The Helmholtz-Kirchhoff Diffraction Integral

Derivation

The Helmholtz-Kirchhoff Diffraction Integral

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Statement

For a monochromatic scalar field \(U(\vec{r})\) obeying the Helmholtz equation \((\nabla^2+k^2)U=0\) in a source-free region \(V\) bounded by a closed surface \(S=\partial V\), the field at any interior point \(P_0\) is fixed by \(U\) and its normal derivative on \(S\) through the integral theorem of Helmholtz and Kirchhoff \[ U(P_0)=\frac{1}{4\pi}\oint_{S}\left[\,G\,\frac{\partial U}{\partial n}-U\,\frac{\partial G}{\partial n}\,\right]dS,\qquad G(\vec{r})=\frac{e^{ikr_{01}}}{r_{01}}, \] where \(r_{01}=|\vec{r}-\vec{r}_0|\) and \(\hat{n}\) is the outward normal. Applying Kirchhoff's aperture conditions to a diffracting screen reduces this to the Fresnel–Kirchhoff diffraction formula, and choosing a Green's function with homogeneous boundary values yields the Rayleigh–Sommerfeld forms.

Why it matters

This is the exact bridge between the wave equation and everything the word "diffraction" names: the single/double slit, the Airy pattern of a telescope, the resolution limit of a microscope, the design of every hologram and Fourier-optical system. Fraunhofer and Fresnel diffraction are not separate theories — they are the far-field and quadratic-phase approximations of this one surface integral.

It also answers the question Huygens could only assert: why a wavefront acts as a set of secondary sources, and with what amplitude, phase, and directional weighting. The obliquity (inclination) factor that drops out of the derivation is exactly the \(\tfrac12(1+\cos\chi)\) that Huygens' construction needed to suppress the backward wave, and which he had no way to justify.

Assumptions

Scalar field: a single complex amplitude \(U\) represents the wave, ignoring the vector coupling of \(\vec{E}\) and \(\vec{B}\).Drop it and the components of \(\vec{E}\) mix at boundaries; one needs the vector (Stratton–Chu / Franz) diffraction formulae, and the pattern becomes polarization-dependent — essential for sub-wavelength apertures and near edges.

Monochromatic, time-harmonic field \(U(\vec{r},t)=U(\vec{r})\,e^{-i\omega t}\), so the wave equation collapses to Helmholtz with \(k=\omega/c\).Drop it and each Fourier component diffracts with its own \(k\); one must superpose over frequency and track temporal coherence.

Source-free homogeneous medium inside \(V\): no charges, scatterers, or index variation between aperture and observation point.Drop it and Green's identity acquires a volume term \(\int_V G\,(\nabla^2+k^2)U\,dV\neq0\); the field is no longer determined by boundary data alone.

Sommerfeld radiation condition: on a large hemisphere at infinity \(\lim_{R\to\infty}R\left(\frac{\partial U}{\partial n}-ikU\right)=0\).Drop it and the contribution of the far spherical cap need not vanish; incoming (advanced) waves are admitted and \(U(P_0)\) is not fixed by the aperture alone.

Kirchhoff boundary conditions on the screen: in the aperture \(U\) and \(\partial U/\partial n\) equal their unobstructed incident values; on the opaque part both are exactly zero.Drop the "both" and you must replace it — as written the two conditions over-determine a Helmholtz solution (a field with \(U=\partial_n U=0\) on any finite patch vanishes everywhere), the internal inconsistency the Rayleigh–Sommerfeld choice repairs.

Observation and source distances large compared with a wavelength, \(kr_{01}\gg1\) and \(kr_{21}\gg1\), so \(ik-1/r\approx ik\).Drop it and the near-field \(1/r\) terms of \(\partial G/\partial n\) survive, correcting the amplitude and phase within a few wavelengths of the aperture.

