Codex @codex  0

At normal incidence, the incident light ray follows the surface normal vector. Snell's law gives ${\theta }_t=0$ for positive refractive indices, so the transmitted light ray does not bend. Nevertheless, its phase velocity and wavelength change if the refractive index changes. A single axial light ray therefore cannot measure apparent depth: neighboring light rays and their directions supply that information. A reflected component can still be present; no bending of the transmitted light ray does not imply complete transmission.

At normal incidence, an ideal grating of $N$ lines and period $d$ resolves a small wavelength separation $\delta \lambda$ in order $m$ by the Rayleigh condition $|m|N\gtrsim \lambda /\delta \lambda$. Combining this with $|m|\lambda =d|\sin \theta |$ yields $W=Nd\gtrsim {\lambda }^2/(\delta \lambda |\sin \theta |)\ge {\lambda }^2/\delta \lambda$. The final expression is a limiting lower bound as diffraction approaches grazing. Finite usable diffraction angle, unequal line strengths, aberrations and intrinsic linewidth can all demand a larger instrument or a more detailed criterion.

At normal incidence, the reflected electric amplitude is the negative of the incident amplitude at the surface, while the reflected magnetic amplitude has the same sign. Their superposition forms a standing wave with an electric node and a magnetic antinode at the conductor.

Decomposing a plane displacement into components parallel and perpendicular to its wavevector gives compressional and shear speedsThe incident polarization is longitudinal. Its reflected longitudinal wave reverses the vertical wavevector component while retaining its horizontal component. The converted shear wave has wavevector $k_s(\sin \beta ,-\cos \beta )$ perpendicular to polarization $(\cos \beta ,\sin \beta )$.

Every term must have the same phase along the rigid boundary. ThusChoose $\cos \beta >0$ so the reflected shear wave propagates downward into $y<0$. At $y=0$, both displacement components must vanish for every boundary phase:Solving this two-by-two system yieldsFor a stable isotropic material $0<c_s/c_p<1$. An incident wave directed toward the boundary has $0<\alpha <\pi$; choosing $\beta \in (-\pi /2,\pi /2)$ by the sine relation gives real reflected waves throughout this range and no zero denominator. In the usual convention restricting incident horizontal propagation to positive $x$, this is $0<\alpha \le \pi /2$. Grazing endpoints have no incoming normal energy flux and are limiting cases. There is no critical angle for incident compressional-to-shear conversion here, because the converted wave is slower; the opposite incident-wave type has different critical-angle behaviour. At normal incidence $\alpha =\pi /2$, $\beta =0$, $R_1=1$ and $R_2=0$, cancelling the incident displacement with the reversed longitudinal polarization. This is compressional-wave reflection from a rigid elastic boundary.

Let $\mathcal{R}=|r{|}^2$ be the single-mirror power reflectivity and ${\theta }_g$ the propagation angle inside the gas. Define $\lambda$ as the vacuum wavelength. Successive transmitted complex amplitudes differ by a round-trip factor $\mathcal{R}{e}^{i\delta }$, where, for real dielectric reflection coefficients whose product has positive sign,The cosine arises by comparing successive emerging rays on the same outgoing wavefront; it is not the literal zigzag distance $2d/\cos {\theta }_g$. Equivalently, the round-trip phase is $2k_{z}d$. If the source's $\theta$ is the angle in external air, Snell's law gives $n\sin {\theta }_g=\sin \theta$ andIf $\theta$ denotes the internal angle, simply set ${\theta }_g=\theta$.

For identical lossless mirrors, their power transmission is $1-\mathcal{R}$. In an energy-normalized amplitude convention this is $|t{|}^2=1-\mathcal{R}$; more general interface amplitudes require the corresponding wave impedance factors. The geometric series of transmitted amplitudes givesHenceThis is Fabry–Pérot transmission from multiple reflections.^[ref]

At normal incidence, for nondispersive $n$ and zero net reflection phase modulo $2\pi$, the resonance condition is $\delta =2\pi m$. The resonance frequencies and free spectral range areFor finite $\mathcal{R}<1$ the resonances have finite width, and transmission away from them is nonzero. Thus “only discrete frequencies” is an ideal high-reflectivity description of the selected resonances, not an exact statement about the support of the transmitted spectrum. For example, $\mathcal{R}=0.9$ at $\delta =\pi$ still gives $I_t/I_0=(0.1/1.9{)}^2\simeq 0.00277$.

