10.08.03 · em-sr / advanced-electrodynamics

Diffraction: Kirchhoff Integral, Fraunhofer and Fresnel Regimes

shipped3 tiersLean: none

Anchor (Master): Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 10; Born & Wolf, Principles of Optics, 7th ed. (1999), Ch. 8-9

Intuition Beginner

Diffraction is the bending of waves around obstacles and through openings. Every wave diffracts -- water waves, sound waves, light, radio waves. The amount of bending depends on one number: the ratio of the wavelength to the size of the obstacle or opening.

Sound waves in air have wavelengths ranging from centimeters (high-pitched sounds) to meters (low bass notes). A typical doorway is about one meter wide. Since the wavelength of everyday sound is comparable to the size of doorways, hallways, and corners, sound waves bend around these obstacles readily. This is why you can hear someone speaking from around a corner, even though you cannot see them. The sound waves spread out around the edge of the wall and reach your ears.

Visible light has wavelengths between about 400 nm (violet) and 700 nm (red). A nanometer is one billionth of a meter. A doorway is roughly one million wavelengths wide for visible light. Because the wavelength is so small compared to everyday objects, light produces sharp shadows -- the diffraction is there, but the bending angle is tiny. You cannot see around a corner because the diffraction angle of visible light around a doorway is about one millionth of a radian, far below what the eye can detect.

Radio waves tell the opposite story. FM radio has wavelengths of about 3 meters, comparable to buildings and hills. This is why radio reception works in valleys and behind obstacles -- the waves diffract around them. AM radio has even longer wavelengths (hundreds of meters), which is why AM signals can be received far beyond the line of sight, bending around the curvature of the earth's surface.

The classic demonstration of diffraction is the single-slit experiment. Shine monochromatic light (one wavelength) through a narrow slit onto a distant screen. If light traveled in perfectly straight lines (geometric optics), you would see a bright stripe exactly as wide as the slit. Instead, you see a pattern: a broad central bright band, flanked by alternating dark and bright bands that get progressively dimmer. This is the diffraction pattern, and it is a direct signature of the wave nature of light.

The central bright region has an angular half-width of approximately $λ / a$ , where $λ$ is the wavelength and $a$ is the slit width. The full angular width of the central maximum is approximately $2 λ / a$ . Narrow the slit and the pattern spreads out. Use a longer wavelength and the pattern spreads out. This inverse relationship -- narrower opening produces wider diffraction -- is a universal feature of wave behavior.

The dark bands (minima) occur at angles $θ$ satisfying $a sin θ = nλ$ where $n = 1, 2, 3, \dots$ . The first dark band sits at $sin θ \approx λ / a$ (for small angles, $θ \approx λ / a$ ). A slit of width $a = 100 λ$ gives a first minimum at about $0.01$ radians, or roughly half a degree. A slit of width $a = 10 λ$ gives a first minimum at about 5.7 degrees. A slit of width $a = λ$ gives a first minimum at 90 degrees -- the light spreads out over the entire forward hemisphere.

The intensity on the screen follows the pattern $I = I_{0} (sin β / β)^{2}$ where $β = π a sin θ / λ$ . The central maximum is twice as wide as each secondary maximum, and the secondary maxima are progressively dimmer: the first carries about $4.7%$ of the central peak intensity, the second about $1.7%$ , and the third about $0.8%$ .

Visual Beginner

Regime	Condition	Approximation	Character
Fraunhofer (far-field)	$L ≫ a^{2} / λ$	Linear phase	Fourier transform of aperture
Fresnel (near-field)	$L \sim a^{2} / λ$	Quadratic phase	Fresnel integrals
Geometric optics	$a ≫ λ$	No phase	Sharp shadow

Worked example Beginner

A helium-neon laser beam ( $λ = 633$ nm) passes through a slit of width $a = 0.1$ mm and strikes a screen $L = 2$ m away. Find the positions of the first three dark bands on the screen.

The minima satisfy $a sin θ = nλ$ , giving $sin θ_{n} = nλ / a$ .

With $a = 0.1$ mm $= 1 0^{- 4}$ m and $λ = 633$ nm $= 6.33 \times 1 0^{- 7}$ m:

sin θ_{n} = n \times \frac{6.33 \times 1 0 ^{- 7}}{1 0 ^{- 4}} = n \times 6.33 \times 1 0^{- 3}

For these small angles, $θ_{n} \approx sin θ_{n}$ , and the position on the screen is $y_{n} \approx L θ_{n}$ :

First minimum ( $n = 1$ ): $y_{1} \approx 2 \times 6.33 \times 1 0^{- 3} = 1.27 \times 1 0^{- 2}$ m $= 12.7$ mm
Second minimum ( $n = 2$ ): $y_{2} \approx 25.3$ mm
Third minimum ( $n = 3$ ): $y_{3} \approx 38.0$ mm

The central maximum extends from $y = - 12.7$ mm to $y = + 12.7$ mm, a total width of 25.4 mm. Each secondary maximum between successive minima is about 12.7 mm wide (half the central width). The central maximum is roughly 25 mm wide on a screen 2 m away -- a dramatic spreading for a slit only 0.1 mm wide.

Check your understanding Beginner

Exercise (easy, multiple choice).