Derivation

\[ \left(\nabla^2+k^2\right)U(\vec{r})=0,\qquad k=\frac{\omega}{c}=\frac{2\pi}{\lambda} \]

Time-harmonic ansatz \(U(\vec{r},t)=U(\vec{r})e^{-i\omega t}\) inserted into the scalar wave equation (prior result) removes the time derivative and leaves the Helmholtz equation for the spatial amplitude. A

\[ G(\vec{r})=\frac{e^{ikr_{01}}}{r_{01}},\qquad \left(\nabla^2+k^2\right)G=-4\pi\,\delta^3(\vec{r}-\vec{r}_0) \]

Take the outgoing spherical (free-space) Green's function of the Helmholtz operator centred on the observation point \(P_0\) (prior result). It is an exact solution away from \(\vec{r}_0\) and has a \(1/r_{01}\) singularity there. A

\[ \int_{V'}\!\left(U\,\nabla^2 G-G\,\nabla^2 U\right)dV=\oint_{S'}\!\left(U\,\frac{\partial G}{\partial n}-G\,\frac{\partial U}{\partial n}\right)dS \]

Green's second identity, valid because \(U,G\) are twice differentiable on the region. To exclude the singularity of \(G\), integrate over \(V'=V\) minus a small ball \(B_\varepsilon\) of radius \(\varepsilon\) about \(P_0\); then \(S'=S\cup S_\varepsilon\). B

\[ U\,\nabla^2G-G\,\nabla^2U=U(-k^2G)-G(-k^2U)=0 \quad\text{on }V' \]

Inside \(V'\) both fields satisfy the source-free Helmholtz equation (Steps 1–2, the delta sits outside \(V'\)), so the volume integrand vanishes identically. The left side of Step 3 is zero. A

\[ \oint_{S}\!\left[U\frac{\partial G}{\partial n}-G\frac{\partial U}{\partial n}\right]dS \;=\; -\oint_{S_\varepsilon}\!\left[U\frac{\partial G}{\partial n}-G\frac{\partial U}{\partial n}\right]dS \]

With the volume integral zero, the total surface integral over \(S\cup S_\varepsilon\) vanishes; move the small-sphere piece to the right. The task is now to evaluate the \(S_\varepsilon\) integral as \(\varepsilon\to0\). B

\[ \left.\frac{\partial G}{\partial n}\right|_{S_\varepsilon}=-\frac{d}{dr}\!\left(\frac{e^{ikr}}{r}\right)_{\!\varepsilon}=\left(\frac{1}{\varepsilon^2}-\frac{ik}{\varepsilon}\right)e^{ik\varepsilon} \]

On \(S_\varepsilon\) the outward normal of \(V'\) points radially inward (toward \(P_0\)), so \(\partial/\partial n=-\partial/\partial r\). Here \(G|_\varepsilon=e^{ik\varepsilon}/\varepsilon\). These are the integrand factors for the shrinking sphere. C

\[ \oint_{S_\varepsilon}\!\left[U\frac{\partial G}{\partial n}-G\frac{\partial U}{\partial n}\right]dS=\int_{4\pi}\!\left[U\!\left(\tfrac{1}{\varepsilon^2}-\tfrac{ik}{\varepsilon}\right)e^{ik\varepsilon}-\tfrac{e^{ik\varepsilon}}{\varepsilon}\frac{\partial U}{\partial n}\right]\varepsilon^2\,d\Omega\;\xrightarrow[\varepsilon\to0]{}\;4\pi\,U(P_0) \]

Write \(dS=\varepsilon^2 d\Omega\). The \(1/\varepsilon^2\) term gives \(4\pi U(P_0)e^{ik\varepsilon}(1-ik\varepsilon)\to4\pi U(P_0)\); the \(1/\varepsilon\) and \(G\,\partial_nU\) terms carry a surviving factor \(\varepsilon\to0\). Continuity of \(U\) at \(P_0\) is used to pull \(U(P_0)\) out. C

\[ \boxed{\,U(P_0)=\frac{1}{4\pi}\oint_{S}\left[\,G\,\frac{\partial U}{\partial n}-U\,\frac{\partial G}{\partial n}\,\right]dS\,} \]

