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Let ${\beta }_0$ be the direction of the optical camera optical axis. The grating equation on that axis isThe central wavelengths in adjacent diffraction orders are therefore ${\lambda }_m=K/m$ and ${\lambda }_{m+1}=K/(m+1)$, with separation$K$ is the fixed order-times-wavelength product directed onto the optical camera axis, or the optical path difference between adjacent grooves in that direction. In a Littrow configuration, $K=2d\sin \alpha$. This adjacent-central-wavelength spacing is a conventional free spectral range of an echelle grating; at high order it is approximately ${\lambda }_m/m$.

There is a qualification in the wording of the printed definition. The exact interval for which order $m$ lies closest to the $y$ axis is not, in general, identical to the spacing of two adjacent order centres. For a concrete counterexample take ${\beta }_0=0$, $0<\alpha <\pi /2$, and a photodetector mapping $x=f_{\mathrm{cam}}\tan \beta$. Equality of the distances from the axis for orders $m,m+1$ means ${\beta }_m=-{\beta }_{m+1}$, so their common boundary is $\lambda =K/(m+1/2)$. The other boundary with order $m-1$ is $K/(m-1/2)$. Consequently the literal closest-order interval has widthfor $m\ge 2$ and accessible non-grazing orders. This differs from the printed $K/[m(m+1)]$. For a nonzero ${\beta }_0$, the exact boundary is determined by ${\beta }_m+{\beta }_{m+1}=2{\beta }_0$ and need not even have the simple half-order form. The two definitions agree to leading order $K/m^2$ when $m$ is large. Thus the printed expression is correct as an adjacent-centre convention or a high-order coverage estimate, rather than an exact consequence of the literal nearest-axis definition. Gap-free coverage should be checked using the actual photodetector mapping and neighbouring order endpoints.

Choose photodetector $x$ along increasing main grating dispersion, and $y$ along increasing wavelength for the cross-disperser. The echellogram shows two order traces and their intersections with the $y$ axis, which passes through the optical camera optical axis.

For fixed incidence, $d\beta /d\lambda =m/(d\cos \beta )>0$, so increasing wavelength goes to the right along each order. The cross-disperser also sends increasing wavelength upward in the chosen orientation; the traces therefore slope upward. At $x=0$, ${\lambda }_m=K/m>{\lambda }_{m+1}=K/(m+1)$, so the trace labeled $m$ lies above the trace labeled $m+1$. Reflecting either plotting axis would reverse its arrow but leave the optical content unchanged. The cross-disperser separates orders that would overlap in the main dispersion direction.

For each beam in vacuum, its magnetic field is ${\mathbf{B}}_j={\hat{\mathbf{k}}}_j\times {\mathbf{E}}_j/c$. Write the two complex scalar contributions, including the beam splitter factor, as $u_1=f(t-t_1)/\sqrt{2}$ and $u_2=f(t-t_2)/\sqrt{2}$. Since their electric fields are both along $z$,The optical-cycle and ensemble-averaged Poynting vector isThe mean individual light intensities are equal, soThus energy flows along the bisector of the two beam directions, crossing a detector plane normal to $x$. Such a sensor can in principle record the variation across its $y$ coordinate. Its actual normal energy flux is $I(y)/\sqrt{2}$ in the normalization used above; this fixed factor does not change interferometric visibility.

There is a practical resolution condition in addition to the direction of energy flow. The interference fringe spacing $\lambda /\sqrt{2}$ is very small, so direct pixels would need spacing no larger than $\lambda /(2\sqrt{2})$ as required by the Nyquist–Shannon sampling theorem, or optical magnification and sufficient angular aperture to preserve both incident directions. An ordinary optical camera with much larger effective pixels averages the interference fringes. The Poynting-vector argument establishes the allowed measurement geometry, not guaranteed resolution with any optical camera.

The $x$ axis lies along the mean Poynting vector. A two-dimensional optical camera sensor intended to sample many $x$ positions at once would have to lie along the energy flow, where it receives no normal flux in the ideal geometry; a side-view optical camera also sees no light from an empty nonscattering propagation region. Putting a detector normal to $x$ measures one transverse section instead, and absorbs the light before a downstream detector can sample the same rays.

One could translate a detector to successive $x$ positions or introduce a scattering screen, but that is a different measurement. Moreover the ideal result here is already the constant $I(x)=2I_0$, so it contains no longitudinal optical coherence structure for the optical camera to resolve. The difficulty is the longitudinal sampling geometry, rather than a hidden $x$-fringe period.