You can hear someone around a corner but not see them because:

A. Sound travels faster than light. B. Sound waves have wavelengths comparable to doorways; visible light wavelengths are far smaller. C. Sound is a longitudinal wave and light is a transverse wave. D. The human eye cannot detect diffracted light.

Hint

The diffraction angle is proportional to $λ / a$ where $a$ is the obstacle size. Compare the wavelengths of sound (centimeters to meters) and visible light (about 500 nm) to the size of a doorway (about 1 m).

Answer

B. Diffraction depends on the ratio of wavelength to obstacle size. Sound wavelengths (centimeters to meters) are comparable to the size of doorways and corridors, so sound diffracts readily around corners. Visible light wavelengths (400--700 nm) are about a million times smaller than a doorway, so the diffraction angle is negligible and light casts sharp shadows. The longitudinal vs transverse distinction (C) does not determine the amount of diffraction.

Formal definition Intermediate+

The Kirchhoff diffraction integral. Consider a monochromatic scalar field $ψ (r)$ satisfying the Helmholtz equation $(\nabla^{2} + k^{2}) ψ = 0$ everywhere outside the sources, with $k = 2 π / λ$ . Choose a closed surface $S$ enclosing the observation point. From Green's theorem 10.08.01 applied to $ψ$ and the Helmholtz Green's function $G = e^{ik R} / (4 π R)$ where $R = ∣ r - r^{'} ∣$ :

ψ (r) = \frac{1}{4 π} \oint_{S} [\frac{e ^{ik R}}{R} \frac{\partial ψ}{\partial n ^{'}} - ψ \frac{\partial}{\partial n ^{'}} (\frac{e ^{ik R}}{R})] d a^{'}

This is the Kirchhoff integral theorem: the field at any point inside $S$ is determined entirely by the field and its normal derivative on $S$ . It is the frequency-domain version of the time-domain Kirchhoff representation derived from the retarded Green's function 10.08.01.

Fresnel-Kirchhoff diffraction formula. For diffraction through an aperture $A$ in an opaque screen, construct $S$ from: (1) the aperture plane, (2) a large hemisphere in the forward direction, and (3) a small sphere around the observation point. Apply the Kirchhoff boundary conditions:

In the aperture $A$ : $ψ$ and $\partial ψ / \partial n^{'}$ equal the values of the incident wave (the field passes through undisturbed).
On the opaque screen: $ψ = 0$ and $\partial ψ / \partial n^{'} = 0$ (the screen blocks the field completely).

These conditions are mutually inconsistent (Sommerfeld showed that $ψ$ and $\partial ψ / \partial n$ cannot both vanish on a physical surface), but the error is small when the aperture is large compared to the wavelength.

For a point source at $r_{s}$ emitting a spherical wave $ψ_{0} = A e^{ik r_{s}} / r_{s}$ , the field at observation point $P$ becomes:

ψ_{P} = - \frac{i A}{λ} \int_{A} \frac{e ^{ik (r + s)}}{r s} K (χ) d a^{'}

where $r$ is the distance from the source to the aperture point, $s$ is the distance from the aperture point to $P$ , and:

K (χ) = \frac{1 + cos χ}{2}

is the obliquity factor (inclination factor), with $χ$ the angle between the source-to-aperture direction and the aperture-to-observer direction. For normal incidence and small observation angles, $K \approx 1$ .

Fraunhofer (far-field) regime. When the observation distance satisfies $L ≫ a^{2} / λ$ (where $a$ is the characteristic aperture dimension), the phase can be linearized and the diffraction integral reduces to:

ψ (θ_{x}, θ_{y}) \propto \int_{A} e^{- ik (s i n θ_{x} x^{'} + s i n θ_{y} y^{'})} d x^{'} d y^{'}

This is the two-dimensional Fourier transform of the aperture function. The Fraunhofer diffraction pattern is the power spectrum of the aperture.

Fresnel (near-field) regime. When the observation distance satisfies $L \sim a^{2} / λ$ , the quadratic phase terms must be retained:

ψ (r_{P}) \propto \int_{A} e^{ik [(x^{'} - x_{P})^{2} + (y^{'} - y_{P})^{2}] / (2 z)} d x^{'} d y^{'}

These integrals involve the Fresnel integrals:

C (u) = \int_{0}^{u} cos (\frac{π t ^{2}}{2}) d t, S (u) = \int_{0}^{u} sin (\frac{π t ^{2}}{2}) d t

The parametric curve $(C (u), S (u))$ is the Cornu spiral, which provides a geometric method for evaluating Fresnel diffraction patterns.

The Fresnel number $N_{F} = a^{2} / (λ L)$ determines the regime:

$N_{F}$	Regime	Character
$N_{F} ≪ 1$	Fraunhofer	Far-field, Fourier optics
$N_{F} \sim 1$	Fresnel	Near-field, Fresnel integrals
$N_{F} ≫ 1$	Geometric	Ray optics, sharp shadows

Fresnel zones. For a point source and a point observer, the wavefront can be divided into concentric annular Fresnel zones such that the path difference from successive zone boundaries to the observer differs by $λ /2$ . The $n$ -th zone boundary has radius $r_{n} \approx nλ L$ for $L$ the distance to the observation point. Adjacent zones contribute with opposite phase (path difference $λ /2$ ), so the total amplitude at the observer is an alternating series that converges to approximately half the contribution from the first zone. This construction underpins both the Fresnel zone plate and the physical explanation of why diffraction decreases with increasing aperture size.

