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The Airy disk is the central bright region, up to the first dark ring, in the scalar diffraction pattern of a uniformly illuminated, unobscured circular aperture. The whole pattern includes the weaker surrounding rings; its radius is distinct from its full width at half maximum.

For a circular optical pupil of diameter $D$ under normal, uniform monochromatic plane wave illumination, Fraunhofer diffraction givesHere $I_0$ is the on-axis light intensity, $\theta$ is the angle from the optical axis, $\lambda$ is the wavelength in the propagation medium, and $J_1$ is the order-one Bessel function of the first kind. The paraxial approximation replaces $\sin \theta$ by $\theta$, measured in radians. The first positive zero $j_{1,1}=3.831705970\dots$ givesThis is the angular radius of the Airy disk; the angular diameter is twice as large. The Airy disk derivation explains the origin of the Bessel function of the first kind and of the numerical coefficient.^[ref]

The image and both radial curves show the same calculated light intensity, divided by its peak. The image's logarithmic scale makes the weak rings visible; its colour is a display choice. The linear curve shows how small the ring peaks are compared with the centre. A dark ring marks a zero of wave amplitude, produced by cancellation across the optical pupil; it is not the edge of all transmitted light. The first Airy disk contains about $83.78\%$ of the total radiant flux, as calculated in Airy disk normalization and encircled energy.

For an original design example, take an ideal optical lens in air with focal length $f=100\mathrm{mm}$, circular optical pupil diameter $D=5.00\mathrm{mm}$ and wavelength $\lambda =550\mathrm{nm}$. In the focal plane, the radial coordinate $r\simeq f\theta$ givesThe first-zero diameter is $26.8327\mu \mathrm{m}$. Solving $P(u_{1/2})=1/2$ gives $u_{1/2}=1.616339948\dots$, so the diameter measured by full width at half maximum is onlyThus a quoted “spot diameter” is incomplete without its definition. Doubling $D$ halves both widths at fixed wavelength and focal length; increasing magnification alone changes the image scale without improving object-space discrimination.

The Airy disk supplies a calculable reference for a point spread function, allowing diffraction to be distinguished from defocus and optical aberrations. Its beauty is that one circular boundary determines an entire sequence of bright and dark rings through a Fourier transform. The ideal formula assumes an unobscured pupil, uniform illumination, ideal optical focus, a single wavelength, and a scalar description with small output angles. A blocked centre, nonuniform pupil transmission or a significant pupil wave phase error changes the response. High numerical aperture may require polarization of an electromagnetic wave and a vector field description. The circular-aperture diffraction laboratory experiment tests the far-field pattern directly; the Rayleigh criterion uses its first zero to define a two-source convention.

An aperture is an opening that admits light into or through an optical instrument. Its shape and size select the transmitted light rays; an aperture alone does not bend them as an optical lens does. The aperture stop is the opening that most restricts the bundle from an on-axis object point. The entrance pupil is the image of that aperture stop formed by the preceding optical lens elements, viewed from the object side. Consequently the effective collecting diameter $D$ is the entrance pupil diameter, which can differ from the physical hole's diameter. For example, preceding optical lens elements that enlarge a $10\mathrm{mm}$ aperture stop by a factor of two give $D=20\mathrm{mm}$; measuring the physical hole alone would misidentify the focal ratio. ^[ref].

For an unobscured circular entrance pupil under a uniform beam normal to its plane, the collecting area and admitted radiant flux arewhere $E_{\mathrm{pupil}}$ is incident irradiance. Doubling $D$ quadruples admitted radiant flux. Obstructions, nonuniform illumination and losses after the entrance pupil require the actual illuminated area and transmission to be included. The same $D$ also sets an angular diffraction-limited resolution scale $\lambda /D$ at wavelength $\lambda$. Collection and diffraction therefore answer different questions from field of view: an aperture selects the bundle from each object point, while the receiving plane's extent helps select which object directions appear in the image.

NASA/JPL's published pinhole demonstration uses two white cards, foil taped over a window in one card, and a small hole pierced in the foil. Sunlight passes through the hole onto the second card; the projected image is viewed on that card. The activity describes a larger image when the card separation increases, improved visibility when the receiving card is shaded, and a separate image from each of several holes. These are qualitative observations of image formation, rather than calibrated measurements of radiant flux or diffraction. ^[ref].

