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Fourier Optics

10.1 Fraunhofer Diffraction as a Fourier Transform

Section titled “10.1 Fraunhofer Diffraction as a Fourier Transform”

The Fraunhofer diffraction pattern of an aperture with transmittance function t(x,y)t(x, y) illuminated by a plane wave is proportional to the 2D Fourier transform of the aperture function:

E(u,v)t(x,y)e2πi(ux+vy)dxdy=F{t(x,y)}(u,v)E(u, v) \propto \iint t(x,y)\, e^{-2\pi i(ux + vy)}\, dx\, dy = \mathcal{F}\{t(x,y)\}(u, v)

Where u=sinθx/λu = \sin\theta_x/\lambda and v=sinθy/λv = \sin\theta_y/\lambda are the spatial frequencies.

Theorem 10.1. The intensity in the Fraunhofer diffraction pattern is

I(u,v)=I0t~(u,v)2I(u,v) = I_0\,|\tilde{t}(u,v)|^2

Where t~(u,v)=F{t(x,y)}(u,v)\tilde{t}(u,v) = \mathcal{F}\{t(x,y)\}(u,v) is the Fourier transform of the aperture function.

Proof. The Huygens-Fresnel principle in the far field gives:

E(x",y)=eikriλrt(x,y)eik(xx+yy)/rdxdyE(x",y') = \frac{e^{ikr}}{i\lambda r}\iint t(x,y)\, e^{-ik(xx' + yy')/r}\, dx\, dy

In the far field, rDr \approx D and the phase factor eik(xx+yy)/re^{-ik(xx' + yy')/r} is exactly the kernel of the Fourier transform. \blacksquare

Theorem 10.2 (Convolution theorem). If an aperture function is the convolution t=t1t2t = t_1 * t_2The diffraction pattern is the product of the individual diffraction patterns:

F{t1t2}=F{t1}F{t2}\mathcal{F}\{t_1 * t_2\} = \mathcal{F}\{t_1\} \cdot \mathcal{F}\{t_2\}

Corollary. If an aperture is the product t=t1t2t = t_1 \cdot t_2The diffraction pattern is the convolution of the individual patterns:

F{t1t2}=F{t1}F{t2}\mathcal{F}\{t_1 \cdot t_2\} = \mathcal{F}\{t_1\} * \mathcal{F}\{t_2\}

10.3 Worked Example: Diffraction Grating via Fourier Transform

Section titled “10.3 Worked Example: Diffraction Grating via Fourier Transform”

Problem. Use the Fourier transform to derive the intensity pattern of a grating with NN slits of width aa and spacing dd.

Solution

The transmittance of a single slit centred at x=0x = 0 is tslit(x)=rect(x/a)t_{\mathrm{slit}(x) = \mathrm{rect}(x/a)}. The full grating is NN slits:

t(x)=n=0N1tslit(xnd)=tslit(x)n=0N1δ(xnd)t(x) = \sum_{n=0}^{N-1} t_{\mathrm{slit}(x - nd) = t_{\mathrm{slit}(x) * \sum_{n=0}^{N-1} \delta(x - nd)}}

The Fourier transform is:

t~(u)=F{tslit}F{n=0N1δ(xnd)}\tilde{t}(u) = \mathcal{F}\{t_{\mathrm{slit}\} \cdot \mathcal{F}\left\{\sum_{n=0}^{N-1}\delta(x - nd)\right\}}

=asinc(πau)n=0N1e2πindu=asinc(πau)sin(Nπdu)sin(πdu)= a\,\mathrm{sinc}(\pi a u) \cdot \sum_{n=0}^{N-1} e^{-2\pi i n d u} = a\,\mathrm{sinc}(\pi a u) \cdot \frac{\sin(N\pi d u)}{\sin(\pi d u)}

The intensity is:

I(u)=I0a2sinc2(πau)sin2(Nπdu)sin2(πdu)I(u) = I_0\,a^2\,\mathrm{sinc}^2(\pi a u)\,\frac{\sin^2(N\pi d u)}{\sin^2(\pi d u)}

The first factor is the single-slit envelope; the second is the NN-slit interference pattern. Principal maxima occur at du=mdu = m (integer mm), giving the grating equation dsinθ=mλd\sin\theta = m\lambda.