Key derivation Intermediate+

Derivation (The Fresnel-Kirchhoff diffraction formula from Green's theorem).

Theorem. The field at an observation point behind a diffracting aperture is given by the Fresnel-Kirchhoff integral, obtained by applying Green's theorem with the Helmholtz Green's function and the Kirchhoff boundary conditions.

Proof. Start with Green's theorem for the Helmholtz equation. Let $ψ$ satisfy $(\nabla^{2} + k^{2}) ψ = 0$ and $G = e^{ik R} / (4 π R)$ satisfy $(\nabla^{'2} + k^{2}) G = - δ^{3} (r - r^{'})$ . Green's theorem gives:

ψ (r) = - \oint_{S} [ψ \frac{\partial G}{\partial n ^{'}} - G \frac{\partial ψ}{\partial n ^{'}}] d a^{'}

Choose $S$ as: (a) the plane of the screen (including the aperture), (b) a large hemisphere of radius $R \to \infty$ on the source side, and (c) a small sphere of radius $ϵ \to 0$ around the observation point. The hemisphere contribution vanishes by the Sommerfeld radiation condition 10.08.01 (only outgoing waves survive at infinity). The small sphere gives $+ ψ (r)$ by Gauss's theorem, which cancels the left-hand side and produces the Kirchhoff integral.

In the aperture, substitute the incident spherical wave $ψ = A e^{ik r_{s}} / r_{s}$ and compute $\partial ψ / \partial n^{'}$ . Noting that $\partial r_{s} / \partial n^{'} = - cos θ_{s}$ where $θ_{s}$ is the angle between the source direction and the outward normal, and $\partial R / \partial n^{'} = cos θ_{P}$ where $θ_{P}$ is the angle between the observer direction and the inward normal:

\frac{\partial G}{\partial n ^{'}} = \frac{e ^{ik R}}{4 π R} (ik - \frac{1}{R}) cos θ_{P} \approx \frac{ik cos θ _{P}}{4 π R} e^{ik R}

where the last step uses $k R ≫ 1$ (the observation distance is many wavelengths, so the $1/ R$ term is negligible compared to $ik$ ). A similar approximation holds for $\partial ψ / \partial n^{'}$ . Combining the two terms and using $k = 2 π / λ$ :

ψ_{P} = - \frac{i A}{λ} \int_{A} \frac{e ^{ik (r + s)}}{r s} \frac{cos θ _{s} + cos θ _{P}}{2} d a^{'}

The factor $(cos θ_{s} + cos θ_{P}) /2 = K (χ)$ is the obliquity factor. For normal incidence ( $θ_{s} = 0$ ) and small observation angles ( $θ_{P} \approx 0$ ): $K \approx 1$ , recovering the simple form of Huygens' principle. $□$

Bridge. The Kirchhoff integral transforms the diffraction problem into an integral over the aperture: each point in the aperture acts as a secondary source (Huygens' principle), and the total field is the coherent sum of all these secondary wavelets. The foundational insight is that the diffraction pattern is determined entirely by the shape of the aperture and the wavelength, with the Kirchhoff boundary conditions providing the bridge from the wave equation to the observed pattern. The central message is that the Fresnel-Kirchhoff formula converts a differential equation (the wave equation) into an integral equation (the diffraction integral), which can then be evaluated in two limiting cases. Putting these together, the two regimes -- Fraunhofer (far-field, Fourier transform) and Fresnel (near-field, Fresnel integrals) -- cover all practical diffraction phenomena, with the Fresnel number $N_{F}$ serving as the control parameter that selects the appropriate approximation.

Exercises Intermediate+

Exercise 1 (easy, symbolic).

A slit of width $a = 0.05$ mm is illuminated by light of wavelength $λ = 500$ nm. Compute the angular positions of the first three minima in the Fraunhofer diffraction pattern. State whether the small-angle approximation is justified.

Hint

The minima satisfy $a sin θ = nλ$ . Check whether $sin θ_{n} ≪ 1$ for the values of $n$ requested.

Answer

The minima occur at $sin θ_{n} = nλ / a$ . With $λ = 5 \times 1 0^{- 7}$ m and $a = 5 \times 1 0^{- 5}$ m:

sin θ_{n} = n \times \frac{5 \times 1 0 ^{- 7}}{5 \times 1 0 ^{- 5}} = n \times 1 0^{- 2}

$n = 1$ : $sin θ_{1} = 0.01$ , so $θ_{1} = 0.57 3^{\circ}$
$n = 2$ : $sin θ_{2} = 0.02$ , so $θ_{2} = 1.14 6^{\circ}$
$n = 3$ : $sin θ_{3} = 0.03$ , so $θ_{3} = 1.71 9^{\circ}$

All angles are small ( $sin θ_{n} \approx θ_{n}$ to better than $0.005%$ ), so the small-angle approximation is well justified. The diffraction pattern is tightly concentrated around the forward direction because the slit is 100 wavelengths wide.

Exercise 2 (easy, symbolic).

The Hubble Space Telescope has a primary mirror diameter of $D = 2.4$ m. For visible light at $λ = 550$ nm, compute the angular resolution (Rayleigh criterion) and the corresponding linear resolution at a distance of 100 km.