A proposed quantitative extension replaces sunlight by a steadily illuminated asymmetric target and puts the foil and a translucent receiving screen in a dark box. Record target distance $u$, hole-to-screen distance $b$, hole diameter $d$, image height, orientation and edge blur. Measure screen brightness with a fixed photodetector, or record the screen with an optical camera whose integration time and gain stay fixed. Exchange centred foils of measured $d$ at fixed $u,b$; then vary $b$ with the same foil. Geometrical optics predicts an inverted image with transverse magnification $m=-b/u$ and a point's geometric blur diameter $d(1+b/u)$. Thus a larger hole admits more radiant flux without changing the central image scale, whereas increasing $b$ enlarges the image. These are predictions to test, not reported data. Keep target illumination and detector position fixed, subtract a hole-covered dark reading, repeat the original foil to check drift, and avoid clipping the detector response. Foil thickness and ragged edges can block oblique light rays; at sufficiently small $d$, diffraction invalidates the claim that ever smaller holes always sharpen the image. A lensless optical camera has no optical lens focal length: $b$ is a propagation distance, not that focal length.

Insert a centred opaque circular stop of diameter $d=6\mathrm{mm}$ directly in the plane of the $D=10\mathrm{mm}$ bundle. It transmits the same $30^{\circ }$ directions at every surviving position, but discards the surrounding annulus. ThereforeWith the uniform radiance of etendue of a circular aperture and cone, the transmitted radiant flux is $2.2207\mathrm{mW}$ and the blocked fraction is $64\%$. Surviving light rays retain their radiance; the stop has not concentrated them. Lossless propagation after the stop preserves the etendue of this surviving subset, not that of the original incident bundle. A stop placed elsewhere can select positions and directions together, so its physical area ratio alone need not give the transmitted radiant flux fraction.

When both endpoint bundles have the filled circular-cone geometry of etendue of a circular aperture and cone, conservation of etendue givesFor the same refractive index, quadrupling the area halves the sine of the cone half-angle. In the $10\mathrm{mm}$, $30^{\circ }$ example, an ideal output of diameter $20\mathrm{mm}$ must have $\sin {\theta }_2=1/4$, so ${\theta }_2\approx 14.4775^{\circ }$. The exact result differs from simply halving $30^{\circ }$. Conversely, reducing the diameter to $5\mathrm{mm}$ requires a complete forward hemisphere, ${\theta }_2=90^{\circ }$, as a limiting case.

Equal bounding area and solid angle do not alone prove equal etendue: an actual output can have unoccupied directions or position-direction correlations. The filled-cone equation is an exact model of the specified bundles and a capacity constraint on an optical instrument, not a guarantee of an aberration-free implementation.

The beam waist is the narrowest cross-section of a freely propagating Gaussian beam. Its radius $w_0$ uses the wave amplitude $1/e$, or light intensity $1/e^2$, convention; its diameter is $2w_0$. Specifying a “spot size” without saying radius or diameter can introduce a factor-of-two error. Within the scalar paraxial approximation, the wavefront at the beam waist is planar. The minimum radius sets both the Rayleigh range and the far-field beam divergence.

Use the balanced cut-vector distance formula with $p=13$. There are $\left(\genfrac{}{}{0ex}{}{51}{25}\right)=247959266474052$ distinct cuts. A subset of strictly smaller diameter contains no pair with side intersection thirteen, so deleting their shared coordinate one and applying the Frankl-Wilson theorem bounds its size by $\left(\genfrac{}{}{0ex}{}{51}{12}\right)=158753389900$. At least 1562 smaller-diameter parts are needed, exceeding $1325+1$. The cut vectors occupy an affine hyperplane in ${\mathbb{R}}^{1326}$, which identifies isometrically with ${\mathbb{R}}^{1325}$ and gives the stated explicit counterexample.

A subset of a metric space has boundedness when it lies in some ball of finite radius, equivalently when its diameter is finite. A function is bounded when its range is a bounded set. For real-valued functions, this means that $|f|$ has a finite upper bound.

A cover of a set $E$ is a family of sets whose union contains $E$. The members may overlap and may include points outside $E$; intersecting each member with $E$ gives a cover by its subsets. An open cover requires its members to be open sets. In Hausdorff measure, countable covers with a bound on each member's diameter determine geometric size. Constructing one inexpensive cover supplies an upper bound, while a lower bound must control all allowed covers.