Insert Step 7 into Step 5: the left side equals \(-4\pi U(P_0)\), so \(U(P_0)=-\frac{1}{4\pi}\oint_S[U\partial_nG-G\partial_nU]\,dS\). This is the Helmholtz–Kirchhoff integral theorem. A

\[ S=\Sigma\cup S_{\text{opaque}}\cup S_\infty,\qquad \left.U\right|_{\text{opaque}}=\left.\frac{\partial U}{\partial n}\right|_{\text{opaque}}=0,\quad \int_{S_\infty}\to0 \]

Take \(S\) to be the screen plane plus a hemisphere \(S_\infty\) of infinite radius closing on the observation side. Kirchhoff's boundary conditions kill the opaque part; the Sommerfeld radiation condition kills \(S_\infty\). Only the aperture \(\Sigma\) survives. B

\[ U(P_1)=A\,\frac{e^{ikr_{21}}}{r_{21}},\qquad \frac{\partial U}{\partial n}\approx ik\cos(\hat{n},\vec{r}_{21})\,U(P_1),\quad \frac{\partial G}{\partial n}\approx ik\cos(\hat{n},\vec{r}_{01})\,G \]

Illuminate the aperture with a spherical wave from a point source \(P_2\) at distance \(r_{21}\). Differentiating and dropping \(1/r\) against \(ik\) (assumption \(kr\gg1\)) gives the normal derivatives, with the cosines the projections of the ray directions on \(\hat{n}\). B

\[ U(P_0)=\frac{A}{i\lambda}\iint_{\Sigma}\frac{e^{ik(r_{21}+r_{01})}}{r_{21}\,r_{01}}\;\underbrace{\frac{\cos(\hat{n},\vec{r}_{01})-\cos(\hat{n},\vec{r}_{21})}{2}}_{\displaystyle K(\chi)}\;dS \]

Insert Step 10 into the boxed theorem over \(\Sigma\). Both surviving terms share \(e^{ik(r_{21}+r_{01})}/(r_{21}r_{01})\); collecting them and using \(\tfrac{ik}{4\pi}=\tfrac{i}{2\lambda}=\tfrac{1}{i\lambda}\cdot\tfrac12\) yields the Fresnel–Kirchhoff formula with the inclination factor \(K\). B

\[ G_{\mp}=\frac{e^{ikr_{01}}}{r_{01}}\mp\frac{e^{ik\tilde r_{01}}}{\tilde r_{01}}\;\Rightarrow\;\begin{cases}G_-=0 &(\text{Dirichlet})\\[2pt]\partial_n G_+=0 &(\text{Neumann})\end{cases}\ \text{on }\Sigma \]

To remove the over-determination, replace \(G\) by the mirror-image combination, where \(\tilde r_{01}\) is measured from the reflection \(P_0'\) of \(P_0\) in the screen plane. On the plane \(r_{01}=\tilde r_{01}\), so \(G_-\) (or \(\partial_nG_+\)) vanishes identically. C

\[ U(P_0)=\frac{1}{i\lambda}\iint_{\Sigma}U(P_1)\,\frac{e^{ikr_{01}}}{r_{01}}\,\cos(\hat n,\vec r_{01})\,dS \]

Using \(G_-\) in the boxed theorem leaves only the \(U\,\partial_nG_-\) term (since \(G_-=0\) on \(\Sigma\)), and \(\partial_nG_-=2\,\partial_nG\) there. This Rayleigh–Sommerfeld first solution needs only \(U\) on the aperture; its inclination factor is \(\cos\chi\), versus Kirchhoff's \(\tfrac12(1+\cos\chi)\) and RS-II's unity. All three agree to first order in the diffraction angle. C

Result

\[ U(P_0)=\frac{1}{4\pi}\oint_{S}\!\left[G\frac{\partial U}{\partial n}-U\frac{\partial G}{\partial n}\right]dS\;\xrightarrow[\text{aperture}]{}\;U(P_0)=\frac{A}{i\lambda}\iint_{\Sigma}\frac{e^{ik(r_{21}+r_{01})}}{r_{21}r_{01}}\,K(\chi)\,dS \]

