## Abstract

We present numerical experiments of light scattering by a circular dielectric
cylinder embedded in a stratified background, using the Green’s
tensor technique. The stratified background consists of two or three dielectric
layers, the latter forming an anti–reflection system. We show movies of the
scattered field as a function of different parameters: polarization, angle of
incidence, and relative position of the cylinder with respect to the background
interfaces.

© 2001 Optical Society of America

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### Equations (16)

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(1)
$$\mathbf{E}\left(\mathbf{r}\right)={\mathbf{E}}^{0}(\mathbf{r})+{\int}_{A}d\mathbf{r}\prime \mathbf{G}(\mathbf{r},\mathbf{r}\prime )\xb7{k}_{0}^{2}\Delta \epsilon \left(\mathbf{r}\prime \right)\mathbf{E}\left(\mathbf{r}\prime \right),$$
(2)
$$\Delta \epsilon \left(\mathbf{r}\right)=\epsilon \left(\mathbf{r}\right)-{\epsilon}_{k},\phantom{\rule{.5em}{0ex}}\mathbf{r}\in \mathrm{layer}\phantom{\rule{.2em}{0ex}}k.$$
(3)
$${\mathbf{E}}^{0}(x,z)={\mathbf{A}}^{0}\mathrm{exp}\left(i\mathbf{kr}\right)={\mathbf{A}}^{0}\mathrm{exp}\left({\mathit{ik}}_{z}z\right),$$
(4)
$${\mathbf{E}}^{0}(x,z)={\mathbf{A}}^{0}\left[\mathrm{exp}\left({\mathit{ik}}_{1z}z\right)+R\mathrm{exp}\left(-{\mathit{ik}}_{1z}z\right)\right],\phantom{\rule{.5em}{0ex}}z\ge 0,$$
(5)
$${\mathbf{E}}^{0}(x,z)={\mathbf{A}}^{0}T\phantom{\rule{.2em}{0ex}}\mathrm{exp}\left({\mathit{ik}}_{2z}z\right),\phantom{\rule{.5em}{0ex}}z<0,$$
(6)
$$R=\frac{\sqrt{{\epsilon}_{1}}-\sqrt{{\epsilon}_{2}}}{\sqrt{{\epsilon}_{1}}+\sqrt{{\epsilon}_{2}}},$$
(7)
$$T=\frac{2\sqrt{{\epsilon}_{1}}}{\sqrt{{\epsilon}_{1}}+\sqrt{{\epsilon}_{2}}}.$$
(8)
$$A={\left({\mathbf{E}}^{0}\xb7{\mathbf{E}}^{{0}^{*}}\right)}^{\frac{1}{2}}=\mid {\mathbf{A}}^{0}\mid {\left[1+{R}^{2}+2R\mathrm{cos}\left({2k}_{1z}z\right)\right]}^{\frac{1}{2}},\phantom{\rule{.5em}{0ex}}z\ge 0,$$
(9)
$${E}_{\alpha}^{0}(x,z)={A}_{\alpha}^{0}\left[\mathrm{exp}\left({\mathit{ik}}_{1z}z\right)+{R}_{\alpha}\mathrm{exp}\left(-{\mathit{ik}}_{1z}z\right)\right]\mathrm{exp}\left({\mathit{ik}}_{x}x\right),\phantom{\rule{.2em}{0ex}}z\ge 0,\phantom{\rule{.2em}{0ex}}\alpha =x,y,z,$$
(10)
$${R}^{s}=\frac{{k}_{1z}-{k}_{2z}}{{k}_{1z}+{k}_{2z}},$$
(11)
$${R}^{P}=\frac{{{\epsilon}_{2}k}_{1z}-{\epsilon}_{1}{k}_{2z}}{{\epsilon}_{2}{k}_{1z}+{\epsilon}_{1}{k}_{2z}}.$$
(12)
$$A=\{{A}_{x}^{{0}^{2}}\left[1+{R}^{{p}^{2}}-2{R}^{p}\mathrm{cos}\left({2k}_{1z}z\right)\right]+{A}_{y}^{{0}^{2}}\left[1+{R}^{{s}^{2}}+{2R}^{s}\mathrm{cos}\left({2k}_{1z}z\right)\right]$$
(12)
$$+{A}_{z}^{{0}^{2}}\left[1+{R}^{{p}^{2}}+{2R}^{p}\mathrm{cos}\left({2k}_{1z}z\right)\right]{\}}^{\frac{1}{2}},\phantom{\rule{.2em}{0ex}}z\ge 0.$$
(13)
$$A=\mid {\mathbf{A}}^{0}\mid {\left(1+{R}^{{p}^{2}}\right)}^{\frac{1}{2}}.$$
(14)
$${\epsilon}_{2}=\sqrt{{\epsilon}_{1}{\epsilon}_{3}}\approx 1.334,$$
(15)
$$d=\left(n+\frac{1}{2}\right)\frac{\lambda}{2\sqrt{{\epsilon}_{2}}},\phantom{\rule{.5em}{0ex}}n=1,2,3,\dots $$