In a normed vector space, the diameter of the convex hull of finitely many points equals their greatest pairwise distance. For $x={\sum }_i{\lambda }_i{v}_i$ and $y={\sum }_j{\mu }_j{v}_j$, with nonnegative coefficients each summing to one,The triangle inequality gives $\Vert x-y\Vert \le {\sum }_{i,j}{\lambda }_i{\mu }_j\Vert v_i-v_j\Vert \le {\max }_{i,j}\Vert v_i-v_j\Vert$. The reverse inequality holds because all the vertices belong to the convex hull. Thus a continuum of distances inside a convex polygon can be controlled by finitely many vertex pairs.

Assume in addition that the probability space is a metric space, continuous integrable functions are dense in $L^1$, and the diameters of its partition cells tend to zero at each point. For a continuous $g$, the oscillation average tends to zero because every point in $M_k(x)$ is within its diameter of $x$. For $h=f-g$, the upper limit of the oscillation of $f$ is bounded by the partition maximal function of $h$ plus $|h(x)|$. The first-crossing maximal inequality for refining partitions and Chebyshev's inequality bound the measure where this exceeds $2\epsilon$ by $2\Vert h{\Vert }_1/\epsilon$. Approximation proves the displayed almost everywhere assertion. Refinement supplies the maximal estimate; shrinking diameter supplies the approximation argument. Without shrinking diameter, a constant partition generally converges only to its cell averages.

Diffraction-limited resolution is a scale for distinguishing details with a specified ideal diffraction response and a specified criterion. For two equally bright, mutually incoherent optical point sources imaged through a uniformly transmitting, unobscured circular optical pupil, the Rayleigh criterion places the centre of one Airy disk at the first zero of the other. At small angles in air this separation isHere $D$ is pupil diameter, $\lambda$ is wavelength in air, and $j_{1,1}$ is the first positive zero of the order-one Bessel function of the first kind. This is the radius to the first zero, not the diameter of the central spot. Fraunhofer diffraction gives the radial peak-normalized light intensityThe removable value at zero follows from $J_1(q)\sim q/2$. The pupil Fourier transform reduces to a radial integral by circular symmetry, which produces the Bessel function of the first kind; its first zero supplies the factor $1.22$.^[ref]

In the usual scalar circular-pupil approximation for a fluorescence microscope, the corresponding object-plane Rayleigh criterion isHere ${\lambda }_0$ is the vacuum emission wavelength, and the object-side numerical aperture already includes the medium's refractive index. Do not divide ${\lambda }_0$ by $n$ again. For small cone angles in air, $\mathrm{NA}\simeq D/(2f)$ and image-plane radius $f{\theta }_{\mathrm{R}}$ agrees with this form. The assumptions include a single wavelength, ideal optical focus, and mutually incoherent emitters. High numerical aperture can require a vector description with polarization of an electromagnetic wave and emitter orientation; confocal microscopy has its own excitation-and-detection response. The same $0.61$ formula therefore does not describe every microscope or illumination method.^[ref]

In this original model, ${\lambda }_0=550\mathrm{nm}$ and numerical apertures $0.25$ and $0.50$ give first-zero radii $1.342\mu \mathrm{m}$ and $0.671\mu \mathrm{m}$. Each point spread function is divided by its own peak to compare shape. The pupil axes are transverse wave vector components multiplied by ${\lambda }_0/(2\pi )$, so pupil radius equals numerical aperture. A wider pupil accepts a wider range of transverse wave vectors, producing a narrower image. Increasing magnification alone enlarges the same spot and does not reduce its object-plane radius.

The Rayleigh criterion is a useful visual convention, not an absolute information barrier. For equal incoherent sources with total detected radiant flux $F$ and separation vector $\mathbf{s}$, the ideal image isBelow the Rayleigh criterion this still differs from $Fp(\mathbf{x})$; for small $\mathbf{s}$, symmetry cancels the linear change and the first change is quadratic in separation. Estimating two unknown source positions is then increasingly sensitive to noise, background, relative brightness and uncertainty in the point spread function. Localizing one isolated source and recovering an arbitrary extended object are different inference tasks. Measurements that separate optical modes can also extract separation information differently from direct image light intensity measurement.^[ref]

