Abbe Theory of Imaging and Spatial Filtering

Derivation

Abbe Theory of Imaging and Spatial Filtering

D-212 Home PU-206 Threads light · waves Depends on Fraunhofer Diffraction as a Fourier Transform, Lensmaker's Formula and the Imaging Equation

Statement

In a coherently illuminated, paraxial optical system, a converging lens performs an exact spatial Fourier transform of the field in its front focal plane onto its back focal plane: the amplitude at back-focal-plane point \((u,v)\) equals the object spectrum evaluated at spatial frequencies \(f_X=u/(\lambda f)\), \(f_Y=v/(\lambda f)\). A second Fourier transform (image formation) reconstructs the object, so the image amplitude is a double Fourier transform of the object field, band-limited by the lens pupil. The pupil therefore acts as a coherent transfer function \(H(f_X,f_Y)=P(\lambda f f_X,\lambda f f_Y)\), an amplitude mask placed in the back focal plane multiplies the object spectrum (spatial filtering), and the finite aperture sets the diffraction-limited resolution \(d_{\min}\sim\lambda/(2\,\mathrm{NA})\).

Why it matters

Abbe's insight reframes imaging as a two-stage diffraction process rather than a ray-tracing exercise: the lens first decomposes the object into plane-wave components (its angular spectrum), physically separating them in the back focal plane by spatial frequency, and then re-superposes them into an image. This is the operational foundation of Fourier optics.

Because the back focal plane is a physical map of the object's spectrum, one can reach in and edit it. Blocking, attenuating, or phase-shifting parts of that plane directly manipulates spatial frequencies — the basis of low-pass "cleaning", edge enhancement, phase contrast, dark-field microscopy, apodisation, and optical pattern recognition. The same analysis fixes the resolution limit of every microscope and lithography tool.

Assumptions

Monochromatic, spatially coherent scalar field.If dropped, one cannot speak of a single complex amplitude that transforms linearly; partially coherent or broadband light requires the mutual coherence function and imaging becomes linear in intensity (the optical transfer function) rather than amplitude, changing the transfer function and the resolution limit.

Paraxial (Fresnel) propagation and a thin lens acting as a pure quadratic phase.Without it the lens no longer imposes \(\exp[-ik(x^2+y^2)/2f]\); aberrations and high-angle terms enter, the back-focal-plane field is a distorted transform, and the clean Fourier relationship holds only near the axis.

Object placed exactly in the front focal plane.If the object sits at a general distance \(d\neq f\), a residual quadratic phase factor \(\exp[i k(1-d/f)(u^2+v^2)/2f]\) multiplies the transform, so the back focal plane carries the correct magnitude spectrum but a curved phase — harmless for intensity measurements, fatal for phase-sensitive filtering.

Aperture large compared with a wavelength; evanescent components neglected.Spatial frequencies with \(f_X^2+f_Y^2>1/\lambda^2\) correspond to evanescent waves that never reach the lens. Ignoring them is what makes the system band-limited; near-field (SNOM) imaging recovers exactly these terms and beats the Abbe limit.

Derivation

\[ U_1(x_1,y_1)=\frac{e^{ikf}}{i\lambda f}\iint U_0(x_0,y_0)\,\exp\!\left(i\frac{k}{2f}\big[(x_1-x_0)^2+(y_1-y_0)^2\big]\right)dx_0\,dy_0 \]

Fresnel diffraction propagates the object field \(U_0\) forward by the focal distance \(f\) to the plane just before the lens; this is the paraxial Huygens–Fresnel integral (a convolution with the Fresnel kernel). A

\[ U_1'(x_1,y_1)=U_1(x_1,y_1)\,P(x_1,y_1)\,\exp\!\left(-i\frac{k}{2f}(x_1^2+y_1^2)\right) \]

A thin lens multiplies the field by a converging quadratic phase; \(P\) is the pupil (unity inside the aperture, zero outside). Multiplication is legal because the lens is thin, so transverse coordinates are unchanged across it. A

\[ U_f(u,v)=\frac{e^{ikf}}{i\lambda f}\iint U_1'(x_1,y_1)\,\exp\!\left(i\frac{k}{2f}\big[(u-x_1)^2+(v-y_1)^2\big]\right)dx_1\,dy_1 \]

