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Elongation is the apparent angular separation of a celestial body from the Sun, measured at the observer. For a lunar phase, the elongation is the Sun–Earth–Moon angle; in the distant-Sun approximation it is supplementary to the phase angle. A given unsigned elongation can occur on either side of the Sun in the sky.

Empty magnification enlarges an image without revealing additional specimen detail. Adequate angular magnification can make existing detail visible to the eye, but it cannot undo the objective's diffraction blur. For equally bright, mutually incoherent optical point sources and an ideal circular optical pupil, a commonly used lateral Rayleigh criterion in object space iswhere $\lambda$ is the vacuum wavelength, $n$ is the specimen-medium refractive index, and $\alpha$ is the admitted half-angle. For this model, $\lambda =550\mathrm{nm}$ and $\mathrm{NA}=0.25$ give $d_{\mathrm{R}}\simeq 1.342\mu \mathrm{m}$; changing only numerical aperture to $0.50$ gives $0.671\mu \mathrm{m}$. Changing only the eyepiece leaves this object-space scale unchanged. Illumination, optical coherence, optical aberrations and specimen visibility matter in real instruments; this criterion is not a universal threshold.^[ref]

The geometric reason is especially simple. If a fixed recorded image is enlarged by a factor $q>0$, a feature separation $s$ and a blur width $b$ become $qs$ and $qb$. Their ratio remainsThis is the lower-panel construction in the diagram. A change to the microscope objective can change numerical aperture and hence the specimen detail captured; a display zoom changes how large that captured detail looks. Microscope scale calibration, diffraction-limited resolution and angular magnification are three complementary properties to report.

For the normalized damped pendulum, the sum of kinetic energy and gravitational potential energy, divided by $mg\ell$, isThe chain rule and the full nonlinear equation give this identity exactly, without a small-angle approximation. Here $\dot{\theta }=d\theta /d\tau$ as in damped pendulum. In physical units the mechanical energy and the power of the resisting torque areThe pendulum's lost mechanical energy is transferred to the damper and surroundings; it does not violate conservation of energy. In an ideal viscous dashpot this loss is frictional heating. The exact cumulative balance is particularly useful for checking a numerical solution:

The figure uses $c=0.20$ and release from rest. Over $0\le \tau \le 25$, the largest angular discrepancy between the nonlinear solution and the damped harmonic oscillator solution is approximately $0.000702\mathrm{rad}$ ($0.0402^{\circ }$) for a $10^{\circ }$ release, but $0.14995\mathrm{rad}$ ($8.592^{\circ }$) for a $60^{\circ }$ release. Their nonlinear initial energies are $1-\cos 10^{\circ }\approx 0.015192$ and $1-\cos 60^{\circ }=0.5$, respectively. At $\tau =25$ the remaining fractions are approximately $0.0064080$ and $0.0063757$. These are illustrative calculations from specified initial conditions, not observations. The generator uses a Runge-Kutta method and integrates $Q$ alongside the motion; halving its step and independently integrating $c{\dot{\theta }}^2$ checks the result.

For $\ell =1\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the first three nonlinear same-side swing intervals after a $60^{\circ }$ release are approximately $2.1008$, $2.0378$, and $2.0221\mathrm{s}$. They approach the linear-model oscillation period $2.0162\mathrm{s}$ as the amplitude shrinks. After a $10^{\circ }$ release the corresponding intervals are already about $2.0185$, $2.0168$, and $2.0164\mathrm{s}$. This makes the experiment's period drift a quantitative test of the small-angle approximation, rather than an automatic reason to change the damping coefficient.

