Fraunhofer Diffraction as a Fourier Transform

Derivation

Fraunhofer Diffraction as a Fourier Transform

D-207 Home PU-206 Threads light · waves Depends on The Helmholtz-Kirchhoff Diffraction Integral, fourier-transform-pairs

Statement

In the far field (Fraunhofer regime), the diffracted scalar field produced by a planar aperture equals, up to a multiplicative amplitude and phase factor, the two-dimensional Fourier transform of the aperture transmission function \(t(x',y')\), evaluated at the spatial frequencies \(f_X = x/(\lambda z)\) and \(f_Y = y/(\lambda z)\) fixed by the diffraction angles. Explicitly, the observed field is \(U(x,y)\propto \tilde{t}(f_X,f_Y)=\iint t(x',y')\,e^{-i2\pi(f_X x' + f_Y y')}\,dx'\,dy'\), so the diffraction pattern is the aperture's angular spectrum.

Why it matters

This is the operational core of physical optics: it turns every diffraction problem for a planar aperture into a Fourier-transform lookup. Slits, gratings, circular pupils, and phase masks all inherit the transform-pair machinery, so a single mathematical fact (a rectangle transforms to a sinc, a Gaussian to a Gaussian, a comb to a comb) delivers the entire catalogue of diffraction patterns without re-doing an integral each time.

It also underpins Fourier optics as an engineering discipline. A lens performs a physical Fourier transform in its back focal plane, spatial filtering removes or enhances chosen frequency bands, and the resolution limits of telescopes and microscopes are read directly off the transform of the pupil. The far-field pattern is the aperture's "frequency content" made visible on a screen.

Assumptions

Scalar diffraction.The field is treated as a single complex scalar \(U\); if dropped, vector (polarization) coupling at the aperture edges matters and the simple transform breaks for features comparable to \(\lambda\).

Monochromatic, coherent illumination of wavelength \(\lambda\) and wavenumber \(k=2\pi/\lambda\).If dropped, each wavelength forms its own scaled pattern and one must integrate over the source spectrum (broadband smearing).

Thin planar aperture with transmission \(t(x',y')\).If dropped, the aperture imposes propagation and multiple internal reflections; \(t\) is no longer a simple multiplicative mask.

Far-field (Fraunhofer) condition: \(z \gg \dfrac{k\,(x'^2+y'^2)_{\max}}{2}\), i.e. \(z \gg D^2/\lambda\) for aperture size \(D\).If dropped, the quadratic phase \(e^{ik(x'^2+y'^2)/2z}\) cannot be neglected and one is in the Fresnel regime; the pattern is then a fractional/chirped transform, not a plain Fourier transform.

Paraxial observation and slowly varying obliquity factor \(K(\chi)\approx 1\).If dropped, large diffraction angles pick up an inclination factor \(\tfrac{1}{2}(1+\cos\chi)\) and the direction-cosine-to-frequency map is no longer linear in the screen coordinate.

Derivation

\[ U(x,y)=\frac{1}{i\lambda}\iint_{A} t(x',y')\,U_{\text{inc}}(x',y')\,\frac{e^{ikr}}{r}\,K(\chi)\,dx'\,dy' \]

Start from the assumed Fresnel–Kirchhoff diffraction integral for a planar aperture; \(r\) is the distance from source point \((x',y')\) in the aperture to the observation point \((x,y,z)\). A

\[ U_{\text{inc}}(x',y')=U_0 \quad\text{(uniform normal-incidence plane wave)},\qquad K(\chi)\approx 1 \]

Take a collimated plane wave illuminating the aperture at normal incidence and use the paraxial obliquity factor. This isolates the geometry-dependent phase in \(e^{ikr}\). A

\[ r=\sqrt{z^2+(x-x')^2+(y-y')^2}=z\sqrt{1+\frac{(x-x')^2+(y-y')^2}{z^2}} \]

Write the exact geometric distance to the observation plane at range \(z\). A

\[ r \approx z + \frac{(x-x')^2+(y-y')^2}{2z} = z + \frac{x^2+y^2}{2z} - \frac{xx'+yy'}{z} + \frac{x'^2+y'^2}{2z} \]

Binomial expansion \(\sqrt{1+\epsilon}\approx 1+\tfrac{\epsilon}{2}\), valid because \(|x-x'|,|y-y'|\ll z\) (paraxial). Expanding the square separates a constant term, an observation-plane term, a bilinear cross term, and an aperture quadratic term. B

\[ \frac{k}{2z}\,(x'^2+y'^2)_{\max}\ll 1 \;\;\Longleftrightarrow\;\; z\gg \frac{D^2}{\lambda} \quad\Rightarrow\quad e^{ik(x'^2+y'^2)/2z}\to 1 \]

