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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.

The following original model isolates the conversion mechanism. Use a periodic image window, unit illumination, ideal unit magnification, and a phase plate that changes only the zero spatial frequency. Write $c_0=⟨e^{i\varphi }⟩$ for the spatial mean. Multiplying that single Fourier coefficient by $a{e}^{i\theta }$, with $0<a\le 1$, gives the exact image field in this model:This formula follows by splitting $e^{i\varphi }=c_0+(e^{i\varphi }-c_0)$: the second term has zero mean and is left unchanged. A real annular, partly coherent instrument alters a finite band of directions, so this calculation is a mechanism model rather than its full image-formation law.

For a weak phase object with $⟨\varphi ⟩=0$, one has $c_0=1+O({\varphi }^2)$ and henceThe cross term is $2\mathrm{Re}(a{e}^{-i\theta }i\varphi )=2a\varphi \sin \theta$. Thus $\theta =\pi /2$ makes positive $\varphi$ brighter and $\theta =-\pi /2$ makes it darker under this stated phase convention for complex amplitudes. The sign comes from this convention and chosen plate; a manufacturer’s positive/negative contrast labels should not be inferred from the sign of $\theta$ alone.

For example, choose $a=0.5$ and a zero-mean periodic pattern whose peak phase is $0.084$. The first-order peak intensities are $0.25\pm 0.084$, compared with the uniform ideal bright-field microscopy value $1$. The relative change about the reduced background is approximately $2\varphi /a$ for a quarter-cycle shift. Attenuating the reference increases relative contrast while reducing the background photon count; it is not light amplification or a universal improvement of signal-to-noise ratio.

Two checks expose what the model can and cannot measure. A spatially constant phase gives $E_{\mathrm{out}}=a{e}^{i\theta }{e}^{i\varphi }$ and $I_{\mathrm{out}}=a^2$, so no absolute phase is recovered. Orthogonality of the constant and zero-mean parts also givesFor $a\le 1$ this cannot exceed the incident mean light intensity. The image redistributes brightness and loses some reference light rather than creating energy.

Opening the pinhole increases accepted in-focus and out-of-focus photons; closing it strengthens confocal optical sectioning while discarding some useful in-focus photons too. In the synthetic model, $\alpha =0.5,1.5,4$ transmits approximately $0.1175,0.6753,0.9997$ of the collected in-focus emission. At $t=2$ the corresponding peak-normalized signals are about $0.0420,0.0597,0.1597$. These are calculated fractions under stated assumptions, not instrument measurements.

An Airy unit conventionally specifies a pinhole diameter equal to the first-zero diameter of the emission Airy disk in that plane. For an ideal circular pupil,where $M$ is the total specimen-to-pinhole magnification, ${\lambda }_{\mathrm{em}}$ the emission wavelength in vacuum, and $\mathrm{NA}$ the objective numerical aperture. A software setting needs its specified wavelength and relay magnification; it is not a universal micrometre diameter. ^[ref]

For ideal independent counting, accepted source and background means $S$ and $B$ give photon shot noise with variance $S+B$ and signal-to-noise ratio in photon counting $S/\sqrt{S+B}$. Closing a pinhole need not improve that ratio: it depends on how both $S$ and $B$ change. Increasing exposure or excitation to replace discarded photons can cause bleaching, an irreversible loss of fluorescent emission capacity, or damage. Detection rejection occurs after excitation and therefore does not undo excitation outside the selected depth. Such material can absorb illumination and emit photons even when their emission is omitted from the recorded image.

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.

Ernst Ruska and Max Knoll developed electron optics while improving the writing spot of a cathode-ray oscilloscope.^[ref] A structured object tests image formation more strongly than a single bright spot: each feature must retain its own image position.

