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By linear scalar diffraction theory, complementary aperture transmissions $t$ and $1-t$ give $U_t+U_{1-t}=U_{\mathrm{open}}$ under identical illumination. In Fraunhofer diffraction, away from the direct forward beam $U_{\mathrm{open}}=0$, so $U_{1-t}=-U_t$: the complementary screens have identical off-axis diffraction intensities. This is Babinet's principle. Their complex amplitudes, rather than their intensities, add. The direct beam is the exception; real conducting-edge vector effects are outside this ideal scalar statement.

The wave impedance is the travelling-wave field ratio $Z=E/H$; for the stated nonmagnetic media it is $Z_0/n$. A reflection coefficient is a reflected-to-incident complex amplitude ratio with an explicitly chosen field orientation. With this reflection-amplitude sign convention, using the question's formula for incidence inside silicon givesWith a common tangential electric-field direction, the usual electric-field Fresnel equations instead give $r_E=(Z_2-Z_1)/(Z_2+Z_1)=+5/9$ for this incidence. The printed formula corresponds to the opposite reflected-field sign convention, for example the magnetic amplitude convention. It gives the same reflectance and the same round-trip factor $r^2$.

Each extra internal round trip multiplies the transmitted amplitude by $r^2{e}^{i\delta }$, with $\delta =4\pi nd/\lambda$. A geometric series therefore givesThis is Fabry–Pérot transmission from multiple reflections. Assuming lossless silicon with wavelength-independent $n$, the full intensity transmittance between air regions isMaxima have $2nd/\lambda =j$, minima $2nd/\lambda =j+\frac{1}{2}$. Here $2nd=7000\mathrm{nm}$, so within the specified tuning range the only maximum is $1400\mathrm{nm}$, and the two minima are $1555.6\mathrm{nm}$ and $1272.7\mathrm{nm}$. The contrast is

Identical highly reflecting coatings increase the finesse: transmission maxima become narrower and the intensity between them falls. Ideal symmetric lossless coatings retain $T_{\max }=1$; absorption or unequal mirrors reduce it. The illustrated maxima stay fixed only because additional reflection wave phases are held unchanged. Real coating wave phases can also shift the resonances.

time average of harmonic power. Expand the two real parts:The terms proportional to $e^{\pm 2i\omega t}$ average to zero over complete periods, while $F{u}^{\ast }$ is time-independent. For real peak amplitudes this is $F_0{u}_0\cos \varphi /2$.

String dynamics. A short element of length $dx$ experiences transverse force $T[{\psi }_x(x+dx)-{\psi }_x(x)]$. Newton's second law therefore gives $\rho dx{\psi }_{tt}=Tdx{\psi }_{xx}$, orThe approximation is a flexible, lossless string with small slope and constant tension; $\rho$ is its linear mass density.

energy transport. Across a cut toward increasing $x$, the transverse force is $F_y=-T{\psi }_x$ and the transverse velocity is $u={\psi }_t$. For a rightgoing wave these have complex amplitudes $ikTA$ and $i\omega A$. HenceThis real mechanical impedance is force divided by transverse velocity for a travelling wave, with units $\mathrm{kg}{\mathrm{s}}^{-1}$.

Matching and power ratios. At an ordinary massless junction without an external transverse force, displacement is continuous and transverse tension is balanced: ${\psi }_1={\psi }_2$, $T_1{\psi }_{1x}=T_2{\psi }_{2x}$. For the same tension, the slope is therefore continuous even if density changes. With real positive impedances,Their sum equals $P_i$; the transmitted power includes the impedance ratio and is not simply $t_{\psi }^2{P}_i$.

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.

Let the first transmitted complex amplitude be $T{E}_{\mathrm{in}}{e}^{ikd}$: two amplitude transmission factors each of magnitude $\sqrt{T}$ give magnitude $T$. Each extra internal round trip adds two amplitude reflections, with combined magnitude $R$, and phase $\delta$. With the mirror phase convention already incorporated in the stipulated $\delta$,Taking the squared modulus givesIt is $R$, not $R^2$, that multiplies the field after a round trip; $R^2$ would be the corresponding round-trip intensity fraction.