At a turning point, $\dot{\theta }=0$ and $\dot{E}=0$ even though the swing is losing mechanical energy over the cycle. Its potential energy there is $E_n=1-\cos A_n$, so measured turning angles give successive energies without estimating angular velocity. In the proposed decay experiment, multiply these values by $mg\ell$ to obtain energies in joules, then subtract adjacent same-side energies to estimate the loss over each full swing. Only for small amplitudes is $E_n\simeq A_n^2/2$. In the linear underdamped model, same-side peak energies consequently satisfy $E_{n+1}/E_n=e^{-\gamma T_d}$, the square of the angular amplitude ratio. For the illustrative parameters above this energy ratio is approximately $0.2828$. Between turning points, kinetic energy and potential energy continue to exchange while their sum decreases; the energy slope vanishes briefly at each extremum and is negative whenever the bob moves.

It is therefore inaccurate to replace the exact identity by $\dot{E}=-cE$. Even for the linear model, $E_{\mathrm{lin}}=({\dot{\theta }}^2+{\theta }^2)/2$ has ${\dot{E}}_{\mathrm{lin}}=-c{\dot{\theta }}^2$. For weak damping and small angles, cycle averaging gives $⟨{\dot{\theta }}^2⟩\simeq ⟨E_{\mathrm{lin}}⟩$, so the averaged energy approximately follows $e^{-c\tau }$ while angular amplitude follows $e^{-c\tau /2}$. This exponential decay is an approximation to the energy trace; the same-side peak ratio is exact within the linear model. The figure distinguishes the actual energy curves from this averaging approximation.

The energy identity also gives a global conclusion without solving the nonlinear equation. If $E$ were constant on a positive-length interval, then $\dot{\theta }$ would vanish throughout it; the equation would force $\sin \theta =0$, making the entire trajectory an equilibrium. Every nonstationary trajectory therefore loses mechanical energy over every positive-length interval, even though its instantaneous derivative can vanish at a turning point. A nonconstant periodic motion would have to return to its initial mechanical energy, so it is impossible for this undriven damped pendulum. Oscillations near a stable focus are successive, shrinking swings, not a repeating cycle. Moreover, if $E(0)<2$, the angle cannot reach an upright equilibrium, whose potential energy is $2$; such motion stays within its original potential well. This is the usefulness of the energy description: one scalar balance controls both the observed decay and which nonlinear motions remain accessible.

For a parent moving at $\beta c$ and decaying into two photons of equal laboratory energies, momentum conservation gives photon directions symmetric about the parent direction with $\cos \theta =\beta$. The opening angle is $2\arccos \beta$. In the parent rest frame the photons are opposite and perpendicular to the boost direction.

Take an illustrative elliptic Kepler orbit with semi-major axis $a=1$, orbital eccentricity $e=0.6$ and gravitational parameter $\mu =1$, in consistent dimensionless units. Its semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, mean motion is $n=\sqrt{\mu /a^3}=1$, and specific angular momentum magnitude is $h=\sqrt{\mu a(1-e^2)}=0.8$. About the attracting focus,The two sectors below each span $\Delta t=0.5$, giving $\Delta A=0.2$. Near periapsis the short radius sweeps a wide angle; near apoapsis the long radius sweeps a narrow angle. Constant areal velocity therefore does not require constant angular velocity or constant speed.

To compute the sectors without mistaking equal angles for equal times, parametrize the ellipse by eccentric anomaly $E$:This follows by integrating $\frac{1}{2}(xdy-ydx)$. Kepler's equation gives $E-e\sin E=M=n(t-t_{\mathrm{p}})$, with mean anomaly $M$ measured from periapsis time $t_{\mathrm{p}}$. Consequently $\Delta A=abn\Delta t/2=h\Delta t/2$. The displayed sectors use mean anomaly intervals $[-0.25,0.25]$ and $[\pi -0.25,\pi +0.25]$, rather than equal eccentric anomaly increments.

An equatorial coordinate system labels a viewing direction on the celestial sphere by right ascension $\alpha$, increasing eastward around the reference celestial equator, and declination $\delta$, increasing northward from that celestial equator. These two angles locate a direction, without specifying the object's Euclidean distance. In the traditional convention, zero right ascension points toward the reference equinox, the northward crossing of the Sun's annual path through the celestial equator. An hour circle through the direction supplies the north-south arc on which declination is measured. ^[ref].