On 7 April 1931, a cold cathode source and two short magnetic coils imaged aperture grids at $50\mathrm{kV}$. A platinum grid preceded the first coil; a bronze grid preceded the second. The fluorescent screen showed one-stage grid images and the first grid reimaged through both stages. Comparing these patterns tested successive image formation, not merely beam concentration. Heating could destroy even metal grids. This prototype did not establish resolution beyond an optical microscope.^[ref]

In 1933, Ruska measured $3\mathrm{mm}$ focal lengths at $75\mathrm{kV}$ with a pole-piece lens, then built an instrument giving $12000$-fold magnification.^[ref] TU Berlin dates images surpassing conventional optical microscope resolution to 1933.^[ref]

A fluorescent screen converts the spatial electron distribution into visible light.^[ref] A collector reporting only total electric current would discard that spatial pattern. In a two-stage image relay, the first lens makes an intermediate real image; the second treats that image as its object. The diagram below illustrates this geometry with invented distances and an ideal beam.

An electron microscope forms a spatially enlarged representation of a specimen using electrons, electron lenses, and a detector. Its appeal is access to structure too fine for an ordinary optical microscope, but a short de Broglie wavelength alone does not guarantee fine electron microscope resolution. Increasing magnification can enlarge an existing blur without revealing any additional structure.

In transmission electron microscopy, an electron gun illuminates a thin specimen; magnetic electron lenses form and enlarge its transmitted image on a fluorescent screen or detector. A vacuum reduces collisions with gas. Lens electric currents control focusing and magnification.^[ref]

Electron microscope resolution concerns distinguishable specimen detail under a specified contrast criterion. It depends on diffraction, lens defects, electron energy spread, specimen motion and detection. A smaller de Broglie wavelength permits finer detail in principle, but neither wavelength nor magnification is the instrument's measured resolution.

For an electron accelerated from negligible initial kinetic energy through a positive electric potential difference $U$, combine $K=eU$, the relativistic energy-momentum relation and the de Broglie wavelength:Here $h$ is the Planck constant, $m_e$ the electron mass, $e$ the elementary charge, and $c$ the speed of light. This includes the relativistic correction omitted by the low-voltage electron accelerating-voltage wavelength approximation. With $m_e\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ and SI defining values of $h,e,c$, calculated wavelengths are approximately $6.979$, $3.701$ and $1.969\mathrm{pm}$ at $30$, $100$ and $300\mathrm{kV}$ respectively. The rounded mass agrees with the recommended value.^[ref]

For a circular aperture admitting a small semi-angle $\alpha$, the Rayleigh criterion suggests a diffraction-limited resolution scale $d_D=0.61\lambda /\alpha$. At $100\mathrm{kV}$ and $\alpha =10\mathrm{mrad}$, this gives $0.226\mathrm{nm}$, already much larger than the $0.00370\mathrm{nm}$ wavelength. This is an ideal aperture estimate. Transmission electron microscopy with coherent illumination and a particular specimen requires its own image-contrast analysis; this scale alone is not a universal two-feature resolution test.

An original illustrative blur model adds competing effects:Define $C_s{\alpha }^3$ as a characteristic transverse blur from the electron-lens analogue of spherical aberration, and $C_c\alpha \Delta K/K$ as one from energy-dependent focusing, the analogue of chromatic aberration. The coefficients $C_s,C_c$ have units of length, $\Delta K$ is a chosen characteristic kinetic energy spread, and $b_0$ is a fixed residual blur. Adding their squares is a declared heuristic, not a measured point spread function or a rigorous resolution theorem. The model omits contrast transfer, specimen scattering and signal statistics, so its minimum is only an estimate within its assumptions.

Choose $C_s=C_c=1\mathrm{mm}$, $\Delta K=0.7\mathrm{eV}$ and $b_0=0.10\mathrm{nm}$ at $K=100\mathrm{keV}$. At $5\mathrm{mrad}$, the four contributions are about $0.452$, $0.125$, $0.035$ and $0.100\mathrm{nm}$, giving $b\simeq 0.480\mathrm{nm}$. At $10\mathrm{mrad}$, the diffraction contribution falls to $0.226\mathrm{nm}$ but the assumed spherical aberration contribution grows to $1.00\mathrm{nm}$. Thus opening the aperture indefinitely worsens this model's image. All these values are calculations with chosen parameters, not observations of any microscope.