At resonance $\delta =2m\pi$, $B_{\max }=T^2/(1-R{)}^2$. If $h$ is the positive phase displacement to half maximum,For $R=0.80$, the stipulated high-reflectivity approximation gives $h=0.2236\mathrm{rad}$, while the exact value is $0.2241\mathrm{rad}$. The full width is twice this value. With $T=0.20$, $B_{\max }=1$, $B_{\min }=T^2/(1+R{)}^2=1/81$ and the pattern repeats every $2\pi$. Its ideal finesse is approximately $\pi \sqrt{R}/(1-R)=14.05$.

A wavelength component resonates at $d=m\lambda /2$. Within the scan interval, the individual resonances areNo neighboring orders of these lines fall within the stated interval. The observed signal is the sum of their two Airy transmissions, since different spectral components add intensities. Individual resonance positions identify the two peaks to the plotted precision; the tails cause very small shifts of the maxima of the sum.

The gap-coordinate full width isThe separation is $0.100\mu \mathrm{m}$, about $2.8$ full widths, so the two components can be resolved by the ideal etalon. The equivalent wavelength resolution near order $1000$ is about $\lambda /(m\mathcal{F})\simeq 0.071\mathrm{nm}$, smaller than their $0.2\mathrm{nm}$ separation.

The added wavelength has an order-$999$ resonance atIt lies between the preceding two peaks; its adjacent orders are outside the scan. Its input strength was not specified, so the third peak's height cannot be uniquely assigned. The original plot uses equal strength only as a declared illustration.

This is an order-identification problem: the free spectral range near $d=0.5\mathrm{mm}$ is $\Delta {\lambda }_{\mathrm{FSR}}\simeq {\lambda }^2/(2d)\simeq 1\mathrm{nm}$. A line more than one free spectral range away can appear among peaks of another order. A lower-resolution prefilter or additional dispersive spectrometer is needed to identify the order; a narrow peak alone does not identify an absolute wavelength. mirror parallelism, beam collimation and detector angular acceptance must also be controlled before attributing a measured width entirely to reflectivity.^[ref]

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.

Use the $e^{i\omega t}$ convention for the harmonic force and displacement complex amplitudes. Substitution in the forced harmonic oscillator equation givesIts magnitude and phase relative to the force areAt zero frequency, $|R|=1/k$ and $\varphi =0$. At ${\omega }_0=\sqrt{k/m}$, $\varphi =-\pi /2$; at high frequency the phase approaches $-\pi$. For light viscous damping, the quality factor $Q=m{\omega }_0/b$ is large and there is a tall narrow displacement peak close to ${\omega }_0$.

For the steady response of a forced harmonic oscillator, write $x=\mathrm{Re}(A{e}^{i\omega t})$ and use the same complex amplitude convention for the driving force. Substitution givesThus the response magnitude and displacement phase relative to the force areThe phase goes from $0$ at low frequency, through $-\pi /2$ at ${\omega }_0$, towards $-\pi$ at high frequency. The static magnitude is $1/(m{\omega }_0^2)=1/k$, while the high-frequency magnitude is approximately $1/(m{\omega }^2)$.

Define the quality factor by $Q={\omega }_0/\gamma$. If $Q>1/\sqrt{2}$, the displacement peak occurs at ${\omega }_r=\sqrt{{\omega }_0^2-{\gamma }^2/2}$; for $Q\le 1/\sqrt{2}$ the magnitude decreases monotonically from its static value. For $Q\gg 1$ there is a tall narrow peak close to ${\omega }_0$, of height approximately $Q/k$ and width approximately $\gamma$. At very small $Q$ the resonant peak disappears and the response is broad. The plotted functions are normalized as $k|R|$, at fixed $m,k$.

The capacitor-charge complex amplitude isFor fixed driving-voltage amplitude, maximizing $|q_0|$ minimizesDifferentiate:The positive-frequency minimum therefore hasThis is the resonance of charge, or capacitor voltage. If the stated inequality fails, their maximum is at zero frequency, and no positive displacement-resonance peak exists. The electric current has the additional factor $\omega$, and instead peaks exactly at ${\omega }_0=1/\sqrt{LC}$ for every $R>0$.