For a concrete construction, choose Cartesian coordinates with $x$ pointing toward the reference equinox, $z$ toward the north reference celestial pole, and positive $y$ given by the cross product of the positive $z$ and $x$ directions. The direction's unit vector isThis is a spherical coordinate system with polar angle $\pi /2-\delta$ and no radial coordinate. The trigonometric functions require consistent angular units: convert right ascension in hours to radians before evaluating them. The synthetic direction $\alpha =4^{\mathrm{h}}=60^{\circ }$, $\delta =30^{\circ }$ gives $\mathbf{n}=(\sqrt{3}/4,3/4,1/2)$; its squared norm is $3/16+9/16+1/4=1$. A change of reference axes applies a rotation to this unit vector before recovering the two angles. Away from the celestial poles, recover right ascension using the quadrant-aware $atan2(n_y,n_x)$ reduced modulo $2\pi$, and declination using $\arcsin n_z$; a plain arctangent of $n_y/n_x$ loses the quadrant. Adding a common offset to every right ascension describes only the special case of a rotation about the same polar axis.

The reference specification matters. For an equatorial coordinate system tied to the Earth, precession changes the reference celestial equator and reference equinox, and nutation distinguishes their mean and true versions. The date specifying those axes is distinct from the astronomical epoch at which a catalogue's source directions apply: advancing a star's proper motion and changing reference axes are separate operations. A modern catalogue may instead use fixed reference frame axes, aligned approximately with the traditional J2000.0 axes, without a moving reference equinox. Thus a pair of numbers labelled merely "J2000" does not settle every convention. State the reference frame, the astronomical epoch, and whether a direction is catalogued or apparent. ^[ref].

During one night's elementary fixed-direction approximation, a star's right ascension and declination remain constant while its altitude, azimuth and hour angle change because the Earth turns. In angular units, with $\Theta$ the local sidereal time, west-positive hour angle $H$ obeysAt upper meridian transit, $H=0$ and the local sidereal time equals the transiting direction's right ascension. Pair apparent sidereal time with apparent right ascension referred to the true celestial equator and reference equinox of date; mean sidereal time refers to the mean reference equinox. Raw fixed-frame catalogue coordinates require conversion before precision pointing. ^[ref].

This separation makes the equatorial coordinate system useful: a catalogue can identify a target independently of the observer's current astronomical horizon, while local sidereal time tells the observer which hour circle is on the celestial meridian. Precision observations also account for proper motion, stellar parallax, aberration of light and atmospheric refraction; the fixed-direction construction above omits those effects. A reference-axis rotation preserves angles between directions, whereas changing the observer or correcting atmospheric refraction is a different operation. ^[ref].

Etendue measures how much position-and-direction space a bundle of light rays occupies. A small spot accepting a wide range of directions can have the same etendue as a large spot accepting a narrow range. This makes etendue useful for deciding whether an optical instrument can deliver an extended source's entire admitted radiant flux to a small receiver. For forward crossings of a surface, the geometrical optics definition iswhere $\mathcal{B}$ is the occupied set of surface positions and directions, $n$ is the local refractive index, $dA$ is actual surface area, $d\Omega$ is solid angle, and $\theta$ is the angle to the forward unit normal. The cosine supplies the projected area. Opposite crossing directions must be counted separately with positive projected area, rather than cancelling them in a signed integral. ^[ref]

The shorthand $G\simeq n^{2}A\Omega$ assumes constant refractive index, a narrow nearly normal bundle, and the same admitted directions across the whole area $A$. At larger angles, use the cosine-weighted integral; a position-dependent angular acceptance cannot generally be replaced by the product of two independent extents.

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.

Euclidean geometry describes flat Euclidean space using lengths, angles, and rigid motions. A coordinate realization uses points $x,y\in {\mathbb{R}}^n$, Euclidean distanceand angles determined by the dot product. The coordinate axes are a convenience: translations and rotations change coordinates while preserving these measurements. This makes Euclidean geometry useful for comparing figures and converting measurements along different directions into a single distance.

