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The polarization generated by the relevant third-order nonlinear optical susceptibility is proportional to $E_1{E}_2{E}_3^{\ast }$. Multiplication of its temporal and spatial wave phase factors givesThe two opposing pump arrows in the original diagram mean ${\mathbf{k}}_2=-{\mathbf{k}}_1$. Thus the mixing source points along $-{\mathbf{k}}_3$, and its conjugate signal amplitude drives a backward field. This is counterpropagating-pump phase matching in four-wave mixing.

There is a phase matching qualification to the initial frequency statement. In the stipulated nondispersive isotropic medium, $|{\mathbf{k}}_j|=n{\omega }_j/c$. Exact equality of the backward free-wave vector to $-{\mathbf{k}}_3$ therefore requires ${\omega }_4={\omega }_3$, hence ${\omega }_3={\omega }_P$. For unequal frequencies the backward direction remains the geometry suggested by the source, but the longitudinal phase mismatch has magnitude $2n|{\omega }_P-{\omega }_3|/c$. Efficient generation over a finite length requires that mismatch times $L$ be small; a sinc function factor suppresses conversion otherwise. Nondispersion does not remove the magnitude mismatch. The remainder of the calculation is explicitly degenerate four-wave mixing, so this difficulty disappears.

At the pump frequency, for the strong field $E_1$, self-phase modulation and cross-phase modulation by $E_2$ dominate the weak signal's influence. They change its propagation wave phase through its own intensity and the other pump's intensity. With continuous uniform plane waves, these are nonlinear wave phase shifts; finite pulses acquire temporal chirp, and finite transverse beams can acquire a lens. No universal factor of two should be assigned between the two shifts without specifying their polarization contraction and the field-amplitude convention.

Write $u={\tilde{A}}_3$ and $v={\tilde{A}}_4$. Differentiating the exponential substitutions cancels the terms proportional to $2\kappa$:Consequently $u^{\prime }=i\kappa v^{\ast }$ and $v^{\prime }=-i\kappa u^{\ast }$. Conjugate the latter before eliminating a field: $(v^{\ast }{)}^{\prime }=i\kappa u$, so $u^{\prime \prime }=-{\kappa }^{2}u$. The backward four-wave-mixing boundary problem requires opposite-end boundary conditions: $u(0)=a$, where $a=A_3(0)$, and $v(L)=0$. The transfer matrix isAt $L$, $0=ia\sin \kappa L+b\cos \kappa L$, so $b=-ia\tan \kappa L$. Substitution givesThe printed target expression for $A_4$ has the opposite overall sign. This is present in the original PDF, not just the transcription. With its stated first-order equations, the derivative at the far face must be $v^{\prime }(L)=-i\kappa a^{\ast }/\cos \kappa L$. The boxed result has that derivative; the printed expression has its negative. Reversing the convention for the mixing coefficient could reverse the generated phase, but would also change the coupled equations. The result above satisfies the equations and both boundary conditions together.

At the entrance, $A_4(0)=i{a}^{\ast }\tan \kappa L$, so the returned amplitude is proportional to the complex conjugate of the signal. An ordinary ideal mirror gives $ra$, with a fixed reflection coefficient $r$, rather than $a^{\ast }$. The mixing pumps supply both the extra energy and a wave phase reference. The ideal conjugate reflectivity and signal transmission areThe constancy of $|u{|}^2+|v{|}^2$ along the medium is another direct check. The relation between the incoming signal and outgoing waves allows net output energy to come from the pumps. As $\kappa L\to \pi /2$, the undepleted-pump approximation fails: the predicted fields would extract an appreciable pump fraction. This is the ideal four-wave-mixing oscillation threshold; saturation, pump depletion and noise seeding replace the divergent linearized solution. It is not an unlimited source of reflected power. Amplified phase-conjugated propagation and parametric oscillation are also the effects identified in the primary analysis by Amnon Yariv and David M. Pepper.^[ref]

For the aberrating linear medium, assume its permittivity is real and stationary at this frequency. Taking the complex conjugate of the forward-envelope equation givesThis is the backward-envelope equation with carrier $e^{-i{k}_{3}z}$. Launching the conjugate boundary field therefore permits retracing of the distorted wave. In a transverse lossless phase screen model, let transmission be $t(x)=e^{i\psi (x)}$. A return proportional to $(tU{)}^{\ast }$ acquires $t$ on its second passage and becomes $t(tU{)}^{\ast }=U^{\ast }$. Thus the aberration phase cancels. A mirror instead gives $r{t}^{2}U$, adding the aberration twice; merely reversing the central propagation direction does not reverse the wavefront phase.

