Optical transfer function and aberration

Journal Pre-proof Optical Transfer Function and Aberration Yu Bai, Jiaqi Chen, Qi Lu, Zhenming Zhao

PII:

S0030-4026(20)30077-2

DOI:

https://doi.org/10.1016/j.ijleo.2020.164243

Reference:

IJLEO 164243

To appear in:

Optik

Received Date:

4 December 2019

Accepted Date:

16 January 2020

Please cite this article as: Bai Y, Chen J, Lu Q, Zhao Z, Optical Transfer Function and Aberration, Optik (2020), doi: https://doi.org/10.1016/j.ijleo.2020.164243

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Optical Transfer Function and Aberration

Yu Bai1, Jiaqi Chen1, Qi Lu2, Zhenming Zhao1 *

1

School of Science, Changchun University of Science and Technology, Jilin 130022,

China 2

Division of Physics, Engineering, Mathematics and Computer Science, Delaware State

*

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University, Dover, Delaware, 19901, USA

Corresponding author at: School of Science, Changchun University of Science and Technology, Jilin 130022, China .

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E-mail address: [email protected] (Z. Zhao).

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Abstract

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The Optical Transfer Function (OTF) is an important part of modern optics and is very useful for optical system design and evaluation. A non-Fourier transform OTF method has been recently proposed. It formulates the OTF based on the propagation of light in the frequency domain, combining the intuitiveness of geometrical optics and the accuracy of wave optics. To advance the non-Fourier transform OTF method, in this paper, we first applied it to a single convex lens and analyzed the influence of the optical pupil function on interference pairs to reveal the mechanism of how aberration affects the non-Fourier transform OTF. It was found that the aperture of the lens affects the utilization efficiency of the interference pair, and the aberration of the lens affects the utilization quality of the interference pair. Then we derived the generalized pupil function based on a two-lens system in applying this new non-Fourier transform OTF method to the analysis of complex optical systems.

Keywords: Optical Transfer Function; Lens aberration; Modulation Transfer Function; Non-Fourier Transform; Evaluation of Imaging Quality; Optical Imaging

Introduction The optical transfer function (OTF) is a method that can be used to evaluate the performance of optical imaging systems comprehensively and objectively [1,2]. Currently, the OTF of an optical system is obtained by Fourier transform in the spatial domain [1-6], and the OTF based on Fourier transform does not track the propagation of light thus theoretically abstract. On the other hand, a recent development of non-Fourier transform OTF studies the image formation in the frequency domain [7,8] , which considers the light emitted from the

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cosine fringe on the object plane as consisting of a large number of interference pairs. This consideration is based on the interference theory of light and the concept of the photon

superposition state. By analyzing the utilization efficiency of the optical system for the

interference pair, we can obtain the modulation of the cosine fringe on the image plane, thus

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the OTF of the optical system.

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Herein, we use the aforementioned non-Fourier transform OTF method to analyze the aberration of a single thin lens and a lens system, which combines the intuitiveness of the

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geometrical optics and the accuracy of wave optics. In this method, the light beams propagation are tracked according to the construction of the optical system. Since the

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propagation of light beams is tracked through every component of an optical system, the non-Fourier transform method offers the advantage of visualizing the light beams in a multi-lens system, which is nonexistent in the traditional OTF theory. To promote the

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non-Fourier transform OTF, we need to further consider the effect of aberrations. The analysis

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of aberrations is important for lens design and image formation analysis. 1.

The states of interference pairs in the object plane

During the lens imaging, the light wave is the carrier of information. In the Fig.1, the light

waves emitted from the object plane carry the brightness distribution of the object plane. In order to fully reveal the lens imaging process, we use the coordinate system x o y o z o , x y z and x i y i z i to represent the object plane, the pupil plane and the image plane respectively.

