## Abstract

We present a generalized holography-based approach with improved spatial resolution for extracting images, viewed through a scattering medium. The various angular directions are encoded either with different wavelengths or by capturing their corresponding images in different time slots. The various encoded images are recorded on a digital hologram with a computer. A digital reconstruction, which includes demodulation of the carrier beam and then a proper decoding algorithm, yields resolved images. The principle is demonstrated by recording image-plane digital holograms. Combining the suggested approach with the first-arriving light technique may further improve the results.

© 2002 Optical Society of America

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### Equations (14)

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(1)
$${u}_{1}\left(x\right)=exp\left(\frac{2\mathrm{\pi}\mathit{ix}}{\mathrm{\lambda}}sin\mathrm{\theta}\right).$$
(2)
$${u}_{2}\left(x\right)={u}_{1}\left(x\right)exp\left(-2\mathrm{\pi}i{\mathit{\nu}}_{0}x\right)=exp\left[2\mathrm{\pi}\mathit{ix}\left({\mathit{\nu}}_{0}-\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)\right],$$
(3)
$${u}_{3}\left(x\right)\propto \int {u}_{2}\left(x\prime \right)exp\left(\frac{\mathrm{\pi}i}{\mathrm{\lambda}z}{\left(x-x\prime \right)}^{2}\right)\mathrm{d}x\prime =\int \mathrm{\delta}\left(\mathit{\nu}-{\mathit{\nu}}_{0}+\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)exp\left(-\mathrm{\pi}i\mathrm{\lambda}z{\mathit{\nu}}^{2}\right)exp\left(2\mathrm{\pi}\mathit{ix}\mathit{\nu}\right)\mathrm{d}\mathit{\nu}=exp\left[-\mathrm{\pi}i\mathrm{\lambda}z{\left({\mathit{\nu}}_{0}-\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)}^{2}\right]exp\left[2\mathrm{\pi}\mathit{ix}\left({\mathit{\nu}}_{0}-\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)\right].$$
(4)
$${u}_{3}\left(x\right)\propto exp\left\{\frac{2\mathrm{\pi}\mathit{ix}}{\mathrm{\lambda}}sin\left[\mathrm{\alpha}\left(\mathrm{\lambda}\right)\right]\right\}.$$
(5)
$${\mathit{\nu}}_{0}=\frac{sin\mathrm{\theta}}{{\mathrm{\lambda}}_{0}}.$$
(6)
$$\mathrm{\alpha}={sin}^{-1}\left[sin\mathrm{\theta}\left(\frac{\mathrm{\lambda}}{{\mathrm{\lambda}}_{0}}-1\right)\right].$$
(7)
$${u}_{4}\left({x}_{0}\right)=\int {u}_{3}\left(x\right)t\left(x\right)exp\left(\frac{-2\mathrm{\pi}{\mathit{ixx}}_{0}}{\mathrm{\lambda}f}\right)\mathrm{d}x=\left\{exp\left[-\mathrm{\pi}i\mathrm{\lambda}z{\left({\mathit{\nu}}_{0}-\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)}^{2}\right]\mathrm{\delta}\left(\frac{{x}_{0}}{\mathrm{\lambda}f}-{\mathit{\nu}}_{0}+\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)\right\}*T\left({x}_{0}\right)=exp\left[-\mathrm{\pi}i\mathrm{\lambda}z{\left({\mathit{\nu}}_{0}-\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)}^{2}\right]T\left({x}_{0}-\mathrm{\lambda}f{\mathit{\nu}}_{0}+fsin\mathrm{\theta}\right),$$
(8)
$$T\left({x}_{0}\right)=\int t\left(x\right)exp\left(-\frac{2\mathrm{\pi}{\mathit{ixx}}_{0}}{\mathrm{\lambda}f}\right)\mathrm{d}x.$$
(9)
$${u}_{5}\left({x}_{0}\right)={u}_{4}\left({x}_{0}\right)\mathrm{\delta}\left({x}_{0}\right)=exp\left[-\mathrm{\pi}i\mathrm{\lambda}z{\left({\mathit{\nu}}_{0}-\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)}^{2}\right]\times T\left[fsin\mathrm{\theta}\left(1-\frac{\mathrm{\lambda}}{{\mathrm{\lambda}}_{0}}\right)\right]\mathrm{\delta}\left({x}_{0}\right).$$
(10)
$${u}_{6}\left(x\right)=exp\left[-\mathrm{\pi}i\mathrm{\lambda}z{\left({\mathit{\nu}}_{0}-\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)}^{2}\right]\times T\left[fsin\mathrm{\theta}\left(1-\frac{\mathrm{\lambda}}{{\mathrm{\lambda}}_{0}}\right)\right],$$
(11)
$${u}_{6}\left(x\right)exp\left[\mathrm{\pi}i\mathrm{\lambda}z{\left({\mathit{\nu}}_{0}-\frac{sin\mathrm{\theta}}{\mathrm{\lambda}}\right)}^{2}\right]=T\left(fsin\mathrm{\alpha}\right)=T\left(\mathit{\nu}\right)t\left(x\right)=\int T\left(\mathit{\nu}\right)exp\left(2\mathrm{\pi}\mathit{ix}\mathit{\nu}\right)\mathrm{d}v.$$
(12)
$$\mathrm{\Delta}x=f{\mathit{\nu}}_{0}\mathrm{\delta}\mathrm{\lambda},$$
(13)
$$\mathrm{\delta}x=f{\mathit{\nu}}_{0}\mathrm{\Delta}\mathrm{\lambda}.$$
(14)
$$\mathrm{\delta}\mathrm{\theta}\approx {sin}^{-4}\left(\frac{\mathrm{\Delta}x}{f}\right),$$