Rayleigh–Sommerfeld diffraction

Remarks

Let $r'$ be the distance to the mirror image of $P$ in the plane. The proof applies Green's representation formula on the half-space with the Dirichlet Green's function $\frac{1}{4\pi}\left(\frac{e^{ikr}}{r} - \frac{e^{ikr'}}{r'}\right)$. For the second formula it uses the Neumann Green's function, with $+$ instead of $-$. Each Green's function kills one of the two boundary terms of Kirchhoff's integral theorem, so only $u$ (or only $\partial u / \partial z$) is needed on the plane. Since $\frac{\partial}{\partial z} \frac{e^{ikr}}{r} = \left(i k - \frac{1}{r}\right) \frac{e^{ikr}}{r} \cos\theta$, where $\cos\theta = z/r$, the first formula becomes, for $r \gg \lambda$, $$u(P) \approx \frac{1}{i\lambda} \int_{\R^2} u(\xi, \eta, 0) \, \frac{e^{ikr}}{r} \cos\theta \, d\xi \, d\eta,$$ the Huygens–Fresnel principle with obliquity factor $\cos\theta$. In diffraction by an aperture $A$ in a screen, one sets $u$ equal to the incident field on $A$ and $0$ on the rest of the plane. Unlike Kirchhoff's conditions, this prescribes only the boundary values, so it is consistent. The radiation condition is needed: the standing wave $\sin(kz)$ solves the Helmholtz equation and vanishes on the plane $z = 0$, so without it the first formula would fail.

Going deeper

For a student

These formulas are exact for outgoing waves in a half-space, and they are the rigorous base of scalar diffraction theory. The first one, with the field given on an aperture, is the usual model for diffraction by a hole in a thin screen. Taking the two-dimensional Fourier transform of the formula gives the angular spectrum method, in which propagation multiplies each plane-wave component by a phase.