## Abstract

The principal maximum of axial irradiance of a focused beam with a low Fresnel number does not lie at its focal point; instead it lies at a point that is closer to the focusing pupil. It has been shown by the numerical example of a weakly truncated Gaussian beam that its value increases and its location moves closer to the pupil when spherical aberration is introduced into the beam. Such an increase has been referred to as “beyond the conventional diffraction limit.” Similarly, an increase in the value and a shift in the location of the principal maximum of axial irradiance of a uniform beam toward the pupil by the introduction of some spherical aberration has been characterized as an unexpected result. We explain why and how such a result comes about and that it neither invalidates any diffraction limit nor is it unexpected. We illustrate this for uniform as well as Gaussian beams of various truncation ratios. Both focused and collimated beams aberrated by spherical aberration or astigmatism are considered.

© 2005 Optical Society of America

Full Article |

PDF Article
### Equations (30)

Equations on this page are rendered with MathJax. Learn more.

(1)
$$A\left(\rho \right)={A}_{0}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma {\rho}^{2}),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}0\u2a7d\rho \u2a7d1,$$
(2)
$$\gamma ={(a\u2215\omega )}^{2}$$
(3)
$${A}_{0}^{2}=\frac{2\gamma}{1-\mathrm{exp}(-2\gamma )}\left(\frac{P}{{S}_{p}}\right).$$
(4)
$$I(z;\gamma )={\left(\frac{R}{z}\right)}^{2}\left(\frac{2\gamma}{{B}_{d}^{2}+{\gamma}^{2}}\right)\frac{1}{\mathrm{sinh}\phantom{\rule{0.2em}{0ex}}\gamma}(\mathrm{cosh}\phantom{\rule{0.2em}{0ex}}\gamma -\mathrm{cos}\phantom{\rule{0.2em}{0ex}}{B}_{d}),$$
(5)
$${B}_{d}\left(z\right)=\pi N(R\u2215z-1)$$
(6)
$$I(R;\gamma )=\left[\mathrm{tanh}(\gamma \u22152)\right]\u2215(\gamma \u22152).$$
(7)
$$2(\frac{\lambda z}{{S}_{p}}-\frac{{B}_{d}}{{B}_{d}^{2}+{\gamma}^{2}})(\mathrm{cosh}\phantom{\rule{0.2em}{0ex}}\gamma -\mathrm{cos}\phantom{\rule{0.2em}{0ex}}{B}_{d})=-\mathrm{sin}\phantom{\rule{0.2em}{0ex}}{B}_{d}.$$
(8)
$$I(z;\gamma )={\left(\frac{R}{z}\right)}^{2}\frac{2\gamma}{{B}_{d}^{2}+{\gamma}^{2}},$$
(9)
$$I\left(R\right)=2\u2215\gamma ,$$
(10)
$$I(0;\gamma )=2\gamma \u2215{\pi}^{2}{N}^{2}.$$
(11)
$$I({z}_{p};\gamma )=(2\u2215\gamma )+(2\gamma \u2215{\pi}^{2}{N}^{2}),$$
(12)
$${z}_{p}\u2215R={[1+{(\gamma \u2215\pi N)}^{2}]}^{-1}.$$
(13)
$$S={[1+{({B}_{d}\u2215\gamma )}^{2}]}^{-1}.$$
(14)
$$I\left(z\right)={(R\u2215z)}^{2}{\{\left[\mathrm{sin}({B}_{d}\u22152)\right]\u2215({B}_{d}\u22152)\}}^{2},$$
(15)
$$\mathrm{tan}({B}_{d}\u22152)=(R\u2215z){B}_{d}\u22152,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}z\ne R,$$
(16)
$$I(z;\gamma )=\frac{2\gamma}{1-\mathrm{exp}(-2\gamma )}{\left(\frac{R}{\pi z}\right)}^{2}{\mid {\int}_{0}^{1}{\int}_{0}^{2\pi}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma {\rho}^{2})\mathrm{exp}\left\{i[\Phi (\rho ,\theta )+{B}_{d}{\rho}^{2}]\right\}\rho \mathrm{d}\rho \mathrm{d}\theta \mid}^{2}.$$
(17)