The one-dimensional printed equation describes longitudinal variation; genuine transverse aberrations require the corresponding transverse Helmholtz equation or paraxial operator, for which the same real-coefficient conjugation argument applies. Exact reconstruction requires adequate aperture, the correct conjugate amplitude and polarization, uniform pump-product phase, and an unchanged reciprocal lossless medium. Absorption, missing scattered modes or a changed aberrator prevent a universal reconstruction. This is phase conjugation and thin phase-plate cancellation by phase conjugation under specified conditions, rather than reversal of every possible propagation loss.

Spherical-particle frequency scaling. For a dielectric sphere of radius $a\ll \lambda$, the quasistatic electric polarizability isThe electric dipole radiation formula and $I_0={ϵ}_{0}c|E_0{|}^2/2$ giveThus Rayleigh scattering scales as ${\omega }^4$, or ${\lambda }^{-4}$, when the polarizability varies slowly with frequency. Material resonances and absorption can change this simple scaling. For the incident $z$ polarization, the dipole is along $z$: intensity is proportional to ${\sin }^2\chi$, where $\chi$ is the angle from $z$. There is no radiation on the dipole axis and a maximum in the $xy$ plane. In a direction $\hat{\mathbf{n}}$ the scattered electric field is along $\hat{\mathbf{z}}-(\hat{\mathbf{z}}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}$, hence linearly polarized wherever nonzero.^[ref]

Needle model used in parts (a)–(d). Let its axis be $\mathbf{u}=(u_x,u_y,u_z)=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. A very slender, high-permittivity needle has dominant longitudinal response, so use the rank-one needle polarizability $\mathbf{\alpha }\simeq {\alpha }_{\parallel }\mathbf{u}{\mathbf{u}}^{\mathsf{T}}$. Its transverse response is neglected at leading order. For observation along $y$, define $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, $s=|\mathbf{v}{|}^2=1-u_y^2$ andFor a unit incident Jones vector $\mathbf{e}$, the common far-field relation is ${\mathbf{E}}_s\propto \mathbf{v}(\mathbf{u}\cdot \mathbf{e})$, and $I_s=K{I}_{0}s|\mathbf{u}\cdot \mathbf{e}{|}^2$. Here $I_s$ is irradiance at distance $R$; multiplying by $R^2$ gives power per solid angle. No absolute value of ${\alpha }_{\parallel }$ is given or needed for the requested relative intensities.

Diffuse random cloud, with the incident polarization of (a). The single-particle amplitudes through $x$ and $z$ analysers are proportional to $u_x{u}_z$ and $u_z^2$. The isotropic orientation fourth moment gives the needle-scattering analyzer momentsFor example, $⟨u_z^4⟩=\frac{1}{2}{\int }_{-1}^1{\mu }^{4}d\mu =1/5$; symmetry and $u_x^2+u_y^2+u_z^2=1$ give $2⟨u_x^2{u}_z^2⟩=1/3-1/5$. Since the particles scatter mutually incoherently, intensities add. For $N$ illuminated particles,The cross correlation vanishes by reflection symmetry. The degree of linear polarization is $(I_z-I_x)/(I_z+I_x)=\boxed{1/2}$: random anisotropic scatterers produce partially depolarized light, even though every individual needle produces a single linear polarization.

Continuous homogeneous dielectric. One must sum amplitudes coherently and include their spatial phases, rather than retain the preceding sum of intensities. Random orientations give $⟨\mathbf{u}{\mathbf{u}}^{\mathsf{T}}⟩=\mathbf{I}/3$, so the averaged response is isotropic and its induced polarization is along $z$. Its coherent radiation toward $y$, when allowed by the sample geometry, is therefore purely $z$ polarized, with $I_x=0$. The independent-needle prefactor is no longer a quantitative dense-medium model; local-field and collective effects determine the effective permittivity.

There is also coherent continuum phase matching: a uniform bulk driven by a plane wave has no arbitrary outgoing $90^{\circ }$ plane-wave component. In an ideal infinite homogeneous medium there is no diffuse side scattering at all; propagation, refraction and reflection are determined by the medium and its interfaces. A finite sample's coherent side signal depends on its shape and boundaries. Thus the old $1:3$ diffuse ratio is replaced by a coherent isotropic-medium response, not by a universal nonzero $N^2$ side-scattering intensity.

