Rayleigh-Sommerfeld Diffraction Theory

This article takes over a story the electromagnetism track begins and hands optics its working end. The rigorous machinery — Green's theorem turning "the field on a surface" into "the field anywhere behind it," the proof that Kirchhoff's boundary conditions describe no field that can exist, and the image-source trick that repairs them — is derived and owned by aq-em-scalar-vector-diffraction-theory, and recalled here by name rather than re-derived. What this article owns is the optical payoff: the one Rayleigh–Sommerfeld formula an undergraduate actually propagates light with, the ladder down from it to the Fresnel and Fraunhofer integrals every slit and telescope pattern rests on, and its rebirth as the angular-spectrum method running inside every digital-holography and diffractive-optics code. The tension at the heart of scalar diffraction is a boundary-value problem in disguise. A field obeying the Helmholtz equation $(\nabla^2+k^2)U=0$ is pinned down by either its value or its normal slope on a surface — never both. Kirchhoff's 1882 integral quietly imposes both on the screen; Poincaré and Sommerfeld later showed the assumptions are self-contradictory, satisfied only by the zero field. Rayleigh and Sommerfeld's repair is an electrostatics friend in new clothes — the method of images, replayed for waves — leaving the integral's structure intact while restoring its logic. The upshot is a formula that can be wrong as an exact statement and indispensable as an asymptotic one.

Key takeaways

Scalar diffraction is a Helmholtz boundary-value problem: the field behind a screen follows from the field on the aperture — but you may specify only its value or its normal derivative, not both.
Kirchhoff imposes both (field and slope) on the screen; that over-determines the problem, so — as the electromagnetism owner proves — no nonzero field satisfies his conditions, though the integral works excellently for apertures many wavelengths wide.
The Rayleigh–Sommerfeld repair builds an image Green's function that vanishes on the aperture plane (RS-I, Dirichlet) or whose normal derivative vanishes there (RS-II, Neumann), so exactly one aperture datum is needed — a well-posed formula.
The three theories differ only in an obliquity factor: RS-I gives $\cos\theta$ , RS-II gives $\cos\theta_0$ , and Kirchhoff gives their average $\tfrac12(\cos\theta+\cos\theta_0)$ ; all $\to 1$ on-axis, so ordinary far-field patterns are formulation-independent.
Key idea: a Helmholtz field is fixed by its value OR its normal slope on a surface, not both. An image Green's function kills one boundary term, so the Rayleigh–Sommerfeld integral needs one consistent aperture datum — curing Kirchhoff's over-determination while reproducing all of Fresnel and Fraunhofer optics in the paraxial limit.

The core insight

A solution of the Helmholtz equation is determined by its value OR its normal derivative on a surface, not both. Choosing a Green's function that vanishes (RS-I) or whose normal derivative vanishes (RS-II) on the aperture plane — built by the method of images — kills one term of the boundary integral, so the Rayleigh–Sommerfeld formula needs only one consistent aperture datum. It cures the over-determination in Kirchhoff's classic theory while reducing to the same Fresnel and Fraunhofer integrals at small angles.

Where this lives today

Rayleigh–Sommerfeld theory is no museum piece; it is the numerical engine of modern wave optics. Because RS-I is a shift-invariant convolution whose transfer function is the exact free-space propagator $\exp[ikz\sqrt{1-(\lambda f_x)^2-(\lambda f_y)^2}]$ , propagating a field reduces to an FFT, a multiply, and an inverse FFT — the angular-spectrum method. Digital holography, computer-generated holograms, diffractive-optical-element design, phase-retrieval imaging, and the freeform-optics pipelines behind AR/VR waveguides all run this loop, and they use the RS/angular-spectrum form precisely because it handles large diffraction angles and the evanescent cut-off correctly where the paraxial Fresnel kernel fails. The theory also anchors the frontier where scalar optics ends. Sub-wavelength apertures — near-field scanning microscopy, extraordinary-optical-transmission plasmonics, metasurface slits — sit exactly where scalar theory breaks down, making Rayleigh–Sommerfeld the natural bridge to the full vector treatment those devices demand.