From Fig.1, the relation can be: x o  x   x i , y o  y   y i , z o  z  d o  z i  d o  d i . If the brightness distribution function of the object plane in Fig 1 can be expressed as: I o ( x o , y o )  Ao  B o c o s  2  ( o x o   o y o ) 

(1)

where I o ( x o , y o ) is the cosine fringe, according to the  o and the  o , we can obtain the fringe spatial frequency and the fringe spatial period. 2

To  f o

1

2

1 2

(2)

 ( o   o ) 2

2

1 2

(3)

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f o  ( o   o )

x xo

ko

ki s y

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Oo

S  ki

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yo

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S

ko s

di

Oi

z zi xi

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do

yi

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Fig.1 The two beams of the interference pair are refracted by the lens which reach the image plane

When we use the OTF theory to study the lens imaging, cosine fringe is an element

function. But the cosine fringe is not eigenstate of the lens, so we have to decompose the light emitted by the cosine fringe into interference pairs. [8,9] In incoherent imaging, cosine fringes on the object surface in Fig. 1 emit a large number of interference pairs. Assuming the two beams of an interference pair be s  ( ro ) and s  ( ro ) , the direction of their propagation can be

represented by the wave vector k o ( k o x , k o y , k o z ) and k o ( k ox , k oy , k oz ) . Sometimes the spatial frequency (  o ,  o ,  o ) and (  o,  o,  o) can be used to represent the direction of their propagation. For the phase of I o ( x o , y o ) , it is zero. Thus, the phase of beams s  ( ro ) and s  ( ro ) at point O o are same. Then s  ( ro ) and s  ( ro ) can be: a e x p ( ik o  ro )  a e x p  i ( k o x x o  k o y y o  k o z z o )    a e x p  i 2  (  o x o   o y o   o z o ) 

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 s  ( ro )        s  ( ro )     

a e x p ( ik o  ro )  a e x p  i ( k ox x o  k oy y o  k oz z o )    a e x p  i 2  (  ox o   o y o   oz o ) 

(4)

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In the Eq. (4), k o x , k o y and k o z are the components of k o respectively, k ox , k oy and k oz are the components of k o respectively. In the Eq. (4), we have used the components of a

k o y  2   o

k o z  2   o

k oy  2   o

k oz  2   o

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 k o x  2   o  k   2   o  ox

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wave vector related to the spatial frequency.

(5)

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When the independent variables of s  and s  are ro , x o and y o , they can represent the space of s  and s  in front of the lens, which is the object space. When the independent

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variables of s  and s  are ri , x i and y i , which can represent the space s  and s  back of the lens.

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Let z o in the Eq. (4) be equal to 0, we can obtain the intensity of the interference pair on

the x o y o plane: d I o  ( s  + s  )( s  + s  )  a  2 a c o s  2  (  o   o) x o  2  ( o   o ) y o  

2

2

(6)

In the photon theory, when the amplitude of light wave is in the form s  + s  , it is called photon superposition state. Comparing the Eq. (6) with the Eq. (1), we have:

  o   o   o    o   o   o

(7)

From the Eq. (7), we find that the difference in frequency between two beams (which is the interference pair) of light is equal to the cosine fringe frequency on the object plane. 2.

The position of the interference pair on the pupil plane Non-Fourier transform OTF method replaces the ray of geometrical optics theory with a

beam whose width is much larger than the wavelength of light but much smaller than the

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diameter of the lens. In Fig.1, we assume the intersection point between the center of the beam s  ( ro ) and the plane x y to be S  ( x  , y  ) ; and the intersection point between the center of

the beam s  ( ro ) and the plane x y to be S  ( x  , y  ) . Since k o and O o S  are collinear,

k o x x  k ox





y



k o y y 



k oy

do

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x

k o z do k oz

(8)

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      

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k o and O o S  are collinear, we have:

have: k o x k o z

 do

k o x k

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  x  d o     x  do  

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For the paraxial rays, we have k z  k , k x   k z , k x   k z . And from the Eq. (8), we

k ox

 do

k

y   d o

k o y k o z k oy k oz

 do

 do

k o y k k oy

(9)

k

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k oz

k ox

y  d o

Given k  2   , then Eq. (5) and Eq. (9) can be used to obtain the coordinates of

S  ( x , y  )

and S  ( x  , y  ) :

 x    o  d o   x    o d o

y    o  d o y    o d o

(10)

From the Eq. (10), we can obtain the coordinate difference between the point S  ( x  , y  ) and the point S  ( x  , y  ) .   x  x   x   (  o   o)  d o    y  y   y   ( o   o )  d o

(11)

Substituting Eq. (7) into Eq. (11), we obtain:  x   o  d o，  y   o  d o

(12)

point S  ( x  , y  ) .