Impose the Fraunhofer condition. The aperture quadratic phase contributes less than a radian across the whole aperture, so it may be dropped. This is the single approximation that converts Fresnel into Fraunhofer. C

\[ r \approx \underbrace{\Big(z+\tfrac{x^2+y^2}{2z}\Big)}_{\equiv\,R(x,y)} - \frac{xx'+yy'}{z} \]

Collect the surviving terms. The first group depends only on the observation point, not on \((x',y')\), so it factors out of the aperture integral. B

\[ \frac{e^{ikr}}{r}\approx \frac{e^{ikR}}{z}\,\exp\!\left(-ik\,\frac{xx'+yy'}{z}\right) \]

Put the expanded phase into the integrand; in the amplitude denominator \(r\approx z\) (slowly varying), while in the phase every term is kept because \(k\) is large. B

\[ U(x,y)=\frac{U_0\,e^{ikR}}{i\lambda z}\iint_{A} t(x',y')\,\exp\!\left(-ik\,\frac{xx'+yy'}{z}\right)dx'\,dy' \]

Substitute step 7 into step 1 and pull the \((x',y')\)-independent prefactor outside the integral. The remaining integral is over the aperture, with \(t=0\) outside it, so the limits extend to \(\pm\infty\). B

\[ f_X\equiv \frac{x}{\lambda z}=\frac{\sin\theta_x}{\lambda},\qquad f_Y\equiv \frac{y}{\lambda z}=\frac{\sin\theta_y}{\lambda},\qquad k=\frac{2\pi}{\lambda} \]

Define spatial frequencies from the diffraction angles via the direction cosines \(\sin\theta_x=x/R\approx x/z\). This is a change of variables that recasts the exponent as \(-i2\pi(f_X x'+f_Y y')\). B

\[ U(f_X,f_Y)=\frac{U_0\,e^{ikR}}{i\lambda z}\iint_{-\infty}^{\infty} t(x',y')\,e^{-i2\pi(f_X x'+f_Y y')}\,dx'\,dy'=\frac{U_0\,e^{ikR}}{i\lambda z}\,\tilde{t}(f_X,f_Y) \]

Recognize the integral as the 2-D Fourier transform \(\tilde t\) of the transmission function (using the assumed Fourier-transform-pair convention). The aperture field maps to its angular spectrum. A

Result

\[ \boxed{\,U(x,y)=\frac{U_0\,e^{ikR}}{i\lambda z}\,\tilde{t}\!\left(\frac{x}{\lambda z},\frac{y}{\lambda z}\right),\qquad I(x,y)=\frac{|U_0|^2}{\lambda^2 z^2}\,\big|\tilde{t}(f_X,f_Y)\big|^2\,}\]

Reading. The complex amplitude on a distant screen is the Fourier transform of the aperture, sampled at the spatial frequency \((x/\lambda z,\,y/\lambda z)\). A point at screen position \((x,y)\) reports how much of the spatial frequency \(f_X=x/\lambda z\) is contained in the aperture. The prefactor \(e^{ikR}/(i\lambda z)\) is a pure phase and \(1/z\) amplitude spreading; it drops out of the measured intensity, which is the squared modulus of the transform. Larger apertures (more high-frequency content) produce narrower patterns — the reciprocal-width property of Fourier pairs.

Units check. \(f_X=x/(\lambda z)\) has units \(\mathrm{m}/(\mathrm{m}\cdot\mathrm{m})=\mathrm{m^{-1}}\), correct for spatial frequency. In \(U\): \([U_0]/[\lambda z]\times[\tilde t]\). With \(t\) dimensionless, \(\tilde t=\iint t\,dx'dy'\) has units \(\mathrm{m^2}\), and \(1/(\lambda z)\) has units \(\mathrm{m^{-2}}\), so \(U\) carries the same units as \(U_0\). Consistent.