Hint

The Rayleigh criterion for a circular aperture is $θ_{min} = 1.22 λ / D$ . Convert the angular resolution to a linear resolution at distance $L$ using $s = θ \cdot L$ .

Answer

θ_{min} = \frac{1.22 λ}{D} = \frac{1.22 \times 5.5 \times 1 0 ^{- 7}}{2.4} = 2.80 \times 1 0^{- 7} rad = 0.058 arcsec

At distance $L = 100$ km $= 1 0^{5}$ m:

s = θ_{min} \times L = 2.80 \times 1 0^{- 7} \times 1 0^{5} = 0.028 m = 2.8 cm

Hubble can resolve objects 2.8 cm apart at a distance of 100 km. At its orbital altitude of about 550 km, the ground resolution is about 15 cm -- sufficient to distinguish large structures. The Rayleigh criterion sets a fundamental limit: larger apertures (or shorter wavelengths) provide finer angular resolution.

Exercise 3 (medium, symbolic).

Derive the Fraunhofer diffraction pattern for a double slit: two identical slits of width $a$ separated by distance $d$ (center-to-center). Show that the intensity is the product of a single-slit envelope and a double-slit interference pattern.

Hint

The aperture function is the convolution of a single slit with two delta functions at $\pm d /2$ . The Fourier transform of a convolution is the product of the Fourier transforms.

Answer

The aperture consists of two slits at positions $x = \pm d /2$ , each of width $a$ . The Fraunhofer amplitude is:

ψ (θ) \propto \int_{- d /2 - a /2}^{- d /2 + a /2} e^{- ik x s i n θ} d x + \int_{d /2 - a /2}^{d /2 + a /2} e^{- ik x s i n θ} d x

Each integral gives the single-slit diffraction amplitude. Define $β = k a sin θ /2 = π a sin θ / λ$ and $δ = k d sin θ /2 = π d sin θ / λ$ :

ψ (θ) \propto \frac{sin β}{β} (e^{i δ} + e^{- i δ}) = 2 \frac{sin β}{β} cos δ

The intensity is:

I (θ) = I_{0} (\frac{sin β}{β})^{2} cos^{2} δ

This is the product of:

The single-slit diffraction envelope $(sin β / β)^{2}$ , with minima at $a sin θ = nλ$ .
The double-slit interference pattern $cos^{2} δ = cos^{2} (π d sin θ / λ)$ , with maxima at $d sin θ = mλ$ (constructive interference) and minima at $d sin θ = (m + 1/2) λ$ .

The ratio $d / a$ controls the number of interference fringes within the central diffraction envelope: there are approximately $2 d / a$ bright fringes in the central maximum. Missing orders occur when a diffraction minimum coincides with an interference maximum, i.e., when $d / a$ is a ratio of integers.

Exercise 4 (medium, symbolic).

A Fresnel zone plate consists of alternating transparent and opaque concentric rings. The $n$ -th zone boundary has radius $r_{n} = nλ f$ where $f$ is the focal length. Show that the zone plate acts as a lens with focal length $f$ , and estimate the intensity gain at the focal point compared to an open aperture of the same outer radius.

Hint

Adjacent Fresnel zones contribute with opposite phase (path difference $λ /2$ ). Blocking every other zone removes the destructive interference, concentrating light at the focal point.

Answer

Zone construction. At the focal point a distance $f$ from the zone plate, the path from the $n$ -th zone boundary to the focal point is $r_{n}^{2} / (2 f) + O (r_{n}^{4} / f^{3})$ . Setting this equal to $nλ /2$ gives $r_{n} = nλ f$ .

The contribution from the $n$ -th zone to the amplitude at the focal point is approximately $a_{n} \approx (- 1)^{n} A_{0}$ where $A_{0}$ is the contribution from a single zone. The alternating sign arises because each zone covers an additional path difference of $λ /2$ , shifting the phase by $π$ .

Open aperture. With all zones present:

ψ_{open} = A_{0} (1 - 1 + 1 - 1 + \dots) \approx \frac{1}{2} A_{0}

The alternating series converges to approximately $A_{0} /2$ (the first zone dominates, and successive cancellations leave about half the first zone's contribution).

Zone plate. Blocking every other zone (say, the even-numbered ones):

ψ_{zone} = A_{0} (1 + 1 + 1 + \dots) = N_{z} A_{0}

where $N_{z}$ is the number of open zones. The intensity gain over the open aperture is:

\frac{I _{zone}}{I _{open}} \approx \frac{N _{z}^{2} A _{0}^{2}}{( A _{0} /2 ) ^{2}} = 4 N_{z}^{2}

A zone plate with 50 open zones concentrates light by a factor of $1 0^{4}$ at the focal point. Unlike a glass lens, a zone plate has multiple focal lengths ( $f, f /3, f /5, \dots$ for odd harmonics) and works for any wavelength (with $f \propto 1/ λ$ ).

Exercise 5 (medium, symbolic).

Compute the Fraunhofer diffraction pattern of a circular aperture of radius $a$ . Identify the Airy disk radius and show that the first zero occurs at $sin θ = 1.22 λ / (2 a)$ .