A pathline records the motion of one identified fluid element: keep its starting position $\mathbf{a}$ and starting time $t_0$ fixed, and let physical time $t$ advance. In a velocity field $\mathbf{u}(\mathbf{x},t)$ its Lagrangian trajectory $\mathbf{X}(t;t_0,\mathbf{a})$ solves the initial value problemA streamline instead freezes the time argument of the velocity field; a streakline fixes the observation time and varies the release time of many fluid elements. A pathline is useful for answering where a particular parcel goes and how long it takes, rather than merely which way the fluid is moving now. ^[ref]

For a concrete analytical model, take spatial coordinates $x,y$ in metres, time $t\ge 0$ in seconds, and the spatially uniform but time-dependent velocity fieldIts numerical components in these units are $(1,t)$. A fluid element starting at the origin at $t_0=0$ has the pathlineAt $t=1\mathrm{s}$ it is at $(1,0.5)\mathrm{m}$; at $t=2\mathrm{s}$ it is at $(2,2)\mathrm{m}$. The parabola records earlier, shallower velocity directions as well as the final direction. Differentiating gives $(\dot{x},\dot{y})=(U,at)$ and acceleration $(0,a)$, so this curved pathline needs no spatial variation of the velocity field. More generally the chain rule along a pathline gives $df(\mathbf{X}(t),t)/dt={\partial }_{t}f+\mathbf{u}\cdot \nabla f$, the material derivative of a transported scalar field.

A motivating tracer demonstration appears in S. J. Kline's film notes: water passes through a contracting channel, and hydrogen bubbles mark a small element at a wire at time $t_0$. The illustrated element is shown again $4.7\mathrm{s}$ later, with its accumulated pathline superposed. Following that same marked element across successive film frames records a history; the entire bubble pattern in one frame has a different meaning. ^[ref]

For a modern version of the readout, use sparse small markers in a transparent water channel, illuminate the viewed plane, and keep an optical camera fixed with a ruler in that plane. Record timestamps and track the same isolated marker between frames; successive positions estimate its pathline and $[\mathbf{X}(t+\Delta t)-\mathbf{X}(t)]/\Delta t$ estimates its velocity. A long exposure records a pathline segment but loses the timing within it, so its total length divided by exposure time gives only mean speed. These are proposed measurement steps based on Jean Hertzberg's teaching guidance, rather than additional measurements from Kline's film. Markers leaving the viewed plane, overlapping tracks and perspective can break particle identification; repeat with a thinner viewed slice and a lower marker concentration. ^[ref]

Radiant flux is the power carried by electromagnetic radiation: radiation energy transferred per unit time, measured in watts, $\mathrm{W}=\mathrm{J}{\mathrm{s}}^{-1}$. Specify what is counted: all radiation emitted by a source, radiation within one beam, or radiation intercepted by a photodetector. Writing $Q(t)$ for the cumulative energy crossing the specified boundary givesThe first expression is an instantaneous derivative; the second is an average over the stated time interval. If $0.020\mathrm{J}$ arrives during $0.010\mathrm{s}$, the mean radiant flux is $2\mathrm{W}$. The photodetector's absorbed power equals this incident radiant flux only if it absorbs all of the incident radiation.

Keeping radiant flux fixed while widening a beam reduces its irradiance, the incident power per receiving area. Radiance additionally resolves the propagation direction and has units $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$; an optical camera can collect different radiant fluxes from sources with the same radiance but different apparent solid angles. Luminous flux, measured in lumens, weights the optical spectrum by the human eye's standardized response instead of counting energy equally at every wavelength. Equal radiant fluxes need not give equal luminous fluxes.^[ref]

Radiant power also differs from the optical power of an optical lens, which is reciprocal focal length in dioptres. Likewise, astronomical radiative flux is power per area. Stating the units prevents these shared words from hiding different physical quantities.

Radiometry measures the energy and power carried by electromagnetic radiation, including light. It asks how much radiation a source supplies, where it travels, and how much reaches a receiving surface. The measurements are physical rather than weighted by the sensitivity of the human eye: equal radiant fluxes at different wavelengths need not look equally bright. ^[ref].

The five quantities below answer different questions about the same radiation. The International System of Units makes the distinctions visible: a watt is a joule per second, while a steradian measures solid angle. ^[ref].

Resolving wavelength or frequency adds a further “per unit interval”, as in spectral radiance. The notation is consistent with NIST's table of spectral quantities. In particular, radiant flux is power, whereas the existing astronomical radiative flux is power per area. ^[ref].

This bookkeeping makes an optical camera's collected signal calculable from a source's radiance and the geometry of the receiving aperture. It also lets radiation measurements from different instruments be compared. For example, NIST describes separate calibrations of spectral radiance and spectral irradiance for instruments used to observe Earth. ^[ref].

Schlieren photography makes spatial changes of refractive index visible by converting deflections of light rays into contrast at an optical camera. In a gas, mass density variations can produce these deflections through refraction. A knife edge in the focused beam is a common implementation.