S S   (  x   y ) 2

1 2

  d 0 ( o   o ) 2

2

1 2

 d0 f

(13)

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2

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According to Eq. (12) and Eq. (2), we can find the distance between point S  ( x  , y  ) and

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From the Eq. (13), we can determine the separation angle between s  ( ro ) and s  ( ro ) to be:

(14)

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 o  S S  d o   f

In other words, the higher the frequency of cosine fringe is, the larger the separation angle of interference between two beams (which is the interference pair) and the longer the distance

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between point S  ( x  , y  ) and point S  ( x  , y  ) is. In addition, the line segment S S  is perpendicular to the cosine fringe. To illustrate this phenomenon, we take the cosine fringe of

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 o  0 as an example. From the Eq. (12), we can get that  y  0 and the line segment

is parallel to the x -axis. At the same time, the cosine fringe is parallel to the y -axis,

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S S 

so the line segment S S  is perpendicular to the cosine fringe.

3. The process of interference pair propagation in image space In the imaging process of Fig.1, after the two beams of the interference pair are refracted by the lens, the directions of both beams change. If the lens in Fig.1 is without aberration, beam

s  ( ro ) will be refracted by the lens and become s  ( ri ) . According to the properties of the

lens, we can know that s  ( ri ) is the spherical wave converging to F  ( x f , y f ) , see Fig.2, and this spherical wave passes through point F  , then it diverges. In order to find the coordinates of point F  , we can draw a line between point O and point F  . And this line is the auxiliary axis which parallel to the beam s  ( ro ) , so O F  is parallel to k o . Then we can have:

k o x



y f k o y



OF 

(15)

k o

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x f

For the paraxial beam, we have that O F   f , then from the Eq. (5) and k  2   , the

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coordinate of F  should be: x f   f  o , y f   f  o

(16)

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According to the Fig.1 and the Fig.2, the coordinate of point F  in the x i y i z i coordinate system is F  (  x f ,  y f , f  d i ) . The amplitude of spherical wave s  ( ri ) is approximately

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equal at each point on the x i y i surface, we use the constant a i to represent it. Taking the point F  as the phase reference point, the expression of spherical wave s  ( ri ) on the surface is:

2 2 2 ( f  d i )  ( x i  x f )  ( y i  y f )  

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s  ( F ; x i , y i , 0 )  a i e x p  ik 

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2 2  ( x i  x f ) ( y i  y f )  a i e x p  ik ( d i  f ) 1   2 2 (d i  f ) (d i  f )  

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2 2   ( x i  x f ) ( y i  y f )   a i e x p  ik ( d i  f ) 1   2 2 2(d i  f ) 2(d i  f )    

   

     

 ik ( x f  y f ) ik ( x f x i  y f y i ) ik ( x 2  y 2 )  i i  a i e x p  ik ( d i  f )  e x p  + +  di  f 2 ( d i  f )   2 ( d i  f ) 2

2

The essence of Eq. (17) is to replace the sphere wave with the ellipsoid wave.

(17)

xy

xy 0

0

di-f

f S

xy i

i

F ( xf , yf )

F Oi

O

Oo

F ( xf , yf )

S 

di

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do

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Fig. 2 Convergence and divergence of the beam on the back focal plane of the lens

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In the Fig. 2, if the reflection and absorption of the lens are ignored, the amplitude of s  ( ri ) and s  ( ro ) on the x y -plane can be considered equal, that is, the amplitude of s  ( ri )

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on the x y -plane is equal to a . According to the relation between the amplitude of the spherical wave and the radius, the amplitude of s  ( ri ) on the x y -plane can be written as U S F 

, the amplitude of s  ( ri ) on x i y i -plane can be written as a i 

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a 

U F O i

. When

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we divide these two equations, we can get:

a

F O i

S F 

(18)

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ai



If s  and s  are the paraxial beams in the Fig.2, the following relationship exists:  F O i  d i  f   S F   f

Then, we can substitute Eq. (19) into Eq. (18) to get:

(19)

a



di  f

ai

f



di

1

(20)

f

Substituting the lens imaging formula 1 f  1 d i  1 d o into Eq. (20) to obtain: ai  a d o d i

(21)

The magnification formula M  d i d o for lens imaging was substituted into Eq. (21) to

ai  a M

(22)

Combining the Eq. (20) and the Eq. (22), we can get: di  f

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M 

ro of

obtain:

f

(23)

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For spherical wave s  ( ri ) , if the phase of the point O i is taken as a reference, according

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to Eq. (17), the expression of s  ( ri ) can be obtained:

(24)

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2 2  ik ( x f x i  y f y i )   ik ( x i  y i )  s ( O i ; x i , y i )  a i e x p   exp   di  f    2(d i  f ) 

In the Fig.2, s  starts from the plane x o y o and passes the point S  to the plane x i y i .