Hint

Use polar coordinates in the aperture plane. The angular integral produces a Bessel function: $\int_{0}^{2 π} e^{- ik r^{'} s i n θ c o s ϕ^{'}} d ϕ^{'} = 2 π J_{0} (k r^{'} sin θ)$ . Then evaluate $\int_{0}^{a} J_{0} (k r^{'} sin θ) r^{'} d r^{'}$ .

Answer

For a circular aperture of radius $a$ with uniform illumination, the Fraunhofer amplitude in polar coordinates is:

ψ (θ) \propto \int_{0}^{2 π} \int_{0}^{a} e^{- ik r^{'} s i n θ c o s ϕ^{'}} r^{'} d r^{'} d ϕ^{'}

The angular integral gives $2 π J_{0} (k r^{'} sin θ)$ . The radial integral evaluates using $\int_{0}^{x} t J_{0} (t) d t = x J_{1} (x)$ :

\int_{0}^{a} J_{0} (k r^{'} sin θ) r^{'} d r^{'} = \frac{a J _{1} ( k a sin θ )}{k sin θ}

The intensity is:

I (θ) = I_{0} [\frac{2 J _{1} ( k a sin θ )}{k a sin θ}]^{2} = I_{0} [\frac{2 J _{1} ( u )}{u}]^{2}

where $u = k a sin θ = (2 π a / λ) sin θ$ . This is the Airy pattern.

The central bright region (the Airy disk) is bounded by the first zero of $J_{1} (u)$ , which occurs at $u_{1} = 3.8317$ . Therefore:

sin θ_{1} = \frac{3.8317}{2 π} \frac{λ}{a} = \frac{1.22 λ}{D}

where $D = 2 a$ is the aperture diameter. The Airy disk contains about $84%$ of the total transmitted intensity. The first bright ring holds about $7%$ , the second about $3%$ , and higher rings carry progressively less.

Exercise 6 (medium, symbolic).

Evaluate the Kirchhoff diffraction integral for a rectangular aperture of dimensions $a \times b$ in the Fraunhofer regime. Express the result as a product of sinc functions and determine the dimensions of the central maximum.

Hint

The Fraunhofer integral separates in $x$ and $y$ . Each direction gives a sinc function: $\int_{- a /2}^{a /2} e^{- ik x s i n θ_{x}} d x = a sinc (π a sin θ_{x} / λ)$ where $sinc (x) = sin (x) / x$ .

Answer

The aperture function is $f (x, y) = 1$ for $∣ x ∣ \leq a /2$ and $∣ y ∣ \leq b /2$ , and zero otherwise. The Fraunhofer integral separates:

ψ (θ_{x}, θ_{y}) = C \int_{- a /2}^{a /2} e^{- ik x s i n θ_{x}} d x \int_{- b /2}^{b /2} e^{- ik y s i n θ_{y}} d y

Evaluating each integral:

\int_{- a /2}^{a /2} e^{- ik x s i n θ_{x}} d x = \frac{2 sin ( π a sin θ _{x} / λ )}{k sin θ _{x}} = a sinc (\frac{π a sin θ _{x}}{λ})

and similarly for $y$ . The intensity is:

I (θ_{x}, θ_{y}) = I_{0} sinc^{2} (\frac{π a sin θ _{x}}{λ}) sinc^{2} (\frac{π b sin θ _{y}}{λ})

where $sinc (x) = sin (x) / x$ .

The central maximum extends to the first zeros at $sin θ_{x} = \pm λ / a$ and $sin θ_{y} = \pm λ / b$ . For a square aperture ( $a = b$ ), the central maximum is square in $(θ_{x}, θ_{y})$ . For a long narrow slit ( $b ≫ a$ ), the pattern is a thin line spread out along the $θ_{x}$ direction -- the diffraction is significant only in the direction perpendicular to the long axis of the slit.

Exercise 7 (hard, symbolic).

State and prove Babinet's principle: if $ψ_{1}$ is the field diffracted by an aperture $A$ in an opaque screen, and $ψ_{2}$ is the field diffracted by the complementary screen (where $A$ and the opaque region are interchanged), then $ψ_{1} + ψ_{2} = ψ_{0}$ where $ψ_{0}$ is the unobstructed incident field. Show that the diffraction patterns of complementary screens have equal intensity outside the forward beam.

Hint

The Kirchhoff integral for screen 1 integrates over the aperture region $A$ . The Kirchhoff integral for the complementary screen integrates over $A^{c}$ . Together these cover the entire plane, reproducing the unobstructed field.

Answer

Babinet's principle. Let screen 1 have aperture $A$ in an opaque screen, and screen 2 have aperture $A^{c}$ (the complement). The Kirchhoff integrals are:

ψ_{1} (r) = - \frac{i}{λ} \int_{A} \frac{e ^{ik s}}{s} K (χ) ψ_{0} d a^{'}

ψ_{2} (r) = - \frac{i}{λ} \int_{A^{c}} \frac{e ^{ik s}}{s} K (χ) ψ_{0} d a^{'}

Adding:

ψ_{1} + ψ_{2} = - \frac{i}{λ} \int_{A \cup A^{c}} \frac{e ^{ik s}}{s} K (χ) ψ_{0} d a^{'}

Since $A \cup A^{c}$ covers the entire plane (no screen present), the integral reproduces the unobstructed incident field $ψ_{0}$ at $P$ (by the Kirchhoff theorem with no screen). Therefore:

ψ_{1} (r) + ψ_{2} (r) = ψ_{0} (r)

Consequence for intensities. Away from the forward direction ( $θ \neq = 0$ ), $ψ_{0} = 0$ for a collimated beam (the unobstructed wave propagates only forward). Therefore $ψ_{2} = - ψ_{1}$ , and:

∣ ψ_{2} ∣^{2} = ∣ ψ_{1} ∣^{2}

The diffraction patterns of complementary screens are identical in intensity outside the forward beam. A thin wire and a thin slit of the same width produce the same diffraction pattern (except for the forward beam). This is useful experimentally: it is sometimes easier to fabricate a wire obstacle than a narrow slit of the same dimension.