$$I(z;\gamma )=\frac{2\gamma}{1-\mathrm{exp}(-2\gamma )}{\left(\frac{R}{z}\right)}^{2}{\mid {\int}_{0}^{1}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma x)\mathrm{exp}\left[i({A}_{s}{x}^{2}+{B}_{d}x)\right]\mathrm{d}x\mid}^{2}.$$
(18)
$$I(z;\gamma )=\frac{2\gamma}{1-\mathrm{exp}(-2\gamma )}{\left(\frac{R}{2}\right)}^{2}{\mid {\int}_{0}^{1}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma x)\mathrm{exp}\left[i(0.5{A}_{a}+{B}_{d})x\right]{J}_{0}\left(0.5{A}_{a}x\right)\mathrm{d}x\mid}^{2},$$
(19)
$${\int}_{0}^{2\pi}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}\left(i{A}_{a}{\rho}^{2}\phantom{\rule{0.2em}{0ex}}{\mathrm{cos}}^{2}\phantom{\rule{0.2em}{0ex}}\theta \right)\mathrm{d}\theta $$
(20)
$$=\mathrm{exp}\left(0.5i{A}_{a}{\rho}^{2}\right){\int}_{0}^{2\pi}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}\left(0.5i{A}_{a}{\rho}^{2}\phantom{\rule{0.2em}{0ex}}\mathrm{cos}\phantom{\rule{0.2em}{0ex}}2\theta \right)\mathrm{d}\theta =2\pi \phantom{\rule{0.2em}{0ex}}\mathrm{exp}\left(0.5i{A}_{a}{\rho}^{2}\right){J}_{0}\left(0.5{A}_{a}{\rho}^{2}\right).$$
(21)
$${\sigma}_{\Phi}^{2}=\u27e8{\Phi}^{2}\u27e9-{\u27e8\Phi \u27e9}^{2},$$
(22)
$$\u27e8{\Phi}^{n}\u27e9=\raisebox{1ex}{${\int}_{0}^{1}{\int}_{0}^{2\pi}A\left(\rho \right){[\Phi (\rho ,\theta )+{B}_{d}{\rho}^{2}]}^{n}\rho \mathrm{d}\rho \mathrm{d}\theta $}\!\left/ \!\raisebox{-1ex}{${\int}_{0}^{1}{\int}_{0}^{2\pi}A\left(\rho \right)\rho \mathrm{d}\rho \mathrm{d}\theta $}\right.=\{\gamma \u2215\pi [1-\mathrm{exp}(-\gamma )]\}{\int}_{0}^{1}{\int}_{0}^{2\pi}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma {\rho}^{2}){[\Phi (\rho ,\theta )+{B}_{d}{\rho}^{2}]}^{n}\rho \mathrm{d}\rho \mathrm{d}\theta ,$$
(23)
$$S={\left[\frac{\gamma}{1-\mathrm{exp}(-\gamma )}\right]}^{2}{\mid {\int}_{0}^{1}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma x)\mathrm{exp}\left[i({B}_{d}x+{A}_{s}{x}^{2})\right]\mathrm{d}x\mid}^{2},$$
(24)
$$I(z;\gamma )=\frac{2\gamma {({B}_{d}\u2215\pi )}^{2}}{1-\mathrm{exp}(-2\gamma )}{\mid {\int}_{0}^{1}{\int}_{0}^{2\pi}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma {\rho}^{2})\mathrm{exp}\left\{i[\Phi (\rho ,\theta )+{B}_{d}{\rho}^{2}]\right\}\rho \mathrm{d}\rho \mathrm{d}\theta \mid}^{2},$$
(25)
$${B}_{d}=\pi \u22154z$$
(26)
$$I(z;\gamma )=\{2\gamma \u2215[1+{(4\gamma z\u2215\pi )}^{2}]\}[\mathrm{coth}\phantom{\rule{0.2em}{0ex}}\gamma -\mathrm{cos}(\pi \u22154z)\u2215\mathrm{sinh}\phantom{\rule{0.2em}{0ex}}\gamma ].$$
(27)
$$I(z;0)=4\phantom{\rule{0.2em}{0ex}}{\mathrm{sin}}^{2}(\pi \u22158z)$$
(28)
$$I(z;\gamma )=2\gamma \u2215[1+{(4\gamma z\u2215\pi )}^{2}],$$
(29)
$$I(z;\gamma )=\frac{2\gamma {B}_{d}^{2}}{1-\mathrm{exp}(-2\gamma )}{\mid {\int}_{0}^{1}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma x)\mathrm{exp}\left[i({A}_{s}{x}^{2}+{B}_{d}x)\right]\mathrm{d}x\mid}^{2},$$
(30)
$$I(z;\gamma )=\frac{2\gamma {B}_{d}^{2}}{1-\mathrm{exp}(-2\gamma )}{\mid {\int}_{0}^{1}\phantom{\rule{0.2em}{0ex}}\mathrm{exp}(-\gamma x)\mathrm{exp}\left[i(0.5{A}_{a}+{B}_{d})x\right]{J}_{0}\left(0.5{A}_{a}x\right)\mathrm{d}x\mid}^{2},$$