## Abstract

The method of obtaining the sine-wave response of a lens, both theoretically and experimentally, is described, and it is shown that the line spread-function can be derived from the sine-wave response alone if the spread function is symmetrical. If the spread-function is not symmetrical, it can be computed from the sine-wave response only when the phase function is also known, and an experimental method of obtaining this function is described. The mathematical procedure for the reciprocal inversion of the spread-function and the sine-wave response is confirmed experimentally.

© 1958 Optical Society of America

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### Equations (25)

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(1)
$$G(\xi )={b}_{0}+{b}_{1}\hspace{0.17em}\text{cos}2\pi \nu \xi ,$$
(2)
$$F(x)={\int}_{-\infty}^{\infty}A(\xi )G(x-\xi )d\xi .$$
(3)
$$F(x)={b}_{0}{\int}_{-\infty}^{\infty}A(\xi )d\xi +{b}_{1}{\int}_{-\infty}^{\infty}A(\xi )\hspace{0.17em}\text{cos}2\pi \nu (x-\xi )d\xi .$$
(4)
$$F(x)={b}_{0}+{b}_{1}{A}^{\#c}\hspace{0.17em}\text{cos}2\pi \nu x+{b}_{1}{A}^{\#s}\hspace{0.17em}\text{sin}2\pi \nu x.$$
(5)
$${A}^{\#c}(\nu )={\int}_{-\infty}^{\infty}A(\xi )\hspace{0.17em}\text{cos}2\pi \nu \xi d\xi ,$$
(6)
$${A}^{\#s}(\nu )={\int}_{-\infty}^{\infty}A(\xi )\hspace{0.17em}\text{sin}2\pi \nu \xi d\xi ,$$
(7)
$$\mid {A}^{\#}\mid =({A}^{\#c2}+{A}^{\#s2}{)}^{{\scriptstyle \frac{1}{2}}}.$$
(8)
$${A}^{\#c}=\mid {A}^{\#}\mid \hspace{0.17em}\text{cos}\varphi $$
(9)
$${A}^{\#s}=\mid {A}^{\#}\mid \hspace{0.17em}\text{sin}\varphi ,$$
(10)
$$F(x)={b}_{0}+{b}_{1}\mid {A}^{\#}\mid \hspace{0.17em}\text{cos}(2\pi \nu x-\varphi ).$$
(11)
$$F(x)+F(-x)=2{b}_{0}+2{b}_{1}{A}^{\#c}\hspace{0.17em}\text{cos}2\pi \nu x.$$
(12)
$$\mathrm{\Delta}\varphi =k\mathrm{\Delta}x/\mathrm{\lambda}=k\nu \mathrm{\Delta}x.$$
(13)
$${A}_{1}(x)={A}_{1}(-x).$$
(14)
$${A}_{1}(x)={\scriptstyle \frac{1}{2}}[A(x)+A(-x)].$$
(15)
$${A}_{2}(x)=-{A}_{2}(-x).$$
(16)
$${A}_{2}(x)={\scriptstyle \frac{1}{2}}[A(x)-A(-x)].$$
(17)
$$A(x)={A}_{1}(x)+{A}_{2}(x).$$
(18)
$$\begin{array}{l}{A}_{1}(x)=2{\int}_{0}^{\infty}\text{cos}2\pi \nu x{\int}_{-\infty}^{\infty}{A}_{1}(\xi )\hspace{0.17em}\text{cos}2\pi \nu \xi d\xi d\nu \\ =2{\int}_{0}^{\infty}{({A}_{1})}^{\#c}\hspace{0.17em}\text{cos}2\pi \nu xd\nu .\end{array}$$
(19)
$${({A}_{1})}^{\#c}={A}^{\#c}.$$
(20)
$${A}_{1}(x)=2{\int}_{0}^{\infty}{A}^{\#c}\hspace{0.17em}\text{cos}2\pi \nu xd\nu .$$
(21)
$${A}_{1}(x)={A}^{\#c\#}.$$
(22)
$$\begin{array}{l}{A}_{2}(x)=2{\int}_{0}^{\infty}\text{sin}2\pi \nu x{\int}_{-\infty}^{\infty}{A}^{\u2033}(\xi )\hspace{0.17em}\text{sin}2\pi \nu \xi d\xi d\nu \\ =2{\int}_{0}^{\infty}{({A}_{2})}^{\#s}\hspace{0.17em}\text{sin}2\pi \nu xd\nu .\end{array}$$
(23)
$${({A}_{2})}^{\#s}={A}^{\#s}.$$
(24)
$${A}_{2}(x)=2{\int}_{0}^{\infty}{A}^{\#s}\hspace{0.17em}\text{sin}2\pi \nu xd\nu ={A}^{\#s\#},$$
(25)
$$A(x)={A}^{\#c\#}+{A}^{\#s\#}.$$