Let the aperture occupy the $z=0$ plane, with coordinates $(x,y)$ and opening $\Sigma$. Write ${\psi }_{\Sigma }(x,y)$ for the incident scalar complex amplitude and $h(x,y)$ for its dimensionless amplitude transmission, including any imposed phase. The observation point has transverse coordinates $(X,Y)$, corresponding to the PDF's $(x_0,y_0)$, and axial distance $R$. Also $k=2\pi /\lambda$ is the wavenumber and $i^2=-1$.

The Huygens-Fresnel principle expresses the propagated field, up to common prefactors and an approximately constant obliquity factor, asIn the paraxial approximation, expand the path asThe last term is negligible in the Fraunhofer diffraction limit, while the first two terms and $1/r\simeq 1/R$ are common across the aperture. Factoring them out leavesIt is the two-dimensional Fourier transform of the transmitted aperture field, evaluated at transverse wavevector $(kX/R,kY/R)$. For appreciable observation angles the corresponding exact directional components are used rather than $X/R,Y/R$.

Represent equal narrow slits at $x=\pm D/2$ by two contributions with equal complex amplitudes. In a direction $\theta$ the path difference is $D\sin \theta$, soTherefore double-slit interference givesThe bright interference fringes satisfy $D\sin \theta =m\lambda$, with a central maximum. In the small-angle region,Dark fringes lie halfway between bright ones. Exactly, spacing is uniform in $\sin \theta$, not in $\theta$ at large angles. The first panel of the pattern figure below shows the narrow-slit result.

Take the two interfering complex amplitudes to have equal intensities $I_0/2$ and relative phase $kx$. The interference cross term then givesThus the mean intensity is $I_0$, the maximum is $2I_0$, and the minima areThe optical path difference period is $\lambda$ and the mirror-displacement period is $\lambda /2$. The assumed equal intensities refer to the two fields at the chosen output, with matching polarization of waves and full optical coherence. They set the normalization; they should not additionally be interpreted as all incident light entering that output port. A physical lossless Michelson interferometer has a complementary second output.

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 $Z$.

Here $x(t)$ is displacement from equilibrium, in metres; $m>0$ is the oscillating mass, in kilograms; $b\ge 0$ is the coefficient of linear viscous damping, in $\mathrm{kg}{\mathrm{s}}^{-1}$; and $k>0$ is the spring constant, in $\mathrm{N}{\mathrm{m}}^{-1}$. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ are respectively the inertial term, the opposing velocity-dependent damping term and the opposing displacement-dependent restoring term when moved to the force side of Newton's second law.

$F_0$ is a possibly complex driving-force complex amplitude, measured in newtons. $\omega$ is the driving angular frequency in $\mathrm{rad}{\mathrm{s}}^{-1}$, and taking the real part of $F_0{e}^{i\omega t}$ gives the physical sinusoidal force. Its frequency in hertz is $f=\omega /(2\pi )$. The natural undamped angular frequency and damping rate areThis damped harmonic oscillator model assumes a linear restoring force and a linear, instantaneous viscous damping law.

For underdamping, the general response is a free transient plus a particular driven response. Use the paper's $e^{i\omega t}$ convention. The steady displacement complex amplitude isThusHere $\gamma =b/(2m)$, ${\omega }_d=\sqrt{k/m-{\gamma }^2}$ and the real constants $C,D$ enforce the initial conditions. For a real positive $F_0$, the steady term is $|x_0|\cos (\omega t-\varphi )$, whereFor light damping, ${\omega }_d\simeq {\omega }_0$ and the transient decays slowly compared with an oscillation. The two-argument phase convention makes the displacement lag evolve from approximately zero below resonance through $\pi /2$ near ${\omega }_0$ to approximately $\pi$ far above it.