In the Euclidean plane, through a point outside a line there is exactly one parallel line. This is a standard equivalent formulation of the parallel postulate. Euclid's fifth postulate in the Elements instead describes when two lines meet using the interior angles made by a crossing line.^[ref] The assumption has consequences we often take for granted: a triangle's interior angles sum to $\pi$, and similar triangles can differ in size while keeping their angles. Comparing with spherical geometry reveals that these are properties of a particular geometry.

As a computed example, the diagonal from $(0,0)$ to $(4,3)$ in a rectangle has length $\sqrt{4^2+3^2}=5$. The Pythagorean theorem explains why perpendicular contributions add as squares. If the directions cease to meet at a right angle, the law of cosines supplies the missing angle-dependent term. The beauty is that a single distance formula unites geometric constructions, coordinate calculations, and orthogonality in linear algebra.

Make the assumptions visible: in a Cartesian coordinate system with orthogonal axes and the same length unit on each axis, take $A=(0,0)$, $B=(4,0)$ and $C=(0,3)$. The vectors $B-A=(4,0)$ and $C-A=(0,3)$ have dot product zero, establishing a right angle at $A$ before using the Pythagorean theorem. A sketch that merely looks like a right triangle supplies no such guarantee. Likewise, an image stretched horizontally and vertically by different amounts can distort both lengths and angles; a displayed shape and the assumptions defining it must be kept separate.

For a less origin-dependent measurement, take $P=(1,2)$ and $Q=(5,5)$. Their displacement vector is $Q-P=(4,3)$, so their Euclidean distance is again $5$. Its Euclidean norm records the direct separation, whereas travelling first horizontally and then vertically covers length $4+3=7$. This connects measuring a diagonal to locating two positions from a Cartesian coordinate system. A translation changes where the positions are recorded without changing their separation; a rotation changes the component values without changing the Euclidean norm.

The planar equilibrium of an inextensible slender beam can have large tangent angle while retaining small elastic strain across its thickness. For constant horizontal compressive force $F$, moment balance and curvature ${\theta }^{\prime }=d\theta /ds$ give the displayed nonlinear equation. Multiplication by ${\theta }^{\prime }$ gives the first integral $EI{\theta }^{\prime 2}/2-F\cos \theta =C$. A tensile load has $F<0$ in this convention.

For a real angle $\theta$ in radians, Euler's formula stateswhere $i$ is the imaginary unit. It connects the complex exponential to the sine and cosine coordinates of the unit circle. In particular, $|e^{i\theta }|=1$, and multiplying by $e^{i\theta }$ rotates a complex number through angle $\theta$ without changing its modulus. At $\theta =\pi /3$ it gives $e^{i\pi /3}=1/2+i\sqrt{3}/2$; at $\theta =\pi$, it gives $e^{i\pi }=-1$.

The identity makes angle addition algebraic: $e^{i\alpha }{e}^{i\beta }=e^{i(\alpha +\beta )}$. Expanding the product and comparing real parts and imaginary parts recovers the sine addition formula and cosine addition formula. Differentiating $e^{i\omega t}$ gives $i\omega e^{i\omega t}$, and differentiating again gives $-{\omega }^2{e}^{i\omega t}$. Its real and imaginary coordinates therefore solve the harmonic oscillator equation. This is why complex exponentials are useful building blocks in Fourier analysis: differentiation acts on each frequency by simple multiplication.

A short justification separates the even and odd powers in the absolutely convergent power series for $e^{i\theta }$. The even powers give ${\sum }_{n\ge 0}(-1{)}^n{\theta }^{2n}/(2n)!=\cos \theta$, and the odd powers give $i{\sum }_{n\ge 0}(-1{)}^n{\theta }^{2n+1}/(2n+1)!=i\sin \theta$. The absolute convergence permits this regrouping. The formula thus links an analytic definition of the complex exponential to a geometric motion rather than merely introducing a convenient notation.