Reading. The field at \(P_0\) is a coherent sum of secondary spherical wavelets \(e^{ikr_{01}}/r_{01}\), one from every aperture point, each carrying the incident amplitude and phase \(A\,e^{ikr_{21}}/r_{21}\), weighted by the directional factor \(K(\chi)\), and advanced in phase by \(90^\circ\) and scaled by \(1/\lambda\) through the \(1/(i\lambda)\) prefactor. This is Huygens' principle made quantitative — the wavelet strength, phase lead, and forward bias are all derived, not assumed. The inclination factor \(K=\tfrac12(1+\cos\chi)=1\) forward and \(=0\) backward, so no wavelet radiates back toward the source.

Units check. In the theorem \([G]=\mathrm{m^{-1}}\), \([\partial_n G]=\mathrm{m^{-2}}\), \([dS]=\mathrm{m^2}\); each term is \([U]\cdot\mathrm{m^{-1}}\cdot\mathrm{m^{-1}}\cdot\mathrm{m^{2}}\)-balanced so \([U(P_0)]=[U]\), and \(1/4\pi\) is dimensionless. In the aperture form \([A]=[U]\cdot\mathrm{m}\) (since \(U=A e^{ikr}/r\)), \([1/(i\lambda)]=\mathrm{m^{-1}}\), \([e^{ik(\dots)}/(r_{21}r_{01})]=\mathrm{m^{-2}}\), \([K]=1\), \([dS]=\mathrm{m^2}\): product \(=[U]\,\mathrm{m}\cdot\mathrm{m^{-1}}\cdot\mathrm{m^{-2}}\cdot\mathrm{m^{2}}=[U]\). Consistent.

Limiting cases

Forward direction \((\chi=0)\): \(K=\tfrac12(1+\cos0)=1\) — full-strength wavelets straight ahead, the Huygens limit.
Backward direction \((\chi=\pi)\): \(K=\tfrac12(1-1)=0\) — the spurious backward wave of naive Huygens is automatically cancelled.
Paraxial / small aperture-angle: \(r_{01}\approx z+\frac{(x-x')^2+(y-y')^2}{2z}\), \(K\approx1\), \(r_{01}\) in the denominator \(\approx z\): the integral becomes the Fresnel (near-field) diffraction transform with its quadratic phase.
Far field \((z\gg ka^2/2)\): the quadratic phase is negligible across the aperture and \(U(P_0)\) becomes the Fraunhofer pattern — the 2D Fourier transform of the aperture field.
Whole plane open, plane-wave input: the integral reproduces the incident plane wave unchanged (self-consistency of the propagator).

Breaks when

Aperture or features approach a wavelength \((a\lesssim\lambda)\). The scalar assumption fails: polarization couples to the boundary, evanescent fields dominate near the rim, and only the vector Stratton–Chu theory (or a full solution of Maxwell's equations, as in Bethe's theory of the small hole) is correct. Scalar Kirchhoff can be off by orders of magnitude.
Very near the aperture \((z\lesssim\lambda)\). Dropping \(1/r\) against \(ik\) (Step 10) is illegitimate; evanescent components that decay within a wavelength carry sub-wavelength structure the far-field integral discards. Near-field microscopy lives precisely where this formula breaks.
Observation on the screen plane itself. Kirchhoff's field violates its own boundary conditions there (\(U\) and \(\partial_nU\) do not both vanish on the geometric shadow edge), giving inconsistencies right at \(z\to0\); RS is exact on the plane but the two disagree in the near zone.
Conducting or non-thin screens. A screen with thickness, conductivity, or index structure supports induced currents and multiple reflection; the "0 on opaque / incident in aperture" idealization no longer holds.