Consider an original, explicitly normalized model of primary coma, rather than a prescription for a particular optical lens. Let $(u,v)$ be Cartesian coordinates in a circular optical pupil divided by its radius $a$, with $u^2+v^2\le 1$. Let $h=\alpha /{\alpha }_0$ be a signed small field angle divided by an arbitrary positive reference angle. With polar coordinates $u=\rho \cos \theta$, $v=\rho \sin \theta$, prescribe the wavefront errorHere $W_0$ is an arbitrary positive optical path difference scale, not a measured error. The cubic optical pupil dependence and linear field dependence are the classical Seidel form.^[ref]

In the paraxial approximation, light ray direction errors follow the transverse gradient of wavefront error. For a reference sphere a distance $L$ from the exit optical pupil in a uniform image medium of refractive index $n$, the corresponding image-error scale is $s=L{W}_0/(na)$. Choose image-error coordinate signs so the normalized displacement is $\nabla w$; reversing the convention reverses the plotted orientation. Image coordinates below are errors relative to the ideal image of this particular object point, rather than positions of different objects:The partial derivatives explain the asymmetry: opposite optical pupil points $(u,v)$ and $(-u,-v)$ land together, rather than on opposite sides of the ideal image. For $h>0$, every $X$ is nonnegative, while $Y$ can have either sign. The figures stipulate this single wavefront error, omit other optical aberrations and diffraction, and assign no instrument dimensions.

For a ring of fixed optical pupil radius $\rho$, the double-angle formulas for trigonometric functions giveThus each ring maps into a circle of radius $|h|{\rho }^2$ whose centre is displaced by $2h{\rho }^2$. Doubling $\rho$ quadruples both sizes; for $h=1$, the edge ring has centre $(2,0)$ and radius $1$, while the half-radius ring has centre $(1/2,0)$ and radius $1/4$. Superposing these displaced circles produces the comet-like envelope.

For uniform irradiance over the circular optical pupil, equal area samples give an equal-weight geometrical image. Its centroid is $(h,0)$ because $⟨u^2⟩=⟨v^2⟩=1/4$ and $⟨uv⟩=0$. Moving the coordinate origin to that centroid translates the image; it does not remove its asymmetric shape. Adding a linear term to $w$ likewise translates all light rays equally. These origin choices must be distinguished from changing the optical aberration itself.

The model's full transverse extent is $3|h|$ along $X$ and $2|h|$ along $Y$. Halving the field angle halves the blur, and reversing it reverses the pattern about the ideal image. This isolates why coma matters for an optical camera's usable field of view: collecting more light does not ensure that off-axis points remain compact. Near diffraction-limited resolution, a physical optics calculation of the point spread function is needed instead of interpreting the geometrical tip as an infinitely sharp image.^[ref]

A streakline at observation time $T$ is the set of present positions of all fluid elements that passed through one fixed release point $\mathbf{a}$ during a specified history of injection. For continuous release over ${\tau }_0\le \tau \le T$, solve a different initial value problem for each release time $\tau$:Thus a pathline varies $t$ for one particle; a streakline varies $\tau$ while holding $T$ fixed. Interrupted release gives only the corresponding subsets of the streakline. A moving nozzle requires a time-dependent release position instead, so the standard fixed-point definition should not silently be applied to it. Continuous release makes streaklines easy to observe and useful for showing where material from one source has reached. ^[ref]

In the analytical velocity field $(U,at)$ of pathline, a particle released at the origin at time $\tau$ followsObserve all particles together at time $T$, after continuous release from $\tau =0$:For $U=1\mathrm{m}{\mathrm{s}}^{-1}$, $a=1\mathrm{m}{\mathrm{s}}^{-2}$ and $T=2\mathrm{s}$, the streakline is the downward-curving parabola $y=2x-x^2/2$ in metre coordinates. Releases at $\tau =0,1,2\mathrm{s}$ occupy $(2,2)$, $(1,1.5)$ and $(0,0)\mathrm{m}$, respectively. At $x=1\mathrm{m}$, the streakline lies above the pathline $y=x^2/2$ and below the origin's streamline $y=2x$. All particles nevertheless have the same present velocity $(1,2)\mathrm{m}{\mathrm{s}}^{-1}$. The streakline slope $dy/dx=a\tau /U$ generally differs from the present velocity slope $aT/U$; they agree at the nozzle. The oldest particle is the downstream endpoint, which explains why the pathline and streakline meet there without being the same curve.

Each panel fixes one observation time. Coloured dots are the present positions of particles with the indicated release times; dashed pathlines show those particles' earlier histories. The black streakline connects present positions of the continuously released family, including release times between the sampled dots. After $T=1\mathrm{s}$, particles already present keep moving and new ones enter at the origin, so the later streakline is not a frozen record of the earlier one. Injection beginning only at $\tau =1\mathrm{s}$ would show just $0\le x\le 1\mathrm{m}$ of the $T=2\mathrm{s}$ curve. Both panels are diffusion-free analytical models, not experimental photographs.

