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Apply the product-to-sum formula $2\cos u\cos v=\cos (u+v)+\cos (u-v)$ to the amplitude modulation:The amplitude modulation sidebands have angular frequencies $\omega +{\omega }_m$ and $\omega -{\omega }_m$, each with amplitude $b/2$ in this real cosine convention. The carrier wave has angular frequency $\omega$ and amplitude $a$.

This identity is exact for the supplied space-time expression. A physical qualification is needed if every term is meant to be a freely propagating electromagnetic wave with speed of light $c$: the dispersion relation requires $k_{\pm }=(\omega \pm {\omega }_m)/c$, whereas the prescribed modulation at every position leaves all three wavenumbers equal to $k=\omega /c$. A freely propagating modulated wave instead has envelope $a+b\cos ({\omega }_m(t-x/c))$; its analogous expansion has $k_{\pm }$. For example, at $x=t=0$ the prescribed expression has $A_{tt}-c^2{A}_{xx}=-b{\omega }_m^2$, which is nonzero for nontrivial modulation. The given formula can describe modulation imposed throughout space or the temporal signal at one position.

Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

Let ${\rho }_i={\rho }_n=\rho$ and $\mu ={\mu }_{in}={\mu }_{ni}$. Use normal modes proportional to $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$, so $\mathrm{Im}\omega <0$ means damping. The linear displacements satisfyFor a nonzero-frequency perturbation, linearizing the ideal magnetohydrodynamic induction equation givesProject both momentum equations with $\mathsf{P}=I-\mathbf{k}{\mathbf{k}}^T/k^2$. This removes the pressure gradients. The magnetic tension is already transverse to $\mathbf{k}$, and, with $a=k_{\parallel }{v}_A$, the projected equations areHere $v_A=B_0/\sqrt{4\pi \rho }$ uses the ion density, as stipulated. For either of the two independent transverse polarizations, the equal-density ion-neutral Alfvén dispersion relation follows fromIts determinant is $\omega [{\omega }^3+2i\mu {\omega }^2-a^2\omega -i\mu a^2]$. Removing the static neutral-displacement factor givesThe removed factor is a time-independent neutral displacement with no velocity or magnetic perturbation, a relabeling of the homogeneous neutral fluid rather than a fourth physical wave. Direct elimination in the first-order system for $(u_i,u_n,\delta B)$ gives exactly the same cubic. This also handles the zero-frequency limits without introducing a displacement-label mode.

The neutral displacement equation gives, for each nonzero-frequency root,The vector displacements are parallel as complex polarization vectors, with this scalar amplitude and phase ratio.

For the rapidly damped root, $\mu -i{\omega }_d=-\mu +a^2/(4\mu )+O(a^4/{\mu }^3)$, and henceThe two fluids move almost equally and oppositely; collisions remove the counterflow at a rate close to $2\mu$.

For either weakly damped Alfvén wave, expand the ratio around unity:Equivalently, the exact slippage relation is ${\mathbf{\xi }}_i-{\mathbf{\xi }}_n=-(i\omega /\mu ){\mathbf{\xi }}_n$. Thus the relative slippage has size $|a|/(\sqrt{2}\mu )\ll 1$ and is primarily a small phase difference. The fluids almost share a common displacement, but magnetic tension acts directly on the ions and drag accelerates the neutrals. This imperfect locking causes ambipolar damping.

The sign of the damping is also checked by the perturbation energy balance. With periodic or vanishing-flux boundaries,It follows by dotting each linear momentum equation with its velocity and the induction equation with $\delta \mathbf{B}/(4\pi )$; magnetic-tension work cancels magnetic-energy change, pressure contributes only boundary flux, and the two drag terms combine into the negative square. Even though spatial diffusion was neglected, ion-neutral friction still dissipates the wave energy.

At fixed number density $n$, the electron current givesThe second equality uses Ampère's law without displacement current and stationary ions. A perturbation with characteristic perpendicular wave number $k_{\perp }$ therefore produces $\delta u_e\sim \alpha k_{\perp }\delta B$. Its nonlinear magnetic-advection rate isFor the anisotropic electron magnetohydrodynamics wave, $k\simeq k_{\perp }$ and $|\omega |\sim v_A{d}_i|k_{\parallel }|k_{\perp }$. Equating this to the nonlinear rate by critical balance givesThis is small-amplitude but strong turbulence: a small perturbation can have order-one nonlinear distortion over one wave period because its perpendicular gradients are much larger than its parallel gradients.

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

Assume $\rho$ is mass per unit equilibrium length, the tension has constant magnitude $T$, the string has negligible bending stiffness, and there is no damping or external transverse force. Assume small slopes $|y_x|\ll 1$ and purely transverse motion. Then a short segment has mass $\rho \Delta x$ to first order. If $\alpha$ is the angle of its tangent, the small-slope approximation gives $\sin \alpha \simeq \tan \alpha =y_x$.