Limiting cases

Single slit \(t=\mathrm{rect}(x'/a)\Rightarrow I\propto \mathrm{sinc}^2(a f_X)=\mathrm{sinc}^2\!\big(\tfrac{a\sin\theta}{\lambda}\big)\); zeros at \(\sin\theta=n\lambda/a\).
Circular aperture diameter \(D\): \(I\propto\big[2J_1(u)/u\big]^2\) with \(u=\pi D f_r\); Airy disc, first dark ring at \(\sin\theta=1.22\lambda/D\).
Aperture shift \(t(x'-x_0)\Rightarrow\) multiply \(\tilde t\) by \(e^{-i2\pi f_X x_0}\): pattern intensity unchanged (shift-invariance of \(|\tilde t|^2\)).
Two identical apertures (convolution with two deltas) \(\Rightarrow\) single-aperture envelope \(\times\) \(\cos^2(\pi d f_X)\): Young's fringes under the diffraction envelope.
Aperture \(D\to\infty\): \(\tilde t\to\delta(f_X)\delta(f_Y)\); no diffraction, a single forward beam (geometric optics limit).

Breaks when

Fresnel (near-field) regime, \(z\lesssim D^2/\lambda\): the dropped quadratic phase \(e^{ik(x'^2+y'^2)/2z}\) is no longer negligible, so the pattern is a chirp-multiplied transform (Fresnel integral), not a plain Fourier transform. A collimated beam through a slit then shows shadow-edge fringes rather than a clean sinc.
Sub-wavelength or high-angle features (\(D\sim\lambda\), \(\sin\theta\to 1\)): the scalar/paraxial approximation and the linear map \(f_X=\sin\theta/\lambda\) fail; polarization, evanescent orders, and the full obliquity factor matter (rigorous vector diffraction required).
Thick or resonant apertures: \(t\) is not a simple multiplicative thin mask — internal propagation, waveguiding, and edge currents make the exit field depend on the incident field non-locally, so no single transform describes it.
Partially coherent or broadband light: the field-level transform must be replaced by an incoherent sum of intensities over wavelength and source points; the sharp \(|\tilde t|^2\) pattern washes out.

Failure modes

Transforming intensity instead of amplitude. The Fourier transform acts on the field \(t(x',y')\); intensity is \(|\tilde t|^2\) taken afterwards. Squaring first destroys the interference structure.
Confusing aperture coordinates with screen coordinates. \(x'\) lives in the aperture, \(x\) on the screen; they are conjugate variables linked only through \(f_X=x/\lambda z\).
Forgetting the \(1/\lambda z\) frequency scale. Writing \(\tilde t(x,y)\) instead of \(\tilde t(x/\lambda z,\,y/\lambda z)\) gives a pattern with the wrong physical size and wrong \(\lambda\)- and \(z\)-dependence.
Using \(\tan\theta\) or \(\theta\) for \(f_X\). The correct direction cosine is \(\sin\theta\); paraxially \(\sin\theta\approx\tan\theta\approx\theta\), but at large angles only \(\sin\theta\) is right.
Applying Fraunhofer in the near field. Students plug in small \(z\) (a benchtop shadow) and expect a sinc; that regime is Fresnel and requires the quadratic phase.
Dropping the aperture finite support. Extending limits to \(\pm\infty\) is legal only because \(t=0\) outside the aperture; forgetting this loses the diffraction entirely.

Discussion

The physical content of this result is that a screen at infinity is a spatial-frequency analyser. Each plane-wave component leaving the aperture travels in a unique direction set by its transverse wavevector \(k_x=k\sin\theta_x=2\pi f_X\); at large range these directions separate spatially, so position on the screen becomes a proxy for angle, and angle is proportional to spatial frequency. The aperture's angular spectrum, defined by decomposing the exit field into plane waves, is literally what the far field displays. Diffraction is thus not a mysterious spreading but the geometric sorting of the aperture's Fourier components.

The reciprocity between aperture size and pattern width is the Fourier uncertainty relation in disguise: an aperture confined to width \(\Delta x'\sim D\) has a transform of width \(\Delta f_X\sim 1/D\), giving an angular spread \(\Delta\theta\sim\lambda/D\). This single scaling governs the resolving power of every optical instrument (the Rayleigh criterion is \(1.22\lambda/D\) for a round pupil) and the beam divergence of every laser. It is the same mathematics that limits the time–bandwidth product of a pulse.

The convolution and shift theorems turn compound apertures into products of simple transforms. A diffraction grating is a single slit convolved with a comb, so its pattern is a sinc envelope multiplied by a sharp comb of orders; a shifted aperture only adds a linear phase and leaves the intensity untouched, which is why a diffraction pattern does not move when the aperture is translated. These are not optical coincidences but direct readouts of Fourier-transform-pair identities.