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Assuming the path from O o to O i of s  can be   . If s  ( ri ) takes the phase of point

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O o as the reference, according to the Eq. (24), the expression of s  ( ri ) on plane x i y i can

be obtained: 2 2  ik ( x f x i  y f y i )   ik ( x i  y i )  s  ( O o ; x i , y i )  a i e x p ( ik   ) e x p  e x p    di  f    2(d i  f ) 

(25)

For the beam s  ( ro ) , it is emitted from the plane x o y o , and refracted at the point S  of the lens, and the refracted beam is s  ( ri ) . After s  ( ri ) converges at point F  ( x f , y f ) ,

diverging spherical waves arrive at the x i y i plane. Assuming the path from O o to O i of s 

can be   . If s  takes the phase of point O o as the reference, the expression of

s  ( ri ) on plane x i y i can be obtained:

2 2  ik ( x f x i  y f y i )   ik ( x i  y i )  s  ( O o ; x i , y i )  a i e x p ( ik   ) e x p  e x p    di  f    2(d i  f ) 

(26)

According to the Fig.2 and by referring to the Eq. (16), we can get: x f   f  o , y f   f  o

ro of

(27)

This paper aims to analyze the interference fringes formed by s  ( ri ) and s  ( ri ) on plane  ik ( x i  y i )  x i y i , and the common phase factors e x p   of Eq.(25) and Eq.(26) can be  2(d i  f )  2

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2

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neglected. From the Eq. (4), we can know that s  ( ro ) and s  ( ro ) have the same phase at the point O o . In addition, the lens in the Fig.1 has no aberration, then we can have that

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     , that is, e x p ( ik   )  e x p ( ik   ) . So we can have that the phase of the s  ( ri ) and

s  ( ri ) are the same at the point O i . When calculating the interference fringe of s  ( ri ) and

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s  ( ri ) on plane x i y i , only the relative phase of two beams of light is taken into

as:

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consideration, then the amplitude of two beams of light on the image plane can be simplified

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  ik ( x f x i  y f y i )   s ( x i , y i )  a i e x p   di  f      ik ( x f x i  y f y i )   s  ( x i , y i )  a i e x p    di  f   

Substituting the parameters of the Eq. (28) as follows:

(28)

x f  k  k ix  di  f   x f   k ix  k  di  f 

k iy 

k iy 

y f di  f y f di  f

k

(29) k

The amplitude of s  ( x i , y i ) and s  ( x i , y i ) on the plane x i y i can be obtained:  s ( x , y )  a e x p  i ( k  x  k  y )  i i i ix i iy i     s  ( x i , y i )  a i e x p  i ( k ix x i  k iy y i )     

(30)

ro of

In the image plane, the relation between spatial frequency and wave vector component is:  k ix  2   i k iy  2   i  k   2   i k iy  2   i  ix

(31)

(32)

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 s  ( x i , y i )  a i e x p  i 2  (  ix i   i y i )    s  ( x i , y i )  a i e x p  i 2  (  ix i   iy i ) 

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Substituting the Eq. (31) into the Eq. (30) to get:

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From the Eq. (30) and the Eq. (32), we can know that s  ( x i , y i ) and s  ( x i , y i ) are the plane wave.

The parameters of the interference pair in the image space

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4.

