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Far-field transform. For transverse aperture scale $b$ and screen distance $z$, the quadratic phase across the aperture is negligible when $k{b}^2/(2z)\ll 1$, equivalently a small Fresnel number up to geometrical factors. Expanding the source-to-screen distance at transverse screen coordinate $X$ givesDropping the final aperture-dependent term and treating the prefactor as constant givesThus intensity is the modulus squared of the aperture Fourier transform. A lens focal plane supplies the same angular transform without a very distant screen. The infinitely extended vertical dimension is treated per unit vertical extent.

half-slit phase-reversal diffraction. Put $h=-1$ on the left half and $h=1$ on the right half. With $q=k\sin \theta$ and $s=qb/4$,ThereforeThe apparent singularity at zero angle is removable: $H\to 0$, so the centre is dark. Normalizing by the uncoated slit's on-axis intensity $I_0$ gives $I/I_0={\sin }^{4}s/s^2$. Noncentral zeros have $s=j\pi$, or $\sin \theta =2j\lambda /b$. The two largest lobes are symmetric, at the first nonzero root of $\tan s=2s$, $s\simeq \pm 1.166$, giving $\sin \theta \simeq \pm 0.742\lambda /b$ when those directions propagate. Smaller lobes lie farther out under a decaying envelope.

Doubled thickness. Neglecting absorption and interface reflection, the phase delay doubles from $\pi$ to $2\pi$. Both halves then transmit with the same phase, recovering the ordinary single-slit patternIt has a bright centre and first zeros at $\sin \theta =\pm \lambda /b$. The phase layer redistributes transmitted power rather than absorbing it in this idealization.

The Debye model replaces the crystal by a harmonic elastic continuum with acoustic dispersion $\omega =v_{s}k$. It counts three polarizations and introduces a cutoff so that the total number of vibrational modes is $3N$, where $N$ is the atom count. Periodic-box counting givesFor an isotropic average speed, $3/v_D^3=1/v_L^3+2/v_T^3$. With cutoff frequency ${\omega }_D$ fixed by the mode count,This density of states is quadratic because a three-dimensional shell has volume proportional to $k^{2}dk$ and the acoustic dispersion is linear. The cutoff is a mode-counting device; it is not a claim that every actual branch stays linear to the zone boundary.

Each phonon mode has mean thermal energy $\hslash \omega /(e^{\beta \hslash \omega }-1)$, using Bose-Einstein statistics with zero phonon chemical potential. The fixed harmonic zero-point contribution has zero temperature derivative. With $x=\hslash \omega /(k_{B}T)$,and differentiating at fixed volume givesFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{j\ge 1}{e}^{-jx}$ gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\sum j^{-4}={\pi }^4/15$. integration by parts makes the heat-capacity integral $4{\pi }^4/15$, soOnly modes with $\hslash \omega \lesssim k_{B}T$ are appreciably excited; their number scales as $T^3$, explaining the power law physically. At high temperature every mode contributes $k_B$, giving the Dulong-Petit limit $C_V\to 3N{k}_B$.

The model assumes the harmonic approximation, a macroscopic three-dimensional solid and acoustic continuum behaviour for the thermally populated modes. Anisotropic sound speeds change the low-temperature coefficient rather than its cubic power. Real optical branches and nonlinear acoustic dispersion limit the full-temperature approximation; electronic excitations can add a linear term in a metal, and anharmonicity produces thermal expansion and distinctions between $C_P$ and $C_V$. At temperatures below the lowest finite-sample mode, the bulk continuum law also fails. Thus the $T^3$ result is the low-temperature lattice contribution, not a universal total heat capacity for every solid.

Two parallel partially reflecting surfaces in a Fabry–Pérot interferometer produce a sequence of coherent transmitted beams. For equal lossless mirror intensity reflectance $\mathcal{R}$ and round-trip phase $\delta =4\pi nd\cos {\theta }_t/\lambda$, summing the geometric series of amplitudes givesresonances have $2nd\cos {\theta }_t=m\lambda$ and unit peak transmission in this symmetric lossless model. Varying angle gives concentric equal-inclination rings; changing spacing scans the resonance wavelengths.

At normal incidence with negligible dispersion, the free spectral range is $\Delta {\nu }_{\mathrm{FSR}}=c/(2nd)$. For high reflectance, the peak full width gives finesse and resolving powerThe coefficient $\mathcal{F}$ in the transmission formula and the finesse $\mathcal{F}$ are different quantities. Multiple round trips sharpen a peak, allowing high-resolution spectroscopy, frequency selection and cavity measurements. Absorption, mirror mismatch, finite angular spread, imperfect parallelism and surface errors reduce usable transmission or finesse. Adjacent orders can overlap in a broadband measurement, requiring order selection or prior spectral information.

