## Abstract

A method to determine *f*-number for an imaging system is discussed. This method uses Ronchi test and is different in that the value of *f*-number of the system can be determined without depending on any particular parameters. Two different sizes of aperture for a common system were investigated and the corresponding *f*-numbers were compared with those calculated by a lens design software. In addition, the determined *f*-numbers are proved to be consistent with other values obtained in this study.

© 2009 Optical Society of America

Full Article |

PDF Article
### Equations (13)

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

(1)
$$\epsilon {\left(i\right)}_{x}=-2\left(F/\#\right)\sum _{k=1}^{35}{a}_{k}\frac{\partial Z{\left(i\right)}_{k}}{\partial x}$$
(2)
$$\epsilon {\left(i\right)}_{y}=-2\left(F/\#\right)\sum _{k=1}^{35}{a}_{k}\frac{\partial Z{\left(i\right)}_{k}}{\partial y}$$
(3)
$$\text{AZ}{\text{Z}}^{T}=-\frac{1}{2\left(F/\#\right)}E{Z}^{T}$$
(4)
$$A=({a}_{1}\phantom{\rule{1em}{0ex}}{a}_{2}\phantom{\rule{1em}{0ex}}\cdots \phantom{\rule{1em}{0ex}}{a}_{35}).$$
(5)
$$E({\epsilon}_{x}\left(1\right)\phantom{\rule{.6em}{0ex}}{\epsilon}_{x}\left(2\right)\phantom{\rule{.6em}{0ex}}\cdots \phantom{\rule{.6em}{0ex}}{\epsilon}_{x}\left({N}_{x}\right)\phantom{\rule{.6em}{0ex}}{\epsilon}_{y}\left(1\right)\phantom{\rule{.6em}{0ex}}{\epsilon}_{y}\left(2\right)\phantom{\rule{.6em}{0ex}}\cdots \phantom{\rule{.6em}{0ex}}{\epsilon}_{y}\left({N}_{y}\right)),$$
(6)
$$Z=\left(\begin{array}{cccccccc}\frac{\partial Z{\left(1\right)}_{1}}{\partial x}& \frac{\partial Z{\left(2\right)}_{1}}{\partial x}& \cdots & \frac{\partial Z{\left({N}_{x}\right)}_{1}}{\partial x}& \frac{\partial Z{\left(1\right)}_{1}}{\partial y}& \frac{\partial Z{\left(2\right)}_{1}}{\partial y}& \cdots & \frac{\partial Z{\left({N}_{y}\right)}_{1}}{\partial y}\\ \frac{\partial Z{\left(1\right)}_{2}}{\partial x}& \frac{\partial Z{\left(2\right)}_{2}}{\partial x}& \cdots & \frac{\partial Z{\left({N}_{x}\right)}_{2}}{\partial x}& \frac{\partial Z{\left(1\right)}_{2}}{\partial y}& \frac{\partial Z{\left(2\right)}_{2}}{\partial y}& \cdots & \frac{\partial Z{\left({N}_{y}\right)}_{2}}{\partial y}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \frac{\partial Z{\left(1\right)}_{34}}{\partial x}& \frac{\partial Z{\left(2\right)}_{34}}{\partial x}& \cdots & \frac{\partial Z{\left({N}_{x}\right)}_{34}}{\partial x}& \frac{\partial Z{\left(1\right)}_{34}}{\partial y}& \frac{\partial Z{\left(2\right)}_{34}}{\partial y}& \cdots & \frac{\partial Z{\left({N}_{y}\right)}_{34}}{\partial x}\\ \frac{\partial Z{\left(1\right)}_{35}}{\partial x}& \frac{\partial Z{\left(2\right)}_{35}}{\partial x}& \cdots & \frac{\partial Z{\left({N}_{x}\right)}_{35}}{\partial x}& \frac{\partial Z{\left(1\right)}_{35}}{\partial y}& \frac{\partial Z{\left(2\right)}_{35}}{\partial y}& \cdots & \frac{\partial Z{\left({N}_{y}\right)}_{35}}{\partial y}\end{array}\right).$$
(7)
$$A=-\frac{1}{2\left(F/\#\right)}E{Z}^{T}{\left(Z{Z}^{T}\right)}^{-1}.$$
(8)
$$\Delta {a}_{3}=\frac{\Delta {a}_{3,F/\#=1}}{\left(F/\#\right)}$$
(9)
$$\Delta {a}_{3}=\frac{<\Delta {a}_{3,F/\#=1}>}{\left(F/\#\right)}.$$
(10)
$${}_{0}{W}_{20}=\frac{\mathrm{\delta z}}{8{\left(F/\#\right)}^{2}},$$
(11)
$$\Delta \left({}_{0}{W}_{20}\right)=\frac{\mathrm{\Delta z}}{8{\left(F/\#\right)}^{2}}.$$
(12)
$$\Delta {a}_{3}=\frac{\Delta z}{16{\left(F/\#\right)}^{2}}.$$
(13)
$$F/\#=\frac{\Delta z}{16<\Delta {a}_{3,F/\#=1}>}.$$