## Abstract

Recently exact and approximate solutions were given for focusing of two-dimensional scalar and electromagnetic waves through a slit aperture in a perfectly reflecting screen. We present numerical comparisons, based on these solutions, between exact results and approximate Rayleigh–Sommerfeld results for a variety of different focusing geometries. These comparisons show that, at sufficiently low Fresnel numbers and large angular apertures, approximate solutions based on the first and the second Rayleigh–Sommerfeld diffraction integrals agree well with the corresponding exact solutions for hard and soft screens, respectively. Inasmuch as these results are contrary to what one would expect from considerations of the boundary conditions, we give an analytical explanation of them through comparisons between exact and approximate near-field solutions for the corresponding half-plane problems.

© 1998 Optical Society of America

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### Equations (39)

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(1)
$${E}_{y}={u}^{s},\hspace{1em}\hspace{1em}{E}_{x}={E}_{z}=0,$$
(2)
$${H}_{x}=-\frac{c}{i\omega \mu}\frac{\partial {E}_{y}}{\partial z},\hspace{1em}\hspace{1em}{H}_{z}=\frac{c}{i\omega \mu}\frac{\partial {E}_{y}}{\partial x},\hspace{1em}\hspace{1em}{H}_{y}=0.$$
(3)
$${H}_{y}={u}^{h},\hspace{1em}\hspace{1em}{H}_{x}={H}_{z}=0,$$
(4)
$${E}_{x}=-\frac{\mathit{ic}}{\omega \u220a}\frac{\partial {H}_{y}}{\partial z},\hspace{1em}\hspace{1em}{E}_{z}=\frac{\mathit{ic}}{\omega \u220a}\frac{\partial {H}_{y}}{\partial x},\hspace{1em}\hspace{1em}{E}_{y}=0.$$
(5)
$${u}^{i}=exp[\mathit{iks}cos(\theta -{\theta}_{0})],$$
(6)
$${u}^{r}=exp[\mathit{iks}cos(\theta +{\theta}_{0})].$$
(7)
$$\left\{\begin{array}{c}{u}^{s}\\ {u}^{h}\end{array}\right\}=F({\xi}^{i}){u}^{i}\mp F({\xi}^{r}){u}^{r},$$
(8)
$$F(x)=\xbd\{1-\sqrt{2}exp(-i\pi /4)[C(x)+\mathit{iS}(x)]\},$$
(9)
$$C(x)=\sqrt{\frac{2}{\pi}}{\int}_{0}^{x}cos({t}^{2})\mathrm{d}t,$$
(10)
$$S(x)=\sqrt{\frac{2}{\pi}}{\int}_{0}^{x}sin({t}^{2})\mathrm{d}t.$$
(11)
$$F(x)-F(-x)=-\sqrt{2}exp(-i\pi /4)[C(x)+\mathit{iS}(x)],$$
(13)
$${\xi}^{i}=-\sqrt{2\mathit{ks}}sin\xbd(\theta -{\theta}_{0}),$$
(14)
$${\xi}^{r}=\sqrt{2\mathit{ks}}sin\xbd(\theta +{\theta}_{0}).$$
(15)
$${\xi}^{i}=-{\xi}^{r}=-\sqrt{2\mathit{ks}}cos\xbd{\theta}_{0}0,$$
(16)
$${u}^{i}={u}^{r}=exp(-\mathit{iks}cos{\theta}_{0}).$$
(17)
$${u}^{h}(x\u2a7d0,z=0)={u}^{i},$$
(18)
$${u}^{s}(x\u2a7d0,z=0)=-{u}^{i}\sqrt{2}exp(-i\pi /4)[C({\xi}^{i})+\mathit{iS}({\xi}^{i})].$$
(19)
$${u}^{s}(x\u2a7d0,z={0}^{+})={u}^{i}-\frac{exp[i(\mathit{ks}+\pi /4)]}{\sqrt{2\pi \mathit{ks}}cos\xbd{\theta}_{0}}.$$
(20)
$${\xi}^{i}={\xi}^{r}=\sqrt{2\mathit{ks}}sin\xbd{\theta}_{0}0,$$
(21)
$${u}^{i}={u}^{r}=exp(\mathit{iks}cos{\theta}_{0}),$$
(22)