For $N$ uniformly illuminated identical slits of a diffraction grating, with width $b$ and period $d$, the far-field amplitude is a single-slit transform times a finite phasor sum. At normal incidence, write $\beta =\pi d\sin \theta /\lambda$. ThenPrincipal maxima obey $d\sin \theta =m\lambda$ and have $N^2$ times the single-slit intensity at that angle. Nearby first minima are separated from a principal maximum by $\delta \theta \simeq \lambda /(Nd\cos {\theta }_m)$, so more illuminated slits sharpen each order. The broad single-slit envelope suppresses high angles; an order is missing if its principal maximum coincides with an envelope zero.

Differentiating the grating equation gives angular dispersion $d\theta /d\lambda =m/(d\cos {\theta }_m)$. Comparing the separation of nearby wavelengths with the first-minimum width yields the Rayleigh resolving powerOnly propagating orders with $|m\lambda |\le d$ exist. Different spectral orders can overlap, and finite source-slit width, limited coherence, aberrations and grating errors worsen ideal resolution. A spectrometer therefore collimates light from an entrance slit and focuses each diffracted direction onto a detector; the slit image width and illuminated grating width must both support the desired spectral separation.

The power linewidth is $\Delta f=f_0/Q\simeq 0.044\mathrm{Hz}$. Fourier frequency resolution is of order $1/T_{\mathrm{obs}}$, so $T_{\mathrm{obs}}\gtrsim Q/f_0\simeq 23\mathrm{s}$, and preferably longer to distinguish the intrinsic linewidth from the observation-window width. zero padding does not improve this physical resolution. A freely decaying signal also needs sufficient microphone dynamic range.

At $50\mathrm{kHz}$ modes $n=1,2,3,4$ can propagate. Their positive $k_x$ values are approximately $769.5,719.8,628.3,471.2{\mathrm{m}}^{-1}$. Mode $n=5$ is a nonpropagating mode at waveguide cutoff: $k_x=0$ and $v_g=0$, so it does not carry longitudinal power as a traveling mode. Higher modes are evanescent. The injected transverse profile determines which permitted modes are actually excited.

Let $T$ be the common tension and ${\mu }_j$ the linear mass densities. The characteristic impedance of a stretched string is $Z_j=T/c_j=\sqrt{T{\mu }_j}$. Continuity of displacement and transverse force at a massless junction gives $1+r=t$ and $Z_1(1-r)=Z_{2}t$. ConsequentlyThe reflected power is $7.18\%$. The minus sign of $r$ represents a phase reversal, not negative power. Both incident and reflected waves travel in the first string, so their power ratio is simply the squared displacement-amplitude ratio. As a check, $\mathcal{T}=(Z_2/Z_1)|t{|}^2=0.92820$ and $\mathcal{R}+\mathcal{T}=1$.

Take the aperture in $z=0$ and the observation point $P=(0,0,b)$, with $b>0$. Under scalar diffraction theory, the Huygens-Fresnel principle sums complex amplitudes from area elements, rather than summing their intensities. Apart from a common normalization,Here $h$ is the complex aperture transmission, $b$ is the axial distance, and $\theta$ is the angle of the secondary ray to the aperture normal. The obliquity factor $K$ describes the direction dependence of the secondary contribution; one Kirchhoff convention is $K=(1+\cos \theta )/2$ for normal illumination. Other scalar boundary approximations have slightly different angular weights, all tending to unity near the axis.

Put $r^2=x^2+y^2$. The paraxial approximation gives $R\simeq b+r^2/(2b)$ in the phase and $1/R\simeq 1/b$ in the amplitude. Removing common $e^{ikb}/b$ yieldsAssumptions are uniform normally incident plane-wave illumination, coherent monochromatic propagation, aperture dimensions large enough for the scalar description, $r/b\ll 1$, and phase accuracy $k{r}^4/(8b^3)\ll 1$ over the relevant aperture. The last condition justifies dropping the quartic distance term; geometric smallness alone does not guarantee phase accuracy.

For a circular aperture, $h=1$ for $r\le a$ and zero elsewhere. Rotational symmetry and $s=r^2$ give $dxdy=rdrd\phi$, $rdr=ds/2$, andThe constant $\pi$ can be absorbed into the proportionality factor, producing the required one-dimensional integral.

