratic-phase Fresnel form $(3)$ , exhibiting the exact defocus coefficient $-\pi\lambda z$ — the seam where Fourier optics hands off to the Fresnel regime.

Step 1. Write the field as a superposition of plane waves and insert one component $e^{i(k_x x+k_y y+k_z z)}$ into $\nabla^2U+k^2U=0$ with $k=2\pi/\lambda$ . Why: the plane waves are eigenfunctions of the transverse Laplacian, so Helmholtz reduces to an algebraic condition on each.

Step 2. The substitution gives $k_x^2+k_y^2+k_z^2=k^2$ , so $k_z=2\pi\sqrt{1/\lambda^2-f_x^2-f_y^2}$ ( $k_x=2\pi f_x$ ), and propagating through $z$ multiplies by $e^{ik_z z}=H$ . Why: once the transverse frequencies are set, the wave equation fixes the axial wavenumber, and hence the propagation phase.

Step 3. Expand for $\lambda^2 f^2\ll1$ : $\sqrt{1/\lambda^2-f^2}\approx\tfrac1\lambda-\tfrac{\lambda}{2}f^2$ , so $2\pi z(\cdot)=kz-\pi\lambda z f^2$ and $H\approx e^{ikz}e^{-i\pi\lambda z f^2}$ — equation $(3)$ . Why: the leading correction turns spherical dispersion into the parabolic one; the $f^2$ coefficient $-\pi\lambda z$ is the Fresnel defocus chirp.

Step 4. For $f^2>1/\lambda^2$ the root is imaginary, $k_z=i|k_z|$ , and $H=e^{-|k_z|z}$ decays — the evanescent branch the next derivations exploit. Why: nothing forbids $f>1/\lambda$ ; the wave equation answers with a damped, non-propagating mode. *Verified: the symbolic $f^2$ -coefficient of $2\pi z\sqrt{1/\lambda^2-f^2}$ is $-\pi\lambda z$ , and the paraxial phase $\pi\lambda z f^2$ evaluates to $1.5708\times10^{5}$ rad for $\lambda=500$ nm, $z=0.1$ m, $f=10^6$ m⁻¹, in verify_aq-optics-fourier-optics-introduction.py.*

Go deeper · Derivation B · 1 minThe lens as a Fourier transformer

Derivation BThe lens as a Fourier transformer

Here is the owned result: a single ideal lens computes an exact two-dimensional Fourier transform of the field in front of it, provided the object sits in the front focal plane. The goal is to track the two quadratic phases — the one Fresnel propagation accumulates and the one the lens imposes — and show they cancel exactly in the $f$ – $f$ geometry, leaving the clean transform of equation $(4)$ with $f_\xi=x/\lambda f$ ; the propagator is recalled from Fresnel Diffraction Regime and Fraunhofer Diffraction.

Step 1. Fresnel-propagate $U_0(\xi,\eta)$ a distance $f$ to just before the lens: $U_1(x_1,y_1)=\dfrac{e^{ikf}}{i\lambda f}\iint U_0\,\exp\!\big[\tfrac{ik}{2f}((x_1-\xi)^2+(y_1-\eta)^2)\big]\,\mathrm{d}\xi\,\mathrm{d}\eta$ . Why: over one focal length the field spreads as a chirp convolution, the near-field propagator from the Fresnel and Rayleigh–Sommerfeld theories.

Step 2. Multiply by the thin-lens transmittance $t_L=e^{-ik(x_1^2+y_1^2)/2f}$ , which exactly annihilates the $\tfrac{k}{2f}(x_1^2+y_1^2)$ term from Step 1. Why: the lens carries the opposite curvature to free space over $f$ , so the two $x_1^2$ phases vanish together.

Step 3. Propagate a further $f$ to the back focal plane. The $x_1$ -integral becomes $\int e^{-i\frac{k}{f}x_1(\xi+x)}\,\mathrm{d}x_1$ , a linear-phase kernel, and for the object at exactly $z=f$ the leftover $\xi^2$ and $x^2$ terms cancel too. Why: only in the symmetric $f$ – $f$ geometry do all residual curvatures disappear, which is what yields a phase-clean transform.