A uniform Lorentz boost preserves the plane wave form, while changing both its frequency and its vector wave amplitude. Decompose vectors parallel and perpendicular to $\mathbf{v}$, so $\mathbf{k}={\mathbf{k}}_{\parallel }+{\mathbf{k}}_{\perp }$. In the observer's primed coordinates,The transformed wave amplitude is specified in part (d). The unprimed “rest frame” in this question is the original coordinate frame, not a rest frame of a vacuum light wave, which has no inertial rest frame.

Use the inverse coordinate transformation,Substitution into the wave phase $\mathbf{k}\cdot \mathbf{r}-\omega t$ and comparison of the coefficients of ${\mathbf{r}}^{\prime }$ and $t^{\prime }$ givesThe first is the relativistic Doppler effect; the changed direction of the wavevector describes relativistic aberration. Equivalently,The electric-field wave amplitude follows from part (b), after rotating the boost axis to $\mathbf{v}$:For a vacuum plane wave, Maxwell equations give ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$ and $\mathbf{k}\cdot {\mathbf{E}}_0=0$. Thus an expression using only the given electric wave amplitude and wavevector isEquivalently, ${\mathbf{E}}_0^{\prime }=\gamma ({\mathbf{E}}_0+\mathbf{v}\times {\mathbf{B}}_0)-[{\gamma }^2/(\gamma +1)]\mathbf{v}(\mathbf{v}\cdot {\mathbf{E}}_0)/c^2$. The wave amplitude is not a scalar carried unchanged by the wave phase transformation; its direction must remain transverse to the transformed wavevector.

In SI units the Maxwell's equations in vacuum, with sources, areTaking the divergence of the last equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot j=0$, the local conservation of electric charge. Thus sources that violate charge conservation cannot satisfy all the Maxwell's equations. Equivalently, with this conservation law the curl evolution preserves the electric divergence constraint if it holds initially; Faraday's law similarly preserves $\nabla \cdot B=0$.

With $\rho =j=0$, use $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-\Delta E$ and substitute the two curl equations. This gives $-\Delta E=-{\mu }_0{ϵ}_0{\partial }_t^{2}E$. Applying the same calculation to $B$ yieldsA monochromatic plane wave has one angular frequency $\omega$ and constant-phase planes $\mathbf{K}\cdot r=\omega t-\text{constant}$, normal to the wavevector $\mathbf{K}$. Its wavelength is $2\pi /|\mathbf{K}|$.

For a nontrivial traveling wave with $\omega \ne 0$, substitution of the displayed electric field into the divergence equation gives $\mathbf{K}\cdot e_1=\mathbf{K}\cdot e_2=0$. Faraday's law is satisfied byIndeed ${\partial }_{t}B=\mathbf{K}\times E_{\varphi }$ and $\nabla \times E=-\mathbf{K}\times E_{\varphi }$. The magnetic divergence is zero, while Ampère's law reduces to $|\mathbf{K}{|}^2{E}_{\varphi }/\omega =(\omega /c^2)E_{\varphi }$, giving the dispersion relation. The amplitudes may be arbitrary transverse vectors; they need not be mutually orthogonal or have equal lengths. Their combination describes the polarization.

For nonzero momentum, physical helicity is the spin projection along the direction of motion,The convention without $1/2$ labels the same two eigenspaces by $\pm 1$. Write $\hat{\mathbf{p}}=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$. Normalized two-component eigenvectors areEach can carry an arbitrary phase. Before imposing an energy equation, the general four-component helicity eigenvector is $(a{\chi }_{\pm },b{\chi }_{\pm }{)}^{\mathsf{T}}$, with arbitrary complex $a,b$; unit Norm imposes $|a{|}^2+|b{|}^2=1$. For a positive-energy free electron, the Dirac equation fixes their ratio and givesMultiplying by an arbitrary overall amplitude and the common plane wave supplies the corresponding free helicity eigenstate. Helicity has no preferred direction at $\mathbf{p}=0$. For a massive particle a boost can overtake its momentum and change its helicity; it should not be identified with Lorentz-invariant chirality.

Let $\chi =(\sqrt{5}/3,2/3{)}^{\mathsf{T}}$, so ${\chi }^{†}\chi =1$. The Pauli identity $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^{2}I$ givesusing $E^2=|\mathbf{p}{|}^2+m^2$. Thus unit spinor probability density is obtained withAn arbitrary constant phase is immaterial. For the alternative Lorentz normalization $u^{†}u=2E$, choose $N=\sqrt{E+m}$. Neither convention makes an infinite plane wave square-integrable; a box-normalized state includes an additional $V^{-1/2}$, or one forms a normalizable wave packet.