The cosine fringes emit interference pairs, which are refracted by the lens, changing the

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direction of both beams. The spatial frequency of the beam in the image space is determined by the spatial frequency of the beam in the object space and the magnification M

of the

Jo

system. Next, we start to analyze the parameters of the beams (the interference pair) in the image space. At first, substituting k o z  k into the Eq. (15) to obtain:  k o x k o x f  f  x f  k o z k   k ox k ox   xf  f  f   k k oz 

y f 

y f 

k o y k o z k oy k oz

f 

k o y

f

k f 

k oy

(33) f

k

When the Eq. (33) is substituted into the Eq. (29), we can get:

f  k ix  k o x  di  f   f  k   k ox ix  di  f

k iy  k iy 

f di  f f di  f

k o y

(34) k oy

Substituting the Eq. (23) into the Eq. (34), we can get:

k iy  k iy 

k o y M

(35)

k oy M

ro of

 k o x  k ix   M  k ox   k ix   M

Eq. (35) give the change in the spatial frequency of the beam after refracted by the lens. From the Eq. (35), we can know that with the magnification M

increasing, k ix , k iy , k ix ,

-p

k iy get smaller. That is, the larger the magnification, the smaller the angle between the beam

and the z-axis, which is consistent with the conclusion of geometric optics.

lP

  o  M  i  o  M  i    o  M  i  o  M  i

re

Substituting the Eq. (31) and the Eq. (5) into the Eq. (35), we can have:

(36)

na

After that, we can substitute magnification: M = d i d o into the Eq. (36), then we can obtain:

d o o  d i i

(37)

Interference pairs are superimposed on the image plane

Jo

5.

d o o  d i i

ur

 d o  o  d i  i,   d o  o  d i  i,

In the Fig.1 and the Fig.2, the interference pair is refracted by the lens and finally

superimposed on the image plane to form cosine fringes. Because of the aberration in the actual lens, the aberration changes the phase of the beam in the image plane, which results in the translation of the interference fringes. In order to study the influence of the aberration on imaging, the optical pupil function of the aberration and aperture of the lens is expressed. If

the pupil diameter of the lens is 2 R and the aberration is W ( x , y ) , then the pupil function of the lens is that[9-13]:  e x p  ik W ( x , y )  P (x, y)   0 

x  y

2

 R

2

x  y

2

 R

2

2

2

(38)

After taking into account the lens aperture and aberration, then according to the Eq. (32), we can get the amplitude of s  ( ri ) and s  ( ri ) on plane x i y i is:

From the Eq. (10) and the Eq. (37), we can know: y   M  i d o

-p

 x   M  i d o   x   M  i d o

y   M  i d o

(39)

ro of

 s  ( x i , y i )  a i P ( x  , y  ) e x p  i 2  (  ix i   i y i )    s  ( x i , y i )  a i P ( x  , y  ) e x p  i 2  (  ix i   iy i ) 

(40)

y    i d i

lP

 x    i d i   x    i d i

re

Substituting d i  M d o into the Eq. (40), we can obtain:

y    i d i

(41)

na

Then, by substituting the Eq. (41) into the Eq. (39), we can know:

(42)

ur

 s  ( x i , y i )  a i P (  d i  i,  d i i ) e x p  i 2  (  ix i   i y i )    s  ( x i , y i )  a i P (  d i  i,  d i i) e x p  i 2  (  ix i   iy i ) 

Jo

According to the Eq. (42), we can obtain the interference fringe of s  ( ri ) and s  ( ri ) on the x i y i plane which can be written down as: d I i ( x i , y i )  s  ( x i , y i )  s  ( x i , y i )

2

dxdy

Substituting the Eq. (42) into the Eq. (43), we can get: 

d I i ( x i , y i )  a i P (  d i  i,  d i i ) P (  d i  i,  d i i ) d x d y 2

(43)



 a i P (  d i  i,  d i i) P (  d i  i,  d i i) d x d y 2

 a i P (  d i  i,  d i i ) P (  d i  i,  d i i) e x p  i 2  (  i   i) x i  i 2  ( i   i) y i  d x d y 

2

 a i P (  d i  i,  d i i ) P (  d i  i,  d i i) e x p   i 2  (  i   i) x i  i 2  ( i   i) y i  d x d y (44) 

2

In Eq. (44), the interference fringe frequency of the image plane depends on the spatial frequency difference between the two beams. Let the spatial frequency difference be:   i   i   i    i   i   i

ro of

(45)

Using Eq. (41) and Eq.(45), we can substitute  d i  i ,  d i i ,  d i  i ,  d i i for x ,

