## Abstract

The backscattering of circularly polarized light at normal incidence to a half-space of scattering particles is studied using the Electric Field Monte Carlo (EMC) method. The spatial distribution of the backscat-tered light intensity is examined for both the time-resolved and continuous wave cases for large particles with anisotropy factor, *g*, in the range 0.8 to 0.97. For the time-resolved case, the backscattered light with the same helicity as that of the incident beam (co-polarized) is found to form a ring centered on the point of incidence. The ring expands and simultaneously grows weak as time increases. The intensity of backscattered light with helicity opposite to that of the incident beam (cross-polarized) is found to exhibit a ring behavior for *g* ≥ 0.85, with significant backscattering at the point of incidence. For the continuous-wave case no such ring pattern is observed in backscattered light for either helicity. The present EMC study suggests that the ring behavior can only be observed in the time domain, in contrast to previous studies of light backscattered from forward scattering media based on the scalar time-independent Fokker-Planck approximation to the radiative transfer equation. The time-dependent ring structure of backscattered light may have potential use in subsurface imaging applications.

©2005 Optical Society of America

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### Equations (9)

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(1)
$$\left({c}_{o}^{-1}{\partial}_{t}+\hat{s}\bullet {\nabla}_{r}+{\sigma}_{t}\right)\mathbf{I}\left(\overrightarrow{r},\hat{s},t\right)={\int}_{\Omega}\mathbf{P}\left(\hat{s}\bullet \hat{s}\prime \right)\bullet \mathbf{I}(\overrightarrow{r},\hat{s}\prime ,t)ds\prime $$
(2)
$$\mathbf{s}\prime =\mathbf{m}\phantom{\rule{.2em}{0ex}}\mathrm{sin}\left(\theta \right)\mathrm{cos}\left(\varphi \right)+\mathbf{n}\phantom{\rule{.2em}{0ex}}\mathrm{sin}\left(\theta \right)\mathrm{sin}\left(\varphi \right)+\mathbf{s}\phantom{\rule{.2em}{0ex}}\mathrm{cos}\left(\theta \right)$$
(3)
$${\mathbf{e}}_{1}=\mathbf{m}\phantom{\rule{.2em}{0ex}}\mathrm{cos}\left(\varphi \right)+\mathbf{n}\phantom{\rule{.2em}{0ex}}\mathrm{sin}\left(\varphi \right)$$
(3)
$${\mathbf{e}}_{2}=-\mathbf{m}\phantom{\rule{.2em}{0ex}}\mathrm{sin}\left(\varphi \right)+\mathbf{n}\phantom{\rule{.2em}{0ex}}\mathrm{cos}\left(\varphi \right)$$
(4)
$${\mathbf{e}}_{1}^{\prime}=\mathbf{m}\phantom{\rule{.2em}{0ex}}\mathrm{cos}\left(\theta \right)\mathrm{cos}\left(\varphi \right)+\mathbf{n}\phantom{\rule{.2em}{0ex}}\mathrm{cos}\left(\theta \right)\mathrm{sin}\left(\varphi \right)-\mathbf{s}\phantom{\rule{.2em}{0ex}}\mathrm{sin}\left(\theta \right)=\mathbf{m}\prime $$
(4)
$${\mathbf{e}}_{2}^{\prime}={\mathbf{e}}_{2}=\mathbf{n}\prime \phantom{\rule{16.5em}{0ex}}$$
(5)
$$\mathbf{E}\prime ={E}_{1}^{\prime}\mathbf{m}\prime +{E}_{2}^{\prime}\mathbf{n}\prime =\left({S}_{2}\mathbf{E}\bullet {\mathbf{e}}_{1}\right)\mathbf{m}\prime +\left({S}_{1}\mathbf{E}\bullet {\mathbf{e}}_{2}\right)\mathbf{n}\prime $$
(6)
$$\left(\begin{array}{c}\mathbf{m}\prime \\ \mathbf{n}\prime \\ \mathbf{s}\prime \end{array}\right)=\left(\begin{array}{ccc}\mathrm{cos}\left(\theta \right)\mathrm{cos}\left(\varphi \right)& \mathrm{cos}\left(\theta \right)\mathrm{sin}\left(\varphi \right)& -\mathrm{sin}\left(\theta \right)\\ -\mathrm{sin}\left(\varphi \right)& \mathrm{cos}\left(\varphi \right)& 0\\ \mathrm{sin}\left(\theta \right)\mathrm{cos}\left(\varphi \right)& \mathrm{sin}\left(\theta \right)\mathrm{sin}\left(\varphi \right)& \mathrm{cos}\left(\theta \right)\end{array}\right)\bullet \left(\begin{array}{c}\mathbf{m}\\ \mathbf{n}\\ \mathbf{s}\end{array}\right)$$
(7)
$$\left(\begin{array}{c}{E}_{1}^{\prime}\\ {E}_{2}^{\prime}\end{array}\right)=\frac{1}{\sqrt{F\left(\theta ,\varphi \right)}}\left(\begin{array}{cc}{S}_{2}\phantom{\rule{.2em}{0ex}}\mathrm{cos}\left(\varphi \right)& {S}_{2}\phantom{\rule{.2em}{0ex}}\mathrm{sin}\left(\varphi \right)\\ -{S}_{1}\phantom{\rule{.2em}{0ex}}\mathrm{sin}\left(\varphi \right)& {S}_{1}\phantom{\rule{.2em}{0ex}}\mathrm{cos}\left(\varphi \right)\end{array}\right)\bullet \left(\begin{array}{c}{E}_{1}\\ {E}_{2}\end{array}\right)$$