The physical picture

Imagine standing behind a lit aperture and asking what the field is at your eye. Huygens' answer — treat every point of the aperture as a little re-emitter and add up the wavelets — is exactly right, but it needs two fixes Huygens never supplied: how strongly each wavelet radiates in each direction, and with what phase. Rayleigh–Sommerfeld theory supplies both from first principles. The trick is to place, behind the screen, a phantom image source that is the mirror twin of your eye. Choose it out of phase and the real-plus-image field cancels on the aperture plane; the plane is then "clamped" to whatever the hole prescribes, and only the aperture field value is needed to propagate forward. That is RS-I, and it comes with a built-in obliquity factor $\cos\theta$ that fades the wavelets to nothing looking backward — the very suppression Fresnel had to postulate by hand.

Figure

The image construction behind RS-I: observation point P behind the screen, its mirror image P′, and the obliquity factor cosθ = z/r that weights each aperture wavelet

At small angles none of this matters: $\cos\theta\approx1$ , the three theories agree to a fraction of a percent, and undergraduate diffraction — single slit, gratings, the Airy disk — never needs to know which one it is using. The differences are a rim effect, showing up only at large diffraction angles, at Fresnel numbers of order one, or when you demand logical rigor rather than a good number. The next figure makes that precise: it plots the three obliquity factors against angle and shows them fanning apart only as the geometry steepens.

Figure

The punchline — the three obliquity factors versus diffraction angle: identical on-axis, fanning apart only at the rim; at θ = 30° RS-I = 0.866, Kirchhoff = 0.933, RS-II = 1.000

The core equations

Everything rests on one identity, and it belongs to electromagnetism. Green's second identity applied to the field $U$ and a Green's function $G$ — both solutions of the Helmholtz equation — converts a volume statement into a surface one, and the delta-source in $G$ sifts out the field at the observation point. This is the Helmholtz–Kirchhoff integral theorem, derived in full in aq-em-scalar-vector-diffraction-theory; here we quote it as the master identity from which every named formula descends.

U(P)=\frac{1}{4\pi}\oint_{S}\left[U\,\frac{\partial G}{\partial n}-G\,\frac{\partial U}{\partial n}\right]\mathrm{d}S \qquad (1)

where: $P$ — observation point enclosed by $S$ ; $U$ — complex scalar field (one Cartesian component of $\mathbf E$ or $\mathbf B$ ); $G$ — Green's function with $(\nabla^2+k^2)G=-4\pi\,\delta(\mathbf r-\mathbf r_P)$ ; $\partial/\partial n$ — derivative along the inward surface normal; $S$ — closed surface (aperture plane plus a large hemisphere whose contribution dies by the radiation condition). The two boundary terms carry the two Cauchy data, $U$ and $\partial U/\partial n$ — and the whole art is choosing $G$ so that you need only one. Kirchhoff's choice is the plain free-space Green's function, the outgoing spherical wave. Feeding it both aperture data — the incident field and its slope on the hole, zero on the screen — gives the workhorse Fresnel–Kirchhoff formula. It is an excellent asymptotic and, as the electromagnetism owner proves by an analyticity argument, an exact impossibility: a Helmholtz field vanishing together with its normal derivative on a finite patch vanishes everywhere.

G_0=\frac{e^{ikr}}{r},\qquad r=|\mathbf r-\mathbf r_P|,\qquad k=\frac{2\pi}{\lambda} \qquad (2)

where: $G_0$ — free-space (Kirchhoff) Green's function, an outgoing spherical wave; $r$ — distance from an aperture point to $P$ ; $k$ — wavenumber. Using $G_0$ forces both data onto the boundary, which is where the inconsistency enters. The Rayleigh–Sommerfeld repair keeps the identity and changes the Green's function. Reflect $P$ across the aperture plane to an image point $P'$ and superpose its wave with a chosen sign. The out-of-phase combination vanishes on the plane (Dirichlet); the in-phase combination has vanishing normal derivative there (Neumann). Each kills one boundary term of $(1)$ .