-p

y , x   d  , y   d  in the Eq. (44). Then, the Eq. (44) can be: i i i i





d I i ( x i , y i )  a i P ( x   d i  i , y   d i i ) P ( x   d i  i , y   d i i ) d x d y  a i P ( x , y ) P ( x , y ) d x d y 2

di  i,

2

y 

di )i



P( ,x )y e x p i 2  i ( i x 

re

 a i P( x  

i

i

y ) d x d y

 a i P ( x   d i  i , y   d i i ) P ( x , y ) e x p   i 2  (  i x i   i y i )  d x d y 

lP

2

na

We can assume:

ur

(47)

Substituting the Eq. (47) into the Eq. (46), we can have: d I i ( x i , y i )  a i P ( x   d i  i , y   d i i ) d x d y  a i P ( x , y ) d x d y

Jo

2

2

2

) e x pi  2i

 x(

i

 i yi

d) x d y

 a i b e x p (  i ) e x p   i 2  (  i x i   i y i )  d x d y 2

 a i P ( x   d i  i , y   d i i ) d x d y  a i P ( x , y ) d x d y 2

(46)



 b e x p ( i )  P ( x   d i  i , y   d i i ) P ( x , y )    b e x p (  i )  P ( x   d i  i , y   d i i ) P ( x , y )

 a i be x p (i

2

2

 2 a i b  c o s ( 2  i xi  2  i y i   )  d x d y 2

(48)

The results of Eq. (48) show that the phase of interference fringe of s  ( ri ) and s  ( ri ) is  , where 

is determined by P ( x   d i  i , y   d i i ) and P ( x , y ) . If s  ( ri ) or

s  ( ri ) does not enter the pupil, assuming s  ( ri ) is the one, then P ( x , y )  0 , P ( x   d i  i , y   d i i )  0 . According to Eq. (47) and Eq. (48), we can obtain that d I i ( x i , y i )  a i d x d y . Then we can conclude that the interference pair s  ( ri ) and s  ( ri ) 2

Using Eq. (45), Eq. (36) and Eq. (7), we can obtain:

M 1 M

(  o   o)  ( o   o ) 

1 M 1 M

o

-p

1

o

(49)

re

    i   i   i          i i i 

ro of

makes no contribution to the image, but increases the background brightness.

6.

lP

Eq. (49) is the relation between imaging magnification and spatial frequency. The OTF of ideal cosine fringe imaging

na

If the object plane in the Fig.1 is an ideal cosine fringe, that is, the darkest part of the cosine fringe is zero, and the quality of the cosine fringe is the best, then A o in the Eq. (1) is equal

ur

to B o , and the brightness of the object plane can be expressed as:

Jo

I o ( x o , y o )  Ao  Ao c o s  2  ( o x o   o y o )   A o 1  c o s  2  (  o x o   o y o ) 

(50)

As can be seen from the Eq. (50), the modulation of the object plane is: M o ( o ,  o )  1

(51)

In the lens imaging process of the Fig.1, we superimposed the cosine fringe formed on the surface of x i y i by all the interference pairs, so as to obtain the light intensity distribution on the surface of x i y i .[1] When the lens has no aberration such as Fig.1, the fringes formed by all the interference pairs overlap. [8] The Fig.3 shows that the coincidence of d I l , d I m and d I n formed by the interference pairs of m, n and l. If the lens in Fig.1 has aberration, the

fringes formed by each interference pair may not be identical. Fig.4 is the situation where the

ro of

fringes formed by the three interference pairs do not coincide. After adding the total fringe modulation reduced, the phase also changed.

Ii

dIn

i xi  i yi

lP

Oi

re

dIl dIm

-p

Ii

dIn

i xi  i yi

Oi

Fig.4 A fringe with an aberration lens

na

Fig.3 The fringes of an ideal lens

dIl

dIm

For the integral of the Eq. (46), the image plane brightness is I i ( x i , y i )  2 a i S  e x p  i 2  (  i x i   i y i )   P ( x   d i  i , y   d i i ) P ( x , y ) d x d y 

ur

2

 a i e x p   i 2  (  i x i   i y i )   P ( x   d i  i , y   d i i ) P ( x , y ) d x d y

Jo

2



In the Eq. (52), S is the pupil area of lens, S  S 



P (x , y ) d x d y .