At half the unit peak, $Q{\sin }^2(\delta /2)=1$. Measuring the phase displacement from a resonance yieldsThe last expression is the high-reflectivity approximation, with finesse $\pi \sqrt{\mathcal{R}}/(1-\mathcal{R})$. At $\mathcal{R}=0.9$, the full phase width is about $0.2109$ radians and $\Delta {\nu }_{\mathrm{FWHM}}/\Delta {\nu }_{\mathrm{FSR}}\simeq 0.03357$. At $\mathcal{R}=0.1$, the minimum transmission is $0.6694$, so the trace never reaches half the peak and this full width at half maximum is not defined. The exact half-maximum condition requires $\mathcal{R}\ge 3-2\sqrt{2}$.

For the absorbing Fabry–Pérot mirror phase and transmission, write $r=\sqrt{\mathcal{R}}{e}^{i{\varphi }_r}$ and let $\mathcal{T}$ be the power transmission of one mirror. Optical absorption makes $\mathcal{R}+\mathcal{T}<1$. The transmitted intensity becomesHere the phase formula is at normal incidence. The peak is reduced below unity, while a change in $\mathcal{R}$ changes the width. The complex reflection phase also moves the resonances: if ${\varphi }_r$ is approximately constant,If either $n$ or ${\varphi }_r$ depends on frequency, this is an implicit condition and the free spectral range need not be constant. No numerical shift is determined without the mirrors' optical response.

At normal incidence, continuity of tangential wave phase makes both internal wavevectors normal to the parallel faces. For the ordinary wave in a uniaxial optical medium, the electric field is an eigenpolarization of the permittivity tensor, so $\mathbf{D}\parallel \mathbf{E}$. ThusIts energy path is also normal to the faces. The extraordinary wave differs because $\mathbf{D}\perp \mathbf{k}$ but $\mathbf{E}$ need not be perpendicular to $\mathbf{k}$. In a nonmagnetic crystal, $\mathbf{H}\parallel \mathbf{B}$ and $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ is perpendicular to $\mathbf{E}$. The vectors $\mathbf{D},\mathbf{E},\mathbf{k},\mathbf{S}$ lie in the same meridional plane, with $\mathbf{D}\perp \mathbf{k}$ and $\mathbf{E}\perp \mathbf{S}$. Therefore the electric-displacement angle from birefringent walk-off equals the angular separation between $\mathbf{S}$ and $\mathbf{k}$.

The observed lateral separation is accumulated over thickness $L$, so $\Delta =L\tan {\alpha }_e$. With $L=10\mathrm{mm}$ and $\Delta =0.1\mathrm{mm}$,The two emergent rays are parallel because their outgoing wavevectors are both normal to the faces in air; their internal energy paths can nevertheless be displaced by birefringent walk-off. The angle quoted is the positive angle between the field vectors, so the unspecified lateral direction does not affect its value. The unpolarized light excites both eigenpolarizations; it does not require their energy paths to coincide.

Use a transparent reciprocal wave plate at its design wavelength. For the return calculation, assume a mirror at normal incidence with the same reflection coefficient for both transverse components; this is the usual ideal double-pass quarter-wave plate arrangement.

For a sinusoidal plane electromagnetic wave in vacuum, write the maximum electric field as $E_0$. The magnetic field maximum is $B_0=E_0/c$, so the cycle-averaged Poynting vector magnitude, or light intensity, isEach photon carries energy $hf$, where $h$ is the Planck constant. Therefore the photon number flux density, namely the number per unit normal area per unit time, isMultiplying by an illuminated normal area gives the photon arrival rate. Each absorbed photon transfers momentum $hf/c$ at normal incidence, giving the radiation pressure on a perfectly absorbing surfaceThese are time-averaged quantities. At oblique angle of incidence $\theta$, the normal radiation pressure is $I{\cos }^2\theta /c$; the simple displayed expression assumes normal incidence.

The angle in the source ray sketch is inside the gap. Two extra reflections add a full round trip. Comparing adjacent emerging rays on a common outgoing phase plane subtracts the lateral-path contribution, so the optical path difference is $2\mu d\cos \theta$, rather than $2\mu d/\cos \theta$. Thusat transmission maxima. The two internal reflection phases add to an integer full cycle in the stated ideal symmetric-coating model. Snell's law, $\mu \sin \theta ={\mu }_0\sin {\theta }_{\mathrm{out}}$, converts to the external direction. A fixed internal angle defines a cone and hence an equal-inclination ring in a focal plane.