Step 4. What survives is $U_f=\dfrac{1}{i\lambda f}\iint U_0\,e^{-i\frac{2\pi}{\lambda f}(x\xi+y\eta)}\,\mathrm{d}\xi\,\mathrm{d}\eta$ — equation $(4)$ , the exact transform with $f_\xi=x/\lambda f$ . Why: comparing with the analysis integral $(1)$ identifies the back-focal-plane coordinate with spatial frequency — angle has become position. *Verified: the Fourier-plane scale $1/\lambda f=1.0\times10^{7}$ cyc/m per m for $\lambda=500$ nm, $f=200$ mm, and the mapping $x=\lambda f\,f_\xi$ (100 cyc/mm → 10 mm), with the residual-phase cancellation confirmed symbolically, in verify_aq-optics-fourier-optics-introduction.py.*

Go deeper · Derivation C · 1 minThe evanescent cutoff and the λ⁄2 limit

Derivation CThe evanescent cutoff and the λ⁄2 limit

The second owned result turns the transfer function into a resolution theorem: only frequencies up to $1/\lambda$ propagate, so the smallest feature reaching a downstream lens is about half a wavelength. The goal is to locate the cutoff in $H$ , show what lies beyond, and convert the highest propagating frequency into a minimum feature size — the limit Abbe Theory of Image Formation sharpens for a real objective.

Step 1. In $H=e^{i2\pi z\sqrt{1/\lambda^2-f^2}}$ the exponent is real, and $H$ a pure phase, while $f^2<1/\lambda^2$ — the homogeneous, propagating waves that reach the far field intact. Why: a real $k_z$ means the component only accumulates path, arriving at a distant screen intact.

Step 2. For $f>1/\lambda$ the root is imaginary, $k_z=i|k_z|$ with $|k_z|=2\pi\sqrt{f^2-1/\lambda^2}$ , so $H=e^{-|k_z|z}$ — evanescent, absent from the far field. Why: sub-wavelength structure lives in these high- $f$ components, and they damp before a distant lens can collect them.

Step 3. The highest surviving frequency is therefore $f_{\max}=1/\lambda$ , so a reconstruction cannot resolve periods finer than $1/f_{\max}=\lambda$ , i.e. half-period detail $\Delta x_{\min}\approx1/2f_{\max}=\lambda/2$ . Why: a band of width $f_{\max}$ sets the narrowest resolvable half-period, by the same reciprocity that ties bandwidth to pulse width.

Step 4. Through a lens of numerical aperture $\mathrm{NA}=n\sin\theta$ the band shrinks to $f_c=\mathrm{NA}/\lambda$ , so the coherent limit becomes $\lambda/2\mathrm{NA}$ ; immersion ( $n>1$ ) or shorter $\lambda$ pushes it out. Why: only components within the acceptance cone enter, so the aperture — not the vacuum cutoff — sets the Abbe resolution. *Verified: the cutoff $f_{\max}=1/\lambda=2000$ cyc/mm and $\Delta x_{\min}=\lambda/2=250$ nm at $\lambda=500$ nm, plus the coherent limit $\lambda/2\mathrm{NA}=275$ nm at $\lambda=550$ nm, NA $=1$ , in verify_aq-optics-fourier-optics-introduction.py.*

Limiting cases

Case	Result	Meaning
Far field / Fraunhofer, $N_F=a^2/\lambda z\ll1$	$U(x,y)\propto\tilde U_0(x/\lambda z,\,y/\lambda z)$	the field is the pure transform of the aperture, scaled by $\lambda z$
Lens back focal plane	$U_f=\tfrac{1}{i\lambda f}\tilde U_0(x/\lambda f,\,y/\lambda f)$	the far-field transform brought to a finite distance $z=f$
Fresnel / near field, $N_F\gtrsim1$	$H\approx e^{ikz}e^{-i\pi\lambda z f^2}$	quadratic-phase transfer; propagation is a chirp convolution
Evanescent cutoff, $f>1/\lambda$	$k_z=i\|k_z\|$ , $H=e^{-\|k_z\|z}$	frequencies above $1/\lambda$ decay; the $\lambda/2$ resolution ceiling
Coherent imaging	$g=h*f$ , $\tilde g=H_{\text{coh}}\tilde f$	pupil is a hard low-pass filter with cutoff $\mathrm{NA}/\lambda$
Gaussian input	transform is Gaussian	self-Fourier beam; a laser focus is the transform of its near field

Common trapThe lens transforms the intensity, and the back-focal-plane pattern is always a clean transform

Two half-truths that misplace phases. First, a lens transforms the complex field amplitude, not the intensity: the whole point of Fourier filtering is that you edit amplitude and phase in the Fourier plane, then let the second lens re-interfere them. Second, the transform is phase-clean only in the $f$ – $f$ geometry, object exactly one focal length in front. For any other distance an extra factor $\exp[ik(x^2+y^2)(1-d/f)/2f]$ multiplies the transform, so the intensity is still a correctly scaled $|\tilde U_0|^2$ but the field carries curvature — which bites the instant you cascade another lens or record a hologram. Ask where the object sits before claiming a clean transform, and never Fourier-filter the intensity when you mean the amplitude.