G_-=\frac{e^{ikr}}{r}-\frac{e^{ikr'}}{r'},\qquad G_+=\frac{e^{ikr}}{r}+\frac{e^{ikr'}}{r'} \qquad (3)

where: $G_-$ — Dirichlet Green's function, $G_-=0$ on the plane (out-of-phase image); $G_+$ — Neumann Green's function, $\partial G_+/\partial n=0$ on the plane (in-phase image); $r'$ — distance from the mirror image $P'$ to the aperture point (equal to $r$ on the plane, which is why the combinations vanish there). This is the method of images of electrostatics, replayed with phases riding along. Inserting $G_-$ into $(1)$ collapses the integral onto the aperture field alone — the datum an illuminated hole actually provides — and, in the far zone $kr\gg1$ , produces the first Rayleigh–Sommerfeld solution: a Huygens sum with the amplitude-phase prefactor $-i/\lambda$ and the obliquity factor $\cos\theta$ that Fresnel could only guess.

U(P)=-\frac{i}{\lambda}\iint_{\Sigma} U(P_1)\,\frac{e^{ikr}}{r}\,\cos\theta\;\mathrm{d}S,\qquad \cos\theta=\frac{z}{r} \qquad (4)

where: $\Sigma$ — aperture area (the screen contributes nothing, since $G_-=0$ there); $P_1$ — an aperture point; $\theta$ — angle between $\mathbf r$ and the propagation axis $z$ ; $\cos\theta=z/r$ — the RS-I obliquity factor, which suppresses wavelets radiated obliquely and forward-only. The complementary RS-II solution uses $G_+$ and needs only the normal derivative $\partial U/\partial n$ , carrying obliquity $\cos\theta_0$ (the incidence angle). Their comparison is the whole taxonomy:

K_{\text{RS-I}}=\cos\theta,\qquad K_{\text{RS-II}}=\cos\theta_0,\qquad K_{\text{Kirchhoff}}=\tfrac12(\cos\theta+\cos\theta_0) \qquad (5)

where: $K$ — the inclination (obliquity) factor multiplying each secondary wavelet; $\theta$ — diffraction angle; $\theta_0$ — incidence angle. Kirchhoff's factor is exactly the arithmetic mean of the two Rayleigh–Sommerfeld factors — which, as the electromagnetism owner shows, is why the Kirchhoff field equals the average $\tfrac12(U_{\rm I}+U_{\rm II})$ of the two consistent theories.

Derivation AThe RS-I obliquity factor from the free-space Green's function

The single most quoted number in this article is the obliquity factor $\cos\theta$ that RS-I attaches to every Huygens wavelet — the factor Fresnel inserted by hand and RS-I inherits cleanly from the image construction. Our aim is to show it is nothing more than the geometric derivative of the outgoing spherical wave along the screen normal, so that " $\cos\theta = z/r$ " is a fact about $e^{ikr}/r$ and not an extra assumption. We differentiate the free-space Green's function along the normal, keep the far-field term, and read the angular factor off.

Step 1. Write the outgoing spherical wave $g(r)=e^{ikr}/r$ and take its derivative with respect to the observation-point coordinate along the aperture normal $\hat n=\hat z$ : $\partial g/\partial n = \dfrac{dg}{dr}\,\dfrac{\partial r}{\partial n}$ . Why: the normal derivative in the boundary integral acts on the Green's function's spatial argument, and the chain rule separates the radial dependence from the geometry.

Step 2. Evaluate the two pieces. The radial derivative is $\dfrac{dg}{dr}=\left(ik-\dfrac1r\right)\dfrac{e^{ikr}}{r}$ , and the geometric factor is $\dfrac{\partial r}{\partial n}=\cos\theta=z/r$ , the cosine of the angle between $\mathbf r$ and the axis. Why: $\partial r/\partial n$ is the projection of the unit vector along $\mathbf r$ onto the normal — pure geometry — and it is what turns a spherical wave into a directional emitter.