Leonhard Euler presents the relation between imaginary exponentials and sines and cosines in his 1748 Introductio in analysin infinitorum, volume 1, chapter VIII, section 138. His derivation uses limiting powers after connecting trigonometric addition to algebra. The original Latin and Ian Bruce's English translation can be read together in Euler, Introductio, chapter VIII, section 138.

Two identical planar simple pendulums of length $l$ and bob mass $m$, with pivots separated horizontally by $r>0$, are coupled by a straight spring of natural length $r$ and spring constant $k$. In the ideal model without contact restrictions, their configuration space is the torus $S^1\times S^1$. Measure both angles from the downward vertical towards the same horizontal direction. Their bob separation isThe exact Lagrangian is the expression in the title. To derive it, use bob positions $(l\sin {\theta }_1,-l\cos {\theta }_1)$ and $(r+l\sin {\theta }_2,-l\cos {\theta }_2)$. Kinetic energy from Cartesian position vectors gives $T=m{l}^2({\dot{\theta }}_1^2+{\dot{\theta }}_2^2)/2$. Gravitational potential energy is $-mgl(\cos {\theta }_1+\cos {\theta }_2)$ and spring potential energy is $k(d-r{)}^2/2$; subtracting these from $T$ proves the formula. Near the downward equilibrium, $d-r=l({\theta }_2-{\theta }_1)+O(|\mathbf{\theta }{|}^3)$ for fixed $r$, giving the familiar normal modes of two spring-coupled equal pendulums.

Consider the symmetric matrix and initial unit vectorIts orthonormal eigenbasis consists of $u_1=(1,1{)}^T/\sqrt{2}$ and $u_2=(1,-1{)}^T/\sqrt{2}$, with eigenvalues $3$ and $1$. Since $x_0=(u_1+u_2)/\sqrt{2}$, direct multiplication givesThe first unnormalized vectors are $(1,0{)}^T$, $(2,1{)}^T$, $(5,4{)}^T$, and $(14,13{)}^T$. If ${\theta }_k$ is the acute angle from the line spanned by the dominant eigenvector, then $\tan {\theta }_k=3^{-k}$. Every application divides the unwanted-to-wanted component ratio by three: normalization changes lengths, while the spectral gap selects the direction.

The right panel compares the power iteration direction error, Rayleigh quotient error, and eigenpair residual. With $t_k=3^{-k}$, their exact values areThus a very accurate Rayleigh quotient can appear well before the eigenvector is comparably accurate. This is an example of quadratic Rayleigh-quotient improvement for the power method.

For a real unit vector $\mathbf{n}$, the Pauli matrix multiplication law gives $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I$. Separating the even and odd powers in the matrix exponential proves the displayed identity. For spin one-half, the physical rotation through angle $\theta$ has $a=\theta /2$.

The extraordinary wave's electric displacement field lies in that plane and is perpendicular to the wavevector; its electric field need not be. For angle $\theta$ between the wavevector and optic axis,Thus $n_{\mathrm{ex}}(0)=n_o$ and $n_{\mathrm{ex}}(\pi /2)=n_e$.^[ref]

A short derivation starts from the existing plane wave in an anisotropic dielectric equation. Put $\mathbf{k}=(k_x,0,k_z)$ and $k_0=\omega /c$, with angular frequency $\omega$ and vacuum speed of light $c$. The $y$ component gives the ordinary wave relation $k_x^2+k_z^2=n_o^2{k}_0^2$. The determinant of the $x,z$ system givesSubstituting $k_x=n_{\mathrm{ex}}{k}_0\sin \theta$ and $k_z=n_{\mathrm{ex}}{k}_0\cos \theta$ yields the displayed directional refractive index.

