## Abstract

Superresolution performance of a plano-convex lens made of absorbing
glass is analyzed numerically. It was found that a reduction of the radius of
the Airy disk depends solely on the center transmittance of the lens (a zero
edge thickness is assumed) for any reasonable ratio of radius of curvature of
the convex surface and the lens radius. The modest decrease in the size of the
central diffraction peak is followed by a large decrease of its energy content
and a rapid brightening of the diffraction rings. The most that can be
achieved with such a lens is a reduction of the radius of the Airy disk to
71% of the corresponding clear aperture value, followed by
approximately 78% of the energy being diverted into diffraction
rings.

© 1997 Optical Society of America

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### Equations (10)

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(1)
$$x=\frac{r}{{r}_{l}},$$
(2)
$$u=\frac{R}{{r}_{l}},$$
(3)
$${t}_{c}=exp\left\{-\mathrm{\kappa}{r}_{l}\left[u-{\left({u}^{2}-1\right)}^{1/2}\right]\right\},$$
(4)
$$t\left(x\right)exp\left\{-\mathrm{\kappa}{r}_{l}\left[{\left({u}^{2}-{x}^{2}\right)}^{1/2}-{\left({u}^{2}-1\right)}^{1/2}\right]\right\}.$$
(5)
$${u}_{1}={u}_{2}\mathrm{and}{\mathrm{\kappa}}_{1}{r}_{l1}={\mathrm{\kappa}}_{2}{r}_{l2}.$$
(7)
$$t\left(x\right)\approx exp\left\{-\mathrm{\kappa}{r}_{l}\left[\left(1-{x}^{2}\right)/2u\right]\right\}.$$
(8)
$$A\left(\mathrm{\vartheta}\right)={\int}_{0}^{2\mathrm{\pi}}{\int}_{0}^{r}t\left(\mathrm{\rho}\right)cos\left(\frac{2\mathrm{\pi}}{\mathrm{\lambda}}\mathrm{\rho}cos\mathrm{\varphi}sin\mathrm{\vartheta}\right)\mathrm{\rho}\mathrm{d}\mathrm{\rho}\mathrm{d}\mathrm{\varphi},$$
(9)
$$I\left(\mathrm{\vartheta}\right)={\left[\frac{A\left(\mathrm{\vartheta}\right)}{A\left(0\right)}\right]}^{2}.$$
(10)
$${\mathrm{\vartheta}}_{\mathrm{o}}=\frac{1.22\mathrm{\lambda}}{r},$$