Step 3. Combine: $\dfrac{\partial}{\partial n}\dfrac{e^{ikr}}{r}=\left(ik-\dfrac1r\right)\dfrac{e^{ikr}}{r}\cos\theta$ . In the far zone $kr\gg1$ the $1/r$ inside the bracket is negligible against $ik$ , leaving $\approx ik\,\dfrac{e^{ikr}}{r}\cos\theta$ . Why: at distances of many wavelengths the wave's phase varies far faster than its amplitude, so the amplitude-gradient term drops and only the phase-gradient survives.

Step 4. The Dirichlet image doubles this normal derivative on the plane ( $\partial G_-/\partial n=2\,\partial G_0/\partial n$ ), and $ik=2\pi i/\lambda$ converts the prefactor to $-i/\lambda$ . Substituting into the boundary integral $(1)$ with $G_-=0$ gives equation $(4)$ , obliquity factor $\cos\theta$ and all. Why: the image contributes an equal normal derivative on the plane, so the surviving field-value term carries a factor of two that cleanly cancels the $1/4\pi$ into the standard $-i/\lambda$ Huygens prefactor. *Verified symbolically ( $\partial_n(e^{ikr}/r)=(ik-1/r)(e^{ikr}/r)\cos\theta$ , $\cos\theta=z/r$ ) and numerically (on-axis $K=1$ ; at $\theta=30^\circ$ , $K_{\rm RS\text{-}I}=0.8660$ ) in verify_rayleigh-sommerfeld-diffraction.py.*

Go deeper · Derivation B · 1 minKirchhoff as the average of the two consistent theories

Derivation BKirchhoff as the average of the two consistent theories

Why does a formula proven to be self-contradictory nonetheless compute every telescope point-spread function correctly? The answer, owned in full by the electromagnetism companion and recalled here, is that Kirchhoff's obliquity factor is the exact arithmetic mean of the two Rayleigh–Sommerfeld factors, so the Kirchhoff field is the average of two well-posed fields — and an average of two good approximations is itself a good approximation. Our aim is to display that averaging cleanly and to see where the three theories must agree.

Step 1. List the three obliquity factors from $(5)$ : RS-I carries $\cos\theta$ (depends only on the diffraction angle), RS-II carries $\cos\theta_0$ (depends only on the incidence angle), and Kirchhoff carries $\tfrac12(\cos\theta+\cos\theta_0)$ . Why: each RS theory trusts one Cauchy datum and inherits the obliquity of that datum's Green's function; Kirchhoff trusts both, so it lands between them.

Step 2. Form the mean: $\tfrac12(K_{\rm RS\text{-}I}+K_{\rm RS\text{-}II})=\tfrac12(\cos\theta+\cos\theta_0)=K_{\text{Kirchhoff}}$ , identically. Because the master identity $(1)$ is linear in $G$ , feeding it $\tfrac12(G_-+G_+)=G_0$ produces exactly $\tfrac12(U_{\rm I}+U_{\rm II})$ . Why: linearity of the boundary integral in the Green's function turns the average of two Green's functions into the average of two fields — the Kirchhoff field is literally that average.

Step 3. Specialise to normal incidence and small diffraction angles: $\cos\theta_0=1$ and $\cos\theta\to1$ , so all three factors $\to1$ and the three fields coincide. Why: on-axis the two data agree and the averaging is invisible — which is exactly why ordinary far-field diffraction never needs to distinguish the theories.

Step 4. Read off the rim behaviour: at $\theta=90^\circ$ the factors are $0$ (RS-I), $1$ (RS-II), $\tfrac12$ (Kirchhoff); at $\theta=180^\circ$ , RS-I gives $-1$ while Kirchhoff gives $0$ — the backward-wave suppression Fresnel postulated. Why: the disagreement is a large-angle, near-edge effect, measurable only for small apertures, high NA, or near-field work. *Verified: $K_{\text{Kirchhoff}}=\tfrac12(\cos\theta+\cos\theta_0)$ symbolically, and numerically at $\theta=30^\circ$ (0.866, 0.933, 1.000) and $\theta=90^\circ$ (0, 0.5, 1) in verify_rayleigh-sommerfeld-diffraction.py.*

