## Abstract

An inverse, Monte Carlo (IMC) technique is developed to solve the electromagnetic inverse-scattering problem from generally complex distributions of dielectric particles. One can verify the technique using simulated scattering data from aerosols composed of spherical dielectrics. The IMC method is found to give accurate inversion results even when the data have a signal-to-noise ratio to as low as 3:1.

© 1996 Optical Society of America

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### Equations (9)

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(1)
$$i\left(\text{\lambda ,}\phantom{\rule{0.2em}{0ex}}\text{\theta}\right)={\displaystyle \sum _{m=1}^{M}{i}_{m}\left(\text{\lambda ,}\phantom{\rule{0.2em}{0ex}}\text{\theta},{\tilde{\text{\zeta}}}^{m}\right),}$$
(2)
$$i\left(\text{\lambda},\text{\theta}\right)={N}_{0}{\displaystyle \int g\left(\tilde{\text{\zeta}}\right)i\left(\text{\lambda},\text{\theta ,}\phantom{\rule{0.2em}{0ex}}\tilde{\text{\zeta}}\right)\mathrm{d}\tilde{\text{\zeta}},}$$
(3)
$${i}_{\text{\mu}}={N}_{0}{\displaystyle \sum _{j}{\tilde{i}}_{\text{\mu}j}{f}_{j}}.$$
(4)
$${\tilde{i}}_{\text{\mu}j}\equiv \tilde{i}\left(\text{\lambda ,}\phantom{\rule{0.2em}{0ex}}\text{\theta ,}\phantom{\rule{0.2em}{0ex}}{a}_{j},{m}^{\prime}\right)=\frac{1}{\text{\delta}a}{\displaystyle {\int}_{{a}_{j}-\text{\delta}a/2}^{{a}_{j}+\text{\delta}a/2}i\left(\text{\lambda ,}\phantom{\rule{0.2em}{0ex}}\text{\theta ,}\phantom{\rule{0.2em}{0ex}}{a}^{\prime},{m}^{\prime}\right)}\mathrm{d}{a}^{\prime}.$$
(5)
$$\begin{array}{cc}\overline{{s}_{\text{\mu}}}=\frac{{s}_{\text{\mu}}}{{s}_{\text{\mu}}},& \overline{{i}_{\text{\mu}}}=\end{array}\frac{{i}_{\text{\mu}}}{{i}_{\text{\mu}}}=\frac{{\displaystyle \sum _{1}^{P}{\tilde{i}}_{\text{\mu}j}{f}_{j}}}{{\displaystyle \sum _{1}^{P}{\tilde{i}}_{0j}{f}_{j}}}.$$
(6)
$${R}^{2}={{\displaystyle \sum _{\text{\mu}}\left(\frac{\overline{{s}_{\text{\mu}}}-\overline{{i}_{\text{\mu}}}}{\overline{{s}_{\text{\mu}}}}\right)}}^{2}.$$
(7)
$${N}_{0}=\frac{1}{n\text{data}}{\displaystyle \sum _{\text{\mu}}^{n\text{data}}\left(\frac{{s}_{\text{\mu}}}{{i}_{\text{\mu}}}\right)}.$$
(8)
$${\text{\chi}}^{2}={\displaystyle \sum _{\text{\mu}}^{n\text{data}}\frac{{\left(\overline{{s}_{\text{\mu}}}-\overline{{i}_{\text{\mu}}}\right)}^{2}}{{\u220a}_{\text{\mu}}}},$$
(9)
$$g\left(a\right)=\frac{1}{2\text{\pi}a{\sigma}_{g}}\phantom{\rule{0.2em}{0ex}}\text{exp}\left[-\frac{{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}a-\text{ln}\phantom{\rule{0.2em}{0ex}}{a}_{m}\right)}^{2}}{2{{\text{\sigma}}_{g}}^{2}}\right],$$