A physicist.^[ref]

The early two-stage electron microscope experiment investigates how magnetic electron lenses can form a magnified image. The electron microscope article distinguishes magnification from electron microscope resolution.

A useful motivating experiment moves a bead emitting fluorescence through the selected optical focus and records the detector signal against position. Measurement of point-spread function (PSF) for confocal fluorescence microscopy by InCheon Song and others (2005, §2 and Fig. 1) describes a $543.8\mathrm{nm}$ laser delivered through a single-mode optical fiber, collimation, beam-size adjustment and a dichroic beam splitter. A $100\times$, numerical aperture $1.3$ oil-immersion objective optical lens excites slide-mounted fluorescent beads and collects their emission. Emission passes a filter and collecting optical lens, then a pinhole, before an avalanche photodiode. A three-dimensional positioning stage moves the specimen. The apparatus also contains a phase plate for deliberately modified responses.^[ref]

The published comparison uses nominal bead diameters $0.17,0.5,1\mu \mathrm{m}$: larger beads give broader measured spots. This is a control on the finite emitter, because the image is the convolution of bead emission with the confocal point spread function. The smallest bead has weaker signal relative to noise; incorrect bead size, defocus, bleaching and optical aberrations limit inference from a restored response. The reported bead-size comparison concerns lateral responses; it is not a measured pinhole-size series or a measured sheet-sectioning curve.^[ref]

To test axial pinhole selection with this kind of apparatus, the following is a proposed extension, not an additional result from that paper. Choose an isolated bead small relative to the response, centre it laterally, then step $z$ through optical focus while recording background-subtracted signal at several pinhole settings. Hold excitation power, dwell interval, detector gain, emission band and sampling fixed. Preserve raw peaks before also plotting peak-normalized traces. A lower peak and faster relative decay with axial displacement would support the paired-selection mechanism; fitting only normalized widths would miss the throughput cost.

Measure blocked-excitation background, include a bead-free region, avoid detector saturation, and repeat scans in reversed order to reveal drift or bleaching. Check pinhole centring separately, since misalignment can mimic signal loss. Record bead diameter, immersion medium, refractive index, numerical aperture, wavelengths and axial step. A finite bead is an approximation to an optical point source; a point trace does not certify confocal optical sectioning of a whole fluorescent sheet. Depth-dependent spherical aberration, scattering and sample motion can broaden or distort the response. Increasing magnification or making a display brighter does not correct those effects.

Magnification compares an image's scale with a specified reference. Transverse magnification compares signed lengths in object and image planes; angular magnification compares viewing angles at an observer. Both are dimensionless quantities, but they answer different questions. A projected real image twice as tall has $|m|=2$; an optical telescope's advertised viewing enlargement normally concerns angular magnification. ^[ref].

Always specify the sign and the reference arrangement. For signed transverse magnification, $m<0$ means inverted and $m>0$ means upright; $|m|>1$ means enlarged and $0<|m|<1$ means reduced. A larger virtual image can subtend a modest viewing angle if it is farther away. At $u=f$, an ideal converging thin lens has no finite image height and hence no finite transverse magnification, while angular magnification can remain finite. Enlargement also does not guarantee extra detail: diffraction-limited resolution and optical aberrations limit the information available to enlarge. ^[ref].

In transmission electron microscopy, the electron objective lens forms an intermediate real-space image and intermediate/projector lenses enlarge it onto a screen or camera. The total optical magnification is the product of the relevant stage magnifications. The user changes imaging-lens excitations to select magnification, while separately controlling illumination and focus. Display resizing introduces an additional enlargement, so the recorded pixel calibration matters.