Go deeper · Derivation C · 1 minThe ladder down to Fresnel and Fraunhofer

Derivation CThe ladder down to Fresnel and Fraunhofer

The reason this rigorous foundation is worth an undergraduate's time is that it collapses, in two controlled steps, into the two integrals that produce every diffraction pattern in the rest of this track. Our aim is to walk RS-I down to the Fresnel propagator by a paraxial expansion of the path length, and then to the Fraunhofer Fourier transform by dropping one more phase term — locating exactly where each approximation is spent.

Step 1. Start from RS-I, equation $(4)$ , and expand the path length for an observation point at $(x,y,z)$ far downstream of an aperture point $(x_1,y_1,0)$ : $r=\sqrt{z^2+(x-x_1)^2+(y-y_1)^2}\approx z+\dfrac{(x-x_1)^2+(y-y_1)^2}{2z}$ . Why: when the transverse separations are small against $z$ , the binomial expansion keeps the leading quadratic phase — the term that actually bends light — and discards higher orders.

Step 2. Freeze the slow factors: $\cos\theta=z/r\approx1$ and $1/r\approx1/z$ , keeping the fast quadratic phase in the exponential only. RS-I becomes the Fresnel propagator, $U(x,y)=\dfrac{e^{ikz}}{i\lambda z}\iint U(x_1,y_1)\exp\!\big[\tfrac{ik}{2z}((x-x_1)^2+(y-y_1)^2)\big]\mathrm{d}x_1\mathrm{d}y_1$ . Why: amplitude factors vary over the scale of $z$ while the phase varies over the scale of a wavelength, so only the phase's quadratic term needs to be kept — the essence of the paraxial approximation.

Step 3. Expand the quadratic and isolate the aperture's own curvature term $\tfrac{k}{2z}(x_1^2+y_1^2)$ . Its largest value across an aperture of half-width $a$ is $\tfrac{k a^2}{2z}=\pi\,\dfrac{a^2}{\lambda z}=\pi N_F$ , with the Fresnel number $N_F=a^2/\lambda z$ . When $\pi N_F\ll1$ this phase is negligible and the integral becomes a pure Fourier transform of the aperture — the Fraunhofer regime. Why: the Fresnel number measures the aperture's self-curvature in half-wave units; once below one, the aperture looks flat to the far field and diffraction becomes a Fourier transform. The single-slit sinc, the rectangular sinc-product, and the Airy pattern of later articles are each one evaluation of this same integral for a particular hole. Verified: the paraxial reduction and the edge-phase bound $\pi N_F$ ( $N_F=a^2/\lambda z=0.25$ for $a=1$ mm, $\lambda=500$ nm, $z=8$ m) in verify_rayleigh-sommerfeld-diffraction.py.

Limiting cases

Case	Result	Meaning
paraxial / small angles	$K\to1$ ; $\cos\theta\approx\cos\theta_0\approx1$	RS-I, RS-II, Kirchhoff coincide — every standard slit/grating/Airy pattern is formulation-independent
RS-I (Dirichlet)	$K=\cos\theta$	uses only the aperture field $U$ ; obliquity set by the diffraction angle; exact for a planar screen
RS-II (Neumann)	$K=\cos\theta_0$	uses only $\partial U/\partial n$ ; obliquity set by the incidence angle
Kirchhoff	$K=\tfrac12(\cos\theta+\cos\theta_0)$	the mean of the RS factors — accurate for $a\gg\lambda$ , but its boundary data are over-specified
backward $\theta\to\pi$	$K_{\rm RS\text{-}I}=-1$ , $K_{\text{Kirch}}=0$	Kirchhoff zeroes the backward wave by construction; RS-I derives an obliquity instead of postulating one

Common trap“Rayleigh–Sommerfeld fixes the numbers Kirchhoff gets wrong.”