There remains a real limit for arbitrary images in this model: the optical transfer function is zero above the transmitted spatial frequency band. Its cutoff for scalar incoherent imaging is ${\nu }_c=2\mathrm{NA}/{\lambda }_0$ in object-plane cycles per unit length. The Nyquist–Shannon sampling theorem therefore suggests object-referred detector pixel pitch $p_{\mathrm{pix}}\le {\lambda }_0/(4\mathrm{NA})$, with finer sampling providing margin. Finite detector pixels also integrate over their area, suppressing contrast; coarse sampling can cause aliasing. Finer sampling cannot alone recover spatial frequencies removed by the optics, and Fourier deconvolution amplifies noise where the optical transfer function is small. Photon shot noise, read noise, optical aberrations and defocus can all worsen practical discrimination. A useful resolution claim therefore reports the object model, criterion, wavelength, numerical aperture, sampling and signal-to-noise ratio in photon counting.

A cylindrical model nucleus of diameter $d$ and height $h$. Its free energy is $\pi d^{2}h\Delta g/4+\pi d^2{\gamma }_f/2+\pi dh{\gamma }_e$, where $\Delta g$ is the signed bulk free-energy change per volume.

A proposed optical bench test uses a steady illuminated diffuse patch, a relay made from two optical lens elements, interchangeable centred circular stops, and a calibrated optical power meter large enough to collect the whole output bundle. Start with small cone angles so an ordinary relay approximates the area-angle tradeoff. Measure the patch's image diameter and the accepted angular extent at both conjugate planes; the transverse magnification gives the spatial scale, but does not by itself measure etendue.

Map the local radiance with a baffled photodiode on translation and rotation stages, using two small aligned holes to select a narrow known solid angle. Keep the sampled position fixed while scanning direction, calibrate the sampling footprint and angular response, and repeat at several positions. Numerically integrate $n^2\cos \theta dAd\Omega$ over the occupied samples, and integrate $(L/n^2)dG$ to predict the collected radiant flux. This checks the filled-cone assumption instead of inferring it from two outer dimensions.

Compare the relay without a stop and with a stop of $0.6$ times the illuminated diameter in the same plane. The uniform-bundle prediction is a radiant flux ratio of $0.36$, with unchanged local radiance in the surviving samples. Cover the input for a dark reading, monitor source drift, repeat the open-stop reading, check detector linearity with calibrated optical attenuators, and enlarge the collecting detector or scan it to check for missed light rays. Lens losses, angular response, illumination gradients, alignment errors and diffraction can spoil the ideal prediction; their effects must be measured separately. This is a procedure to try, with predicted readouts, not purported experimental data.

Suppose an extended source fills every position of the entrance aperture with a cone of half-angle ${\theta }_{\mathrm{in}}$, and all these light rays must reach a one-sided planar receiver. If the receiver accepts at most a cone of half-angle ${\theta }_{\mathrm{out}}$, its available etendue bounds the admitted etendue:ConsequentlyThis original derivation combines conservation of etendue with the exact cone integral; the second inequality allows the complete output hemisphere. It bounds capacity and does not prove that a chosen shape attains the bound.

For the $30^{\circ }$ input cone with $n_{\mathrm{in}}=n_{\mathrm{out}}=1$, $C\le 4$: a $10\mathrm{mm}$ entrance cannot deliver the entire filled bundle to a circular receiver smaller than $5\mathrm{mm}$ in diameter. If the output half-angle is restricted to $60^{\circ }$, $C\le 3$ and the receiver diameter must be at least $10/\sqrt{3}\approx 5.7735\mathrm{mm}$. For a receiver medium with model refractive index $n_{\mathrm{out}}=1.5$, the full-hemisphere capacity bound instead gives $C\le 9$; that larger value also depends on transmitting the bundle into that medium.

Attempting $C=10$ in the air-to-air example forces rejection of some input. For the uniform radiance already specified, even filling the entire output hemisphere givesThe maximum mean irradiance gain remains $\eta C\le 4$. With nonuniform radiance, rejecting a given fraction of geometric etendue need not reject the same fraction of radiant flux.

These bounds use a filled extended-source bundle in passive reversible geometrical optics, comparing the same wavelength and the same radiation channels. An ideal optical point source does not fill that four-dimensional set. Diffraction, optical coherence, or coupling a Gaussian beam into an optical fiber require a physical optics calculation; the cone bound is not a universal guarantee for a single wave mode. Likewise, optical gain or fluorescence changes the propagation assumptions.