Put $r=p/(E+m)$ and assume $p>0$. The common plane wave has unit modulus. The component norm isFor unit spinor probability density choosewhere $E^2=p^2+m^2$ was used. This is spinor normalization, not square normalization of a plane wave over infinite space; a box or wavepacket gives the corresponding spatial normalization. If the convention is instead ${\psi }^{†}\psi =2E$, use $N=\sqrt{(E+m)/10}$.

For momentum along $+z$, the positive-helicity components are the first and third, while the second and fourth have negative helicity. HenceBoth are independent of speed. In the source's chiral block order the exact left-chiral weight is $1/2+3p/(10E)$, tending to $4/5$ at high energy; helicity and chirality probabilities agree only in that limit.

There are two distinct explanatory questions. One asks why a proposed system of equations has a particular dynamical symmetry. The other asks why the common spacetime geometry is the right geometry for the behavior of physical rods, clocks and fields. A geometric account and a dynamical account give different directions of explanation, and their merits cannot be decided by a change of notation alone.

For the geometric account, Minkowski spacetime has a fixed Lorentzian metric; its spacetime isometries form the Poincaré group. If the matter equations use that metric tensor and no additional symmetry-breaking background structure, with couplings and boundary assumptions appropriate to the transformation, these spacetime isometries take solutions into solutions. For example, the Klein-Gordon equationin metric signature $(-,+,+,+)$ is invariant under the Poincaré group. This provides one common reason for the same transformation rules appearing in many laws. An individual solution, such as a particular plane wave, need not possess every symmetry of the equation.

Geometry alone is insufficient unless the assumed coupling to geometry is included. In the covariant equation with a preferred timelike vector,the fixed timelike $n$ selects a preferred rest frame. All the quantities can be transformed consistently under a coordinate transformation, but boosts that change $n$ are not symmetries with the physical background fixed. Thus general covariance, tensor notation and the mere presence of the Minkowski metric do not by themselves establish Einstein's principle of relativity.

A dynamical account instead starts with the common invariance of the actual interactions and asks why their composite rods and clocks measure a common interval. If their equations and suitable equilibrium solutions transform consistently under Lorentz transformations, their moving configurations can exhibit length contraction and their periodic processes time dilation. This is the explanatory strategy defended by Harvey R. Brown and Oliver Pooley, who deny that merely redescribing these effects geometrically supplies their microscopic explanation.^[ref] Such an account must still explain why independent interactions have the same relevant symmetry and why a particular clock is a good clock; it cannot simply infer those facts from the word “dynamics”.

The positions therefore need not disagree about experimental predictions. The geometric presentation makes a shared constraint explicit and permits economical deductions. The constructive presentation investigates why material systems realize that constraint. Defining geometry from universal dynamical behavior is a substantive proposal; deriving behavior from geometry also requires substantive matter-coupling assumptions. Neither direction is made compulsory by the formal equivalence of the descriptions.

In vacuum, with possible sources $\rho$ and $\mathbf{J}$, Maxwell's equations areTheir integral forms, with compatible orientations and fixed surfaces, areCharge- and current-free space additionally sets $\rho =0$, $\mathbf{J}=0$. Vacuum constitutive constants alone do not imply that no charges exist.

Take the curl of the source-free Ampere-Maxwell law, use Faraday's law of induction and $\nabla \cdot \mathbf{B}=0$:ConsequentlyFor the specified complex plane wave, the wavevector is $k\hat{\mathbf{z}}$ and $\omega =ck$. Faraday's law of induction fixesThe physical fields are the real parts. They are in phase and their Poynting vector points along $+z$.

The instantaneous electric and magnetic energy densities are $u_E={ϵ}_0{E}^2/2$ and $u_B=B^2/(2{\mu }_0)$. Since $E=cB$, they are equal for this ideal travelling plane wave, and so are their cycle averages: $⟨u_E⟩/⟨u_B⟩=1$. For real peak amplitude $E_0$, each average is ${ϵ}_0{E}_0^2/4$. The Poynting vector and mean intensity satisfyIt represents transported energy per area per time, in $\mathrm{W}{\mathrm{m}}^{-2}$.