Failure modes

Forgetting the obliquity factor. Writing Huygens wavelets with \(K=1\) everywhere re-introduces a backward wave and mis-normalizes the forward amplitude; \(K=\tfrac12(1+\cos\chi)\) is not optional.
Dropping the \(1/(i\lambda)\) prefactor. The \(90^\circ\) phase lead and the \(1/\lambda\) scaling are physical (they make the wavelet sum reconstruct the incident wave); omitting them corrupts phase in interferometric and holographic calculations.
Using both Kirchhoff conditions as if independent. Specifying \(U\) and \(\partial_nU\) freely over-determines a Helmholtz solution; students who "verify" the field on the plane find it contradicts the assumed values. This is the very inconsistency RS removes.
Confusing \(r_{01}\) in the phase with \(r_{01}\) in the amplitude. The exponent needs \(r_{01}\) to \(\sim\lambda\) accuracy (quadratic terms matter); the \(1/r_{01}\) prefactor tolerates \(r_{01}\approx z\). Approximating them at the same order breaks Fresnel diffraction.
Applying the scalar formula to polarized sub-wavelength gratings. A very common error in metamaterial/photonic contexts; scalar diffraction cannot see TE/TM splitting.
Sign of \(k\) / \(e^{-i\omega t}\) convention. An \(e^{+i\omega t}\) convention flips the outgoing Green's function to \(e^{-ikr}/r\) and the prefactor to \(-1/(i\lambda)\); mixing conventions inverts every phase.

Discussion

The derivation's deepest content is that it turns Huygens' heuristic into a theorem. Huygens (1678) posited that each point of a wavefront is a source of secondary spherical wavelets; Fresnel added interference by hand to explain diffraction fringes; but the construction needed an unexplained obliquity factor to kill the backward wave and an unexplained \(90^\circ\) phase shift to make the wavelets rebuild a plane wave. The Helmholtz–Kirchhoff integral produces all three — wavelet form \(e^{ikr}/r\), inclination factor \(K(\chi)\), and prefactor \(1/(i\lambda)\) — as forced consequences of the wave equation plus Green's theorem. Nothing is inserted by hand except the boundary values.

The mathematical engine is Green's second identity applied to two Helmholtz solutions, one of which is singular at the observation point. The bulk integrand cancels (both obey the same equation) and the entire field is squeezed out of the infinitesimal sphere surrounding \(P_0\) — the \(4\pi\) from the solid angle and the \(1/\varepsilon^2\) from the Green's function conspire to reproduce \(U(P_0)\) exactly. This "sifting" is structurally identical to how the Coulomb Green's function extracts a source in electrostatics or how the retarded Green's function builds Jefimenko's fields: it is the general machinery of inverting a linear differential operator against a point response.

The Kirchhoff and Rayleigh–Sommerfeld formulations expose a genuine tension in boundary-value theory. A solution of the Helmholtz equation is fixed by either its value (Dirichlet) or its normal derivative (Neumann) on a closed surface — not both. Kirchhoff imposes both on the aperture, so his field is mathematically inconsistent (it does not reproduce the assumed data on the screen), yet it agrees with experiment because the inconsistency is confined to within a wavelength of the rim, a region of negligible weight in the far field. Sommerfeld's fix is elegant: build a Green's function that vanishes (or whose normal derivative vanishes) on the whole plane by the method of images, so that only one datum — \(U\) alone, or \(\partial_nU\) alone — is needed and the problem is well-posed. RS-I and RS-II bracket Kirchhoff: their inclination factors are \(\cos\chi\) and \(1\), while Kirchhoff's \(\tfrac12(1+\cos\chi)\) is exactly their average. In the paraxial regime that dominates optics, the three collapse to the same Fresnel/Fraunhofer integral, which is why the distinction rarely surfaces in practice yet matters foundationally.

Common misconceptions. The formula does not say light "spreads because it is a wave" in a vague sense — it gives the precise amplitude and phase of the spreading, and reduces to rectilinear rays only in the \(\lambda\to0\) stationary-phase limit. The secondary wavelets are a calculational device, not literal re-emitters: nothing physically absorbs and re-radiates at the aperture. And "Kirchhoff is exact" is false — it is a consistent-to-first-order approximation whose success in the far field masks a boundary inconsistency that RS was invented to cure.

Worked examples

On-axis field behind a circular aperture, radius \(a\), plane-wave illumination (amplitude \(U_0\)), observation distance \(z\). Find the on-axis intensity.