Propagate a second focal distance \(f\) from the lens to the back focal plane \((u,v)\), again by the Fresnel integral. A

\[ U_f(u,v)\propto\iint\!\!\iint U_0(x_0,y_0)\,P(x_1,y_1)\,\exp\!\left(i\frac{k}{2f}\big[(x_1-x_0)^2-x_1^2+(u-x_1)^2\big]+\{y\}\right)dx_0dy_0dx_1dy_1 \]

Substitute steps 1–2 into step 3 and collect the exponent. I display only the \(x\)-part; the \(y\)-part is identical. This is legal reordering of a convergent multiple integral. B

\[ (x_1-x_0)^2-x_1^2+(u-x_1)^2 = x_0^2+u^2-2x_1(x_0+u)\;+\;x_1^2 \]

Expand and combine the three squares in \(x_1\). The crucial cancellation: the lens phase \(-x_1^2\) removes one of the two \(+x_1^2\) terms, leaving a single \(x_1^2\) rather than two. This survival of exactly one quadratic term is the whole reason the front-focal-plane geometry yields a clean transform. B

\[ \iint P(x_1,y_1)\,\exp\!\left(i\frac{k}{2f}(x_1^2+y_1^2)\right)\exp\!\left(-i\frac{k}{f}(x_1 u+y_1 v)+\cdots\right)dx_1dy_1 \]

For an ideal lens the pupil is effectively infinite, \(P\to1\); completing the square in \(x_1\) turns the inner integral into a Gaussian/Fresnel integral whose value \(\propto\exp[-i k u^2/2f\cdot(\text{coeff})]\) exactly cancels the leftover \(u^2\) term from step 5. Only the cross term \(x_0 u\) survives. C

\[ U_f(u,v)=\frac{1}{i\lambda f}\iint U_0(x_0,y_0)\,\exp\!\left(-i\frac{2\pi}{\lambda f}(x_0 u+y_0 v)\right)dx_0\,dy_0 \]

Using \(k=2\pi/\lambda\), all residual quadratic phases have cancelled and the back-focal-plane field is the exact Fourier transform of the object. Constant phase prefactors \(e^{2ikf}\) are dropped as they do not affect intensity or filtering. A

\[ U_f(u,v)=\frac{1}{i\lambda f}\,\tilde U_0(f_X,f_Y),\qquad f_X=\frac{u}{\lambda f},\quad f_Y=\frac{v}{\lambda f} \]

Identify the transform variable: position \((u,v)\) in the back focal plane is spatial frequency, scaled by \(\lambda f\). The back focal plane is the Fourier (diffraction) plane. Each object spatial frequency has been mapped to a distinct point. A

\[ U_f'(u,v)=U_f(u,v)\,M(u,v),\qquad H(f_X,f_Y)\equiv M(\lambda f f_X,\lambda f f_Y) \]

Because distinct frequencies are physically separated, a mask \(M\) placed in the back focal plane multiplies the spectrum point-by-point. Defining the coherent transfer function \(H\) as the mask expressed in frequency coordinates makes filtering a simple multiplication. Setting \(M=P\) (the lens's own pupil) gives \(H(f_X,f_Y)=P(\lambda f f_X,\lambda f f_Y)\). A

\[ U_i(x_i,y_i)=\frac{1}{i\lambda f}\iint U_f'(u,v)\,\exp\!\left(-i\frac{2\pi}{\lambda f}(u\,x_i+v\,y_i)\right)du\,dv \]

A second identical lens (the 4f configuration) Fourier-transforms the filtered back-focal-plane field onto the image plane by the same argument as steps 1–7. Two forward transforms compose. A

\[ U_i(x_i,y_i)=\big(U_0\ast h\big)(-x_i,-y_i),\qquad h=\mathcal{F}^{-1}\{H\} \]

Substituting step 9 into step 10, the double transform reproduces the object with inverted coordinates, convolved with the amplitude point-spread function \(h\), the inverse transform of the transfer function. Multiplication in the Fourier plane becomes convolution in the image plane (convolution theorem). This is a linear, space-invariant, amplitude map — the defining statement of coherent imaging. C

\[ H(f_X,f_Y)=1\ \text{for}\ \sqrt{f_X^2+f_Y^2}\le\frac{a}{\lambda f}\equiv f_c,\qquad f_c=\frac{\mathrm{NA}}{\lambda} \]