Faraday rotation exhibits optical nonreciprocity: with the bias magnetic field held fixed, a retracing beam gains another rotation in the same laboratory sense. Natural optical activity in a reciprocal medium instead unwinds on retracing the path.^[ref]

To reconcile this with the signed $B_{\parallel }$ formula, retain the laboratory $x,y,z$ axes for comparing physical lines. For the backward ray choose a right-handed ray basis $x^{\prime }=x$, $y^{\prime }=-y$, $z^{\prime }=-z$. Its magnetic flux density component is $B_{\parallel }^{\prime }=-B_z$, so its local rotation increment is $-\theta$. But positive $y^{\prime }$ is negative laboratory $y$: the backward increment is therefore $+\theta$ in the laboratory orientation plot. Reversing the ray reverses both the longitudinal component and the basis used for its signed angle. Reversing only one creates the false prediction of cancellation.

In fixed laboratory transverse components, define the rotation matrixFor an ideal mirror at normal incidence that contributes the same complex amplitude factor $r$ to both transverse components, the round-trip map is $rR(\theta )R(\theta )=rR(2\theta )$. The common mirror factor does not change the oscillation line. A reciprocal natural rotator instead gives $rR(-\alpha )R(\alpha )=rI$. This comparison assumes a retraced straight path and no additional polarization-changing elements.

For example, a chosen single-pass $30^{\circ }$ Faraday rotation produces a returned line at $60^{\circ }$ to the original laboratory $x$ axis. A chosen reciprocal $30^{\circ }$ natural rotator returns it to $0^{\circ }$. The same addition makes an ideal optical isolator possible: polarizers at $0^{\circ }$ and $45^{\circ }$ surround a $45^{\circ }$ Faraday rotator. Forward light follows $0^{\circ }\to 45^{\circ }$ and passes. Returning light selected at $45^{\circ }$ reaches $90^{\circ }$ at the entrance polarizer and is rejected by Malus's law. Real transmission loss and incomplete extinction limit the isolation.

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 finesse of a Fabry–Pérot interferometer is its free spectral range divided by a transmission peak's full width at half maximum. Large finesse means a narrow accepted frequency interval compared with the spacing between repeated orders. Here half maximum means $T=1/2$ of the ideal unit peak, with zero detector background subtracted; it does not mean halfway between the transmission minimum and maximum.

Write $A=4R/(1-R{)}^2$ for the coefficient multiplying ${\sin }^2(\delta /2)$ in Fabry–Pérot transmission from multiple reflections. This coefficient is sometimes called the coefficient of finesse; it is not the finesse $\mathcal{F}$. About a resonance $\delta =2\pi m$, put $ϵ=\delta -2\pi m$. Solving $T=1/2$ givesand, since one free spectral range advances the wave phase by $2\pi$,These are exact for the ideal nondispersive model with constant $R$ and linear frequency dependence of $\delta$. The inverse sine requires $(1-R)/(2\sqrt{R})\le 1$, or $R\ge 3-2\sqrt{2}\simeq 0.171573$. At equality, neighbouring peaks merely touch half maximum at their shared minima: the nominal width is one full free spectral range and $\mathcal{F}=1$. Isolated half-maximum peaks require the strict inequality. Below this threshold the transmission never falls to $1/2$, so this peak-width definition supplies no finesse; at $R=0$ the transmission is flat.

For large reflectivity, the small-angle approximation $\arcsin x\simeq x$ givesThis is a high-reflectivity approximation, not an identity for every $R$. The original dimensionless examples below have exact full width at half maximum values obtained from the inverse sine formula.

The beauty of the narrowing is visible directly in the geometric series: large $R$ preserves contributions from many round trips, so a small phase difference accumulates across many terms and spoils their addition. For the stipulated $10\mathrm{mm}$ vacuum gap, $R=0.95$ gives an ideal full width at half maximum of about $244.792\mathrm{MHz}$. This instrumental width is not a measured spectral line width. A real scan also depends on the source profile, accepted angles, mirror shape, alignment, and scan stability. The University of Washington handout separately discusses mirror-reflectivity, mirror-flatness, and detector-pinhole contributions; increasing $R$ alone cannot remove the others.^[ref]