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Fourier Optics, Imaging and Resolution

A numerical 4f imaging system: the first lens Fourier-transforms the object, a pupil and filter act on its spectrum, the second lens forms the image. Compare coherent (complex amplitude) and incoherent (intensity) imaging, watch the PSF and MTF change with NA and aberrations, and measure resolution.

Every slider has a number box for typed values • Hover the plots to read values • Settings are kept in the URL
⚙ Incoherent Imaging mode
NA 0.50 Numerical aperture
rR — Rayleigh 0.61λ₀/NA
fc — Incoherent cutoff 2NA/λ₀
S 1.000 Strehl ratio
DOF — Depth of focus λ₀/NA²

1 · Object |t|²

Object intensity |t(x, y)|² over the full periodic field (image-space coordinates, magnification 1).

2 · Fourier plane |Ĩ|

Spectrum at the back focal plane of lens 1. Solid circle: pupil edge fc = NA/λ₀; dashed: 2NA/λ₀.

3 · Pupil × filter W, F

Wavefront error W(f) in waves inside the aperture (piston included). Hatched: blocked by the filter; ringed: +π/2 phase dot.

4 · Image incoherent

Image intensity (same coordinates as the object; the 4f inversion x → −x is undone for comparison).

5 · Point-spread function |h|²

System intensity PSF (aberrations and filter) relative to the unaberrated peak. Dashed circle: Airy first zero 0.61λ₀/NA.

6 · Transfer functions MTF active

Cuts along fx (fy = 0). MTF = |OTF| (incoherent); |CTF| = |P·F| (coherent amplitude). ★ marks the transfer function of the active mode. Dashed/dotted: diffraction-limited circular-pupil references.

7 · Line cut through y = 0 absolute

Object intensity (dashed) and image intensity (solid) along x. Hover to read values.

8 · Through focus on-axis

On-axis intensity versus defocus Δz from the unaberrated focus. Solid: exact angular-spectrum calculation with the current pupil (aberrations included). Dashed: paraxial sinc²(πΔz NA²/2λ₀) for a perfect pupil. Markers: ±λ₀/(2NA²) and the current Δz.

9 · Aperture trade-offs analytic

Top: Rayleigh radius 0.61λ₀/NA and depth of focus λ₀/NA² (log scale). Bottom: fraction of an isotropic emitter's power collected, (1 − cos θ)/2 ≈ NA²/4.

💡 How to use

Learn with this tool

Learning objectives

  • Explain why a lens produces the Fourier transform of an object and how a pupil acts as a filter on spatial frequencies.
  • Predict the coherent cutoff NA/λ₀ and the incoherent cutoff 2NA/λ₀, and use the MTF to predict image contrast.
  • Quantify how NA trades resolution against depth of focus and light collection, and how aberrations reduce the Strehl ratio.

Prerequisites

Model

An on-axis unit plane wave illuminates a thin object with complex transmittance t(x, y). In a 4f system lens 1 (focal length f) puts the angular spectrum T(fx, fy) at its back focal plane, at position u = λ₀ f · fx. A pupil and filter there multiply the spectrum by P(f) = A(f) F(f) exp[iΦ(f)], and lens 2 transforms back. Coordinates are referred to image space (magnification 1, the 4f inversion removed), the medium is air, and the calculation is scalar.

Coherent: Uimg = F⁻¹{ T · P }, I = |Uimg|², cutoff fc = NA/λ₀

Incoherent: Iimg = Iobj ⊛ |h|², h = F⁻¹{P}, OTF = (P ⋆ P)/∫|P|², cutoff 2NA/λ₀

Pupil phase: Φ = 2π Σ cj Zj(ρ, θ) + 2πΔz[√(λ₀⁻² − f²) − λ₀⁻¹], ρ = |f|/fc

Strehl: S = max|h|² / max|h0|² ≈ exp[−(2πω)²] (Maréchal, ω = RMS wavefront in waves)

λ₀
vacuum wavelength (the medium is air, n = 1)
NA
image-side numerical aperture, sin of the marginal-ray angle
fc
coherent cutoff NA/λ₀ (cycles per metre); pupil radius in frequency space
t, T
object amplitude transmittance and its Fourier transform
P, F
pupil (aperture A and phase Φ) and Fourier-plane filter
h, |h|²
amplitude and intensity point-spread functions
OTF, MTF
optical transfer function (normalised, OTF(0) = 1) and its modulus
Zj, cj
Noll-normalised Zernike polynomial (unit RMS on the unit disk) and its coefficient in waves RMS
S, ω
Strehl ratio and RMS wavefront error (waves, piston removed)

Numerics. Fields live on an N × N periodic grid of spacing dx = fov/N, so frequencies are spaced 1/fov. FFTs use the numpy sign convention. Pupil edge samples are area-weighted. Point objects use exact, sub-pixel analytic spectra. The incoherent image is scaled by the collected fraction ∫|P|²/∫|A|², so a uniform object gives I = 1 in both modes with an open pupil.