Exercise 8 (hard, symbolic).

The Sommerfeld half-plane solution (1896) gives the exact diffracted field for a plane wave incident on a perfectly conducting semi-infinite screen. Describe the structure of the solution and derive the asymptotic behavior in three regions: (a) the deep illuminated region, (b) the deep shadow, and (c) near the shadow boundary.

Hint

The solution is expressed in terms of Fresnel integrals. The shadow boundary is the line of geometric optics. Far from the boundary, the Fresnel integrals approach their asymptotic limits (0 or 1). Near the boundary, they are order unity.

Answer

Setup. A plane wave $e^{ik x}$ propagates in the $+ x$ direction. A semi-infinite perfectly conducting screen occupies $y = 0$ , $x \leq 0$ . The shadow boundary is the half-line $y = 0$ , $x > 0$ (the geometric optics continuation of the screen edge).

Sommerfeld's solution (for the scalar case, TE polarization). The total field can be written as:

ψ (x, y) = ψ_{geom} (x, y) + ψ_{diff} (x, y)

where $ψ_{geom}$ is the geometric optics field (present only in the illuminated region $y > 0$ ) and $ψ_{diff}$ is the diffracted field expressed using the Fresnel integral $F (w) = \int_{0}^{w} e^{iπ t^{2} /2} d t$ . Define the Fresnel parameter:

w = \frac{2 k}{π} \frac{y}{2 x} = y \frac{k}{π x}

which measures the transverse displacement from the shadow boundary in units of the Fresnel scale $λ x$ .

(a) Deep illuminated region ( $w ≫ 1$ ): $F (w) \to (1 + i) /2$ and $ψ \to e^{ik x}$ (the undisturbed plane wave). Diffraction is negligible: the geometric optics limit is recovered.

(b) Deep shadow ( $w ≪ - 1$ ): $ψ_{geom} = 0$ and $F (w) \to 0$ . The diffracted field behaves as:

ψ_{diff} \sim \frac{e ^{i (k x + π /4)}}{2 π k} \frac{1}{w} = \frac{e ^{ik x}}{y} \frac{x}{k} \cdot e^{iπ /4} \frac{1}{2 π}

The field decays as $1/∣ w ∣$ into the shadow, meaning it falls off as $λ x / y$ -- the wave penetrates a distance of order $λ x$ into the shadow. The penetration depth grows with distance from the edge and shrinks with frequency.

(c) Near the shadow boundary ( $∣ w ∣ ≲ 1$ ): The Fresnel integrals are order unity and the field transitions smoothly from illuminated to shadow. At the boundary itself ( $w = 0$ ): $F (0) = 0$ , and $ψ = e^{ik x} /2$ -- the field is exactly half the incident value. The transition region has width $Δ y \sim λ x$ , which grows with distance from the edge.

This exact solution validates the Fresnel-Kirchhoff theory in the appropriate limit and reveals where the scalar approximation breaks down (near the edge, where polarization-dependent corrections matter).

Lean formalization Intermediate+

Mathlib has no diffraction theory and no framework for the oscillatory integrals involved. The Kirchhoff integral, Fresnel integrals, the Airy pattern, Babinet's principle, and the Sommerfeld half-plane solution are all absent. Formalising the Fresnel integrals alone would require asymptotic analysis of oscillatory integrals (stationary phase, steepest descent), which is well beyond Mathlib's current scope. lean_status: none.

Advanced results Master

Vector diffraction theory. The scalar Kirchhoff theory treats only one component of the field and cannot handle polarization. For electromagnetic waves, a proper treatment must respect the full vector nature of $E$ and $B$ . The vector Kirchhoff integral is obtained by applying the Kirchhoff theorem to each Cartesian component of the electric field and then enforcing the Maxwell constraint $\nabla \cdot E = 0$ . The result replaces the scalar obliquity factor with a dyadic (tensor) kernel:

E (r) = \frac{i}{λ} \int_{A} \frac{e ^{ik s}}{s} [\hat{s} \times (\hat{n} \times E_{0})] d a^{'}

where $\hat{s}$ is the unit vector from the aperture point to the observer, $\hat{n}$ is the outward normal to the aperture, and $E_{0}$ is the incident field. The vector nature introduces polarization-dependent corrections absent from the scalar theory. For large apertures ( $a ≫ λ$ ) and small observation angles, the scalar theory is an excellent approximation, but near edges and for small apertures, the vector corrections become significant.