For the numerical moonlight estimate, adopt $L_{\odot }=3.828\times 10^{26}\mathrm{W}$ and lunar mean radius $R_M=1.7374\times 10^6\mathrm{m}$. These additional astronomical inputs give solar irradiance at the Moon of ${\mathcal{I}}_{\odot }=L_{\odot }/(4\pi d_{\odot }^2)\simeq 1361\mathrm{W}{\mathrm{m}}^{-2}$.^[ref][ref]

Under the simple isotropic point-reflector approximation, the intercepted power is $\pi R_M^2{\mathcal{I}}_{\odot }$, of which fraction $A=0.12$ is redistributed over $4\pi$ steradians. ThusThese are equivalent sinusoidal peak amplitudes; the corresponding root mean square fields are smaller by $\sqrt{2}$. Actual moonlight is broadband and incoherent.

A total reflected fraction alone does not specify the directional brightness of a full Moon. For a Lambertian sphere at full phase, with the same total reflectance $A$, integrating its radiance gives ${\mathcal{I}}_M=(2A/3){\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, instead giving $E_0\simeq 1.30\mathrm{V}{\mathrm{m}}^{-1}$ and $B_0\simeq 4.32\mathrm{nT}$. If 0.12 is instead the geometric albedo, the full-phase formula is ${\mathcal{I}}_M=0.12{\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, giving $E_0\simeq 1.59\mathrm{V}{\mathrm{m}}^{-1}$. NASA's table itself distinguishes Bond albedo from geometric albedo.^[ref] The numerical estimate therefore needs an explicit reflection convention; the first boxed values use the stated isotropic approximation.

In source-free vacuum, Maxwell's equations readTake the curl of the third equation and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. ThenThe prescribed plane wave travels in the $+z$ direction. Its polarization is along $x$, and Faraday's law requires $B_y=(k/\omega )E_x$ with $\omega =ck$. The wave's magnetic flux density is thereforeThe direction is also fixed by $\mathbf{E}\times \mathbf{B}$ pointing along the propagation direction. A superposed static uniform magnetic field is not fixed by the oscillatory electric field; the expression above is the magnetic field belonging to the wave itself.

In a source-free vacuum, the Maxwell equations in SI arewhere ${ϵ}_0$ and ${\mu }_0$ are the vacuum permittivity and vacuum permeability. Take the curl of the last equation, substitute Faraday's law of induction, and use the curl of the curl identity:The zero divergence therefore gives the wave equationFor a propagating plane wave, write $\mathbf{B}=\mathrm{Re}[{\mathbf{B}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. The wave equation gives ${\omega }^2=c^2|\mathbf{k}{|}^2$, and Gauss's law for magnetism gives $\mathbf{k}\cdot {\mathbf{B}}_0=0$. Thus the oscillating magnetic field is a transverse wave: it is perpendicular to the propagation wavevector. A static background magnetic field, if present, is separate from this propagating component.

Using $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ directly with the specified plane wave givesThe electric field, magnetic flux density and propagation direction form a right-handed triad, and for a nonzero wave amplitudeThe electric field and magnetic field have the same wave phase in vacuum.

In charge-free, current-free vacuum, Maxwell's equations areTaking the curl of the third equation and using the fourth, the curl of the curl identity, and the zero divergence of the electric field givesThe magnetic flux density obeys the same wave equation. For the plane wave factor $e^{i(\omega t-kx)}$, the spatial Laplacian contributes $-k^2$ and the second time derivative contributes $-{\omega }^2$. Thusfor propagation in the positive $x$ direction. The zero-divergence conditions require $E_{0x}=B_{0x}=0$, while the curl equations require ${\mathbf{B}}_0=(k/\omega )\hat{\mathbf{x}}\times {\mathbf{E}}_0$. These conditions show that the proposed plane waves satisfy all four Maxwell's equations, rather than only their derived wave equations. Physical electric fields and magnetic fields are the real parts of the complex expressions.

The free particle dispersion relation is $\hslash \omega ={\hslash }^2{k}^2/(2m)$. HenceThe phase velocity follows constant-phase surfaces of a plane wave. The group velocity follows the envelope of a narrow-band wave packet and equals its classical velocity $p/m$. Dispersion can spread the packet; the phase speed is not the particle's transport velocity. In three dimensions ${\mathbf{v}}_g=\hslash \mathbf{k}/m$ and the phase speed quoted is along $\mathbf{k}$.

Single travelling wave. Differentiation of the plane wave gives ${\partial }_x\psi =ik\psi$, so its probability current isThis is probability density times the free-particle velocity $\hslash k/m$.

