## Abstract

A Fourier transform digital holographic adaptive optics imaging system and its basic principles are proposed. The CCD is put at the exact Fourier transform plane of the pupil of the eye lens. The spherical curvature introduced by the optics except the eye lens itself is eliminated. The CCD is also at image plane of the target. The point-spread function of the system is directly recorded, making it easier to determine the correct guide-star hologram. Also, the light signal will be stronger at the CCD, especially for phase-aberration sensing. Numerical propagation is avoided. The sensor aperture has nothing to do with the resolution and the possibility of using low coherence or incoherent illumination is opened. The system becomes more efficient and flexible. Although it is intended for ophthalmic use, it also shows potential application in microscopy. The robustness and feasibility of this compact system are demonstrated by simulations and experiments using scattering objects.

© 2012 Optical Society of America

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### Equations (8)

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(1)
$$O({x}_{2},{y}_{2})=\frac{1}{j\lambda {f}_{2}}\iint P({x}_{1},{y}_{1})\phantom{\rule{0ex}{0ex}}\mathrm{exp}[-j2\pi \frac{1}{\lambda {f}_{2}}({x}_{2}{x}_{1}+{y}_{2}{y}_{1})]\mathrm{d}{x}_{1}\mathrm{d}{y}_{1},$$
(2)
$$O({f}_{x},{f}_{y})=\mathrm{FT}\{P({x}_{1},{y}_{1})\}({f}_{x},{f}_{y}),$$
(3)
$${f}_{x}=\frac{{x}_{2}}{\lambda {f}_{2}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}{f}_{y}=\frac{{y}_{2}}{\lambda {f}_{2}}.$$
(4)
$$P({x}_{1},{y}_{1})=\mathrm{IFT}\{O({f}_{x},{f}_{y})\}({x}_{1},{y}_{1}).$$
(5)
$${O}_{c}({x}_{2},{y}_{2})=\mathrm{FT}\{P({x}_{1},{y}_{1})\mathrm{exp}(-j\varphi ({x}_{1},{y}_{1}))\}({f}_{x},{f}_{y}).$$
(6)
$$\mathrm{\Delta}{f}_{x}=\frac{\mathrm{\Delta}{x}_{2}}{\lambda {f}_{2}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}\mathrm{\Delta}{f}_{y}=\frac{\mathrm{\Delta}{y}_{2}}{\lambda {f}_{2}}.$$
(7)
$$\mathrm{\Delta}{x}_{1}=\frac{\lambda {f}_{2}}{M\mathrm{\Delta}{x}_{2}}\phantom{\rule[-0.0ex]{2.0em}{0.0ex}}\mathrm{\Delta}{y}_{1}=\frac{\lambda {f}_{2}}{N\mathrm{\Delta}{x}_{2}}.$$
(8)
$$D\le \frac{\sqrt{2}\lambda {f}_{2}}{4\mathrm{\Delta}{x}_{2}}.$$