Consider a planar circular aperture of diameter $D$ in constant refractive index $n$. Suppose every position in its area $A=\pi D^2/4$ is filled with directions inside the same rotationally symmetric cone about the unit normal, with half-angle ${\theta }_{\max }$. Using spherical coordinates, the exact etendue iswhere $\mathrm{NA}=n\sin {\theta }_{\max }$ is the appropriate side's numerical aperture. The ordinary cone solid angle is $2\pi (1-\cos {\theta }_{\max })$; its cosine-weighted value is $\pi {\sin }^2{\theta }_{\max }$. A full forward hemisphere therefore gives $G=n^{2}A\pi$, rather than $2\pi n^{2}A$. ^[ref]

For an original model example, choose $n=1$, $D=10\mathrm{mm}$, and ${\theta }_{\max }=30^{\circ }$. ThenIf the radiance is uniform across this filled bundle and equals $L=100\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, its radiant flux is $\Phi =LG/n^2\approx 6.1685\mathrm{mW}$. Neither the geometry nor the chosen radiance is an experimental measurement.

The field of view is the region of object directions sampled by an optical instrument in one pointing. State whether it is horizontal, vertical or diagonal, and whether it is a full angle or a half-angle. An object-plane width in millimetres also describes a field of view, but requires a specified object distance. Neither that width nor the angular field of view is the aperture diameter or the diffraction-limited resolution.

For a centred flat rectangular photodetector of width $w$ in the focal plane, with positive effective focal length $f$, infinity focus and ideal rectilinear imaging without optical distortion, the full horizontal field of view isThe model maps an object direction $\theta$ to image height of magnitude $|y|=f|\tan \theta |$; set $|y|=w/2$ and use the inverse tangent. This retains exact right triangle geometry within the stipulated imaging model; a paraxial focal length alone does not guarantee that a real wide-field optical lens obeys it. With detector height $h$, replace $w$ by $h$ for the vertical field of view, and by $\sqrt{w^2+h^2}$ for the diagonal field of view. The small-angle approximation gives ${\Theta }_x\simeq w/f$ in radians. ^[ref].

For an original worked example, a $36\mathrm{mm}$ detector width gives ${\Theta }_x\approx 39.60^{\circ }$ at $f=50\mathrm{mm}$, and $20.41^{\circ }$ at $f=100\mathrm{mm}$. Keeping $D=25\mathrm{mm}$ makes these f/2 and f/4 respectively: the field of view changes because $f$ changes. By contrast, closing $D$ from $25$ to $12.5\mathrm{mm}$ at $f=100\mathrm{mm}$ changes f/4 to f/8 while retaining $20.41^{\circ }$ in this model. The clean separation is useful: entrance pupil area controls collection, focal ratio controls the paraxial cone, and $w/f$ controls angular image coverage. Edge clipping, optical distortion, field curvature and optical aberrations can restrict the practically usable field of view.

A proposed angular readout on an optical bench illuminates a small asymmetric target at the optical focus of a collimator, providing a bundle with a known direction. Rotate the optical camera about the centre of its entrance pupil, reading a graduated rotation stage, and record the target image position on its fixed photodetector. Centre the target to establish zero; turn left and right until its centre reaches each detector edge. The difference between stage readings measures the full horizontal field of view. Repeat with a smaller aperture stop: unchanged edge directions but reduced signal support the distinction between collection and angular coverage. Keep the same active detector width, infinity focus, illumination, integration time and gain; approach each reading from the same direction to check mechanical backlash. A finite target blurs the edge criterion, detector thresholds can imitate lost coverage, and a wide bundle must illuminate the whole entrance pupil at every tested angle. These checks distinguish a geometrical boundary from insufficient signal. The analytical diagram is the prediction for this protocol, not a record of its execution.

The focal ratio, or f-number, of a converging optical instrument focused at infinity is the dimensionless quantitywhere $f>0$ is effective focal length and $D$ is the on-axis entrance pupil diameter. The notation f/4 means $N=4$, not $f=4$ in a particular unit. A $100\mathrm{mm}$ focal length with $D=25\mathrm{mm}$ gives f/4. Closing the aperture stop until $D=12.5\mathrm{mm}$ gives f/8 while retaining the same focal length. The mechanical opening need not equal $D$. ^[ref].