Kline's contracting-channel demonstration switches from marking an element to continuous bubble production at a marking wire: the resulting trace is a streakline. The apparatus therefore changes the curve being displayed by changing the marking history, even with the same channel. A wire produces a family of traces; conceptually each fixed emitting location supplies its own streakline. Individual bubbles are markers rather than perfect fluid elements: Jean Hertzberg's teaching notes describe hydrogen production by electrolysis at a thin cathode wire and warn that larger bubbles have more buoyancy. ^{[ref] [ref]}

A dye version uses a small fixed injection port upstream in a transparent water channel, a dilute premix of dye with the channel water, illumination and a fixed optical camera. Start continuous injection at a recorded time ${\tau }_0$ and take a short exposure at $T$: the visible filament approximates a streakline, not the accumulated pathline of one particle. Repeat with the same port and imaging geometry after changing the flow with time; compare the filament against local directions estimated from short displacements of sparse markers at $T$. Agreement in a steady flow and possible disagreement when the flow changes are the predictions being tested; no measured curves are claimed here. The protocol follows Hertzberg's advice to use small ports, little injected volume and a dilute premix with matched density and viscosity. Check for nozzle disturbance by reducing injection rate, for buoyancy by watching the premix in still water, and for diffusion by comparing filament widths at different ages. Keep exposure short enough to suppress motion blur. Dye concentration approximates a passive scalar only when it has negligible feedback on the velocity field; finite port width, diffusion and out-of-plane motion limit how literally a visible filament can be read as a geometrical streakline. ^[ref]

A thin lens models an optical lens by collapsing its refracting surfaces into one plane, neglecting thickness relative to the relevant radius of curvature values and object and image distances. Here use an ideal converging thin lens in air and the paraxial approximation: light rays make small angles with the optical axis, so Snell's law can be linearized using the small-angle approximation. Equal surrounding refractive indices on both sides give the familiar symmetric equation.^[ref]

Put the thin lens at $x=0$, with incident light travelling to the right. An object at $x=-u$ has $u>0$. The image coordinate is $x=v$: $v>0$ for a real image and $v<0$ for a virtual image. The converging thin lens has focal length $f>0$, and heights $h_o,h_i$ are signed relative to the optical axis. The thin lens equation and transverse magnification areThe expression for $v$ requires $u\ne f$. Negative transverse magnification means inverted; positive transverse magnification means upright. Remember that the positive distance $u$ differs in sign from the object's coordinate $-u$. ^[ref].

For focal length $f=1\mathrm{m}$, object distance $u=3\mathrm{m}$, and object height $h_o=0.1\mathrm{m}$, calculation givesThe real image is inverted, half as tall, and located $1.5\mathrm{m}$ to the right of the thin lens. A screen at that position receives the image because the outgoing light rays actually meet there.

For the same focal length $f=1\mathrm{m}$ and object height $h_o=0.1\mathrm{m}$, move the object inside the front optical focus, to $u=0.5\mathrm{m}$. ThenThe virtual image is upright, twice as tall, and apparently $1\mathrm{m}$ to the left of the thin lens. Outgoing light rays still travel to the right; only their backward extensions intersect at the virtual image. A screen there cannot collect that image from the outgoing bundle. This geometry explains why a converging optical lens can act as a magnifier when held close to an object. ^[ref].

Both diagrams use two rules for ray tracing. A light ray from the object's tip parallel to the optical axis reaches the thin lens at $(0,h_o)$ and exits along the straight line through the rear optical focus $(f,0)$. Another light ray through the centre $(0,0)$ continues undeviated in the ideal thin lens model. Their intersection, or the intersection of their backward extensions, locates the image tip. ^[ref].

A short geometric proof outline makes the signs transparent. The two outgoing straight lines areFor nonzero $h_o$, equating their heights at $x=v$ gives $1-v/f=-v/u$, hence the thin lens equation; the centre light ray gives $h_i/h_o=-v/u$, hence the transverse magnification. This derives the image equations from the ideal ray tracing rules, whose physical justification comes from Snell's law in the paraxial approximation.

At $u=f$, the two outgoing straight lines are parallel: there is no finite image position. For $u>f$ the real image has $m<0$; for $0<u<f$ the virtual image has $m>0$. These cases let an optical camera designer choose a receiving position and let a user understand a magnifier. An actual optical lens has finite apertures, which cause diffraction and impose diffraction-limited resolution; optical aberrations can further blur the image. The diagrams are ideal geometrical optics, with exact coordinates within that model, rather than exact predictions of Maxwell's equations. ^[ref].