advanced30 min studyLesson 212 in Physics

Fraunhofer diffraction and aperture transforms

Derive the far-field Fourier relation and separate normalized diffraction shape from absolute field and transmitted power.

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What you’ll understand

  • Derive the aperture transform using cycles per metre.
  • Compute the first zeros and normalization of a rectangular aperture.
  • Separate a lateral shift from a phase tilt in the aperture.

A far-field diffraction pattern is a map of the spatial frequencies present in an aperture field. A narrow opening produces a broad angular pattern because rapid transverse localization needs a broad range of plane-wave directions. This statement becomes quantitative only after declaring a Fourier-transform convention and the field being transformed. The mask transmission alone is insufficient when illumination across the mask has a varying amplitude or phase.

Discard one quadratic phase deliberately

Start with the Fresnel integral and expand the transverse squared distance into an observation-only term, a cross term, and an aperture-only term. For an aperture of radius or half-width aa, the aperture phase has scale πa2/(λz)\pi a^2/(\lambda z). When this is much smaller than one radian, replace its exponential by unity. The observation-only quadratic phase may be retained: it has unit modulus and does not alter the intensity pattern.

Define the Fourier transform with a negative exponential and spatial frequencies in cycles per metre. The physical observation coordinates select frequencies fx=x/(λz)f_x=x/(\lambda z) and fy=y/(λz)f_y=y/(\lambda z) in the paraxial limit. Frequencies specified as angular wavenumbers instead require factors of 2π2\pi in a different place. Mixing those conventions shifts every predicted zero, even when the qualitative pattern still looks convincing.

U~0(fx,fy)=∬U0(ξ,η)e−i2π(fxξ+fyη)dξ dη,U≃eikzeik(x2+y2)/(2z)iλzU~0 ⁣(xλz,yλz)\widetilde U_0(f_x,f_y)=\iint U_0(\xi,\eta)e^{-i2\pi(f_x\xi+f_y\eta)}d\xi\,d\eta,\qquad U\simeq\frac{e^{ikz}e^{ik(x^2+y^2)/(2z)}}{i\lambda z}\widetilde U_0\!\left(\frac{x}{\lambda z},\frac{y}{\lambda z}\right)
The transform of a field U0 in V/m has units V·m, and division by lambda z restores V/m. This scalar paraxial far-field formula assumes negligible aperture quadratic phase; the prefactor is not part of a normalized pattern.↔ Scroll wide equations horizontally.

A rectangle makes the normalization visible

For constant incident phasor UaU_a across a centred rectangular opening of full widths bb and hh, integrate each coordinate independently. Each integral is its width multiplied by the normalized sinc function. At zero frequency both sinc factors approach one, so the on-axis field magnitude is ∣Ua∣bh/(λz)|U_a|bh/(\lambda z). This provides a useful dimensional and limiting check on the transform calculation.

Dividing the intensity by its on-axis value removes the area factor and yields a shape function. It does not imply equal throughput for different aperture widths. With fixed incident irradiance, widening the aperture increases transmitted power, narrows the angular pattern, and raises the central intensity. A calculation that rescales every pattern to the same peak can hide two of those three effects, so always state which quantity has been normalized.

sinc⁡(u)=sin⁡(πu)πu,sinc⁡(0)=1,I(x,y)I(0,0)=sinc⁡2 ⁣(bxλz)sinc⁡2 ⁣(hyλz)\operatorname{sinc}(u)=\frac{\sin(\pi u)}{\pi u},\quad \operatorname{sinc}(0)=1,\qquad \frac{I(x,y)}{I(0,0)}=\operatorname{sinc}^2\!\left(\frac{bx}{\lambda z}\right)\operatorname{sinc}^2\!\left(\frac{hy}{\lambda z}\right)
b and h are full aperture widths in metres; x and y are screen offsets. The first x zero is lambda z/b for uniform coherent illumination. The normalized sinc here differs from the alternative sin(u)/u convention.↔ Scroll wide equations horizontally.

Shifts change phase; tilts move patterns

Translating an aperture field by a fixed amount multiplies its Fourier transform by a phase ramp. Its isolated far-field intensity is therefore unchanged when the illumination translates with it. Tilting the incident plane wave multiplies the aperture field by a phase ramp instead; the shift theorem then moves the transform in spatial frequency. These are reciprocal operations, not two descriptions of the same displacement.

A lens can produce a Fourier-transform plane at its back focal plane within paraxial optics, because its quadratic phase cancels the relevant propagation phase. Thus a laboratory need not put a screen kilometres away to inspect the far-field angular distribution. Finite lens aperture, aberrations, and polarization still matter. The statement concerns a suitably arranged coherent optical system, not every camera image or every plane behind a lens.

First dark fringe from a narrow rectangle

A uniformly illuminated slit has full width b=0.20 mmb=0.20\,\mathrm{mm}, wavelength 500 nm500\,\mathrm{nm}, and screen distance z=2.0 mz=2.0\,\mathrm m. Treat the vertical dimension separately. Find the first horizontal zeros and the central-lobe full width; check the far-field scale using half-width.

  1. The first zeros of the normalized sinc occur when its argument is plus or minus one. Therefore the horizontal offsets satisfy x=±λz/bx=\pm\lambda z/b rather than an argument equal to pi.

  2. Insert SI lengths: (5.0×10−7)(2.0)/(2.0×10−4)=5.0×10−3 m(5.0\times10^{-7})(2.0)/(2.0\times10^{-4})=5.0\times10^{-3}\,\mathrm m. The zeros are 5.0 mm on either side of the optical axis.

  3. The central lobe extends between those zeros, giving 10.0 mm full width. With a=b/2=10−4 ma=b/2=10^{-4}\,\mathrm m, the aperture phase scale is πa2/(λz)=0.0314\pi a^2/(\lambda z)=0.0314 rad, making the far-field shape approximation reasonable.

The result

The first zeros are at ±5.0 mm and the central lobe is 10.0 mm wide; the omitted aperture phase is about 0.031 rad.

Explore the model

Interactive lab

Interactive lab · spectral

Spatial scale and Fourier content

For each run, Reset and set BOTH Mode count N and Coefficient decay p. Compare N=1 and N=8 with p=1, predicting finer spatial structure for eight modes; inspect the curves with Step. The model is exactly a finite sum sin(nx−t)/n^p, not the aperture transform or its intensity.

Spatial scale and Fourier contentA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forMore spatial frequencies can support finer structure, but this sine-sum lab does not calculate slit widths, Fourier-plane coordinates, or transmitted power.

Check your understanding

A centred aperture field is translated while its illumination translates with it. What happens to its isolated Fraunhofer intensity as a function of angle?

Try it yourself

At fixed incident irradiance and screen distance, a rectangular aperture doubles only its horizontal width. Predict the ratios of the on-axis intensity, first horizontal zero offset, and total transmitted power. Assume the far-field and paraxial conditions still hold.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Track the area factor before normalizing. Total transmitted power counts the open area, whereas peak intensity squares the integrated field.

Compare with the solution

The aperture area doubles, so the on-axis field doubles and its intensity becomes four times as large. The first-zero offset is inversely proportional to the horizontal full width, so it halves. The open area doubles and the incident irradiance is fixed, so total transmitted power doubles. The taller peak and narrower horizontal lobe are consistent with this power change. Normalizing both plots to their own maximum would show only the width change and conceal the absolute intensity ratio.

Take it with you

  • Transform the illuminated aperture field with a stated frequency convention.
  • Normalized sinc patterns discard physical area and propagation prefactors.
  • Aperture translation and incident phase tilt have different Fourier consequences.

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