Counterpropagating waves. Remove the common time phase and set $z=B^{\ast }A{e}^{2ikx}$. Direct multiplication givesConsequentlyThe interference term $z-z^{\ast }$ is purely imaginary in the last real-part operation. The probability density can have interference fringes even though this net probability current is spatially constant.

Scattering amplitudes. Locate the step at $x=0$ and first take $E>V$. With $k_1=\sqrt{2mE}/\hslash$ and $k_2=\sqrt{2m(E-V)}/\hslash$, useThe common factor $e^{-iEt/\hslash }$ has been suppressed. A finite step with the same mass on both sides requires continuity of $\psi$ and ${\psi }^{\prime }$, giving $1+r_a=t_a$ and $k_1(1-r_a)=k_2{t}_a$. HenceIf $0<E<V$, choose $k_2=i\kappa$, $\kappa =\sqrt{2m(V-E)}/\hslash$, so the transmitted wave decays. The same amplitude formulae continue to hold, but the semi-infinite forbidden region carries no outgoing probability current: $R=1$ and $T=0$.

Flux balance. In the propagating case,In particular,The velocity ratio in $T$ is essential: the squared transmission amplitude alone is not the transmission probability when the two velocities differ.

Numerical transmission. The mass cancels from the ratio, and the energies can remain in the same electronvolt units:Thus about $92.8\%$ is transmitted and $7.18\%$ reflected, despite the incident energy exceeding the step.

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.

A Fresnel zone plate replaces a conventional refracting surface by concentric alternating transmitting and absorbing annuli. Successive annuli span half a wavelength of optical-path excess to the chosen focus; blocking alternate zones prevents their approximately opposite wave phases from cancelling. For a normally incident plane wave,The paraxial radii grow as $\sqrt{j}$, so the zones become narrower outwards.

Using the photon energy relation $\lambda =hc/E$ gives $\lambda =4.133\times 10^{-11}\mathrm{m}=0.04133\mathrm{nm}$. For $f=10^{-2}\mathrm{m}$,The first three boundaries are approximately $0.6429$, $0.9092$ and $1.1135\mu \mathrm{m}$. This fixes the zone pattern but does not fix a unique outside diameter: the number of zones is a design choice. For example, $N=100$ gives radius $6.429\mu \mathrm{m}$, diameter $12.86\mu \mathrm{m}$ and outermost zone width about $32.2\mathrm{nm}$. A larger usable diameter requires finer outer zones.

One could pattern absorbing metal annuli on a thin X-ray-transparent support, choosing absorber thickness from its attenuation at the selected photon energy; that thickness cannot be determined from energy and focal length alone. The design is chromatic: for fixed radii $f\propto 1/\lambda$, so a narrow energy band is needed for a sharp common focus.

The Fourier expression is the scalar Fraunhofer diffraction approximation for coherent, effectively monochromatic illumination with a uniform incident plane wave over the aperture. At a distant screen the aperture-dependent quadratic propagation phase must be negligible, roughly $D^2/(\lambda z)\ll 1$ for transverse size $D$. Alternatively, a thin lens forms the angular Fourier transform in its back focal plane. That focal-plane construction is the appropriate one here; the aperture-to-lens distance need not itself be an enormous free-propagation distance. aperture features should be large compared with wavelength for this simple scalar transmission-mask model.

If complementary transmissions are $h$ and $1-h$, linearity gives $U_h+U_{1-h}=U_{\mathrm{free}}$. Away from the undiffracted forward beam, the uniform plane-wave contribution $U_{\mathrm{free}}$ vanishes in angular space, so $U_{1-h}=-U_h$ and their intensities agree. This is Babinet's principle. It equates the off-forward diffracted intensities, not the transmitted power or the on-axis intensity including the direct beam.

Let $X,Y$ be screen coordinates conjugate to aperture coordinates $x,y$, so $p\simeq kX/f$, $q\simeq kY/f$. With width $a=0.2\mathrm{mm}$ along $x$ and height $b=0.5\mathrm{mm}$ along $y$, define $\mathrm{sinc}u=\sin u/u$. The complementary rectangle has transformThus the obstruction's off-forward scattered component hasIts dark lines occur atThe central-lobe zero-to-zero size is $1.00\mathrm{mm}$ by $0.400\mathrm{mm}$; it is wider in the direction conjugate to the narrower obstruction. The central forward image of the illuminating beam must be distinguished from this scattered envelope in a real screen observation.