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Angular magnification is the ratio of the signed viewing angle with an optical instrument to the signed viewing angle in a stated reference arrangement. Measure both as apparent directions from the eye, positive toward the same side of the optical axis; then an inverted view has negative $M_{\theta }$. If only positive angular sizes are compared, the result is $|M_{\theta }|$ and carries no orientation information. ^[ref].

For the upright virtual image of a converging thin lens, let the eye be at $x=e\ge 0$ and take the unaided reference to be the same small object at a distance $D$ from the eye. Within the small-angle approximation,Specify $D$: keeping the object in its original position means $D=e+u$, whereas the customary magnifier rating uses a near-point reference, often $D=25\mathrm{cm}$. A reference distance is a convention, not a claim that every eye has that near point. ^[ref].

For the virtual image diagram, $m=4$, $v=-30\mathrm{cm}$ and $e=5\mathrm{cm}$. With $D=25\mathrm{cm}$, $M_{\theta }\simeq 20/7$, whereas keeping the object at $D=e+u=12.5\mathrm{cm}$ gives $M_{\theta }\simeq 10/7$. Thus neither viewing comparison gives the transverse magnification four. A person may also be unable to focus on the unaided object at the shorter reference distance.

At $u=f$, each outgoing bundle is parallel with slope $-h_o/f$; its apparent viewing direction has angle ${\theta }_i\simeq h_o/f$. Consequently $M_{\theta }\simeq D/f$ remains finite even though $m$ is undefined at a finite image plane. Equivalently, in the limit $u\to f^{-}$, the factors $m$ and $e-v$ both grow without bound but their ratio tends to $1/f$. For $f=10\mathrm{cm}$ and $D=25\mathrm{cm}$, the relaxed-view prediction is $M_{\theta }\simeq 2.5$. ^[ref].

A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

A double pendulum consists of two pendula coupled because the second pivot moves with the first mass. It has two degrees of freedom: even when the upper pivot is fixed, moving the upper arm carries the lower pivot along. This makes the double pendulum a compact example of how mechanical constraints couple motion, and how Lagrangian mechanics converts geometry into dynamics.

Use two equal point masses $m>0$ and two massless rigid rods of equal length $\ell >0$, with joints that allow free rotation in one vertical plane. The upper pivot is fixed; gravity is uniform, with gravitational acceleration $g>0$. Neglect joint friction and drag force. Both angles ${\theta }_1,{\theta }_2$ are measured from the downward vertical, positive toward the right, in radians. In particular, ${\theta }_2$ is an absolute angle, not the bend between the rods: that bend is ${\theta }_2-{\theta }_1$.

The double pendulum Lagrangian below gives the exact dynamics of this ideal model. Near the downward stable equilibrium, normal modes of a double pendulum replace the coupled motion with two independent harmonic oscillators. Their fixed amplitude ratios explain which releases make the rods swing together and which make them swing in opposite directions.

Write $\Delta ={\theta }_1-{\theta }_2$. Applying the Euler-Lagrange equations to the double pendulum Lagrangian, then dividing by $m{\ell }^2$, givesFor example, the upper generalized momentum is $\partial L/\partial {\dot{\theta }}_1=m{\ell }^2(2{\dot{\theta }}_1+\cos \Delta {\dot{\theta }}_2)$. Its time derivative contains $-m{\ell }^2\sin \Delta ({\dot{\theta }}_1-{\dot{\theta }}_2){\dot{\theta }}_2$. Subtracting $\partial L/\partial {\theta }_1=-m{\ell }^2\sin \Delta {\dot{\theta }}_1{\dot{\theta }}_2-2mg\ell \sin {\theta }_1$ cancels the mixed product and leaves the positive $\sin \Delta {\dot{\theta }}_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.

The matrix multiplying the two angular accelerations isIt is invertible for every configuration, so specified angles and angular velocities determine both angular accelerations. For the same reason $T>0$ whenever the angular velocities are not both zero. The mass cancels from the motion equations, while the timescale is set by $\sqrt{\ell /g}$.

As an exact constructed release, put ${\theta }_1(0)=\alpha$, ${\theta }_2(0)=0$, and both angular velocities zero. Solving the two equations at release yieldsFor $0<\alpha <\pi /2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's acceleration makes this possible even though the lower rod starts vertical. The small-angle approximation gives ${\ddot{\theta }}_1(0)\simeq -2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)\simeq 2g\alpha /\ell$.

The conserved mechanical energy is $E=T+V$, because the Lagrangian has no explicit time dependence. Directly, $dE/dt$ is $m{\ell }^2$ times the first equation multiplied by ${\dot{\theta }}_1$ plus the second multiplied by ${\dot{\theta }}_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a trajectory.

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.

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]

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.

For a rapidly spinning gyrocompass with axle constrained horizontally, take azimuth $\theta$ positive from north towards west and positive spin angular momentum $J_s=I_1\omega$. In the terrestrial rotating reference frame, the vertical component of the torque equation, with zero external vertical torque, is $I_2\ddot{\theta }+J_s{\Omega }_E\cos \lambda \sin \theta \simeq 0$. Here $I_2$ is the vertical moment of inertia and terms of order ${\Omega }_E^2$ have been neglected. The small-angle approximation about $\theta =0$ gives simple harmonic motion with the displayed squared angular frequency. Linearizing about $\theta =\pi$ reverses its sign and gives an unstable equilibrium. At the poles this restoring effect vanishes. For a thin rotationally symmetric flywheel, the perpendicular axis theorem gives $I_1=2I_2$.