For a thin lens in air with a circular aperture stop at the thin lens, the converging bundle from a distant on-axis point has half-angle $\alpha$ satisfyingThe last step uses the small-angle approximation. For a compound optical lens, the cone is set by the image-side optical pupil and its distance from the receiving plane; the simple triangle is not an exact high-angle rule. At finite object distance a thin lens instead gives $v=f(1+|m|)$, by its imaging equation and transverse magnification, so its working cone corresponds to $v/D=N(1+|m|)$. Nominal focal ratio therefore does not alone specify close-range image irradiance. ^[ref].

Two useful scalings follow under stated conditions. For a fixed uniform incident beam and unchanged transmission, collected radiant flux scales as $D^2$. For an extended source with fixed radiance, the central image irradiance scales approximately as $N^{-2}$ in the paraxial approximation: the solid angle of the cone scales as $(D/f{)}^2$. For an ideal uniformly illuminated unobscured circular entrance pupil, the first dark ring of the point spread function has angular radius and focal plane radiusHere $\lambda$ is wavelength. These are first-zero radii, not diameters or a universal rule for resolving arbitrary objects. A proof outline evaluates the circular aperture's Fraunhofer diffraction using its Fourier transform, then uses the first zero of the resulting Bessel function. ^[ref].

At $\lambda =0.55\mu \mathrm{m}$, the example's f/4 and f/8 first-ring radii are $2.684$ and $5.368\mu \mathrm{m}$. Closing the aperture stop halves $D$, quarters collected radiant flux and doubles this diffraction radius. It can also reduce optical aberrations by excluding outer light rays; actual sharpness reflects both effects. This explains why a smaller aperture need not produce a sharper image. ^[ref].

A proposed optical bench check puts a fixed collimator before a converging thin lens and a photodetector large enough to collect the entire image behind it. Compare two centred aperture stops of known diameters while fixing source output, detector gain and integration time. The integrated signal tests the $D^2$ collection law; a spatially resolved optical camera at the focal plane can separately measure the point spread function. Repeat a dark reading and the first aperture stop, and check that neither the incoming beam nor detector clips either bundle. A resolved pinhole source, imperfect collimation, optical aberrations and detector pixel size can dominate the apparent spot width; a single peak reading does not measure total collected radiant flux.

Consider an ideal coherent monochromatic wave from an unresolved source on the optical axis, a uniform circular entrance pupil of diameter $D$, and ideal relay optics without wavefront errors. In the scalar Fraunhofer diffraction approximation, a Fourier transform carries the pupil's complex amplitude into the image plane. Choose dimensionless pupil coordinates $\mathbf{x}$ in units of $D$ and angular coordinates $\mathbf{u}=D\mathbf{\theta }/\lambda$, where $\lambda$ is the wavelength. DefineThe coronagraphic focal plane mask has transmission $M=1-W$, with $W=1$ inside the occulting disk and $W=0$ outside. Thus the field immediately after the mask is $MA$. Outside the disk $M=1$, so the local wave intensity is still $|A{|}^2$: hiding the core has left those rings intact. Reimaging an unlimited pupil giveswhere $\ast$ is convolution, $L$ is the Lyot stop transmission, and the proportionality constant in the irradiance is omitted. Coordinate reversals from physical relay optics are absorbed into the coordinate choice. These equations use wave amplitudes throughout propagation; applying a Fourier transform to wave intensity would give a different and incorrect model. This is the pupil/image calculation described by Sivaramakrishnan et al.; the circular disk and numerical choices below are our illustrative specialization. ^[ref].

The subtraction explains the beauty of the arrangement: in much of the interior, the original pupil field and the convolution term nearly cancel. Near a boundary, that cancellation is incomplete. The Lyot stop can therefore block a region containing disproportionately much unwanted light. A hard-edged coronagraphic focal plane mask has a transform with extended tails, so the interior need not be perfectly dark. With an unlimited relay, $L=1$ gives $A_{\mathrm{final}}=MA$ exactly: no further ring suppression occurs. A stop covering the full original optical pupil instead uses $L=P$ and already clips field outside the optical pupil; an undersized Lyot stop clips more of the remaining rim.

