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An optical microscope whose microscope objective is designed to form a real image at a specified finite distance. The intended image distance and optical aberration corrections belong together; mechanical tube length is a mounting specification, not automatically the image distance measured from a principal plane.^[ref]

Why can a small displacement produce a large image? For a converging thin lens, write $u_o=f_o+ϵ$ with $epsilon>0$. Substitution into the thin lens equation givesThe similar triangles construction supplies the minus sign. The example has $epsilon=2\mathrm{mm}$ and hence $m_o=-5$. Letting $epsilon$ decrease moves the real image farther away; it does not keep the microscope eyepiece in focus at a fixed position. Large transverse magnification therefore requires the correct conjugate geometry, not merely a short focal length.

The focal length $f$ specifies the signed distance from an ideal thin lens to its optical focus for incident light rays parallel to the optical axis. A converging thin lens has $f>0$, with a rear optical focus beyond the thin lens; a diverging thin lens has $f<0$, with a virtual optical focus on the incoming side. The focal length is a property of the optical lens in its specified surroundings, not the image distance for every object. For example, a converging thin lens with $f=1\mathrm{m}$ images an object $3\mathrm{m}$ away at $1.5\mathrm{m}$, as computed below. ^[ref].

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.

At fixed incidence angle, differentiating the grating equation gives $d\beta /d\lambda =m/(d\cos \beta )$. A camera with focal length $f$ therefore has local focal-plane dispersion $q=fm/(d\cos \beta )$ near its optical axis. With coordinate $X=f\tan (\beta -{\beta }_0)$, the full derivative is $q=f{\sec }^2(\beta -{\beta }_0)m/(d\cos \beta )$.

Image formation assigns each object point a corresponding image point: outgoing light rays from that object point either converge there, giving a real image, or diverge as though they originated there, giving a virtual image. The requirement applies separately to each point. Bringing unrelated light rays together makes a bright patch, but does not preserve the object's pattern. This point-by-point correspondence explains how an optical camera records a scene. ^[ref].

Use the same convention as thin lens: place the optical lens at $x=0$, with incident light travelling to the right and a real object at $x=-u$, $u>0$. Write the image coordinate as $x=v$, so $v>0$ for a real image and $v<0$ for a virtual image. Heights $h_o,h_i$ are positive above the optical axis and negative below it. In air, within the paraxial approximation, a thin lens obeysHere $f$ is the signed focal length, positive for a converging optical lens and negative for a diverging optical lens; the expression for $v$ assumes $u\ne f$. Object distance $u$ is positive even though the object's coordinate is negative. These coordinate rules describe transmission through the optical lens; a mirror reverses the outgoing direction, so do not identify its image distance with this same rightward coordinate. ^[ref].

The beauty of geometrical optics is that two convenient light rays can locate a point, while a whole admitted bundle forms its image. For a converging thin lens, an incident light ray parallel to the optical axis exits through the rear optical focus; a light ray through the centre continues undeviated. Real optical lens elements do not focus every admitted light ray exactly: optical aberrations and diffraction limit the point correspondence. The diagrams below are original ideal-model constructions, not measured data. ^[ref].

An optical microscope whose microscope objective, with the specimen at its designed front focal plane, sends out a parallel bundle of light rays from each specimen point. Different specimen points generally give different bundle directions. A tube lens then forms the finite intermediate microscope image; the microscope eyepiece views it. This parallel region permits suitable optical accessories to be inserted. Finite-conjugate and infinity-corrected components must be matched to their intended system.^[ref]

In the ideal paraxial approximation, an object height $h_o$ at objective focal length $f_o$ gives bundle slope $-h_o/f_o$. A tube lens of focal length $f_t$ maps that slope to $h_i=-f_t{h}_o/f_o$, soThis is a two-assembly result. Substituting infinity into the single-objective transverse magnification $-v_o/u_o$ would not describe the finite recorded image. Infinity correction concerns the region after the objective; relaxed-eye viewing concerns the region after the eyepiece. They are distinct conditions.

With fixed spectral resolving power, wavelength interval, diffraction grating geometry, photodetector sampling and collimator focal ratio, a beam diameter scales as the spaxel width $\theta$, but the optical camera focal length stays fixed. A simple on-axis beam-envelope model therefore gives collimator and disperser volumes proportional to ${\theta }^3$ and optical camera volume proportional to ${\theta }^2$. With a fixed number of spaxels per photodetector, replication multiplies these volumes by $M\propto {\theta }^{-2}$. A fixed photodetector field adds transverse clearances and changes the small-$\theta$ limit; these exponents are not universal mechanical-volume laws.