Sommerfeld's exact half-plane solution. The Sommerfeld solution (1896) for diffraction by a semi-infinite conducting screen is one of the few exact solutions in diffraction theory. It uses the method of contour integration in the complex plane [06.01] -- specifically, separation of variables in parabolic coordinates (equivalent to the Wiener-Hopf technique). The scalar solutions for TE and TM polarization differ by their boundary conditions (Dirichlet: $E_{z} = 0$ on the screen vs Neumann: $\partial B_{z} / \partial n = 0$ ), and this polarization dependence is absent from the scalar Fresnel-Kirchhoff theory. The Sommerfeld solution serves as the benchmark for testing approximate theories and reveals the structure of edge diffraction: the edge acts as a line source of cylindrical waves, producing a characteristic $1/ r$ amplitude decay.

Fourier optics. In the Fraunhofer regime, the diffraction pattern is the Fourier transform of the aperture function:

\tilde{f} (k_{⊥}) = \int_{A} f (r^{'}) e^{- i k_{⊥} \cdot r^{'}} d^{2} r^{'}

where $k_{⊥} = k (sin θ_{x}, sin θ_{y})$ is the transverse wavevector. This identification makes diffraction a linear, shift-invariant system analyzable with the full machinery of Fourier analysis. A thin lens of focal length $f$ performs a Fourier transform in hardware: the field at the back focal plane is the Fourier transform of the field at the front focal plane, with spatial frequency $k_{x}$ mapped to transverse position $x = f k_{x} / k$ . This is the basis for:

Spatial filtering (the $4 f$ system): an object at the front focal plane of the first lens produces its Fourier transform at the intermediate plane, where a filter can block or modify selected spatial frequencies. The second lens performs the inverse transform, producing the filtered image.
Optical correlation and pattern recognition: the Fourier transform enables convolution and correlation operations at the speed of light.
Abbe theory of imaging: in a microscope, the objective lens forms the Fourier transform of the object, the condenser aperture selects which spatial frequencies pass, and the eyepiece reconstructs the image. The resolution limit $δ x \approx λ / (2 NA)$ where NA is the numerical aperture follows directly from the band-limiting imposed by the finite lens aperture.

Synthesis. Diffraction theory connects the wave equation to the observed optical patterns through the Kirchhoff integral, which expresses the field as a coherent sum of secondary wavelets from the aperture. The foundational insight is that the far-field diffraction pattern is the Fourier transform of the aperture function, making diffraction and Fourier optics inseparable. The central message is that the Fresnel number $N_{F}$ controls the transition from near-field (Fresnel) to far-field (Fraunhofer) behavior, and the Rayleigh criterion quantifies the ultimate resolution limit imposed by diffraction. Putting these together, diffraction theory provides the bridge between wave physics and practical optics, explaining why no optical instrument can resolve features smaller than about $λ /2$ and why larger apertures produce sharper images.

Full proof set Master

Proposition (Fraunhofer limit of the Kirchhoff integral). For an observation point at distance $z$ satisfying $z ≫ a^{2} / λ$ (where $a$ is the maximum aperture dimension), the Fresnel-Kirchhoff integral reduces to the Fourier transform of the aperture function.

Proof. Start with the Fresnel-Kirchhoff formula for a plane wave $e^{ik z}$ incident along the $z$ -axis. The diffracted field at point $(x_{P}, y_{P}, z)$ is:

ψ (x_{P}, y_{P}, z) = \frac{e ^{ik z}}{iλ z} \int_{A} e^{ik [(x^{'} - x_{P})^{2} + (y^{'} - y_{P})^{2}] / (2 z)} d x^{'} d y^{'}

Expand the exponent:

\frac{( x ^{'} - x _{P} ) ^{2} + ( y ^{'} - y _{P} ) ^{2}}{2 z} = \frac{x _{P}^{2} + y _{P}^{2}}{2 z} - \frac{x ^{'} x _{P} + y ^{'} y _{P}}{z} + \frac{x ^{'2} + y ^{'2}}{2 z}

The first term depends only on the observation point and factors out of the integral. The second term gives the Fourier transform kernel. The third term is the Fresnel phase $k (x^{'2} + y^{'2}) / (2 z)$ .

In the Fraunhofer limit $z ≫ a^{2} / λ$ , the Fresnel phase satisfies:

\frac{k a ^{2}}{2 z} = \frac{π a ^{2}}{λ z} = π N_{F} ≪ π

The maximum phase variation across the aperture due to this term is much less than $π$ , so $e^{ik (x^{'2} + y^{'2}) / (2 z)} \approx 1$ and the integral reduces to:

ψ (x_{P}, y_{P}, z) = \frac{e ^{ik z}}{iλ z} e^{ik (x_{P}^{2} + y_{P}^{2}) / (2 z)} \int_{A} e^{- ik (x^{'} x_{P} + y^{'} y_{P}) / z} d x^{'} d y^{'}

The remaining integral is the 2D Fourier transform of the aperture function evaluated at spatial frequencies $k_{x} = k x_{P} / z$ , $k_{y} = k y_{P} / z$ . For small angles: $k_{x} \approx k sin θ_{x}$ and $k_{y} \approx k sin θ_{y}$ . The prefactor $e^{ik (x_{P}^{2} + y_{P}^{2}) / (2 z)}$ is a quadratic phase that does not affect the intensity pattern. $□$

Connections Master

Green's functions 10.08.01 provide the Helmholtz Green's function from which the Kirchhoff integral is derived via Green's theorem.
Wave equation 10.04.02 is the governing PDE; the Helmholtz equation is its frequency-domain form, and the Kirchhoff integral is its integral representation.
Complex analysis [06.01] provides the contour integration techniques used in the Sommerfeld half-plane solution and the asymptotic evaluation of diffraction integrals via the method of steepest descent.
Multipole radiation 10.07.02 is the radiation analogue: the far-field radiation pattern is the angular distribution of the multipole expansion, just as the Fraunhofer pattern is the Fourier transform of the aperture.
Babinet's principle connects diffraction by obstacles to diffraction by complementary apertures, halving the number of independent diffraction problems one must solve.