At the deepest level the Fraunhofer transform is the stationary-phase / far-field limit of the angular-spectrum propagator: writing the exit field as \(U(x',y')=\iint A(k_x,k_y)e^{i(k_xx'+k_yy')}\,dk_xdk_y\) and propagating each plane wave by \(e^{ik_z z}\), the method of stationary phase selects, at observation point \((x,y,z)\), exactly the component whose direction points there, \((k_x,k_y)=k(x,y)/R\). The far field is therefore \(A\) evaluated on the Ewald-sphere direction toward the observation point — the same object as \(\tilde t\) when the aperture field is \(U_0 t\). This viewpoint makes the linear frequency map \(f_X=\sin\theta_x/\lambda\) exact (it is the true direction cosine), with paraxiality only entering when one replaces \(\sin\theta\) by \(x/z\).

Common misconceptions. The pattern is not "the shadow of the aperture blurred out": it is the transform, so a wide slit gives a narrow central peak, the opposite of a shadow. Also, the far field does not require the light to physically travel an infinite distance — a converging lens brings "infinity" to its back focal plane, so the Fraunhofer transform is realised on a benchtop, with \(z\) replaced by the focal length \(f\).

Worked examples

Single slit, position of the first minimum. Width \(a=0.10~\mathrm{mm}\), \(\lambda=633~\mathrm{nm}\) (HeNe), screen at \(L=2.0~\mathrm{m}\).

Aperture \(t=\mathrm{rect}(x'/a)\); its transform is \(\tilde t\propto a\,\mathrm{sinc}(a f_X)\) with \(\mathrm{sinc}(u)=\sin(\pi u)/(\pi u)\). A

\[ I(x)\propto \mathrm{sinc}^2\!\big(a f_X\big),\qquad f_X=\frac{x}{\lambda L}\;\Rightarrow\; \text{zeros at } a f_X=n \]

Intensity is \(|\tilde t|^2\); the first zero is \(n=1\), i.e. \(a\sin\theta=\lambda\) with \(\sin\theta=x/L\). A

\[ x_1=\frac{\lambda L}{a}=\frac{(633\times10^{-9}~\mathrm{m})(2.0~\mathrm{m})}{1.0\times10^{-4}~\mathrm{m}} \]

Solve \(x_1/(\lambda L)=1/a\) for \(x_1\); symbols first, then insert numbers in SI. A

\[ x_1=1.27\times10^{-2}~\mathrm{m}=12.7~\mathrm{mm} \]

Reading. The first dark fringe sits \(12.7~\mathrm{mm}\) from centre; the central bright band is \(2x_1\approx 25~\mathrm{mm}\) wide. Check Fraunhofer validity: \(D^2/\lambda=(10^{-4})^2/633\times10^{-9}\approx 1.6\times10^{-2}~\mathrm{m}\), and \(L=2~\mathrm{m}\gg 0.016~\mathrm{m}\), so the far-field transform applies.

Circular pupil (telescope), angular resolution. Aperture diameter \(D=0.20~\mathrm{m}\), \(\lambda=550~\mathrm{nm}\).

A uniformly illuminated disc transforms to \(\tilde t\propto 2J_1(u)/u\), \(u=\pi D f_r\), \(f_r=\sin\theta/\lambda\) — the Airy pattern. B

\[ I(\theta)\propto\left[\frac{2J_1(u)}{u}\right]^2,\qquad \text{first zero at } u=3.832=\pi D\,\frac{\sin\theta}{\lambda} \]

The first zero of \(J_1\) is at \(3.832\); set \(u=3.832\) and solve for the angle. B

\[ \sin\theta_{\min}=\frac{3.832\,\lambda}{\pi D}=\frac{1.22\,\lambda}{D}=\frac{1.22\,(550\times10^{-9}~\mathrm{m})}{0.20~\mathrm{m}} \]

Rewrite \(3.832/\pi=1.22\) (Rayleigh criterion), then substitute numbers. A

\[ \theta_{\min}\approx \sin\theta_{\min}=3.36\times10^{-6}~\mathrm{rad}\approx 0.69~\text{arcsec} \]

Reading. Two stars closer than \(0.69''\) merge into one Airy disc for this \(20~\mathrm{cm}\) aperture. Bigger \(D\) sharpens the transform (\(\theta_{\min}\propto 1/D\)); this reciprocal scaling is exactly the Fourier width relation of the pupil.