The lensmaker's equation connects focal length to the refractive index and shape of a thin lens. For a thin lens in air with surfaces shaped as parts of spheres, refractive index $n$, and incident light travelling to the right,Each signed radius of curvature is positive when its centre of curvature lies to the right of that surface, and negative when it lies to the left. A symmetric biconvex thin lens with $n=1.5$, $R_a=1\mathrm{m}$, and $R_b=-1\mathrm{m}$ therefore has $1/f=0.5(1+1){\mathrm{m}}^{-1}=1{\mathrm{m}}^{-1}$, giving $f=1\mathrm{m}$ and optical power $+1\mathrm{D}$. This is an ideal shape example consistent with the thin lens diagrams, not a specification of their finite thickness.

In a uniform surrounding medium of refractive index $n_s$ on both sides, replace $n-1$ by $n/n_s-1$. Thus the same shape reverses its focusing sign if its refractive index is lower than its surroundings: for $n=1.5$ and $n_s=2$, the same radii give $1/f=-0.5{\mathrm{m}}^{-1}$, or $f=-2\mathrm{m}$. Positive focal length therefore cannot be inferred from convex shape in every environment. A proof outline applies Snell's law with the small-angle approximation at each surface, adds their direction changes, and neglects the distance between them. ^[ref].

An optical lens is a transparent element, or assembly of elements, that redirects a bundle of light rays to form images or change how the bundle spreads. In an ordinary glass optical lens, refraction at the surfaces provides the redirection: their changing normal vectors make Snell's law bend different light rays by different amounts. A converging optical lens brings incoming light rays parallel to its optical axis toward an optical focus; a diverging optical lens makes them spread as if from an optical focus on the incoming side. ^[ref].

For a usual glass thin lens in air, a form thicker at the centre than at the edges normally converges, with positive focal length; one thinner at the centre normally diverges. Shape alone is insufficient: the refractive index relative to the surroundings also matters, as the lensmaker's equation shows. An aperture selects which light rays pass; an optical lens also changes their directions. The distinction explains how an optical camera can gather light from one object point over many incoming directions and deliver it to one image location.

The beautiful simplification is that an entire bundle can be characterized by a focal length, and two constructed light rays can locate an image. The thin lens examples below turn that geometry into a prediction of image position, orientation and transverse magnification. A real optical lens retains optical aberrations and diffraction, so the construction is a useful leading model rather than an exact solution of Maxwell's equations.^[ref]

Early mathematical treatments include Johannes Kepler's 1611 Dioptrice, whose title explicitly concerns vision through newly invented instruments using transparent glass, and René Descartes's 1637 La Dioptrique, whose seventh discourse discusses improving vision with shaped glass. These are landmarks in explaining and designing optical instruments, rather than dates for the invention of the optical lens. ^[ref]; ^[ref].

A magnifying glass uses a converging optical lens to present an upright enlarged virtual image. For focal length $f$ and eye near point distance $D$, the near-eye angular magnification is $D/f$ with a relaxed eye, or $1+D/f$ with the virtual image at the near point. These formulas assume negligible eyepiece-to-eye distance. With eye-to-lens distance $e$, object distance $u$, and the virtual image at the eye's near point, the angular magnification is $(D-e)/u$.

For a converging thin lens forming a real image of a real object, let $u,v$ be positive distances from the lens and $f$ the focal length. Set $u^{\prime }=u-f$ and $v^{\prime }=v-f$, measuring from the neighboring focal points. Substituting into $1/u+1/v=1/f$ gives $(f+u^{\prime })(f+v^{\prime })=f(2f+u^{\prime }+v^{\prime })$, so $u^{\prime }{v}^{\prime }=f^2$. Other object-image geometries require an explicitly consistent signed-distance convention.

An objective aperture selects outgoing electron directions in the objective lens's back focal plane. Passing the central transmitted beam gives bright-field transmission electron microscopy; selecting a separated diffracted beam gives dark-field transmission electron microscopy.^[ref]

In a paraxial approximation, an outgoing angle $\theta$ maps approximately to position $r=f\theta$ in this focal plane, where $f$ is the objective focal length. A centred opening of radius $a$ therefore admits $|\theta |\lesssim a/f$. Selecting a direction selects a component of the outgoing wavefunction; the subsequent image still assigns intensity to specimen positions. The opening is not a hole cut into a real-space specimen image. Its angular selection must be distinguished from selecting a specimen region for electron diffraction.

The institution-issued AY 105 Lab Experiment 3: Optical aberrations by California Institute of Technology (Spring 2011 offering, Part A, pp. 2–3, 7–8) specifies a concrete optical bench: a lamp, opal diffuser and condensing optical lens illuminate a $25$ or $30\mu \mathrm{m}$ optical pinhole; a photographic telephoto optical lens acts as a collimator, producing an approximately $75\mathrm{mm}$ beam. A test optical lens sits on a rotation stage, followed by an adjustable optical microscope. The procedure establishes the on-axis image, then rotates the test optical lens a few degrees and examines the off-axis image through optical focus. It compares this readout with through-focus ray tracing at the corresponding field angles and offers a Ronchi ruling to calibrate the measuring eyepiece.