Derivation: why coherent and incoherent imaging differ

A space-invariant system maps each object point to a shifted copy of the amplitude PSF h. With coherent light the contributions from different points keep fixed relative phases, so amplitudes add: Uimg(x) = ∫ t(x′) h(x − x′) dx′ = t ⊛ h. Taking Fourier transforms gives Ũimg = T · P, because h = F⁻¹{P}. The pupil is a hard stop at fc = NA/λ₀, so no amplitude frequency above NA/λ₀ reaches the image.

With incoherent light the relative phases fluctuate randomly and the cross terms average to zero: ⟨|Σ tk hk|²⟩ = Σ |tk|² |hk|². Intensities add, so Iimg = Iobj ⊛ |h|². The transform of |h|² = h h* is the autocorrelation of P, which extends to twice the pupil radius: 2NA/λ₀. The incoherent system passes finer periods than the coherent one, but with falling contrast, given by the MTF. The coherent transfer function is flat up to fc, but it acts on amplitude. The image intensity |U|² then contains cross terms, so coherent imaging is not linear in intensity.

For small aberrations, S = |⟨eiΦ⟩|² ≈ |1 + i⟨Φ⟩ − ⟨Φ²⟩/2|² ≈ 1 − (⟨Φ²⟩ − ⟨Φ⟩²) = 1 − (2πω)², which Maréchal wrote as exp[−(2πω)²]. Defocus by Δz adds W ≈ Δz NA² ρ²/2. At WP-V = λ₀/4, the Rayleigh quarter-wave limit, Δz = ±λ₀/(2NA²) and S = sinc²(π/4) = 0.81, which gives the depth of focus λ₀/NA².

Exercise 1: the incoherent cutoff

  1. With λ₀ = 550 nm and NA = 0.5, predict the finest sinusoidal period an incoherent system can image, and its contrast at ν = 1.2 cycles/µm.
  2. Choose Sinusoidal intensity target, incoherent mode, and scan ν from 0.5 to 2 cycles/µm.
  3. Read the image modulation from the readouts and compare it with the MTF curve at the ν marker.
  4. Why does the contrast fall smoothly to zero rather than stop abruptly?
Show answer

The cutoff is 2NA/λ₀ = 1.82 cycles/µm, so the finest period is 0.55 µm. At ν = 1.1875 cycles/µm (after snapping), v = ν/(2NA/λ₀) = 0.653 and MTF = (2/π)[acos v − v√(1 − v²)] ≈ 0.23. The tool measures 0.237 on the discrete pupil. The OTF is the overlap area of two pupils shifted by λ₀ν. That area shrinks continuously to zero at a shift of one full diameter.

Exercise 2: coherent versus incoherent two-point resolution

  1. Two points are separated by the Rayleigh distance. Predict the midpoint/peak ratio for incoherent light and for coherent in-phase light.
  2. Load Rayleigh limit, then switch the mode to coherent. Then set the relative phase to 180°.
  3. Record the "two-point dip" readout in each case.
  4. Why can the same two points be "resolved" or not, depending on the illumination?
Show answer

Incoherent: 0.735 (a 26 % dip). Coherent in-phase: the amplitudes add at the midpoint, the ratio exceeds 1 and there is no dip, so the points are unresolved. Coherent in antiphase: the midpoint is exactly dark, so the points look resolved at any separation. The Rayleigh criterion is a convention for incoherent point sources. With coherent light the answer depends on the relative phase.