In Fresnel diffraction, the propagation wave phase varies quadratically across the aperture and that curvature must be retained. For a transverse scale $w$ at propagation distance $L$, its size is of order $\pi w^2/(\lambda L)$. In Fraunhofer diffraction, that quadratic term is negligible, approximately $w^2/(\lambda L)\ll 1$, while the linear phase encoding the observation angle remains. A lens can also form the far-field angular pattern in its focal plane.

For a normally incident plane wave and aperture transmission $h(x,y)$, the far-field complex amplitude is its Fourier transform:The detected light intensity is proportional to $|\psi {|}^2$. For identical parallel slits, write the transmission as a sum of shifted copies of one slit. The Fourier shift theorem gives one single-slit transform multiplied by the sum of positional phase factors:The single-slit diffraction envelope sets the broad scale, while coherent interference between slit centres produces the finer array pattern. One sums complex amplitudes before squaring, rather than adding independent slit intensities.

Put the slit centres at $x_j=\pm D/2,\pm 3D/2$. Each slit has the same width $a$ and illumination. Its contribution to the Fraunhofer diffraction complex amplitude isThe four phase factors pair into cosines:Therefore the four-slit Fraunhofer diffraction field isThe common factor $2a$ can be absorbed in the proportionality constant, giving the requested form. In particular $\psi (0)\propto 4a$, as expected from four equal in-phase slits. Normalizing by the central intensity givesThe last form follows from $\cos x+\cos 3x=2\cos 2x\cos x$ and avoids apparent singularities in the usual sine-ratio array formula.

Place the junction at $x=0$ and use the complex amplitude convention $e^{-i\omega t}$. With incident displacement wave amplitude $A_i$, writeHere $r$ and $t$ are displacement wave reflection and transmission coefficients, and $k_j=\omega /c_j$. The junction is assumed massless, with no externally applied transverse force, and both strings have the common equilibrium tension $T$. Continuity of displacement and transverse force givesSince $T{k}_j=\omega Z_j$, these become $1+r=t$ and $Z_L(1-r)=Z_{R}t$. Solving,A larger right-hand mechanical impedance produces a reflected displacement with reversed sign, as expected in the fixed-end limit $Z_R\to \infty$. Equal mechanical impedances give $r=0$ and $t=1$.

Let the middle segment occupy $0<x<l$. Normalize the incident displacement complex amplitude to one, suppress the common time factor $e^{-i\omega t}$, and chooseHere $r$ is the reflected complex amplitude at $0$, $a$ and $b$ are the rightgoing and leftgoing middle-segment complex amplitudes referenced to $0$, and $t$ is the outgoing complex amplitude referenced to $l$. The middle wavenumber is $k_M=\omega /c_M$, andis the propagation phase factor across the lossless middle segment.

At each junction, continuity of displacement and transverse force means continuity of $y$ and $T{y}_x$. Using $T{k}_j=\omega Z_j$ givesThese boundary conditions also include all multiple reflections inside the middle segment: $a$ and $b$ are the total travelling components, not just the first waves to cross a junction.

Put $\delta =k_{M}l$. Eliminating the middle-segment complex amplitudes using the right-hand boundary conditions givesHence the displacement reflection coefficient isFor finite positive real mechanical impedances, its denominator cannot vanish at real $\delta$. The real and imaginary parts of the numerator must therefore vanish separately:This gives all three possibilities.

If the exterior strings differ, $Z_L\ne Z_R$, then $\cos \delta =0$, so $\sin \delta \ne 0$ andThe middle segment is a quarter-wave impedance transformer. Its two reflected contributions cancel at these odd quarter-wavelength thicknesses. Here ${\lambda }_M=2\pi /k_M$ is specifically the wavelength in the middle string; the matching is not generally valid at every frequency.

If $Z_L=Z_R$ but $Z_M$ differs, the second condition requires $\sin \delta =0$, givingA whole number of half-wavelengths reproduces the exterior mechanical impedance at the input. The formal $n=0$ limit is zero frequency or zero segment length, rather than a finite-wavelength resonance.

Finally, if $Z_L=Z_M=Z_R$, both conditions hold identically: $\boxed{r=0\text{ at every wavelength}}$. Since the system is lossless, each of these cases also gives unit transmitted power.