It usually changes no number you care about. For normal incidence and small diffraction angles all three obliquity factors approach one, so RS-I, RS-II, and Kirchhoff predict the same single-slit, grating, and Airy patterns — the Rayleigh resolution limit included. What Rayleigh–Sommerfeld buys is logical consistency (one boundary datum, no self-contradiction) and a cleaner large-angle obliquity factor, not a correction to textbook far-field diffraction. Two further cautions: RS-I and RS-II are exact only for a planar screen with the image construction — they are not general-purpose boundary conditions — and scalar theory of any flavour fails for holes near or below a wavelength, where the vector (polarization) boundary conditions take over and transmission collapses far faster than any scalar estimate. Different diseases, different cures: images repair Kirchhoff's logic; the full vector theory repairs sub-wavelength physics.

Worked examples

1 · How far apart are the three theories? (confidence builder)

An aperture is illuminated at normal incidence ( $\theta_0=0$ ). Compare the RS-I, RS-II, and Kirchhoff obliquity factors for light diffracted to $\theta=30^\circ$ , and again on-axis, and say when the difference could matter.

Step 1. Normal incidence fixes RS-II: $K_{\rm RS\text{-}II}=\cos\theta_0=\cos0^\circ=\mathbf{1.000}$ at every diffraction angle.

Step 2. At $\theta=30^\circ$ , RS-I gives $K_{\rm RS\text{-}I}=\cos30^\circ=\mathbf{0.8660}$ , and Kirchhoff gives the mean $K=\tfrac12(\cos30^\circ+1)=\tfrac12(1.8660)=\mathbf{0.9330}$ . The spread from smallest to largest is $0.134$ , about a $\pm7\%$ variation in the wavelet weighting.

Step 3. On-axis ( $\theta=0$ ) all three collapse: $\cos0^\circ=1$ for each, so $K_{\rm RS\text{-}I}=K_{\rm RS\text{-}II}=K_{\text{Kirch}}=\mathbf{1.000}$ — no difference at all. This is why the central peak of any far-field pattern is theory-independent.

Step 4. That spread reaches intensity only through $K^2$ , and only matters where $30^\circ$ -scale angles carry real power — small apertures ( $N_F\sim1$ ), high-NA optics, the near field. For a telescope diffracting to arc-seconds it is negligible. RS-I 0.866, Kirchhoff 0.933, RS-II 1.000 at 30°; all 1.000 on-axis.

2 · Where does the far field begin? (exam level)

A circular aperture of radius $a=1$ mm is illuminated by $\lambda=500$ nm light. Using the Fresnel-number bookkeeping of the ladder derivation, find the distance beyond which the Fraunhofer (Fourier-transform) approximation holds, and evaluate $N_F$ at a screen $8$ m away.

Step 1. The aperture's self-curvature phase across its edge is $\pi N_F$ with $N_F=a^2/\lambda z$ . Fraunhofer requires $\pi N_F\ll1$ , i.e. $z\gg a^2/\lambda$ .

Step 2. Compute the scale distance: $a^2/\lambda=(10^{-3}\,\text{m})^2/(500\times10^{-9}\,\text{m})=10^{-6}/5\times10^{-7}=\mathbf{2.0\ m}$ . The true far field is many times this — which is why a lens is used in the lab to bring Fraunhofer diffraction into the focal plane.

Step 3. At $z=8$ m, $N_F=a^2/\lambda z=2.0/8=\mathbf{0.25}$ , so the edge phase is $\pi N_F=0.785$ rad — not yet $\ll1$ . At $8$ m you are on the threshold of Fraunhofer, not deep inside it.

Step 4. Sanity check with RS-I's descendants: at $N_F=0.25$ the on-axis Fresnel result $I/I_0=4\sin^2(\pi N_F/2)=4\sin^2(0.393)=0.585$ still differs from the geometric value, confirming residual near-field structure. Only for $z\gtrsim10\,a^2/\lambda\approx20$ m is the pattern a clean Fourier transform. $a^2/\lambda=2.0$ m; $N_F=0.25$ at 8 m — the far field has barely begun.