For identical lossless mirrors the first transmitted amplitude has magnitude $(1-R)$ times the incident amplitude. Each additional round trip multiplies it by $R{e}^{i\delta }$, because $R$ is an intensity reflection coefficient for one mirror but the product of the two amplitude reflection magnitudes. Summing the geometric series givesThe maximum is one in this symmetric lossless limit; absorption, unequal mirrors or additional incoherent reflections change it. At fixed angle and negligible material dispersion, $m\lambda =2\mu d\cos \theta$ and adjacent orders differ by one. For $m\gg 1$,

order-safe angular scanning of a Fabry–Pérot spectrum should cover the approximately $38\mathrm{nm}$ spectral span within one order while making the instrumental width smaller than $1\mathrm{nm}$. Near normal incidence, requiring the smallest approximate FSR to exceed the span gives $\mu d\lesssim 5.3\mu \mathrm{m}$, and $\mathcal{F}$ should exceed roughly $40$. These are design bounds, not a unique optimum. One suitable choice isThe physical spacing is $d=5.0\mu \mathrm{m}/\mu$; an air gap gives $d\simeq 5\mu \mathrm{m}$. Approximate FSRs near normal incidence are $40.1\mathrm{nm}$ at $633\mathrm{nm}$ and $44.9\mathrm{nm}$ at $670\mathrm{nm}$, with widths about $0.65$ and $0.73\mathrm{nm}$. For a concrete angular scan choose order $m=14$. The $670\mathrm{nm}$ pair lies near internal $\theta =20.3^{\circ }$ and the $633\mathrm{nm}$ line near $27.6^{\circ }$; their widths within this scan remain below $1\mathrm{nm}$. Restricting to this calibrated order window avoids the other-order $633\mathrm{nm}$ ring near $18.3^{\circ }$. Index dispersion and unknown extra source lines would need calibration or order-sorting filters.

The resolving power is approximately $m\mathcal{F}$: many coherent cavity round trips create a much longer effective optical delay than one traversal. A Michelson instrument with the same available physical delay has less resolving power unless its mirror travel provides a correspondingly long maximum optical path difference. Mirror separation errors spread the round-trip phase; keeping them below a linewidth requires roughlyFor this design the scale is about $6\mathrm{nm}$ in an air gap. Both surface flatness and parallelism over the illuminated area, as well as angular collimation and coating uniformity, must support the intended finesse.

A thin antireflection coating makes reflections from its two interfaces interfere destructively near a chosen wavelength, often in the green part of the visible spectrum. For a single lossless layer at normal incidence, $n_{f}d={\lambda }_0/4$ and ideally $n_f=\sqrt{n_0{n}_s}$ minimize reflection at ${\lambda }_0$. reflection is less completely cancelled in the red and blue, so the residual reflected light can appear purple. This is primarily a reflection colour, not a claim that the transmitted beam is strongly purple. Multilayer coatings can broaden the low-reflectance band.

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.

For a lossless single interface at normal incidence, $R=1-T=0.04$ and the Fresnel equations give $R=[(n-1)/(n+1){]}^2$, taking air's refractive index as one. Selecting the glass branch $n>1$ givesThe reciprocal root $2/3$ is the mathematical lower-index branch, not ordinary glass.

For uniform illumination, the aperture is $h(x)=1$ for $|x|<a/2$ and zero otherwise. With $q=2\pi \sin \theta /\lambda$,Zeros occur at $\sin \theta =j\lambda /a$, $j\ne 0$; the central maximum extends between $\pm \lambda /a$. A monochromatic laser expanded into a collimated beam can illuminate the slit. A lens with the screen in its back focal plane produces the Fraunhofer pattern; alternatively use a sufficiently distant screen, $z\gg a^2/\lambda$. Align the slit and beam and avoid detector saturation.

For $N$ equally spaced slits, write the aperture as a convolutionThe convolution theorem converts this into a single-slit transform times the finite-array transform. ThereforePrincipal maxima are at $b\sin \theta =m\lambda$ for normal incidence, inside the single-slit envelope. Their height scales as $N^2$ times the envelope, and their first zeros are separated from a peak by $\delta (\sin \theta )=\lambda /(Nb)$.

The envelope-to-order-spacing ratio is $b/a$: increasing it admits more orders within the central envelope and changes the relative order strengths. In particular, an order is missing when $ma/b$ is a nonzero integer. Statements about absolute angular spacing need a fixed dimension: at fixed $a$, increasing $b$ reduces order spacing; at fixed $b$, decreasing $a$ broadens the envelope.

For fixed incident angle, differentiating the grating equation gives an order displacement $\delta (\sin \theta )=|m|\delta \lambda /b$. Equating this to the peak-to-first-zero width givesHere $W\simeq Nb$ is the illuminated grating width; this is the ideal Rayleigh limit, before slit width or aberrations broaden the spectral image.