The vertical components of tension at the two ends give net transverse force $T[y_x(x+\Delta x,t)-y_x(x,t)]$. Newton's second law therefore givesto first order. Dividing by $\Delta x$ and taking its limit to zero yieldsThus $c$ is the propagation speed of small transverse waves.

Write $y(x,t)=Y(\xi ,\eta )$ with $\xi =x-ct$, $\eta =x+ct$. The chain rule gives the differential operatorsTheir squared difference isConsequently the wave equation becomes $Y_{\xi \eta }=0$. Integrating first in $\eta$ gives $Y_{\xi }=F^{\prime }(\xi )$; integrating in $\xi$ gives the general solutionThe arbitrary twice differentiable functions describe right-moving and left-moving waves, respectively. Conversely the chain rule shows that each such sum satisfies the wave equation.

Enforce the fixed-end Dirichlet boundary condition by taking the odd extension of the initial velocity to the whole line:This agrees with the specified initial condition for $s\ge 0$. To derive the relevant form of D'Alembert's formula, the zero initial displacement in b gives $G=-F$, up to a cancelling constant. The initial velocity then gives $-2c{F}^{\prime }=v_o$. HenceAn antiderivative, valid across $s=0$, isEvaluation therefore gives the solution throughout $x>0,t>0$:For $ct\le x$, replace $|x-ct|$ by $x-ct$; for $ct\ge x$, replace it by $ct-x$. Thus the same expression includes the reflected wave after it reaches the fixed end.

At $x=0$ the two terms agree, so $y(0,t)=0$. At $t=0$ they again agree, so $y(x,0)=0$. Differentiating the integral with respect to $t$ giveswhich recovers $y_t(x,0)=\kappa x{e}^{-x/L}$. Since $v_o$ is a continuously differentiable function, the formula is twice continuously differentiable even across $x=ct$, and satisfies the wave equation there as well.

Since the initial velocity vanishes, D'Alembert's formula reduces toEach translated copy has half the original amplitude. To identify each requested cross-section, track the translated support intervals and add the values where they overlap.

At $t=\pi$, $f(x-\pi )$ is supported on $[-\pi ,0]\cup [2\pi ,3\pi ]$ and $f(x+\pi )$ on $[-3\pi ,-2\pi ]\cup [0,\pi ]$. On these intervals $\sin (x\pm \pi )=-\sin x$, soThe pieces meet continuously at zero, where both contributions vanish.

At $t=3\pi /2$, the support of $f(x-3\pi /2)$ is $[-\pi /2,\pi /2]\cup [5\pi /2,7\pi /2]$; that of $f(x+3\pi /2)$ is $[-7\pi /2,-5\pi /2]\cup [-\pi /2,\pi /2]$. In the central overlap,so the two contributions cancel. The remaining wave profile isFor the time cross-section at the origin, the initial profile is an odd function, soFor the time cross-section at $x=\pi /2$, the left-travelling contribution is nonzero for $\pi /2\le t\le 3\pi /2$ and equals $\cos t/2$. The right-travelling contribution is nonzero for $3\pi /2\le t\le 5\pi /2$ and also equals $\cos t/2$. At the common endpoint both vanish. HenceThe following graphs show all four cross-sections. In particular, the last trace reaches $-1/2$ at $t=\pi$ and $1/2$ at $t=2\pi$.

Because $u(0,t)=0$, restriction to the half-line solves the following initial-boundary value problem with a homogeneous Dirichlet boundary condition:This is wave reflection by odd extension. If a condition at infinity is desired, take the solution with finite wave energy; here the profile has bounded support at every finite time.

The initial profile is continuous and piecewise continuously differentiable, but its first derivative jumps at the support endpoints. Thus the formula gives a weak solution globally and a piecewise classical solution away from the characteristic lines for the one-dimensional wave equation issuing from those endpoints. The initial velocity is zero in the square-integrable function sense and pointwise wherever the initial profile is differentiable; at the kink points a pointwise right time derivative need not equal zero. This regularity qualification is needed for the given data.

The Huygens-Fresnel principle treats every point of an advancing wavefront as a source of secondary spherical waves; their forward envelope constructs the next wavefront. Consider two points where an oblique plane wavefront meets a plane interface a distance $AB$ apart. After its first point reaches $A$, let time $t$ elapse until the second reaches $B$. In medium 1, the incident travel distance is $v_{1}t=AB\sin i$. Meanwhile, a secondary wave from $A$ has travelled distance $v_{2}t$ in medium 2. The tangent from $B$ to that secondary wavefront gives $v_{2}t=AB\sin r$. ThereforeHere $i,r$ are the angle of incidence and angle of refraction, measured from the interface normal vector, and $n_j=c/v_j$ are the refractive indices at the common frequency. This is Snell's law.

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.