Worked examples

1 · Reading the Fourier plane of a lens (confidence builder)

A transparency carrying spatial frequencies up to $100$ cyc/mm sits one focal length in front of an $f=200$ mm lens at $\lambda=500$ nm. Where does the $100$ cyc/mm component land in the back focal plane, and what is the plane-to-frequency scale?

Step 1. The Fourier-plane scale is $1/\lambda f=1/(500\times10^{-9}\times0.200)=\mathbf{1.0\times10^{7}\ \text{cyc/m per m}}$ — $10$ cyc/mm per millimetre of radius. Position maps to frequency by $x=\lambda f\,f_x$ .

Step 2. For $f_x=100$ cyc/mm $=1.0\times10^{5}$ cyc/m, $x=\lambda f\,f_x=(500\times10^{-9})(0.200)(1.0\times10^{5})=\mathbf{10\ \text{mm}}$ from the axis. The dc light stays on axis; higher frequencies fan out linearly.

Step 3. The component leaves at $\sin\theta=\lambda f_x=0.05$ , so $\theta=2.87^\circ$ — safely paraxial, $x\approx f\sin\theta$ . A $100$ cyc/mm feature focuses to $10$ mm off axis; each mm of Fourier-plane radius reads $10$ cyc/mm.

Step 4. Large angles bend the map: a $1000$ cyc/mm feature needs $\sin\theta=\lambda f_x=0.5$ ( $\theta=30^\circ$ ), where the paraxial $x=\lambda f\,f_x$ ( $100$ mm) understates the true $f\tan\theta=115$ mm.

2 · The diffraction limit from the cutoff (exam level)

At $\lambda=500$ nm, what is the finest object detail that can propagate to a distant lens, and how does an immersion objective change the answer?

Step 1. The highest propagating frequency is the cutoff $f_{\max}=1/\lambda=1/(500\times10^{-9})=2.0\times10^{6}\ \text{cyc/m}=\mathbf{2000\ \text{cyc/mm}}$ . Above it, components are evanescent and vanish before the lens.

Step 2. The narrowest surviving half-period is $\Delta x_{\min}\approx1/2f_{\max}=\lambda/2=500/2=\mathbf{250\ \text{nm}}$ . No far-field instrument recovers finer vacuum-propagated detail — the sub-wavelength information is simply not there.

Step 3. A real objective collects only up to $f_c=\mathrm{NA}/\lambda$ . With immersion at $\mathrm{NA}=1.0$ the coherent limit is $\lambda/2\mathrm{NA}=550/2=\mathbf{275\ \text{nm}}$ at $\lambda=550$ nm; higher NA or shorter $\lambda$ sharpens it — the strategy quantified by Abbe Theory of Image Formation. Vacuum cutoff $2000$ cyc/mm; $\Delta x_{\min}=250$ nm; a lens caps it at $\lambda/2\mathrm{NA}$ .

Physics of the portrait

The portrait here is not a person but a point of view: Fourier optics, describing light by the spatial frequencies it carries rather than the rays it follows. It is the lens through which a mesh becomes a constellation of dots and a microscope a two-stage transform.

Origins · 1 minFourier optics and how this was discovered

Fourier optics

the linear-systems view of light · optics · imaging · signal processing

Fourier optics applies linear-systems theory and the two-dimensional Fourier transform to the propagation, diffraction, and imaging of coherent light. Its founding move is that free-space propagation is linear and shift-invariant, hence diagonal in the basis of complex exponentials — the tilted plane waves labelled by spatial frequency. In that basis propagation is multiplication by the transfer function, diffraction is filtering, and a lens evaluates a transform in the time light takes to cross it. The viewpoint grew from Ernst Abbe's diffraction theory of the microscope, was cast in explicit Fourier language by Pierre-Michel Duffieux, and became the dialect of modern optics through transfer-function analysis and mid-century spatial filtering. Its reach runs from coherent imaging and holography to the $4f$ correlator, phase contrast, beam shaping, and the optical transfer function. To think in Fourier optics is to stop asking where each ray goes and ask which spatial frequencies survive — the shift that makes resolution, filtering, and image formation one idea.

History

1873

— Ernst Abbe publishes his diffraction theory of image formation, treating the microscope as a two-stage diffraction process

1935

— Frits Zernike introduces phase-contrast microscopy, editing the zero-frequency light in the Fourier plane to reveal transparent structure

1946

— Pierre-Michel Duffieux publishes L'intégrale de Fourier et ses applications à l'optique, founding Fourier optics as a discipline

1953

— André Maréchal and Paul Croce demonstrate spatial filtering by double diffraction, editing images in the Fourier plane

1968

— Joseph Goodman publishes Introduction to Fourier Optics, the text that standardised the transfer-function language

Going further

Texts. Goodman, Introduction to Fourier Optics, 4th ed., Chs. 3–6 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 4 · Hecht, Optics, 5th ed., Ch. 11 · Born & Wolf, Principles of Optics, 7th ed., §8 (angular-spectrum representation) · Loudon, The Quantum Theory of Light, 3rd ed. (mode decompositions).