$$\left\{\begin{array}{c}{u}^{s}(x>0,z={0}^{+})\\ \\ {u}^{h}(x0,z={0}^{+})\end{array}\right\}=\left\{\begin{array}{l}0\\ \\ 2F({\xi}^{i}){u}^{i}\end{array}\right..$$
(23)
$$2F({\xi}^{i}){u}^{i}\sim \frac{exp[i(\mathit{ks}+\pi /4)]}{\sqrt{2\pi \mathit{ks}}sin\xbd{\theta}_{0}}.$$
(24)
$${u}^{s}(x,{0}^{+})=\left\{\begin{array}{ll}-{u}^{i}\sqrt{2}exp\left(\frac{-i\pi}{4}\right)[C({\xi}^{i})+\mathit{iS}({\xi}^{i})]\sim {u}^{i}-\frac{1}{cos\xbd{\theta}_{0}}\frac{exp\left[i\left(\mathit{ks}+\frac{\pi}{4}\right)\right]}{\sqrt{2\pi \mathit{ks}}}& x0\\ 0& x0\end{array},\right.$$
(25)
$${u}^{h}(x,{0}^{+})=\left\{\begin{array}{ll}{u}^{i}& x0\\ 2{u}^{i}F({\xi}^{i})\sim \frac{1}{sin\xbd{\theta}_{0}}\frac{exp[i(\mathit{ks}+\pi /4)]}{\sqrt{2\pi \mathit{ks}}}& x0\end{array}.\right.$$
(26)
$${u}_{\mathrm{I}}=\frac{1}{2i}{\int}_{-\infty}^{0}{u}^{i}(x,0)\frac{\partial}{\partial {z}_{2}}{H}_{0}^{(1)}({\mathit{kR}}_{2})\mathrm{d}x=\frac{\mathit{ik}}{2}{\int}_{-\infty}^{0}{u}^{i}(x,0)\frac{{z}_{2}}{{R}_{2}}{H}_{1}^{(1)}({\mathit{kR}}_{2})\mathrm{d}x,$$
(27)
$${u}_{\mathrm{II}}=\frac{1}{2i}{\int}_{-\infty}^{0}{\left.\frac{\partial {u}^{i}(x,z)}{\partial z}\right|}_{z=0}{H}_{0}^{(1)}({\mathit{kR}}_{2})\mathrm{d}x,$$
(28)
$${R}_{2}=[(x-{x}_{2}{)}^{2}+z_{2}{}^{2}{]}^{1/2},$$
(29)
$${u}^{i}(x,0)=exp(\mathit{ikx}cos{\theta}_{0}),$$
(30)
$${\left.\frac{\partial {u}^{i}(x,z)}{\partial x}\right|}_{z=0}={\mathit{iku}}^{i}sin{\theta}_{0}.$$
(31)
$${u}_{j}={\int}_{-\infty}^{0}{g}_{j}(x)exp[\mathit{ikf}(x)]\mathrm{d}x\hspace{1em}\hspace{1em}(j=\mathrm{I},\mathrm{II}),$$
(32)
$${g}_{\mathrm{I}}(x)=\frac{exp(-i\pi /4)}{\sqrt{\mathrm{\lambda}{z}_{2}}}{\left(\frac{{z}_{2}}{{R}_{2}}\right)}^{3/2},$$
(33)
$${g}_{\mathrm{II}}(x)=\frac{exp(-i\pi /4)}{\sqrt{\mathrm{\lambda}{z}_{2}}}{\left(\frac{{z}_{2}}{{R}_{2}}\right)}^{1/2}sin{\theta}_{0},$$
(34)
$$f(x)={R}_{2}+xcos{\theta}_{0}.$$
(35)
$${u}_{\mathrm{I}}(s,\theta )\sim {u}^{i}H(\theta -{\theta}_{0})-\frac{sin\theta}{cos{\theta}_{0}-cos\theta}\frac{exp[i(\mathit{ks}+\pi /4)]}{\sqrt{2\pi \mathit{ks}}},$$
(36)
$${u}_{\mathrm{II}}(s,\theta )\sim {u}^{i}H(\theta -{\theta}_{0})-\frac{sin{\theta}_{0}}{cos{\theta}_{0}-cos\theta}\frac{exp[i(\mathit{ks}+\pi /4)]}{\sqrt{2\pi \mathit{ks}}},$$
(37)
$$s=\sqrt{x_{2}{}^{2}+z_{2}{}^{2}},\hspace{1em}\hspace{1em}\theta =\mathrm{arctan}({x}_{2}/{z}_{2}).$$
(38)
$${u}_{\mathrm{I}}(x,{0}^{+})=\left\{\begin{array}{ll}{u}^{i}& x0\\ 0& x0\end{array}\right.,$$
(39)
$${u}_{\mathrm{II}}(x,{0}^{+})\sim \left\{\begin{array}{ll}{u}^{i}-\frac{sin\xbd{\theta}_{0}}{cos\xbd{\theta}_{0}}\frac{exp\left[i\left(\mathit{ks}+\frac{\pi}{4}\right)\right]}{\sqrt{2\pi \mathit{ks}}}& x0\\ \frac{cos\xbd{\theta}_{0}}{sin\xbd{\theta}_{0}}\frac{exp\left[i\left(\mathit{ks}+\frac{\pi}{4}\right)\right]}{\sqrt{2\pi \mathit{ks}}}& x0\end{array}\right..$$