Physics of the portrait

The formula that repaired scalar diffraction, and the first exact diffraction solution of any kind, both carry the name of a Munich theorist who trained a generation of quantum physicists.

Origins · 1 minArnold Sommerfeld and how this was discovered

Arnold Sommerfeld

1868–1951 · Munich — the great teacher of theoretical physics

Arnold Sommerfeld built, at the University of Munich, the most productive school of theoretical physics of the early twentieth century — Heisenberg, Pauli, Debye, and Bethe were among his students, and he was nominated for the Nobel Prize a record number of times without ever winning it. In 1896 he solved the diffraction of a wave by a perfectly conducting half-plane exactly, using a branched Riemann-surface construction — the first rigorous solution of a diffraction problem in the history of the subject, against which all approximate theories are still checked. In the same period he and Lord Rayleigh independently repaired Kirchhoff's scalar theory by replacing its Green's function with an image-built one, curing the inconsistency Poincaré had exposed in 1892. His four-volume Lectures on Theoretical Physics — its optics volume laying out exactly this diffraction theory — remained a standard for decades. He also introduced the fine-structure constant and the azimuthal and magnetic quantum numbers into atomic theory, and died in 1951 after a traffic accident, one of the field's most consequential mentors.

History

1882

— Kirchhoff derives the diffraction integral from Green's theorem, obliquity factor included

1892

— Poincaré exposes the inconsistency of Kirchhoff's boundary conditions

1897

— Rayleigh's Philosophical Magazine paper treats wave passage through plane-screen apertures with image constructions

1896

— Sommerfeld solves the perfectly conducting half-plane exactly — the first rigorous diffraction solution

1954

— Sommerfeld's Optics (Lectures on Theoretical Physics, Vol. IV) codifies the Rayleigh–Sommerfeld theory

1968

— Goodman's Introduction to Fourier Optics makes RS-I and the angular-spectrum method the standard computational tools

Going further

References. Hecht, Optics, 5th ed., Ch. 10 (Diffraction) and the Kirchhoff/RS integral appendix · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 4 (Fourier Optics) · Goodman, Introduction to Fourier Optics, 4th ed., Ch. 3 (Foundations of scalar diffraction; RS-I, RS-II) and §3.10 (angular spectrum) · Born & Wolf, Principles of Optics, 7th ed., §8.3–8.4 · Sommerfeld, Optics (Lectures on Theoretical Physics, Vol. IV).

Provenance. Equations are standard: the Helmholtz–Kirchhoff integral theorem (Green's second identity + Helmholtz equation), the image Green's functions $G_\pm$ , the RS-I/RS-II integrals and obliquity factors $\cos\theta$ , $\cos\theta_0$ , $\tfrac12(\cos\theta_0+\cos\theta)$ , the Fresnel integral, and the angular-spectrum transfer function. A verifier should: (1) confirm the RS-I obliquity factor $\cos\theta=z/r$ by differentiating $e^{ikr}/r$ along the normal (Goodman §3.5); (2) verify Kirchhoff's factor is the mean of the RS factors (Goodman §3.5.3, Born & Wolf §8.4); (3) check the over-determination argument (Born & Wolf §8.3.2); (4) confirm the angular-spectrum transfer function (Goodman §3.10) and the Airy first-zero constant $1.22$ (Born & Wolf §8.5.2). The rigorous Green's-theorem derivation, the Kirchhoff-inconsistency proof, and the vector corrections are owned by aq-em-scalar-vector-diffraction-theory and recalled here by name. All numbers standard textbook values; none fabricated.

Builds on. Fresnel–Kirchhoff Diffraction Integral · Huygens–Fresnel Principle · Helmholtz equation and Green's functions. Leads to. Fresnel Diffraction Regime · Fraunhofer Diffraction Regime · Circular Aperture Diffraction and the Airy Pattern · Spatial Frequency and the Angular Spectrum.

If Kirchhoff's theory is provably inconsistent, why is it still taught and used everywhere?