At a focus a distance $f$ away, half-period-zone boundaries obeyWith slowly varying obliquity and illumination, each zone has almost the same area and its integrated phasor has the opposite sign to its neighbour. A zone plate blocks alternate zones, so the remaining phasors add instead of nearly cancelling. Choosing an open central zone, the first three open regions have outer boundaries $n=1,3,5$:These are outer radii, not zone widths; the second open region is the annulus between $r_2$ and $r_3$.

For the specified outer radius, $R^2/(\lambda f)=49$: there are approximately 49 half-period zones, of which 25 odd zones are open. Normalize to the field ${\psi }_0$ of the unobstructed plane wave. The normalized Fresnel aperture integral isFor any odd open annulus this gives $2$, so the whole plate gives $\psi /{\psi }_0\simeq 50$ andThe axial intensity gain of an amplitude zone plate grows as the square of the number of open zones. Treating the number of zones continuously gives $49^2\simeq 2.40\times 10^3$, the same accuracy as an estimate. This large local intensity gain is consistent with conservation of energy because light is concentrated into a small focal spot. It is not a power efficiency; the ideal first-order power efficiency of a binary amplitude zone plate is only $1/{\pi }^2$.^[ref]

“Without the zone plate” has here meant removing the finite plate and its outer stop, giving the unobstructed incident wave. If instead a clear circular aperture of radius $R$ is retained, its axial field is ${\psi }_{\mathrm{clear}}/{\psi }_0=1-e^{i49\pi }=2$, so the ratio to that finite-aperture reference is $2500/4=625$. The two reference experiments differ; neither ratio should be confused with total transmitted power.

The Q factor measures stored energy relative to loss: $Q=2\pi E/(\text{energy lost per cycle})$, or $Q={\omega }_{0}E/P_{\mathrm{loss}}$ in the weak-loss limit. For viscous damping $m\ddot{x}+b\dot{x}+kx=0$, write $\gamma =b/(2m)$, ${\omega }_0=\sqrt{k/m}$ and ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The conventional oscillator parameter isIn free ringdown, fit the amplitude envelope $A(t)=A_0{e}^{-\gamma t}$. Its decay time is ${\tau }_A=1/\gamma$, whereas the cycle-averaged energy envelope decays on ${\tau }_E=1/(2\gamma )$; thus $Q={\omega }_0{\tau }_A/2={\omega }_0{\tau }_E$. Equivalently the logarithmic decrement between same-sign peaks is $\delta =\ln (A_n/A_{n+1})=2\pi \gamma /{\omega }_d$. It gives $Q\simeq \pi /\delta$ for weak damping, or the exact parameter relation $Q=\sqrt{1/4+(\pi /\delta {)}^2}$. Fit multiple peaks above the noise floor, measure the damped period, and verify an exponential envelope rather than assuming all amplitude loss is viscous.

For a driven measurement, allow transients to settle and scan frequency at fixed force amplitude. Near resonance and at high $Q$, the response is Lorentzian in displacement power, with half-power points at displacement amplitudes $1/\sqrt{2}$ of the maximum. Their angular-frequency separation gives $Q\simeq {\omega }_0/\Delta \omega$. Halving the amplitude is not a half-power criterion.

The absorbed-power bandwidth of a viscously damped oscillator can be found exactly. Sinceits maximum is at $\Omega ={\omega }_0$, and the half-power frequencies satisfy $|k-m{\Omega }^2|=b\Omega$. HenceFor broad resonances, displacement and absorbed-power peaks differ, so the measured observable must be specified. Frequency-dependent drive, backgrounds or several overlapping modes can otherwise masquerade as a changed Q factor.

The characteristic impedance of a stretched string is $Z=T/c=\rho c$, since its wave speed obeys $c^2=T/\rho$. Put the endpoint at $x=0$ and use displacement complex amplitudes $A$ incident and $B$ reflected:At the massless end, the string's transverse force $-T{\Psi }_x$ balances the damper's force $-\alpha {\Psi }_t$. Thus $T{\Psi }_x=-\alpha {\Psi }_t$, or $Z(A-B)=\alpha (A+B)$. This viscous termination of a stretched string hasHalf the incident mean power is reflected when $r=\pm 1/\sqrt{2}$, giving two positive solutions:They are approximately $0.1716\rho c$ and $5.828\rho c$, in $\mathrm{kg}{\mathrm{s}}^{-1}$. The smaller coefficient produces no reflected displacement-phase reversal; the larger produces a $\pi$ reversal. Both dissipate the remaining half of the incident power. Choosing the impedance-matched value $\alpha =\rho c$ would instead absorb all of it.