For a circular pupil of radius \(a\) the transfer function is a hard low-pass: frequencies beyond the cutoff \(f_c=a/(\lambda f)=\mathrm{NA}/\lambda\) (with \(\mathrm{NA}=n\sin\theta\approx n a/f\)) are simply not collected by the lens and are absent from the image. This band limit is the diffraction limit. B

Result

\[ U_i(x_i,y_i)=\Big(U_0\ast h\Big)(-x_i,-y_i),\quad h=\mathcal{F}^{-1}\{H\},\quad H(f_X,f_Y)=P(\lambda f f_X,\lambda f f_Y),\quad f_c=\frac{\mathrm{NA}}{\lambda} \]

Reading. Imaging is a Fourier transform to the back focal plane followed by an inverse transform to the image plane. In between, the object's spatial spectrum lies exposed and separated by frequency, and the lens pupil (or any inserted mask) acts as a multiplicative transfer function \(H\). Whatever \(H\) leaves out never appears in the image; the finite pupil low-passes the object with cutoff \(f_c=\mathrm{NA}/\lambda\), blurring it by the amplitude point-spread function \(h\) (an Airy pattern for a circular aperture). Coherent imaging is linear and shift-invariant in complex amplitude, not intensity.

Units check. \(f_X=u/(\lambda f)\) has dimensions \(\mathrm{m}/(\mathrm{m}\cdot\mathrm{m})=\mathrm{m^{-1}}\) (cycles per metre), correct for a spatial frequency. The prefactor \(1/(i\lambda f)\) has units \(\mathrm{m^{-2}}\); acting on the field integrated over \(dx_0\,dy_0\) (units of field \(\times\,\mathrm{m^2}\)) returns a field of the original units. The cutoff \(f_c=\mathrm{NA}/\lambda\) is dimensionless/metre \(=\mathrm{m^{-1}}\), and \(d_{\min}=\lambda/(2\,\mathrm{NA})\) is a length.

Limiting cases

Infinite aperture (\(a\to\infty\), \(H\to1\)): \(h\to\delta\), the image is a perfect inverted copy \(U_i=U_0(-x_i,-y_i)\) — geometrical imaging with no diffraction blur.
Pinhole in the Fourier plane (\(H\to\delta\) at the origin): only the DC term survives, the image is a uniform field carrying the object's average transmission — the ultimate low-pass.
DC blocked (\(H=0\) at origin only): the mean is removed and only edges/fine structure remain — high-pass edge enhancement; a dark background shows bright object outlines.
Point object (\(U_0=\delta\)): its spectrum is flat, uniformly filling the pupil, and the image is the point-spread function \(h\) itself — a direct read-out of \(H\).
Grating finer than \(\lambda/\mathrm{NA}\): its \(\pm1\) orders fall outside the pupil, only the DC order passes, and the grating vanishes from the image — the Abbe resolution limit made visible.

Breaks when

Incoherent or partially coherent illumination. The amplitude transfer function is replaced by the intensity-domain optical transfer function \(\mathrm{OTF}=\mathcal{F}\{|h|^2\}\), which is the autocorrelation of the pupil and extends to twice the coherent cutoff. Amplitude linearity fails; adding two point sources adds intensities, not fields, so all coherent-filtering intuition (edge ringing, phase contrast phases) breaks down.
Large angles / high NA / thick or aberrated lenses. The quadratic-phase (paraxial) lens model fails; the pupil phase carries aberration terms \(W(f_X,f_Y)\), the back-focal-plane field is no longer a clean Fourier transform, and vector (polarisation) effects appear as \(\mathrm{NA}\to1\). The scalar theory and the exact FT relation are only leading-order.
Object not in the front focal plane, phase filtering attempted. A residual quadratic phase multiplies the spectrum, so magnitude filtering still works but any phase-sensitive operation (Zernike phase contrast, complex matched filtering) is corrupted.
Sub-wavelength / near-field structure. Spatial frequencies above \(1/\lambda\) are evanescent, never propagate to the lens, and cannot be imaged by any far-field system regardless of aperture — the fundamental ceiling behind the Abbe limit.