Because being wrong as an exact statement and indispensable as an asymptotic one are different things. The inconsistency is confined to a boundary layer of order one wavelength around the aperture rim; for an aperture many wavelengths wide, that rim contributes a negligible fraction of the integral, so the Kirchhoff formula reproduces measured far-field patterns to high accuracy. It is also the arithmetic mean of the two consistent Rayleigh–Sommerfeld theories, so it inherits their accuracy where they agree. Treat it as a controlled asymptotic — excellent for $a\gg\lambda$ — and switch to Rayleigh–Sommerfeld when you need consistency: back-propagation, near-aperture fields, or careful phase work.

What is physically different between RS-I and RS-II, and when would I choose one?

They trust different data. RS-I propagates the aperture field $U$ (a Dirichlet datum, obliquity $\cos\theta$ ); RS-II propagates its normal slope $\partial U/\partial n$ (a Neumann datum, obliquity $\cos\theta_0$ ). Each is exact given its datum exactly, but in practice you know neither perfectly and pick the more defensible guess. Behind a thin illuminated aperture, "field ≈ incident wave" is natural and standard in Fourier optics, so RS-I dominates. RS-II suits problems where the slope is the clean boundary condition — a pressure-release surface in the acoustic analogue, for instance. Their difference shrinks as $a/\lambda\to\infty$ , and where they disagree materially, no scalar theory deserves trust.

Where does the mysterious −i/λ factor in the Huygens sum actually come from?

It falls out of the derivation rather than being postulated. Differentiating the outgoing spherical wave $e^{ikr}/r$ along the normal brings down a factor $ik$ in the far zone; the Dirichlet image doubles the normal derivative on the plane; and $ik=2\pi i/\lambda$ combined with the $1/4\pi$ of the boundary integral assembles into $-i/\lambda = e^{-i\pi/2}/\lambda$ . Each secondary wavelet is thus scaled by $1/\lambda$ and advanced a quarter-cycle relative to the primary wave — exactly the amplitude-and-phase factor Fresnel inserted by hand, now derived.

Is the obliquity factor cosθ or ½(1+cosθ) — different books show different things?

Both are correct — for different theories. $\cos\theta$ is the RS-I factor; $\tfrac12(\cos\theta+\cos\theta_0)$ is the Kirchhoff factor, which for normal incidence ( $\cos\theta_0=1$ ) becomes the familiar $\tfrac12(1+\cos\theta)$ . They are the same to first order at small angles and differ only at large diffraction angles, where the two theories genuinely disagree. If a text writes $\tfrac12(1+\cos\theta)$ it is using Kirchhoff at normal incidence; if it writes $\cos\theta$ it is using RS-I. Quote whichever obliquity factor matches the theory you are actually propagating with, and note that off-axis they are not interchangeable.

How can propagating light really be "just" a Fourier transform?

Through the angular-spectrum picture. Decompose the aperture field into plane waves by a transverse Fourier transform; each plane wave travels $z$ and accumulates the phase $k_z z$ with $k_z=k\sqrt{1-(\lambda f_x)^2-(\lambda f_y)^2}$ ; an inverse transform reassembles the propagated field. RS-I, written as a convolution, has exactly this transfer function, so "propagate the field" and "transform, multiply by a phase, inverse transform" are the same operation. Spatial frequencies with $(\lambda f_x)^2+(\lambda f_y)^2>1$ make $k_z$ imaginary — the evanescent waves that decay within a wavelength and never reach the far field, the deep origin of the diffraction limit.

Does Rayleigh–Sommerfeld theory account for polarization?

No — it is a scalar theory, tracking one component $U$ and ignoring the coupling Maxwell's equations impose between components at a boundary. That is fine for apertures large compared with a wavelength, where scalar patterns are accurate and polarization enters weakly, but it fails badly for sub-wavelength holes, where the true transmission is throttled by the vector boundary conditions and can fall below a scalar estimate by orders of magnitude. That regime needs the full vector theory — Bethe's small-hole result and the electromagnetic boundary conditions — owned by aq-em-scalar-vector-diffraction-theory. Rayleigh–Sommerfeld is the last word in scalar diffraction and the launch point into the vector story.