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].

A mathematical model represents selected features of a system with mathematical objects and relations so that its behavior can be analyzed or predicted. Choosing the representation is part of the work: specify the quantities tracked, the equations or probability distributions connecting them, and the conditions under which those connections are expected to apply. A mathematical proof can establish a consequence of the model; comparing that consequence with observations tests its suitability for the intended use.

For a simple pendulum, the angle $\theta (t)$ and gravitational acceleration $g$ give the ideal equation $\ddot{\theta }+(g/\ell )\sin \theta =0$. Its assumptions include a point mass, fixed length $\ell$, uniform gravitational acceleration, and no damping. The small-angle approximation gives the simpler harmonic oscillator equation and oscillation period $T_0=2\pi \sqrt{\ell /g}$. A larger amplitude changes the full model's period; drag requires a damped pendulum. The simpler model remains useful when its omitted effects are small for the question being asked.

A model need not track every detail. A Bernoulli process models repeated success/failure observations using one probability and an independence assumption; a Gaussian process specifies a mean and a covariance function to model related measurements. Each representation enables calculations by retaining some distinctions and omitting others. Useful simplification therefore comes with a stated domain of use and quantities against which its predictions can be checked.

The viewing optical lens assembly that acts as a magnifier of the intermediate microscope image. With that image at its front focal plane, the outgoing light rays from each point are parallel. In the small-angle approximation, the viewed angle is $h_i/f_e$, versus $h_i/D$ for the same image-sized object at reference distance $D$, giving $M_e=D/f_e$.^[ref]

A finite-distance virtual image requires a different eyepiece conjugate. In the ideal thin lens model, with the eye close to the eyepiece and the virtual image at distance $D$, the angular magnification becomes $1+D/f_e$. State the viewing condition before multiplying by the objective's transverse magnification; the relaxed-eye example above uses $D/f_e$.

The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

Parallax localization of a virtual image matches the apparent depth of a virtual image to a directly viewed physical marker. A sideways change of viewpoint changes their relative apparent positions if their depths differ. Adjust the marker until that relative motion vanishes, and use its coordinate to estimate image position. The marker must be viewed directly; viewing it through the same optical lens compares two images rather than an image and a known physical depth. ^[ref].

For two nearly aligned points at distances $L_i,L_m$ from an eye displaced sideways by $\delta$, the small-angle approximation gives a change of relative viewing angle proportional to $\delta (1/L_i-1/L_m)$. It vanishes when $L_i=L_m$, and changes sign across that match. When both distances are large, the difference is small, explaining the poor depth precision near the thin lens condition $u=f$.

The parsec is a distance unit defined by the exact conventionThe astronomical unit is exactly $149597870700\mathrm{m}$; one arcsecond is exactly $\pi /648000$ radians. The IAU resolution explicitly records the parsec convention. A megaparsec is one million parsecs. ^[ref]. ^[ref].

The familiar description "the distance giving one arcsecond of stellar parallax" uses the small-angle approximation. Exact right triangle geometry instead gives $\mathrm{au}/\tan (\pi /648000)$, slightly shorter than the conventional parsec. The relative difference is approximately $(\pi /648000{)}^2/3\simeq 7.83\times 10^{-12}$. This is negligible for the examples here, but distinguishing an exact unit convention from an approximate geometric interpretation prevents a misleading equality.

The small-angle approximation replaces $\sin \theta$ by $\theta$, giving the damped harmonic oscillator equation $\ddot{\theta }+c\dot{\theta }+\theta =0$. Its characteristic roots satisfy $r^2+cr+1=0$.

For $0<c<2$, put $\nu =\sqrt{1-c^2/4}$; the motion is underdamped:It oscillates with decaying amplitude. For $c=2$, the repeated root gives critical damping:For $c>2$, the motion is overdamped:Both real exponents are negative, so there is no oscillatory factor. The two constants are arbitrary in each case; these approximations require the displacement to remain small.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.

For the nearly edge-on orbit $I=\pi /2-I^{\prime }$ with $I^{\prime }\ll 1$, the small-angle approximation gives $\cos I=\sin I^{\prime }=I^{\prime }+O(I^{\prime 3})$. HenceAlso,Thus, on a branch where $\cos f>0$ and with separation measured in units of $a$,as stated. The absolute value is required when $R_{\mathrm{sky}}$ denotes the nonnegative separation over the whole orbit.

Take the pivot as origin, with $x$ horizontal and $y$ vertically upward. For the simple pendulum, putThe velocity and acceleration areThe forces are the tension $-T{\mathbf{e}}_r$ and gravity $(0,-mg)$. Taking the dot product of Newton's second law with ${\mathbf{e}}_{\theta }$ eliminates the tension and givesNear the downward stable equilibrium $\theta =0$, the small-angle approximation $\sin \theta \simeq \theta$ gives $\ddot{\theta }=-(g/\ell )\theta$ to first order. Since $x=\ell \sin \theta \simeq \ell \theta$, its linearization isThis is simple harmonic motion with angular frequency $\sqrt{g/\ell }$; the restoring sign confirms the stable equilibrium.