By symmetry the integral over \(\Sigma\) is a sum over annular Fresnel zones; on axis the wavelet from radius \(\rho\) has extra path \(\sqrt{\rho^2+z^2}-z\).

\[ U(z)=U_0\!\left(1-e^{ik\delta}\right),\quad \delta=\sqrt{a^2+z^2}-z\approx\frac{a^2}{2z},\quad k\delta=\pi\frac{a^2}{\lambda z}\equiv\pi N_F \]

Symbols first: evaluating the on-axis diffraction integral gives the incident wave minus the rim contribution; the phase across the aperture is \(\pi N_F\) with the Fresnel number \(N_F=a^2/(\lambda z)\).

\[ I(z)=|U|^2=I_0\,\bigl|1-e^{i\pi N_F}\bigr|^2=4I_0\sin^2\!\left(\frac{\pi N_F}{2}\right) \]

Take the squared modulus; \(|1-e^{i\theta}|^2=2(1-\cos\theta)=4\sin^2(\theta/2)\).

\[ a=1.0\ \mathrm{mm},\ \lambda=500\ \mathrm{nm},\ z=0.50\ \mathrm{m}:\quad N_F=\frac{(1.0\times10^{-3})^2}{(500\times10^{-9})(0.50)}=\frac{1.0\times10^{-6}}{2.5\times10^{-7}}=4.0 \]

Insert numbers. \(N_F=4\) is even, so \(\sin^2(\pi N_F/2)=\sin^2(2\pi)=0\).

\[ I=4I_0\sin^2(2\pi)=0\quad(\text{dark on axis});\qquad N_F=1\Rightarrow I=4I_0\ (\text{four-fold bright}) \]

Reading. Even Fresnel-zone numbers give an on-axis null, odd numbers a bright spot up to four times the unobstructed intensity — the Poisson/Arago alternation that follows directly from the diffraction integral. A "bright spot in the centre of a shadow" is the same mechanism with a disc instead of a hole.

Fraunhofer single slit of width \(b\), normal plane-wave input. Reduce the diffraction integral to the far-field amplitude and find the first minimum angle.

Far field: \(K\to1\), \(r_{01}\) in the denominator \(\to z\), and \(r_{01}\approx R-x'\sin\theta\) in the phase (Fraunhofer approximation). The 1D aperture integral over \(x'\in[-b/2,b/2]\) remains.

\[ U(\theta)\propto\int_{-b/2}^{b/2}e^{-ikx'\sin\theta}\,dx'=b\,\frac{\sin\!\left(\tfrac{kb\sin\theta}{2}\right)}{\tfrac{kb\sin\theta}{2}}=b\,\mathrm{sinc}\!\left(\frac{\pi b\sin\theta}{\lambda}\right) \]

Symbols first: the far-field field is the Fourier transform of the slit, a sinc; intensity \(\propto\mathrm{sinc}^2\).

\[ \text{first minimum: }\ \frac{\pi b\sin\theta}{\lambda}=\pi\ \Rightarrow\ \sin\theta=\frac{\lambda}{b};\qquad b=0.10\ \mathrm{mm},\ \lambda=633\ \mathrm{nm} \]

The sinc vanishes when its argument is \(\pi\). Insert numbers: \(\sin\theta=633\times10^{-9}/1.0\times10^{-4}=6.33\times10^{-3}\).

\[ \theta_1=\arcsin(6.33\times10^{-3})\approx6.3\times10^{-3}\ \mathrm{rad}\approx0.36^\circ \]

Reading. The familiar \(\sin\theta=\lambda/b\) single-slit result is not a separate law — it is the Fraunhofer limit of the Helmholtz–Kirchhoff integral, obtained by keeping only the linear phase term across the aperture.