In scanning transmission electron microscopy, a focused probe is rastered across the specimen and a detector signal is assigned to each probe position. The displayed magnification is display width divided by scanned specimen width. The user increases magnification by reducing raster-deflection amplitude, independently of choosing a smaller incident probe with the probe-forming lenses. TEM changes image-forming lens magnification; STEM changes the raster field of view. A smaller field or more pixels does not create finer physical resolution than the probe, specimen scattering and detection permit. A calibrated real-space scale is more transferable than the apparent size on a monitor.^[ref]

For a displayed width $D$ representing a scanned specimen width $d$, the linear image magnification isThe user changes the scan-coil excursion, hence $d$. A smaller raster displayed at the same width gives larger magnification. Changing the display or print size also changes displayed magnification, so the calibrated scale bar is more portable than a magnification label. This is the scan/display principle described in the manufacturer's microscope guide.^[ref]

At the low-magnification extreme, the probe scans the widest accessible field. Coil range, apertures, scan distortion and maintaining focus constrain the actual usable width. Formally increasing $d$ without bound would make $M$ arbitrarily small, but a real column has a finite accessible field.

At the high-magnification extreme, the scan covers a very small region. Formally $d\to 0$ makes $D/d$ diverge, but exactly zero scan width samples one location and supplies no spatial image. Electron microscope resolution, finite probe size, pixel spacing, drift, noise and specimen damage set the useful limit. Enlarging a field already sampled more finely than its resolvable information produces empty magnification, not new structural detail. For $N$ pixels across a field, the specimen sampling interval is $d/N$; this is distinct from the probe's spatial resolution.

A scanning transmission electron microscope makes an image by assigning a transmitted-electron signal to each position of a scanned probe. Its essential chain is electron source, probe-forming optics, scan system, thin specimen, transmitted-electron detectors and synchronized acquisition.

An electron gun and accelerating supply provide a stable beam of the chosen kinetic energy. Condenser lens optics and an aperture control probe current and convergence; the probe-forming objective lens focuses a small spot on the specimen. Alignment, focusing and astigmatism correction keep the probe usable across the field.

Scan coils deflect the probe in a raster. A stable specimen holder positions an electron-transparent region, and a high-vacuum column limits gas scattering and contamination. A bright-field transmission electron microscopy detector collects the central transmitted beam; an annular detector can select annular dark-field electron imaging, including high-angle annular dark-field imaging. Different angular selections give different contrasts, rather than a universal map of height.

Detector electronics, scan timing and a computer assign the collected signal to the corresponding image pixel. Spectrometers are useful additions but not essential to basic image formation. The post-specimen optics can adjust detector collection geometry; they do not set scan-image magnification in the same way as the projection lenses of conventional transmission imaging.

Treat the magnetic electron lens as a paraxial thin lens. A point in the specimen sends rays through different lens positions; the lens brings these to a conjugate point in the image plane, with $1/u+1/v=1/f$ and transverse magnification $-v/u$. Conversely, waves leaving different specimen points at the same angle are focused to the same point in the back focal plane. This plane therefore contains the electron diffraction pattern, with radius $r\simeq f\theta$ for small scattering angle $\theta$. The specimen is just beyond the front focal plane, giving a magnified real image farther downstream.

The sketch assumes a thin specimen, coherent approximately parallel illumination, small angles and an ideal lens without aberrations. Quantitative diffraction intensities may require dynamical scattering; the ray sketch only establishes the plane conjugacies.^[ref]

Adopt positive object and real-image distances $u$ and $v$ measured from a converging thin lens, with the object on the left. A ray through the lens centre is undeviated in the thin-lens approximation. For object height $h$ and signed image height $h^{\prime }$, similar triangles give $h^{\prime }/h=-v/u$.