\blacksquare

10.4 Worked Example: Circular Aperture and the Airy Pattern

Section titled “10.4 Worked Example: Circular Aperture and the Airy Pattern”

Problem. Compute the Fraunhofer diffraction pattern of a circular aperture of radius aa.

Solution

The aperture function is t(r)=1t(r) = 1 for rar \leq a and t(r)=0t(r) = 0 for r>ar > a. By circular symmetry, the Fourier transform in polar coordinates is:

t~(q)=2π0aJ0(2πqr)rdr\tilde{t}(q) = 2\pi\int_0^a J_0(2\pi q r)\, r\, dr

Where J0J_0 is the Bessel function of the first kind and q=sinθ/λq = \sin\theta/\lambda is the radial spatial frequency. Using the identity:

0aJ0(2πqr)rdr=a2πqJ1(2πqa)\int_0^a J_0(2\pi q r)\, r\, dr = \frac{a}{2\pi q}J_1(2\pi q a)

t~(q)=πa22J1(α)α\tilde{t}(q) = \pi a^2 \cdot \frac{2J_1(\alpha)}{\alpha}

Where α=2πaq=2πasinθ/λ\alpha = 2\pi a q = 2\pi a\sin\theta/\lambda. The intensity is:

I(θ)=I0(2J1(α)α)2I(\theta) = I_0\left(\frac{2J_1(\alpha)}{\alpha}\right)^2

This is the Airy pattern. The first zero occurs at α=3.832\alpha = 3.832Giving the angular radius of the first dark ring:

sinθ1=1.22λd\sin\theta_1 = 1.22\,\frac{\lambda}{d}

Where d=2ad = 2a is the diameter.

\blacksquare

  • Rayleigh criterion: Two point sources are just resolved when the centre of the Airy disc of one coincides with the first dark ring of the other, giving the minimum resolvable angle θmin=1.22λ/d\theta_{\min} = 1.22\lambda/d.
  • Fourier scaling property: If t(x,y)t(x,y) is scaled by aa, i.e., t(x/a,y/a)t(x/a, y/a), then t~(u,v)\tilde{t}(u,v) scales as a2t~(au,av)|a|^2\tilde{t}(au, av). A larger aperture produces a narrower diffraction pattern.
  • Parseval’s theorem: t(x,y)2dxdy=t~(u,v)2dudv\iint |t(x,y)|^2\,dx\,dy = \iint |\tilde{t}(u,v)|^2\,du\,dv. The total power in the aperture equals the total power in the diffraction pattern.
  • Uncertainty principle analogy: A narrow aperture (small Δx\Delta x) produces a wide diffraction pattern (large Δu\Delta u), and vice versa. Quantitatively, ΔxΔu1\Delta x \cdot \Delta u \gtrsim 1.
  • Confusing Fraunhofer with Fresnel diffraction: Fraunhofer diffraction requires the far-field condition Da2/λD \gg a^2/\lambda. At shorter distances, Fresnel (near-field) diffraction must be used, and the pattern is not a simple Fourier transform.
  • Forgetting the intensity is the squared modulus: The diffraction pattern is I(u,v)=I0t~(u,v)2I(u,v) = I_0|\tilde{t}(u,v)|^2, not t~(u,v)\tilde{t}(u,v). Phase information is lost in the intensity measurement.
  • Neglecting the obliquity factor: The Huygens-Fresnel principle includes a directional cosine factor. For small angles this is approximately constant, but at large angles it modifies the pattern.
  • Assuming the Fourier transform of a real function is real: Even if t(x,y)t(x,y) is real and non-negative, t~(u,v)\tilde{t}(u,v) is generally complex. The phase of t~\tilde{t} carries information about the spatial structure of the aperture.
  • Telescope resolution: The Airy pattern sets the fundamental resolution limit of any circular-aperture optical system. The diameter of the primary mirror determines the smallest detail that can be resolved.
  • Spectrometre design: A diffraction grating disperses light according to the grating equation dsinθ=mλd\sin\theta = m\lambda. The resolving power R=mNR = mN depends on the order mm and the number of illuminated slits NN.
  • Spatial filtering: By placing masks in the Fourier plane (at the focal length of a lens), specific spatial frequencies can be blocked or attenuated. This enables edge enhancement, noise removal, and pattern recognition.
  • Holography: A hologram records both the amplitude and phase of the diffracted field. Reconstruction involves illuminating the hologram, which acts as a complex transmittance function whose Fourier transform reproduces the original wavefront.