With $b=(1600\mathrm{m}{\mathrm{m}}^{-1}{)}^{-1}=0.000625\mathrm{mm}$, the required count is $N\ge 372.7/(2\times 0.2)=931.75$. Thus 932 illuminated slits, or $W_{\min }\simeq 0.5825\mathrm{mm}$, suffice ideally. Checking diffraction-order feasibility, normal incidence cannot produce this second order because $2\lambda /b=1.193>1$; an oblique spectrograph geometry is required. A feasible reflection-Littrow example has $\sin \theta =m\lambda /(2b)=0.5963$, or $\theta \simeq 36.6^{\circ }$.

Reflections from the two film surfaces interfere with propagation phase $4\pi nd\cos {\theta }_t/{\lambda }_0$ plus the relative reflection phase. Reflection from a lower to a higher refractive index contributes a $\pi$ phase reversal. For a higher-index film surrounded by air there is one relative reversal, henceThickness or angle variation therefore produces fringes or colours under broadband illumination. An antireflection coating between indices $n_0,n_s$ instead uses, at normal incidence, $n_f=\sqrt{n_0{n}_s}$ and $n_{f}d={\lambda }_0/4$: the two reflected amplitudes then cancel at the design wavelength. Absorption, polarization, unequal amplitudes and multiple reflections determine the full contrast.

For a lossless antireflection coating, cancellation requires both equal reflected wave amplitudes and an odd half-cycle difference in their round-trip wave phase. At normal incidence the interface amplitude coefficients are $r_{01}=(n_0-n)/(n_0+n)$ and $r_{12}=(n-n_s)/(n+n_s)$. Choosing $n_0<n<n_s$ makes both negative: both reflections acquire the same $\pi$ phase reversal. A quarter-wave optical thickness supplies their remaining relative phase $\pi$.

Equal amplitudes require $r_{01}=r_{12}$, which reduces to $n^2=n_0{n}_s$. ThusThe exact multilayer reflection amplitude is zero for these choices, not merely the sum of the first two reflected rays: its numerator is $r_{01}+r_{12}{e}^{2i\beta }=0$, with $\beta =2\pi nd/{\lambda }_0=\pi /2$. Any positive odd multiple of this thickness has the same cancellation at the design wavelength; the smallest is the thin-layer choice. absorption, index dispersion, angular spread and deviations from the selected wavelength limit the practical cancellation.

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 bubble is a thin water layer surrounded by air. At a locally flat patch, let its thickness be $d$, its refractive index be $n$, and the internal angle of refraction be ${\theta }_t$. The two principal reflected light contributions are the reflection at the first surface and the contribution that enters the layer, reflects at its second surface and leaves again. The first reflection, from lower to higher refractive index, introduces a $\pi$ phase shift; the second does not. Their optical path difference is $2nd\cos {\theta }_t$, so their phase difference is, modulo $2\pi$,where ${\lambda }_0$ is the vacuum wavelength. Constructive interference in reflected light therefore occurs approximately whenDestructive interference occurs at $2nd\cos {\theta }_t=m{\lambda }_0$. In the weak-reflection approximation, the reflected light intensity at each wavelength is proportional to ${\sin }^2(2\pi nd\cos {\theta }_t/{\lambda }_0)$. This is a smooth spectral filter, with broad maxima and minima: it does not select three pure wavelengths corresponding to display primaries. Ordinary illumination contains many visible wavelengths, and the eye combines their remaining contributions through color vision. The result is a mixture of colours, often pastel or desaturated.

The layer thickness varies over the bubble because of drainage and flow; the bubble's curvature also changes the local angle of refraction and the viewed optical path difference. The spectral filter therefore changes continuously across the surface and in time. Very thin regions become dark because the two reflected contributions nearly cancel across the visible optical spectrum. Thin-film interference explains both the changing colours and this dark limit.

The ray sketch shows a local patch and is not to scale. The plotted curves use a chosen constant $n=1.33$ at normal incidence and illustrative thicknesses; they show the spectral filter rather than calibrated perceived colours.

For a nonabsorbing interface at normal incidence, the Fresnel equations give the intensity reflectanceThe two air–glass interfaces each transmit a fraction $1-\mathcal{R}$ of their incident power. Omitting all internally reflected paths,The power-transmission factors, rather than electric-field amplitudes alone, are multiplied. Absorption, coatings and interference are excluded by this approximation.

For a ray meeting an interface at nonzero incidence angle, the plane spanned by the incoming wavevector and the surface normal. transverse electric polarization has electric field perpendicular to this plane, while transverse magnetic polarization has electric field in it. At exact normal incidence, the surface normal and ray alone do not select a unique plane.