Provenance. Equations follow Saleh & Teich Ch. 4 and Goodman, Introduction to Fourier Optics (Chs. 3–6). The lens-transform derivation reproduces Saleh & Teich Eq. (4.2-8); the Fraunhofer reduction matches Hecht Ch. 10–11; the angular-spectrum transfer function and evanescent cutoff are standard (Goodman Ch. 3). A verifier should confirm quadratic-phase cancellation when the object sits at exactly $z=f$ , the sinc far field of a slit, the Fourier-plane scale $1/\lambda f=10^7$ cyc/m per m at $\lambda=500$ nm and $f=200$ mm, the cutoff $f_{\max}=1/\lambda=2000$ cyc/mm with $\Delta x_{\min}=\lambda/2=250$ nm, and the coherent limit $\lambda/2\mathrm{NA}\approx275$ nm at $\lambda=550$ nm, NA $=1$ . No non-standard numbers are introduced.

Prerequisites. Huygens–Fresnel Principle · Fresnel–Kirchhoff Diffraction Integral · Fraunhofer Diffraction · Fresnel Diffraction Regime. Leads to. Fraunhofer Diffraction · Rayleigh Resolution Criterion · Circular Aperture Diffraction and the Airy Pattern · Spatial Frequency and the Angular Spectrum · Abbe Theory of Image Formation.

Is Fourier optics the same thing as Fraunhofer diffraction?

No — Fraunhofer diffraction is one theorem inside Fourier optics. Fraunhofer says the far-field pattern of an aperture is its transform; Fourier optics generalises that to any field over any distance, to lenses, and to imaging. The single transfer function $H=e^{i2\pi z\sqrt{1/\lambda^2-f^2}}$ contains Fraunhofer (far-field limit), Fresnel (paraxial expansion), and the evanescent cutoff as special cases. So Fraunhofer Diffraction is the far-field corner of the same framework, and a lens brings it to the finite distance $z=f$ .

Why does the lens transform the amplitude and not the intensity?

Because the lens is a linear phase element acting on the complex field, not its modulus squared. Propagation and the lens phase give the transform of the amplitude $U_0$ , and Fourier filtering depends on it: a mask changes amplitude and phase, and only the second lens re-interferes those components into a new intensity. If a lens transformed intensity you could never do phase contrast, since a pure phase object has uniform intensity but a richly structured amplitude spectrum.

What is the difference between the coherent transfer function and the OTF?

It is the difference between adding amplitudes and adding intensities. Coherent illumination is linear in amplitude, and the transfer function equals the scaled pupil $P$ — a hard low-pass filter with cutoff $\mathrm{NA}/\lambda$ . Incoherent illumination is linear in intensity, and the transfer function is the OTF, the pupil autocorrelation, with cutoff $2\mathrm{NA}/\lambda$ and a gradual roll-off. So incoherent imaging passes twice the bandwidth but with declining contrast; the full treatment belongs to the Optical Transfer Function.

How can spatial frequency have units of cycles per millimetre — frequency of what?

Of position, not time. A grating of 100 line pairs per millimetre has spatial frequency 100 cyc/mm; the field oscillates 100 times per millimetre transversely. The transform $(1)$ is with respect to the transverse coordinates $x,y$ , so its conjugate variable is cycles per length — exactly as a time signal's transform lives in cycles per second. A steeper tilt is a higher spatial frequency, through $f_x=\sin\theta_x/\lambda$ .

If the object is not exactly one focal length in front, is the focal plane still a Fourier transform?

The intensity is, but the field is not phase-clean. Move the object off the front focal plane by $d\ne f$ and a factor $\exp[ik(x^2+y^2)(1-d/f)/2f]$ multiplies the transform. The intensity $|\tilde U_0|^2$ is still correctly scaled — which is why the mesh experiment tolerates sloppy placement — but the curvature bites the moment you cascade a lens or record phase, so use the $f$ – $f$ geometry when phase counts.

Does treating light as a scalar field limit where Fourier optics applies?

It sets the boundaries, and they are generous. Scalar theory ignores polarisation coupling and the vector boundary conditions at sub-wavelength edges, so it is superb for apertures many wavelengths across and softens only at very high NA or feature sizes near $\lambda$ . For most imaging, filtering, and beam-shaping at visible wavelengths it is accurate to measurement precision; the rigorous vector ancestry is reached through Rayleigh–Sommerfeld Diffraction.