Historical & philosophical context Master

Huygens (1678, published 1690) proposed that every point on a wavefront acts as a source of secondary spherical wavelets, and the envelope of these wavelets forms the new wavefront. This was a geometric principle, not derived from any equation, and it could not account for diffraction patterns -- it predicted only wavefront propagation, not amplitudes or intensities.

Fresnel (1818) placed Huygens' principle on a quantitative foundation by adding interference: the secondary wavelets superpose coherently, with definite phases, and the resulting amplitude determines the intensity. His submission to the French Academy prize competition on diffraction (1818) was a landmark. Poisson, one of the judges, derived a consequence that he considered absurd: a bright spot should appear at the center of the shadow of a circular obstacle (the Poisson-Arago spot). Arago immediately verified this experimentally, converting Poisson's attempted refutation into decisive evidence for the wave theory of light.

Fraunhofer (1821) observed and classified the far-field diffraction patterns of single slits, double slits, and gratings. His work was primarily experimental and phenomenological, but the regularity of the patterns he observed -- particularly the grating spectra -- established spectroscopy as a quantitative science. The patterns are now understood as the Fourier transforms of the respective aperture functions.

Kirchhoff (1882) derived the integral theorem from the scalar wave equation and Green's theorem, showing that Huygens' principle and the Fresnel-Kirchhoff formula follow as rigorous consequences rather than heuristic postulates. He formulated the boundary conditions (field and normal derivative specified on the screen) that bear his name, although their self-consistency was later questioned by Poincare and others. Sommerfeld demonstrated that the Kirchhoff conditions are mutually incompatible on physical grounds (one cannot independently specify both $ψ$ and $\partial ψ / \partial n$ ), but the resulting errors are small for apertures much larger than the wavelength.

Sommerfeld (1896) obtained the first exact solution of a diffraction problem: the half-plane. This achievement required the full apparatus of complex analysis [06.01] and demonstrated that exact diffraction theory is far richer than the scalar Kirchhoff approximation. The Sommerfeld solution showed that the edge acts as a line source of cylindrical waves, a feature that the Kirchhoff theory captures only approximately. This work established the mathematical framework -- parabolic coordinates, contour integration, Wiener-Hopf technique -- that underpins modern diffraction theory.

Bibliography Master

Foundations of wave optics

Huygens, C., Traite de la Lumiere (Leiden, 1690).
Fresnel, A., "Memoire sur la diffraction de la lumiere," Mem. Acad. Sci. Inst. Fr. 5, 339 (1826).
Fraunhofer, J., "Kurzer Bericht von den Resultaten neuerer Versuche uber die Gesetze des Lichtes," Denkschr. K. Akad. Wiss. Munchen 8, 1 (1822).

Kirchhoff and rigorous diffraction theory

Kirchhoff, G., "Zur Theorie der Lichtstrahlen," Sitzungsber. K. Preuss. Akad. Wiss. 22, 641 (1882).
Sommerfeld, A., "Mathematische Theorie der Diffraction," Math. Ann. 47, 317 (1896).
Born, M. and Wolf, E., Principles of Optics, 7th ed. (Cambridge, 1999).

Modern references

Jackson, J. D., Classical Electrodynamics, 3rd ed. (Wiley, 1999).
Goodman, J. W., Introduction to Fourier Optics, 3rd ed. (Roberts, 2005).
Hecht, E., Optics, 5th ed. (Pearson, 2017).

Prerequisites

10.04.02
10.08.01

Tier anchors

beginner: Griffiths, Introduction to Electrodynamics, 4th ed. (2017), Ch. 10.2-10.3; Hecht, Optics, 5th ed. (2017), Ch. 10
intermediate: Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 10.5-10.9
master: Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 10; Born & Wolf, Principles of Optics, 7th ed. (1999), Ch. 8-9

References

Griffiths — Introduction to Electrodynamics, 4th ed. (Cambridge, 2017) · Ch. 10.2-10.3 Diffraction; Kirchhoff's theory; Fraunhofer and Fresnel diffraction
Jackson — Classical Electrodynamics, 3rd ed. (Wiley, 1999) · Ch. 10.5-10.9 Scalar diffraction theory; Kirchhoff integral; Fraunhofer and Fresnel diffraction; Babinet's principle
Born & Wolf — Principles of Optics, 7th ed. (Cambridge, 1999) · Ch. 8-9 Diffraction theory; Fresnel and Fraunhofer diffraction; rigorous diffraction theory

Estimated time

beginner: 20m
intermediate: 45m
master: 70m