A second ray leaves the object parallel to the axis, reaches the lens at height $h$ and then passes through the far focal point. Its height at distance $v$ is $h^{\prime }=h(1-v/f)$. Equating the two descriptions givesThe signed transverse magnification is $\boxed{m=h^{\prime }/h=-v/u}$. Since a real image has $v>0$ for $u>0$, $m<0$: it is inverted. Its size relative to the object is $|m|=v/u$.

The derivation assumes the paraxial approximation, a negligible lens thickness and a focal length defined in the surrounding medium. The ray diagrams show actual paths as solid lines and backward extrapolations as dashed lines.

For a real image, $u>f$ and $v=fu/(u-f)>0$. The object-to-image separation isIts excess above $4f$ factors into a nonnegative expression:ThereforeAt this minimum the image is inverted and has unit absolute magnification. The restriction to a real object and a real image matters: the positive separation used here is not the same quantity as a signed distance to a virtual image.

For a lens between an object and a screen, $u+v=D$ and the thin lens relation requires $uv=fD$. Eliminating $v$ gives the quadratic equationFor $D>4f$, both roots lie between $0$ and $D$, and they exchange the object and image distances. Both therefore focus the image on the same screen, with reciprocal absolute magnifications. The distance between lens positions isThe existence qualification is essential: for $D=4f$ the two roots coincide and there is only one distinct position; for $D<4f$ there is none. For example, $D=3f$ makes the discriminant negative, providing a concrete counterexample to an unconditional assertion of two positions. This is the basis of the Bessel lens displacement method, with $f=(D^2-l^2)/(4D)$ when two distinct positions can be measured.

TODO: the source glass-rod diagram is unavailable. The recovered prose gives a surface curvature and object distance, but the diagram's orientation, labels and any further optical surfaces cannot be verified. The following solves the surviving request for a single convex entrance surface; it is conditional on that interpretation and does not reconstruct the missing diagram or supply a rod length.

For refraction at a spherical surface, take incident light travelling rightwards from refractive index $n_0$ to $n_g$, object distance $u>0$ to the left of the vertex, image distance $v$ positive to the right, and signed curvature $R_s$ positive when the centre is inside the rod. In the paraxial approximation, applying Snell's law to a light ray at height $y$ givesAfter cancelling $y$, the image position and transverse magnification areThe magnification follows by considering a paraxial light ray through the vertex: Snell's law there gives $n_0(-y_o/u)=n_g(y_i/v)$.

For a convex entrance, $R_s=+20\mathrm{mm}$ and $u=60\mathrm{mm}$. In air, the denominator $(n_g-n_0)/R_s-n_0/u$ is $1/75\mathrm{m}{\mathrm{m}}^{-1}$, givingThis is a real, inverted image within a sufficiently long glass rod, before any subsequent surface.

For water with refractive index $1.33$, the denominator is $-13/1500\mathrm{m}{\mathrm{m}}^{-1}$, andThe emerging light rays diverge as though from an upright virtual image on the object side. The change of sign is consistent with $v$ diverging when $n_0=n_{g}u/(u+R_s)=1.20$. A reversed curvature or a second exit surface changes these answers; the unavailable diagram prevents a definitive answer for the full illustrated arrangement.

Use positive real object and image distances $u,v$ for a thin lens. The required absolute magnification is $M=1000/15=200/3$, and the specified LCD-to-screen separation gives $u+v=2000\mathrm{mm}$, not $v=2000\mathrm{mm}$. Since $v/u=M$,The thin lens equation yieldsPlace the projection lens $29.56\mathrm{mm}$ beyond the LCD towards the screen. Its real image is inverted; the signed magnification is $-200/3$.

Use the thin lens equation with positive object and real-image distances $u,v$:The transverse magnification is $M=-v/u=-f/(u-f)=-4/296$. For an object dimension $500\mathrm{mm}$, the corresponding image dimension isThe minus sign of the magnification means that the image is inverted. This is the ideal thin-lens prediction; it assumes the relevant field is admitted by the camera and ignores lens distortion.