Part B (pp. 9–10) substitutes a spherical mirror of $1\mathrm{m}$ radius of curvature, with the optical microscope beside the incoming beam so the image is necessarily off-axis. It calls for measuring the field angle, calibrating the eyepiece and recording the image's length and width at minimum size. These are specified apparatus dimensions and instructions, not supplied measurement results or dimensions of the normalized figures.

A controlled extension would record images at zero and equal positive and negative field angles, repeat zero afterwards, and scan through optical focus at every setting. Keep the illuminated optical pupil and illumination fixed, recenter the optical microscope without clipping the bundle, and use its calibrated scale to separate a mere image displacement from a change in blur shape. The pure Seidel coma pupil mapping predicts reversal and linear growth; a stationary asymmetric central image instead calls for checking alignment. A focus-dependent change between perpendicular elongations suggests an astigmatism contribution. These comparisons test the model rather than assuming every off-axis flare is pure coma.

The optical pinhole has finite angular size, approximately its diameter divided by the collimator's focal length; the collimator and observing optical microscope also contribute their own optical aberrations. A broadband lamp can mix chromatic aberration with the geometric blur, and diffraction limits the smallest visible structure. A narrow spectral band, repeated field settings and a reference image help separate these effects. Consequently a measured length and width are useful checks on off-axis image formation, but they cannot by themselves determine a unique coma coefficient or recover the full point spread function.

An optical microscope uses light and optical lens assemblies to reveal small structures, such as the boundaries of biological cells or defects in a manufactured surface. In a compound optical microscope, the microscope objective supplies an enlarged real image; the microscope eyepiece makes that intermediate microscope image subtend a larger viewing angle. The transverse magnification of the first stage and the angular magnification of the second answer different questions.^[ref]

Consider an original ideal finite-conjugate microscope in air, with both optical lens assemblies treated as converging thin lens elements in the paraxial approximation. Choose objective focal length $f_o=10\mathrm{mm}$ and object distance $u_o=12\mathrm{mm}$. The thin lens equation and signed transverse magnification givePlace a microscope eyepiece of focal length $f_e=25\mathrm{mm}$ a further $25\mathrm{mm}$ beyond the intermediate microscope image. The objective-to-eyepiece separation is therefore $85\mathrm{mm}$. Each object point then supplies a parallel outgoing bundle of light rays, appropriate to a relaxed eye focused at infinity. Define the unaided reference as the same object viewed at distance $D=250\mathrm{mm}$; this is a rating convention, not a universal property of eyesight. The eyepiece angular magnification and total signed angular magnification areFor instance, a hypothetical object height $h_o=0.020\mathrm{mm}$ becomes $h_i=-0.100\mathrm{mm}$ at the intermediate microscope image. The signed viewing angles are approximately $h_i/f_e=-0.004\mathrm{rad}$ with the instrument and $h_o/D=0.00008\mathrm{rad}$ unaided. Their ratio is $-50$: enlarged and inverted. These are stipulated model dimensions, not measurements or specifications of a modern commercial optical microscope.

The attraction of an optical microscope is that seeing can become measuring: a stage micrometer supplies a known object-space length, and microscope scale calibration converts an image readout into a specimen length. This also separates a large-looking image from a trustworthy size estimate. Meanwhile, diffraction-limited resolution depends on numerical aperture and wavelength; empty magnification enlarges an existing blur without revealing additional structure.

The optical power is reciprocal signed focal length, $\Phi =1/f$. With $f$ expressed in metres, optical power is expressed in dioptres: $f=1\mathrm{m}$ gives $\Phi =+1\mathrm{D}$, while $f=-0.5\mathrm{m}$ gives $\Phi =-2\mathrm{D}$. The sign distinguishes converging and diverging optical lens elements; the magnitude measures how strongly they change the direction of a light ray at a given height. ^[ref].

For two ideal thin lens elements in contact in air, optical powers add: ${\Phi }_{\mathrm{total}}={\Phi }_1+{\Phi }_2$. Thus $+2\mathrm{D}$ and $-1\mathrm{D}$ combine to give $+1\mathrm{D}$, or $f_{\mathrm{total}}=1\mathrm{m}$. Separation changes the rule: for two ideal thin lens elements separated by $d$ in air, their equivalent optical power is ${\Phi }_1+{\Phi }_2-d{\Phi }_1{\Phi }_2$. The contact rule is the $d=0$ case.^[ref]

Use the thin lens sign convention in which a real object on the incident side has positive object distance, a real image on the outgoing side has positive image distance, and a converging lens has positive focal length. The first image distance isFor the image distance for two separated thin lenses, the second lens has object distance $u_2=L-v_1$, which can be negative when its intermediate object is virtual. Applying the thin lens equation again givesThe image is $d_2$ beyond the second lens when $d_2>0$; $d_2<0$ denotes a virtual image on its incident side. A vanishing denominator with nonzero numerator denotes an image at infinity. The rational expression also supplies appropriate limits when the first image is at infinity; for example, $d_1=f_1$ gives $d_2=f_2$ for finite nonzero focal lengths.