The toy uses a coronagraphic focal plane mask of angular radius $3\lambda /D$ and a Lyot stop of diameter $0.8D$. The discrete Fourier transform grid has $1024\times 1024$ samples, with 128 across $D$ and angular sampling $0.125\lambda /D$; it approximates the continuous equations on a finite periodic domain. The plot averages wave intensity in annuli of width $0.25\lambda /D$. Over the annulus $4\le |\mathbf{u}|\le 10$, the mean wave intensity divided by the direct stellar peak is approximately $1.40\times 10^{-4}$ without a coronagraph, the same with only the focal mask and an unlimited relay, $4.44\times 10^{-5}$ with the mask and full pupil, and $1.66\times 10^{-6}$ with the mask and $0.8D$ Lyot stop. Thus the undersized opening reduces this particular annular mean about 27-fold relative to the full-pupil comparison. It does not suppress every direction by that factor, and these values do not predict a real instrument's detection limit.

The ideal toy excludes a central obstruction, support vanes, scattering of light, wavefront errors, atmospheric effects, a spread of wavelengths, pointing drift and detector noise. It also contains only the star: no simulated planet or solar corona. For a far off-axis source whose image misses the coronagraphic focal plane mask, the continuous geometric coronagraph throughput of the $0.8D$ opening tends to $0.8^2=0.64$ relative to the full disk. Rejecting stellar light therefore costs some desired light as well. Enlarging the coronagraphic focal plane mask narrows the characteristic scale of its pupil-plane transform by the scaling property of the Fourier transform, but also hides a larger neighborhood of the source.

The Hausdorff dimension is the critical exponent at which Hausdorff measure changes from infinite to zero:with the standard endpoint convention for an empty supremum. This measures fine-scale geometric size rather than the number of coordinates used to describe the set. A line segment has Hausdorff dimension one; a square has dimension two; the middle-third Cantor set has dimension $\log 2/\log 3\approx 0.63093$.

Why can a dimension be fractional? At a scale reduced by a factor of three, the Cantor set needs twice as many construction intervals. The balance $2(1/3{)}^d=1$ gives $d=\log 2/\log 3$, between the growth rates of finitely many points and a whole interval. Making this argument rigorous requires a cover upper bound and the mass distribution principle lower bound, as in Hausdorff dimension of the Cantor set.

At the critical exponent itself, Hausdorff measure can be zero, finite and positive, or infinite; dimension does not specify that endpoint value. For example, finite sets and countably infinite sets both have dimension zero, although their counting measures differ. Dimension is also different from a finite-resolution estimate: box-counting dimension is another asymptotic definition, and fitting a scaling slope over a handful of observed scales does not establish either dimension.

The metric matters. Multiplying every distance by a fixed positive constant multiplies ${\mathcal{H}}^s$ by that constant to the power $s$ and preserves Hausdorff dimension. A Lipschitz map cannot increase it, because mapped cover diameters increase by at most a fixed factor. A bi-Lipschitz map therefore preserves it by applying this fact in both directions. These properties make Hausdorff dimension useful for comparing geometric models while controlling how their measurements change.

The middle-third Cantor set has Hausdorff dimensionIts Lebesgue measure is zero, yet $0<{\mathcal{H}}^d(C)<\infty$. These assertions measure different aspects of size: interval length is too coarse to distinguish this set from a null set, while the critical fractional measuring rule captures its remaining structure.

At level $n$, the construction covers $C$ by $2^n$ intervals of diameter $3^{-n}$. Their total $s$-cost isFor $s>d$ it tends to zero, proving the upper dimension bound. At $s=d$ it equals one, giving ${\mathcal{H}}^d(C)\le 1$. For $s<d$ this particular cover's cost grows, but that alone would not prove a lower bound on Hausdorff measure.

For the matching lower bound, use the Cantor distribution, which assigns mass $2^{-n}$ to each level-$n$ Cantor cylinder. An interval of length $r$ with $3^{-(n+1)}<r\le 3^{-n}$ meets at most two level-$n$ construction intervals, since their gaps are at least $3^{-n}$. Its mass is therefore at mostEvery covering set of diameter $r$ in the line lies in an interval of length $r$, so any small cover must have total $d$-cost at least $1/4$. This is the mass distribution principle in its interval form: $1/4\le {\mathcal{H}}^d(C)\le 1$, sufficient to prove the claimed dimension. The constants are convenient bounds, rather than an assertion of the exact critical Hausdorff measure. In the diagram, the right panel is a finite-stage approximation to the Cantor function, showing how probability is shared while the middle third receives none.

The reusable pattern is to obtain an upper bound by constructing covers and a lower bound by distributing mass. The same strategy underlies Cartesian powers of the Cantor set and the dimension formula for a finite similarity iterated function system satisfying the open set condition.