Failure modes

Confusing the coherent cutoff with the incoherent one. The amplitude cutoff is \(f_c=\mathrm{NA}/\lambda\); the intensity (OTF) cutoff is \(2\mathrm{NA}/\lambda\). Students quote one resolution number for both illumination conditions.
Applying \(|h|^2\)-based (incoherent) resolution formulas to a laser (coherent) system, or vice versa — e.g. using the Rayleigh two-point criterion for coherent illumination, where the two-point response depends on the relative phase of the sources and the "resolvable" separation shifts.
Forgetting the image inversion and magnification. A general 4f uses two focal lengths \(f_1,f_2\) giving magnification \(-f_2/f_1\) and frequency scale \(\lambda f_1\) in the Fourier plane; only \(f_1=f_2\) gives unit magnitude.
Placing the filter at the lens instead of the back focal plane. Only in the back focal plane are spatial frequencies fully separated; a mask elsewhere mixes position and frequency and does not perform clean spectral filtering.
Treating the Fourier-plane coordinate as a length without the \(\lambda f\) scale, so predicted diffraction-order positions are off by orders of magnitude.
Ignoring the DC term when high-pass filtering, then being surprised the "edge image" is dim — most of the energy lived in the blocked zero order.

Discussion

Abbe's theory dissolves the apparent opposition between "ray" imaging and "wave" diffraction: they are the same process viewed at the two ends of the system. Geometrical optics describes the mapping from object plane to image plane; Fourier optics describes what happens in the plane between them, where the object has been decomposed into its plane-wave (angular-spectrum) components. A plane wave leaving the object at angle \(\theta\) is a single spatial frequency \(f_X=\sin\theta/\lambda\), and the lens focuses it to one point \(u=\lambda f f_X\) in the back focal plane. The back focal plane is therefore quite literally a picture of the object's diffraction pattern — the same pattern one would see far away in Fraunhofer diffraction, brought to a finite distance by the lens.

The power of the picture is that it makes imaging editable. Because each frequency occupies its own location, physical hardware placed there — an opaque stop, a graded filter, a phase plate, a spatial light modulator — performs an analogue Fourier-domain computation at the speed of light and in parallel over the whole field. Zernike's phase contrast (retarding the undiffracted zero order by a quarter wave to turn invisible phase objects into visible intensity), dark-field microscopy (blocking the zero order so only scattered light forms the image), schlieren imaging, apodisation to suppress diffraction rings, and optical correlators for pattern recognition are all one-line prescriptions for \(H\).

The resolution limit follows without any extra physics. To reproduce a periodic object of period \(d\) the image must contain at least its zeroth and first diffraction orders; the first order emerges at \(\sin\theta=\lambda/d\) and must fall within the collection cone, \(\sin\theta\le\mathrm{NA}/n\). Axial coherent illumination then requires \(d\ge\lambda/\mathrm{NA}\), but tilting the illumination so the zeroth and one first order sit at opposite pupil edges doubles the captured bandwidth to give Abbe's celebrated \(d_{\min}=\lambda/(2\,\mathrm{NA})\). Every route to higher resolution is a route to larger effective NA: immersion oil raises \(n\), structured illumination folds high frequencies into the passband, and near-field probes capture the evanescent tail outright.

At the deepest level the coherent transfer function is the pupil, and the amplitude point-spread function is its Fourier transform — a statement that the achievable spot is set entirely by the aperture's shape and phase. This is the optical face of the uncertainty relation between aperture size and angular spread, and it connects directly to antenna theory (the far-field pattern is the transform of the aperture illumination) and to the diffraction-limited spot of a focused laser. Aberrations are not a separate phenomenon but simply a non-flat phase \(P=|P|e^{iW}\) across the pupil; the entire Zernike-polynomial machinery of lens design lives inside \(H\). Common misconceptions. The back focal plane does not contain a small inverted image — it contains the object's spectrum; the image forms only after the second transform. "Coherent" resolution is not simply "better" than incoherent — the coherent amplitude cutoff is half the incoherent intensity cutoff, and coherent images suffer speckle and edge ringing that incoherent images do not. Finally, a bigger lens does not help once \(\mathrm{NA}\) already fills the propagating cone: the ceiling is \(1/\lambda\), set by the wave, not the glass.