Near order $m$, the first angular zero from a principal maximum is $\delta \theta =\lambda /(ND\cos \theta )$. Two wavelengths have peak separation $\Delta \theta =m\Delta \lambda /(D\cos \theta )$. Equating them applies the Rayleigh criterion and gives the resolving power of a diffraction grating,Here the required resolving power is $400/0.2=2000$, so in second order $N\ge 1000$. The illuminated width is thereforeWhether one counts the full illuminated window or the first-to-last slit-centre distance changes the width by one pitch and is immaterial at this precision. The result assumes negligible aberrations and adequate slit illumination and coherence.

For a slit of width $a$, the single-slit diffraction amplitude is proportional to $\mathrm{sinc}(\pi a\sin \theta /\lambda )$. At a grating maximum, $D\sin \theta =m\lambda$, so the envelope factor isThe specified $a=1\mu \mathrm{m}=D/2$ gives $ma/D=1$ at $m=2$. ThusThe grating maximum coincides exactly with the first single-slit minimum for each wavelength. The preceding width and detector calculations describe the ideal array factor, but this particular rectangular-aperture grating cannot deliver the assumed second-order signal. Pixel refinement cannot repair the missing optical peak. One must change the aperture duty cycle or grating profile, or use an available order and recalculate its resolving power.

Circular apertures and circular obstructions illustrate both the usefulness and the limits of geometric shadow arguments. In scalar diffraction theory, each point of a transmitted wavefront contributes a complex secondary field. Their coherent sum, described by the Huygens-Fresnel principle, determines the intensity. Circular symmetry simplifies the integral, but the answer depends on whether propagation is in the Fraunhofer diffraction or Fresnel diffraction regime.

For aperture radius $a$ and axial propagation distance $z$, the Fresnel number $N_F=a^2/(\lambda z)$ measures the importance of quadratic phase across the aperture. The far-field condition is $N_F\ll 1$; a lens can produce the same angular spectrum in its back focal plane. Fresnel propagation retains that quadratic phase. For an axial point source a distance $s$ before the object and an observer a distance $r$ after it, the phase uses the effective distance $z_{\mathrm{eff}}=sr/(s+r)$, because both incident and outgoing spherical paths contribute curvature. Scalar, paraxial propagation and an opening large compared with wavelength are assumed throughout.

A circular hole in the far field gives an Airy pattern. For uniform plane illumination the field is the two-dimensional Fourier transform of the aperture. In polar coordinates, let $Q=k\sin \theta$; the angular integral gives the Bessel function $J_0$:ThusThe limiting central value is one because $J_1(u)\simeq u/2$ near zero. A bright central Airy disk is surrounded by progressively weaker concentric rings. The first dark ring has $ka\sin \theta =3.8317$, or $\sin \theta \simeq 1.22\lambda /D$ with diameter $D=2a$. This scale gives the Rayleigh criterion for a circular optical instrument. Larger apertures narrow the angular pattern; a lens of focal length $f$ maps the first dark ring to radius $1.22\lambda f/D$ in its focal plane.

In the Fresnel regime, the axial intensity oscillates as aperture size or distance changes. Fresnel half-period zones have boundaries at path excesses $m\lambda /2$ and approximately opposite phases in successive zones. For a plane wave, their radii obey $r_m^2=m\lambda z$. A circular opening can therefore expose alternating constructive and cancelling zone contributions. More precisely, Fresnel diffraction by a circular aperture gives, normalized by the unobstructed field,An odd integer number of zones gives a bright axial point with ideal intensity $4I_0$; an even number gives an axial zero. Intensity can locally exceed the unobstructed value because interference redistributes energy; total transmitted power still obeys conservation of energy. The relative normalisation here differs from the Airy expression, which normalises each far-field pattern by its own central intensity.

Off axis, the circular Fresnel field isIts concentric structure evolves with $N_F$ and includes fringes near the geometrical edge. The axial zone argument does not determine the complete off-axis pattern.

Complementary holes and opaque disks are related at the field level. Babinet's principle states $U_a+U_d=U_0$ for complementary aperture and disk transmissions under the same illumination. In the Fraunhofer region away from the direct transmitted beam, the background field is absent and the diffracted fields are opposite in sign. Their scattered intensity profiles therefore have the same Airy angular dependence. Exactly in the forward direction the unobstructed field must be retained; subtracting intensities is incorrect. Finite incident beams also give a finite-width direct beam rather than an ideal angular delta function.