For a uniform taut sheet in the small-displacement approximation, the two-dimensional wave equation isSeek a harmonic solution $Y(y)e^{i\omega t-i{k}_{x}x}$. The clamped edges impose Dirichlet boundary conditions $Y(0)=Y(b)=0$. Hence $Y^{\prime \prime }+k_y^{2}Y=0$ has nontrivial solutions only forTaking the real part gives the stated sinusoidal transverse profile. There is no nonzero $n=0$ mode satisfying both clamped boundaries.

Substitution gives the dispersion relationThe three branches have cutoffs ${\omega }_{c,n}=n\pi v/b$ at $k_x=0$ and approach $\omega =v|k_x|$ at large $|k_x|$. Only frequencies above a branch's cutoff give traveling waves along an infinitely long guide; below it, the axial dependence is evanescent. For a forward propagating mode,The phase velocity exceeds $v$, while the group velocity is smaller: the carrier phase speed is not the signal speed.

Here the first two cutoff frequencies are $50$ and $100\mathrm{Hz}$, so both modes propagate at $200\mathrm{Hz}$. With $\omega =400\pi {\mathrm{s}}^{-1}$,Each transverse mode independently acquires its axial phase. The second mode first changes sign relative to the first when $(k_1-k_2)x_1=\pi$, givingAt that distance a common temporal phase is $\varphi =-k_1{x}_1$ modulo $2\pi$; the second term is negative relative to the first for every $y$ and every time, not just at one selected instant.

In an optical fibre, different guided modes have different group velocities, so an input pulse becomes a sum of differently delayed pulses. This intermodal dispersion broadens pulses and can make adjacent digital symbols overlap. A single-mode optical fiber removes this particular mode-delay mechanism; a graded-index optical fiber can reduce it in a multimode design. Single-mode propagation still permits wavelength-dependent group delay, so it does not remove every kind of optical dispersion.

The reciprocal lattice consists of all wavevectors $\mathbf{G}$ whose plane-wave phase is unchanged under every direct-lattice translation $\mathbf{R}$:It expresses periodicity in wavevector space, with dimensions of inverse length.

Choose incident field convention $e^{i{\mathbf{k}}_i\cdot \mathbf{r}-i\omega t}$. The wave at the second atom acquires incident phase ${\mathbf{k}}_i\cdot {\mathbf{r}}_n$ relative to the atom at the origin. Its path to a distant detector is shorter by approximately ${\hat{\mathbf{k}}}_f\cdot {\mathbf{r}}_n$, contributing outgoing phase $-{\mathbf{k}}_f\cdot {\mathbf{r}}_n$. ThereforeNeglecting variation of detector distance across the specimen, identical atoms have a common scattering amplitude factor andThe original geometry diagram shows the incident direction, outgoing direction and direct-space displacement separately; ${\mathbf{k}}_s$ is a wavevector difference, not an extra ray path.

The primitive reciprocal vectors obeywith cyclic versions for $\mathbf{B},\mathbf{C}$. If ${\mathbf{k}}_s=h\mathbf{A}+k\mathbf{B}+l\mathbf{C}$, every lattice term has phase $2\pi$ times an integer, so all terms add constructively. Conversely, full translational reinforcement requires ${\mathbf{k}}_s\cdot \mathbf{a}$, ${\mathbf{k}}_s\cdot \mathbf{b}$ and ${\mathbf{k}}_s\cdot \mathbf{c}$ to be integer multiples of $2\pi$, hence ${\mathbf{k}}_s$ is reciprocal. These are the Laue equations. A finite crystal gives finite-width maxima rather than ideal reciprocal-space delta peaks; elastic scattering must also satisfy $|{\mathbf{k}}_i|=|{\mathbf{k}}_f|$.

For the two-dimensional Bravais lattice interpretation, with one atom per primitive cell, takeSolving the duality conditions givessoThe reciprocal primitive vectors have equal length $4\pi /(\sqrt{3}a)$ and angle $120^{\circ }$; the array itself is triangular. A different primitive basis can use a $60^{\circ }$ angle without changing the lattice. The direct primitive-cell area is $\sqrt{3}{a}^2/2$ and its reciprocal-cell area is $(2\pi {)}^2$ divided by that area.

For nonzero integer pairs, $h^2-hk+k^2$ has minimum one, attained for six shortest vectors. Therefore the first nonforward maximum has $|{\mathbf{k}}_s|=4\pi /(\sqrt{3}a)$. If the forward transmitted maximum is included literally, the smallest value is $|{\mathbf{k}}_s|=0$; the nonzero value is the first diffraction shell. Reaching that shell in elastic scattering additionally requires $4\pi /(\sqrt{3}a)\le 2|{\mathbf{k}}_i|$.

Fresnel diffraction is interference during propagation when curvature of the optical path across an aperture matters. It provides a continuous account of fringes near geometric edges and their evolution into the far-field pattern. A screen that is many aperture widths away is not automatically in the Fraunhofer diffraction limit; wavelength also enters the comparison.