Worked examples

Example 1 — Fourier-plane geometry and low-pass removal of a grating (4f, coherent).

\[ \lambda=633\ \text{nm},\quad f=200\ \text{mm},\quad \text{grating period } d=100\ \mu\text{m},\quad \text{object in front focal plane} \]

A He–Ne laser (633 nm) illuminates an amplitude grating; we ask where its diffraction orders land in the back focal plane. A

\[ f_{X,m}=\frac{m}{d}\ \Longrightarrow\ u_m=\lambda f\,f_{X,m}=\frac{m\,\lambda f}{d} \]

The grating's spectrum is a comb at \(f_X=m/d\); using the frequency-to-position map \(u=\lambda f f_X\) gives each order's location. Symbols first. A

\[ u_{\pm1}=\pm\frac{(633\times10^{-9})(0.200)}{100\times10^{-6}}=\pm\frac{1.266\times10^{-7}}{1.0\times10^{-4}}\ \text{m}=\pm1.27\ \text{mm} \]

Insert numbers: the \(\pm1\) orders sit \(1.27\) mm off axis, the \(\pm2\) orders at \(\pm2.53\) mm, the DC order on axis. A

\[ \text{Pinhole radius } r_{\text{stop}}<1.27\ \text{mm}\ \Rightarrow\ \text{only } m=0 \text{ passes} \]

A central pinhole smaller than the first-order spacing blocks every diffracted order, leaving only the uniform DC term. By the result, the image is then featureless. B

\[ u_{\pm1}=\pm1.27\ \text{mm};\quad \text{a pinhole of }r_{\text{stop}}\lesssim1\ \text{mm removes the grating, giving a uniform image of transmission }\langle t\rangle. \]

Reading. The lens has spread the grating's spectrum across millimetres of physical space, so a modest stop can select or reject individual orders. Passing only DC erases the periodic structure — a textbook low-pass filter.

Units check. \((\text{m})(\text{m})/(\text{m})=\text{m}\); \(1.266\times10^{-7}/10^{-4}=1.27\times10^{-3}\) m ✓.

Example 2 — Abbe resolution of a microscope objective and the gain from oil immersion.

\[ \lambda=550\ \text{nm (green)},\quad \text{dry objective } \mathrm{NA}=0.65,\quad \text{oil objective } n=1.515,\ \sin\theta=0.90 \]

Compare a dry and an oil-immersion objective at the same collection angle regime, using Abbe's two-beam limit with optimally tilted illumination. A

\[ d_{\min}=\frac{\lambda}{2\,\mathrm{NA}},\qquad \mathrm{NA}=n\sin\theta \]

Abbe's resolution with symmetric (tilted) coherent illumination. Rearranged symbolically before substitution. A

\[ d_{\min}^{\text{dry}}=\frac{550\times10^{-9}}{2(0.65)}=\frac{5.50\times10^{-7}}{1.30}=4.23\times10^{-7}\ \text{m}=423\ \text{nm} \]

Dry objective. A

\[ \mathrm{NA}_{\text{oil}}=1.515\times0.90=1.36,\qquad d_{\min}^{\text{oil}}=\frac{550\times10^{-9}}{2(1.36)}=2.02\times10^{-7}\ \text{m}=202\ \text{nm} \]

Immersion in oil raises \(n\), hence NA, hence resolution — the same \(\sin\theta\) now buys a smaller \(d_{\min}\). B

\[ d_{\min}^{\text{dry}}\approx423\ \text{nm},\qquad d_{\min}^{\text{oil}}\approx202\ \text{nm}\quad(\text{a factor }\approx2.1\text{ improvement}). \]

Reading. Resolution is set entirely by \(\mathrm{NA}/\lambda\). Oil immersion works purely by increasing \(n\) in \(\mathrm{NA}=n\sin\theta\), letting the objective collect the same high diffraction orders that would otherwise be lost. Shorter wavelength would help identically.

Units check. \(\text{m}/(\text{dimensionless})=\text{m}\); NA is dimensionless ✓.