For Fresnel diffraction by an opaque disk, the axial complement is particularly striking:The ideal circular disk has an Arago spot at the centre of its geometric shadow. Circular edge contributions reach the axial point with the same phase. This does not mean the whole shadow is illuminated: off-axis cancellation produces the surrounding shadow and rings, and the axial spot becomes narrower for larger obstruction or shorter propagation distance. An aperture's axial intensity oscillates, while its complementary disk's axial intensity stays constant in this ideal paraxial model.

A practical observation uses a compact coherent source, an accurately circular hole or disk, and an aligned screen or camera. Varying the propagation distance or aperture radius reveals the transition between regimes. A point source, or spatial filtering, prevents different source positions from washing out the fine structure. Finite source size, incomplete coherence, edge roughness, misalignment, vector effects for wavelength-sized objects and lens aberrations reduce the ideal contrast. These experimental limitations explain why a geometrically unexpected axial bright point requires a suitable setup even though its scalar derivation is simple.

A characteristic wave impedance relates an effort-like quantity to its conjugate flow-like quantity in a single travelling wave: for a mechanical wave these are force and velocity, and their product is power. The sign depends on the propagation and force conventions. Use the transverse force transmitted towards increasing $x$, $F=-T{\Psi }_x$. A rightgoing travelling wave $\Psi =f(x-vt)$ has transverse velocity $u={\Psi }_t=-v{\Psi }_x$, soThis is the positive characteristic impedance of a stretched string. A leftgoing component has $F/u=-Z$ with the same force orientation. Its units differ from pressure-based acoustic impedance because it uses force, rather than force per area.

For a rightgoing harmonic travelling wave in the lossless string, its force and transverse-velocity complex amplitudes satisfy $F=Zu$ with real $Z=\sqrt{T\rho }$. Thus the supplied mean-power expression giveswhere $A$ is the displacement complex amplitude and $u=-i\omega A$ for convention $e^{i(kx-\omega t)}$. The same result follows directly from instantaneous string power $P=-T{\Psi }_x{\Psi }_t$ averaged over a cycle. The amplitudes here are peak amplitudes; with root mean square force or velocity the factor $1/2$ is absent. If reflections return to the driver, its net mean input is still $\mathrm{Re}(F{u}^{\ast })/2$, but its input mechanical impedance need not equal the travelling-wave value $Z$.

The reflected and transmitted mean-power fractions areFor strictly positive finite linear mass densities, $\mathcal{T}>0$; a finite density step cannot produce exact total reflection. The ideal limits areThe first behaves as a fixed end; the second as a free end. In the free-end limit the transmitted displacement tends to $2A$, but its mechanical impedance tends to zero, so it carries vanishing power. Thus total reflected energy does not require a vanishing boundary displacement. “Entirely reflected” at a merely large density contrast is an approximation.

In thin-film interference, coherent beams reflected at the two faces of a transparent film overlap. Let the film have refractive index $n_f$, thickness $t$, and internal ray angle ${\theta }_f$, determined by Snell's law. For vacuum wavelength ${\lambda }_0$, the propagation part of the relative phase isAlthough the internal round-trip ray length is $2t/\cos {\theta }_f$, the two emergent rays must be compared on the same outgoing wavefront. Subtracting their external lateral path contribution gives the effective optical path difference $2n_{f}t\cos {\theta }_f$, explaining the cosine rather than its reciprocal.

Reflection also changes phase. At normal incidence in lossless positive-index media, the electric-field Fresnel equations give $r_{ij}=(n_i-n_j)/(n_i+n_j)$: reflection towards higher index reverses sign, whereas reflection towards lower index does not. If the two interfering reflections differ by one such phase reversal, their total relative phase is ${\delta }_{\mathrm{prop}}+\pi$ modulo $2\pi$. Then reflected maxima and minima obeyIf there are zero or two reversals, the conditions interchange. These bright/dark conditions assume overlapping beams with adequate temporal optical coherence; complete cancellation further requires equal amplitudes. In transmitted light, a lossless complete film has complementary intensity variations because reflected plus transmitted power is conserved.