Problems

An aperture of half-width \(a=0.50\ \mathrm{mm}\) is illuminated at \(\lambda=600\ \mathrm{nm}\) and observed at \(z=2.0\ \mathrm{m}\). Compute the Fresnel number \(N_F=a^2/(\lambda z)\) and state whether Fresnel or Fraunhofer diffraction applies.
Solution
\(N_F=\dfrac{(0.50\times10^{-3})^2}{(600\times10^{-9})(2.0)}=\dfrac{2.5\times10^{-7}}{1.2\times10^{-6}}=0.21\). Since \(N_F\ll1\), the quadratic phase across the aperture is negligible and the Fraunhofer (far-field) regime applies. (Fresnel diffraction is the regime \(N_F\sim1\); geometric optics is \(N_F\gg1\).)
For a circular aperture \(a=1.0\ \mathrm{mm}\), \(\lambda=500\ \mathrm{nm}\), find the largest distance \(z\) at which the on-axis point is bright with the maximum \(4I_0\) (i.e. \(N_F=1\)).
Solution
Maximum on-axis brightness needs the first Fresnel zone only, \(N_F=1\): \(z=\dfrac{a^2}{\lambda N_F}=\dfrac{(1.0\times10^{-3})^2}{(500\times10^{-9})(1)}=\dfrac{1.0\times10^{-6}}{5.0\times10^{-7}}=2.0\ \mathrm{m}\). At \(z=2.0\ \mathrm{m}\), \(I=4I_0\sin^2(\pi/2)=4I_0\). For \(z>2.0\ \mathrm{m}\), \(N_F<1\) and \(I<4I_0\), so this is the farthest four-fold-bright point.
Evaluate the Kirchhoff inclination factor \(K(\chi)=\tfrac12(1+\cos\chi)\) at \(\chi=0^\circ,\ 90^\circ,\ 180^\circ\) and explain physically what the \(\chi=180^\circ\) value guarantees.
Solution
\(K(0)=\tfrac12(1+1)=1\); \(K(90^\circ)=\tfrac12(1+0)=0.5\); \(K(180^\circ)=\tfrac12(1-1)=0\). The vanishing at \(\chi=180^\circ\) means no secondary wavelet radiates back toward the source, so the diffraction integral produces a purely forward-propagating field — the cancellation of the backward wave that Huygens' original construction could not justify.
The step \(ik-1/r\approx ik\) requires \(kr\gg1\). Compute \(kr\) for \(r=10\lambda\) and \(r=100\lambda\) and comment on the validity of the approximation.
Solution
\(k=2\pi/\lambda\), so \(kr=2\pi(r/\lambda)\). For \(r=10\lambda\): \(kr=2\pi(10)=62.8\), and \(1/r\) is smaller than \(k\) by a factor \(kr\approx63\) (a \(\sim1.6\%\) correction). For \(r=100\lambda\): \(kr=628\), a \(\sim0.16\%\) correction. The approximation is excellent beyond a few tens of wavelengths and degrades only in the deep near field \(r\lesssim\lambda\), where \(kr\lesssim2\pi\) and the \(1/r\) term is no longer negligible.
The Fresnel–Kirchhoff prefactor is \(1/(i\lambda)\). Write it in magnitude-and-phase form and state, for \(\lambda=633\ \mathrm{nm}\) and \(\lambda=450\ \mathrm{nm}\), the magnitude \(1/\lambda\) and the phase. Why must the wavelets carry this phase?
Solution
\(\dfrac{1}{i\lambda}=\dfrac{1}{\lambda}e^{-i\pi/2}\): magnitude \(1/\lambda\), phase \(-90^\circ\) (a \(90^\circ\) phase advance of the wavelet relative to a naive \(e^{ikr}/r\) source). For \(\lambda=633\ \mathrm{nm}\), \(1/\lambda=1.58\times10^{6}\ \mathrm{m^{-1}}\); for \(\lambda=450\ \mathrm{nm}\), \(1/\lambda=2.22\times10^{6}\ \mathrm{m^{-1}}\). The \(90^\circ\) phase and \(1/\lambda\) scaling are exactly what make the coherent sum of secondary wavelets reconstruct the incident wave when the whole plane is open; without them Huygens' construction reproduces neither the correct amplitude nor the correct phase of free propagation.