Problems

Fourier-plane scale. A 4f system uses \(\lambda=532\) nm and \(f=150\) mm. What length in the back focal plane corresponds to a spatial frequency of \(50\) cycles/mm?
Solution
Position maps as \(u=\lambda f f_X\). With \(f_X=50\ \text{mm}^{-1}=5.0\times10^{4}\ \text{m}^{-1}\): \(u=(532\times10^{-9})(0.150)(5.0\times10^{4})=(7.98\times10^{-8})(5.0\times10^{4})=3.99\times10^{-3}\) m \(\approx4.0\) mm. So each 50 cycles/mm of object detail sits 4.0 mm off axis.
Designing a low-pass stop. An amplitude object contains a wanted image band up to \(20\) cycles/mm and unwanted fine noise near \(120\) cycles/mm, imaged with \(\lambda=633\) nm, \(f=250\) mm. Give the radius range of a circular aperture in the Fourier plane that passes the signal but blocks the noise.
Solution
\(u=\lambda f f_X\), \(\lambda f=(633\times10^{-9})(0.250)=1.583\times10^{-7}\) m·m. Signal edge: \(u_{\text{sig}}=(1.583\times10^{-7})(2.0\times10^{4})=3.17\times10^{-3}\) m \(=3.17\) mm. Noise: \(u_{\text{noise}}=(1.583\times10^{-7})(1.2\times10^{5})=1.90\times10^{-2}\) m \(=19.0\) mm. Any aperture radius \(r\) with \(3.17\ \text{mm}<r<19.0\ \text{mm}\) passes all signal frequencies while rejecting the noise; e.g. \(r\approx8\)–\(10\) mm is a safe choice.
Coherent cutoff and spot size. A lens of clear-aperture radius \(a=10\) mm and focal length \(f=100\) mm images at \(\lambda=500\) nm in a coherent 4f system. Find the coherent cutoff frequency \(f_c\), the effective NA, and estimate the amplitude point-spread-function radius (first Airy zero) \(\Delta r\approx0.61\lambda/\mathrm{NA}\).
Solution
\(\mathrm{NA}\approx a/f=10/100=0.10\). Cutoff \(f_c=\mathrm{NA}/\lambda=0.10/(500\times10^{-9})=2.0\times10^{5}\ \text{m}^{-1}=200\) cycles/mm. PSF radius \(\Delta r=0.61\lambda/\mathrm{NA}=0.61(500\times10^{-9})/0.10=3.05\times10^{-6}\) m \(=3.05\ \mu\text{m}\). The system passes spatial frequencies up to 200 cycles/mm and blurs points to about a 3 µm radius.
Axial vs. oblique illumination. For \(\mathrm{NA}=0.80\), \(\lambda=488\) nm, compute the finest resolvable grating period under (a) axial coherent illumination (needs orders \(0,\pm1\)) and (b) optimally tilted illumination (needs \(0,+1\)). Comment.
Solution
(a) Axial: the \(\pm1\) orders must both fit, \(\lambda/d\le\mathrm{NA}\Rightarrow d_{\min}=\lambda/\mathrm{NA}=488\times10^{-9}/0.80=6.10\times10^{-7}\) m \(=610\) nm. (b) Tilted so the 0 and +1 orders straddle the pupil, the captured angle doubles: \(d_{\min}=\lambda/(2\mathrm{NA})=488\times10^{-9}/1.60=3.05\times10^{-7}\) m \(=305\) nm. Oblique illumination doubles the resolution — this is Abbe's original argument and the reason condensers are set to fill the aperture.
High-pass edge enhancement (energy accounting). A coherent 4f images a transparency whose amplitude is \(t(x)=0.5+0.3\cos(2\pi x/\Lambda)\) with period \(\Lambda=50\ \mu\text{m}\), \(\lambda=633\) nm, \(f=200\) mm. (i) Where are the spectral components in the back focal plane? (ii) A small opaque dot blocks only the DC order. Write the resulting image amplitude and intensity, and state qualitatively what is seen.
Solution
(i) The spectrum has a DC term (amplitude 0.5) at \(u=0\) and two sidebands (each amplitude 0.15) at \(f_X=\pm1/\Lambda\), i.e. \(u=\pm\lambda f/\Lambda=\pm(633\times10^{-9})(0.200)/(50\times10^{-6})=\pm2.53\times10^{-3}\) m \(=\pm2.53\) mm. (ii) Blocking DC leaves the two sidebands: the image amplitude becomes \(U_i(x)\propto0.3\cos(2\pi x/\Lambda)\) (coordinate-inverted, but even so unchanged in form). The intensity is \(|U_i|^2\propto0.09\cos^2(2\pi x/\Lambda)=0.045\,[1+\cos(4\pi x/\Lambda)]\) — a pattern at half the original period with the uniform background removed. Physically, the bright field turns dark and the edges/structure of the modulation stand out (dark-field / edge enhancement), but the image is dim because the 0.5 DC amplitude, carrying most of the energy, was discarded.