For multiple internal reflections at normal incidence, summing the geometric series gives the exact film reflection amplitudeThis includes both the amplitude mismatch and repeated-beam interference. A normal-incidence antireflection coating on substrate index $n_s>n_0$ can cancel reflection at one wavelength when $n_f=\sqrt{n_0{n}_s}$ and $t={\lambda }_0/(4n_f)$: both interface reflections reverse sign and the film's round-trip propagation adds $\pi$, with equal interface amplitudes. A varying thickness or angle selects different maxima for different wavelengths, producing the colours of transparent films in white light. Broad bandwidth and large incidence angles generally prevent perfect cancellation across all wavelengths.

Assume a massless junction and common tension $T$. The characteristic impedance of a stretched string is $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ is its linear mass density. The displacement reflection coefficient is $r=(Z_1-Z_2)/(Z_1+Z_2)$, so the reflected fraction of power isSet $\mathcal{R}=1/2$. The positive solutions are $z=3\pm 2\sqrt{2}$, which are reciprocals. Thus the string density ratio for prescribed reflected power isThe impedance ratio is about $5.83$, rather than the required density ratio. Reversing which string is on the incident side changes the reflected displacement sign but not the reflected power fraction.

Let ${\lambda }_1=589.0\mathrm{nm}$, ${\lambda }_2=589.6\mathrm{nm}$ and $\Delta \lambda =0.6\mathrm{nm}$. Requiring the separation to be at least $100$ times the measured full width at half maximum means a width no larger than $0.006\mathrm{nm}$, or a resolving power of approximately $589.3/0.006\simeq 9.8\times 10^4$.

The etalon resolving power and order overlap constraints are particularly clear in frequency. Put $\Delta \nu =c|1/{\lambda }_1-1/{\lambda }_2|$. At normal incidence, the free spectral range is ${\nu }_{\mathrm{FSR}}=c/(2d)$ and the instrumental width is $\delta \nu ={\nu }_{\mathrm{FSR}}/\mathcal{F}$. ThusAvoiding ambiguity between the doublet and adjacent interference orders requires a free spectral range exceeding the doublet separation:Together these giveUsing the narrow-peak finesse from the previous part, $\pi \sqrt{R}/(1-R)>100$ gives $\boxed{R\gtrsim 0.9691}$, with a practical margin above that limiting value. At the exact limit, the gap between orders vanishes, so equality is not a sensible operating point.

One feasible ideal design is $R=0.99$, $\mathcal{F}\simeq 313$, and $d=0.14018\mathrm{mm}$. It has an approximate wavelength free spectral range of $1.24\mathrm{nm}$ and instrumental width $0.0040\mathrm{nm}$ near the doublet. The doublet spacing is about $151$ widths. Even the spacing to the nearest replica of the other line exceeds $100$ widths in this example. If that latter separation is also required in general, impose the stronger condition $\min (\Delta \nu ,{\nu }_{\mathrm{FSR}}-\Delta \nu )\ge 100{\nu }_{\mathrm{FSR}}/\mathcal{F}$, rather than just the two basic inequalities.

Such mirror reflectivities and submillimetre gaps are plausible for a high-quality Fabry–Pérot etalon. Flatness, parallelism, coating loss, angular spread, mechanical stability and source width limit the achieved finesse. A wavelength width of $0.004\mathrm{nm}$ corresponds to a gap variation of order $d\delta \lambda /\lambda \simeq 1\mathrm{nm}$, so the ideal coating-limited estimate demands stringent uniformity. The paper expressly excludes intrinsic source broadening.

The first panel is the fixed etalon's transmission versus wavelength, a comb of resonances. It does not by itself have one peak at each sodium line. The second panel shows the two peaks obtained by scanning the etalon phase or gap and calibrating the scan to wavelength. A fixed-gap output spectrum is instead the incident spectrum multiplied by the transmission function. This distinction is needed to interpret a doublet measurement correctly.

A high-finesse Fabry–Pérot etalon obtains very narrow transmission peaks by summing many coherent round trips, allowing high resolving power with a compact optical path. It can transmit a substantial fraction of resonant light when coating absorption is small, and works well for precise relative line separations and narrow-line measurements. A diffraction grating may require many illuminated grooves to achieve the same resolving power.

The etalon's free spectral range is limited, and repeated orders cause ambiguity across a broad optical spectrum. It commonly needs scanning and an order-selecting optical filter or another dispersive element. Mirror alignment, flatness and illumination angle strongly affect peak positions and widths, and absorption lowers throughput. A diffraction grating instead separates a broad range of wavelengths spatially for simultaneous detection, although grating orders can also overlap. The preferred instrument depends on whether narrow-band precision or broad spectral coverage is the priority.