10.8 Worked Example: Double-Slit via Fourier Transform

Section titled “10.8 Worked Example: Double-Slit via Fourier Transform”

Problem. Use the convolution theorem to derive the double-slit diffraction pattern.

The aperture is the product of a double-slit function t1(x)=rect((xd/2)/a)+rect((x+d/2)/a)t_1(x) = \mathrm{rect}((x - d/2)/a) + \mathrm{rect}((x + d/2)/a) and a wide rectangular window. However, it is simpler to view the double slit as a single slit convolved with two delta functions:

t(x)=rect(x/a)[δ(xd/2)+δ(x+d/2)]t(x) = \mathrm{rect}(x/a) \cdot [\delta(x - d/2) + \delta(x + d/2)]

Wait, the double slit is a product (two slits cut from an opaque screen). The transmittance is:

t(x)=[rect((xd/2)/a)+rect((x+d/2)/a)]t(x) = [\mathrm{rect}((x - d/2)/a) + \mathrm{rect}((x + d/2)/a)]

The Fourier transform is:

t~(u)=asinc(πau)eiπdu+asinc(πau)eiπdu=2asinc(πau)cos(πdu)\tilde{t}(u) = a\,\mathrm{sinc}(\pi a u)\,e^{-i\pi d u} + a\,\mathrm{sinc}(\pi a u)\,e^{i\pi d u} = 2a\,\mathrm{sinc}(\pi a u)\cos(\pi d u)

The intensity is:

I(u)=4I0a2sinc2(πau)cos2(πdu)I(u) = 4I_0 a^2\,\mathrm{sinc}^2(\pi a u)\cos^2(\pi d u)

The cos2\cos^2 factor produces the double-slit interference fringes with spacing Δu=1/d\Delta u = 1/d, and the sinc2^2 factor provides the single-slit envelope.

\blacksquare

Problem. Find the Fraunhofer diffraction pattern of a rectangular aperture of width aa and height bb.

The aperture function is separable: t(x,y)=rect(x/a)rect(y/b)t(x,y) = \mathrm{rect}(x/a)\,\mathrm{rect}(y/b). By the separability of the 2D Fourier transform:

t~(u,v)=F{rect(x/a)}(u)F{rect(y/b)}(v)=absinc(πau)sinc(πbv)\tilde{t}(u,v) = \mathcal{F}\{\mathrm{rect}(x/a)\}(u) \cdot \mathcal{F}\{\mathrm{rect}(y/b)\}(v) = ab\,\mathrm{sinc}(\pi a u)\,\mathrm{sinc}(\pi b v)

The intensity is:

I(u,v)=I0a2b2sinc2(πau)sinc2(πbv)I(u,v) = I_0\,a^2 b^2\,\mathrm{sinc}^2(\pi a u)\,\mathrm{sinc}^2(\pi b v)

The pattern is a product of two sinc2^2 functions. The first zero along uu occurs at u=1/au = 1/a (angular position sinθx=λ/a\sin\theta_x = \lambda/a), and along vv at v=1/bv = 1/b (sinθy=λ/b\sin\theta_y = \lambda/b). A wider aperture produces a narrower diffraction pattern in that direction.