Problems

A double slit has slit width \(a=20~\mu\mathrm{m}\) and centre-to-centre separation \(d=100~\mu\mathrm{m}\), illuminated at \(\lambda=633~\mathrm{nm}\), screen at \(L=1.5~\mathrm{m}\). How many bright interference fringes fit inside the central single-slit diffraction maximum?
Solution
The pattern is the single-slit envelope \(\mathrm{sinc}^2(af_X)\) times interference \(\cos^2(\pi d f_X)\). Envelope first zero: \(a f_X=1\). Interference maxima: \(d f_X=m\). Missing-order ratio \(d/a=100/20=5\). The envelope zero coincides with the \(m=5\) order, so orders \(m=0,\pm1,\pm2,\pm3,\pm4\) lie strictly inside (order \(\pm5\) is suppressed). That is \(9\) bright fringes inside the central maximum. (The \(\pm5\) fringes are "missing orders.")
Show that translating an aperture by \(x_0\) in its own plane leaves the Fraunhofer intensity pattern unchanged, and find the phase acquired by the field.
Solution
Let \(t'(x',y')=t(x'-x_0,y')\). By the Fourier shift theorem, \(\tilde t'(f_X,f_Y)=e^{-i2\pi f_X x_0}\,\tilde t(f_X,f_Y)\). The field picks up a linear phase \(\phi=-2\pi f_X x_0=-k x x_0/z\). The intensity \(|\tilde t'|^2=|e^{-i2\pi f_X x_0}|^2|\tilde t|^2=|\tilde t|^2\) is unchanged because the modulus of a pure phase is 1. Hence the diffraction pattern does not move when the aperture is slid sideways — only its phase distribution changes.
A slit of width \(a=50~\mu\mathrm{m}\) is used with \(\lambda=500~\mathrm{nm}\). Estimate the minimum distance \(z\) beyond which the Fraunhofer approximation is good to better than a \(\pi/8\) phase error across the aperture.
Solution
The dropped quadratic phase at the aperture edge is \(\Delta\phi=\dfrac{k}{2z}\left(\dfrac{a}{2}\right)^2=\dfrac{\pi a^2}{4\lambda z}\). Require \(\Delta\phi\le \pi/8\): \(z\ge \dfrac{2a^2}{\lambda}=\dfrac{2(50\times10^{-6})^2}{500\times10^{-9}}=\dfrac{2(2.5\times10^{-9})}{5\times10^{-7}}=1.0\times10^{-2}~\mathrm{m}\). So \(z\gtrsim 1~\mathrm{cm}\). (The looser criterion \(z\gg a^2/\lambda=5~\mathrm{mm}\) is consistent.)
A Gaussian amplitude aperture \(t(x')=e^{-x'^2/w^2}\) is illuminated by a plane wave. Find the far-field intensity profile and its \(1/e^2\) angular half-width.
Solution
The 1-D Fourier transform of a Gaussian is a Gaussian: \(\tilde t(f_X)=\int e^{-x'^2/w^2}e^{-i2\pi f_X x'}dx'=w\sqrt{\pi}\,e^{-\pi^2 w^2 f_X^2}\). Intensity \(I\propto|\tilde t|^2\propto e^{-2\pi^2 w^2 f_X^2}\). With \(f_X=\sin\theta/\lambda\approx\theta/\lambda\), \(I\propto e^{-2\pi^2 w^2\theta^2/\lambda^2}\). The \(1/e^2\) half-width (where the exponent \(=-2\)) is \(2\pi^2 w^2\theta^2/\lambda^2=2\Rightarrow \theta_{1/e^2}=\dfrac{\lambda}{\pi w}\). A wider beam diverges less, \(\theta\propto 1/w\), the standard Gaussian-beam divergence.
A transmission grating has \(N=600\) lines/mm. Monochromatic light at \(\lambda=589~\mathrm{nm}\) illuminates it at normal incidence. Find the diffraction angle of the first order, and the highest order observable.
Solution
Grating period \(d=1/(600~\mathrm{mm^{-1}})=1.667~\mu\mathrm{m}=1.667\times10^{-6}~\mathrm{m}\). Grating equation (from the comb transform, maxima at \(d\sin\theta=m\lambda\)): \(\sin\theta_1=\lambda/d=589\times10^{-9}/1.667\times10^{-6}=0.3534\Rightarrow\theta_1=20.7^\circ\). Highest order: \(\sin\theta\le 1\Rightarrow m\le d/\lambda=1.667\times10^{-6}/589\times10^{-9}=2.83\), so \(m_{\max}=2\) (orders \(0,\pm1,\pm2\) appear; \(\theta_2=\arcsin(2\times0.3534)=\arcsin(0.7069)=44.97^\circ\)).