As $L\to 0$, the effective focal length is $F=f_1{f}_2/(f_1+f_2)$ andFor a real incident object, $d_1>0$, and positive $F$, the real-image condition is exactly $d_1>F$, yielding the requested inequality. This includes two converging lenses. The positivity assumption matters: with $f_1=f_2=-1\mathrm{m}$ and $d_1=1\mathrm{m}$, one has $F=-0.5\mathrm{m}$ and $d_1>F$, but $d_2=-1/3\mathrm{m}$ is virtual. If $f_1+f_2=0$, the touching pair has zero net optical power and produces no real image of this real object in the ideal thin lens model.

Place the input plane one focal length before a single positive lens and the output plane one focal length after it. The lens maps each incident transverse spatial frequency to its corresponding point in the back focal plane. Referencing the input to the front focal plane removes the otherwise present output phase curvature, giving an exact scaled Fourier transform within the same paraxial model.

A short ray-matrix check makes the geometry explicit. With free propagation $P(f)$ and lens matrix $L(f)$,The vanishing diagonal entries are the Fourier-plane condition; the output is proportional to ${\tilde{E}}_{\mathrm{in}}(kx/f)$. The final panel of the original diagram in part a shows the two focal distances. A lens with input immediately on its surface and output at the back focal plane would generally leave a quadratic output phase; the front-focal-plane choice avoids it.

Use the unfolded propagation convention after reflection. Free-space propagation and a concave spherical mirror have matricesThe mirror is equivalent in this paraxial convention to a thin lens of focal length $R/2$. The excursion from the reference plane to mirror 1 and back isLikewise $S_2=T(L-L_1)F(R_2)T(L-L_1)$. Multiplying $S_2{S}_1$, or using cyclic invariance of the trace to move the reference plane, givesConsequently the strict criterion of a isThe trace is independent of $L_1$, as a change of reference plane only conjugates the round-trip matrix. For the concrete confocal counterexample $R_1=R_2=L$, $L_1=L/2$, each excursion is $\left(\begin{array}{cc}0& L/2\\ -2/L& 0\end{array}\right)$ and its square is $-I$. This gives $g_1{g}_2=0$ and the degenerate boundary exception already explained.

A thin lens has ray transfer matrixThe complex beam parameter transformation gives $q_2=q_1/(1-q_1/f)$, henceThe sign is also obtained by multiplying the incident wave amplitude by the thin lens phase $\exp [-ik{\rho }^2/(2f)]$.

Choose the thin lens plane as $z=0$ and put $a=\pi w_0^2/\lambda$, the incident Rayleigh range. Since the incident beam waist lies there, $q_1=-ia$. Immediately after the thin lens,Subsequent free propagation adds $z$, so write $q(z)=z-z_f-i{z}_R^{\prime }$. Comparing real and imaginary parts givesFor a positive focal length, the new beam waist lies downstream but closer than the geometrical focus. At fixed $f,\lambda$, as $w_0\to \infty$,This is the formal unbounded-aperture scalar limit; eventually finite aperture and the paraxial approximation limit the extrapolation.

Place the optical lens at $z=0$. For a sufficiently broad incident Gaussian beam, the beam waist positions approach the focal lengths frequency by frequency:Add the group delay at the optical lens to propagation through vacuum:Here an arbitrary common time origin has been omitted. Differentiating both equations with respect to angular frequency yieldsFor $\beta >0$, define $u=2\eta c/\beta$. The velocity graph in the preceding figure has horizontal asymptote zero and a vertical asymptote at $u=-1$. For $\eta >0$, $0<v_f<c$: the higher-frequency components both enter later and travel farther. At $\eta =0$, $v_f=c$. For $-\beta /(2c)<\eta <0$, $v_f>c$: the farther-focusing components leave earlier, partly compensating their extra travel. At $\eta =-\beta /(2c)$ every component focuses at the same time, so the trajectory is not a finite-velocity function of time. For $\eta <-\beta /(2c)$ the farther positions focus first, giving $v_f<0$; it tends to zero from below as $\eta \to -\infty$.

A flying focus is a moving pattern assembled from different portions of an optical pulse. Each portion propagates causally at $c$ in this model. The superluminal or backward motion of the focus therefore does not describe a material object or information overtaking light. The calculation assumes negligible additional frequency-dependent group delay from the optical lens; any such delay must be included in $\psi$.