Exercise 3 (limiting case): very small and very large NA

  1. What should the image of the letters become as NA → 0? And as the pupil grows beyond every frequency on the grid, in coherent mode?
  2. Choose Letters, coherent mode, and set NA to 0.05, then to 0.95 with λ₀ = 400 nm and fov = 80 µm.
  3. Compare the image mean with the object mean, and the line cut shapes.
  4. Which conservation law does the large-NA limit test?
Show answer

As NA → 0 only the DC term passes, and the image tends to a uniform intensity |⟨t⟩|² (the squared mean transmittance, not the mean intensity). When the pupil covers the whole sampled spectrum (fc above the Nyquist frequency 1/(2dx)), P = 1 on every sample, the image equals the object, and ∫|U|² is conserved (Parseval). A warning notes that the pupil is clipped by the grid.

Exercise 4: seeing a transparent object

  1. For a weak phase object t = eiφ with φ = 0.1 rad, predict the bright-field contrast and the phase-contrast intensity step.
  2. Load Phase contrast, then change the filter to None, Dark field and Knife edge.
  3. Read the image min/max and the line cut in each case.
  4. Why does incoherent mode show nothing whatever the filter?
Show answer

Bright field: |1 + iφ|² = 1 + φ² ≈ 1, so there is almost no contrast. Retarding the undiffracted light by π/2 gives |i(1 + φ)|² ≈ 1 + 2φ, a 0.2 step. Dark field removes the DC term, so the edges glow on black with intensity ∝ φ². The knife edge keeps one sideband and gives a gradient (Schlieren) shading. An incoherent object is described by its intensity alone, which is uniform, so a phase object carries no image information in that mode.

Worked example: a microscope objective

A 0.65 NA dry objective images a fluorescent sample (incoherent) at λ₀ = 520 nm. (a) Rayleigh resolution: r = 0.61 × 520 nm / 0.65 = 488 nm. (b) Incoherent cutoff: 2NA/λ₀ = 2.50 cycles/µm, so the finest transferable period is 400 nm. At a period of 800 nm (ν = 1.25 cycles/µm, v = 0.5), MTF = (2/π)[acos 0.5 − 0.5√0.75] = 0.39. (c) Depth of focus: λ₀/NA² = 1.23 µm (±0.62 µm). (d) Collected fraction of an isotropic emitter: (1 − √(1 − 0.4225))/2 = 0.12. (e) A residual spherical aberration of 0.07 waves RMS gives S ≈ exp[−(2π·0.07)²] = 0.82, just above the 0.8 "diffraction-limited" criterion. To check with the tool, set λ₀ = 520, NA = 0.65 and Z₁₁ = 0.07 with a single point, and read the Rayleigh, cutoff, DOF and Strehl readouts. The computed Strehl is about 0.82.

When the model fails

  • High NA (≳ 0.6): the scalar model ignores polarisation. Vector (Richards–Wolf) focusing gives an elongated, polarisation-dependent PSF.
  • Partial coherence: real microscopes use Köhler illumination with a finite condenser NA. Imaging is then neither fully coherent nor fully incoherent (Hopkins' transmission cross-coefficient). This tool models only the two limits. A real phase-contrast objective uses an annular phase ring matched to an annular condenser. Here the ring reduces to a phase dot at DC because the illumination is a single on-axis plane wave.
  • Thin-object approximation: t(x, y) is multiplicative. Thick or strongly scattering samples need beam propagation or rigorous solvers.
  • Space invariance: the PSF is the same across the field. Field-dependent aberrations (off-axis coma, field curvature, distortion) are not modelled.
  • Sampling: the field is periodic, so objects wrap around, the edge object has a second edge at ±fov/2, and PSF tails alias. Keep the pupil radius ≥ 4 samples and 2NA/λ₀ below the Nyquist frequency 1/(2dx).
  • Depth of focus: λ₀/NA² is a paraxial quarter-wave criterion. At high NA the exact angular-spectrum defocus used here departs from it.

References

  • J. W. Goodman, Introduction to Fourier Optics, 4th ed., W. H. Freeman (2017), ch. 5–7.
  • M. Born and E. Wolf, Principles of Optics, 7th ed., Cambridge (1999), §8.5 (Airy pattern), §9.1–9.3 (aberrations, Strehl, Maréchal).
  • R. J. Noll, "Zernike polynomials and atmospheric turbulence", J. Opt. Soc. Am. 66, 207 (1976).
  • F. Zernike, "Phase contrast, a new method for the microscopic observation of transparent objects", Physica 9, 686 (1942).
  • E. Hecht, Optics, 5th ed., Pearson (2017), §11.3 (Fourier optics and spatial filtering).