In the Eq. (52), we can assume:



P ( x   d i  i , y   d i i ) d x d y ,

(52)

  P ( x   d i  i , y   d i i ) P ( x , y ) d x d y  m (  i ,  i ) e x p  i (  i ,  i )        P ( x   d i  i , y   d i i ) P ( x , y ) d x d y  m (  i ,  i ) e x p   i (  i ,  i ) 

(53)

Substituting the Eq. (53) into the Eq. (52) to get: I i ( x i , y i )  2 a i S  a i m (  i ,  i ) e x p  i (  i ,  i )  e x p  i 2  (  i x i   i y i )  2

2

 a i m (  i ,  i ) e x p   i (  i ,  i )  e x p   i 2  (  i x i   i y i )  2

 2 a i S  2 a i m ( i ,  i ) c o s  2  ( i x i   i y i )   ( i ,  i )  2

ro of

2

1   2  2 a i S 1  m ( i ,  i ) c o s  2  ( i x i   i y i )   ( i ,  i )   S  

(54)

-p

As can be seen from the Eq. (54), due to the limited aperture of the lens and the aberration

of the lens, part of the light energy is transformed into the background brightness of the image

re

plane, resulting in that the modulation of the image plane fringe is not 1. From the Eq. (54), it

lP

can also be seen that the phase of cosine fringe of image plane is  (  i ,  i ) . By comparing it with Eq. (54) and (50), the phase transfer function ( P T F ) of the lens can be obtained: P T F (  i ,  i ) = e x p  i (  i ,  i ) 

na

(55)

From the Eq. (54), it can also be concluded that the modulation of image plane fringe is: m ( i ,  i )

(56)

ur

M i ( i ,  i ) 

S

Jo

According to Eq. (51) and Eq. (56), the modulation transfer function ( M T F ) of the lens is:

M T F ( i ,  i ) 

M i ( i ,  i ) M o ( o ,  o )



m ( i ,  i )

(57)

S

According to Eq. (55) and Eq. (57), the OTF of the lens is [4] O T F ( i , i ) = M T F ( i , i )P T F ( i , i ) 

1 S

m (  i ,  i ) e x p  i (  i ,  i ) 

(58)

Substituting the Eq. (53) into the Eq. (58) to obtain the expression of pupil function, O T F ( i , i ) 

7.

1 S

 P ( x   d  i



i

, y   d i i ) P ( x , y ) d x d y

(59)

The OTF of actual cosine fringe imaging If the object plane in the Fig.1 is not an ideal cosine fringe, we can know that B o  A o . I o ( x o , y o )  Ao  B o c o s  2  ( o x o   o y o ) 

ro of

 B o  B o c o s  2  ( o x o   o y o )   Ao  B o

 I o  I o

(60)

Where I o  B o  B o c o s  2  (  o x o   o y o )  , I o  A o  B o . According to the Eq. (50)

-p

and the Eq. (54), the brightness distribution of I o formed on the x i y i -plane can be

I i  2 a i S

Bo  1  m ( i ,  i ) c o s  2  ( i x i   i y i )   ( i ,  i )   1  Ao  S 

(61)

lP

2

re

obtained:

In order to obtain the brightness distribution of I o formed on the x i y i -plane, which is I i ,

na

we can assume  o   o  0 . Thus, according to the Eq. (50), we can get:

(62)

ur

I o ( x o , y o )  2 Ao

Jo

When  o   o  0 , from Eq. (49), we can get that  i   i  0 , and substitute it into Eq. (53), we can get m ( 0 , 0 )  S , then use Eq. (54), we can know:

I i ( xi , yi )  4 ai S 2

(63)

According to I o  A o  B o and Eq. (62) and Eq. (63), the brightness distribution of I o formed on the x i y i -plane can be obtained:

I i  4 a i S 2

Ao  B o

(64)

2 Ao

From Eq. (61) and Eq. (64), the brightness distribution on the x i y i -plane can be: Ao  B o

I i  I i + I i  4 a i S 2

2 Ao

2 a i B o m( i,

 2 ai S 2

2

 2 ai S  2

i

Ao

Bo  1  m ( i ,  i ) c o s  2  ( i x i   i y i )   ( i ,  i )   1  Ao  S 

) c o s 2  i( x i 

i

, i

y i )  ( i

)

ro of

  B o m ( i ,  i ) 2  2 a i S 1  c o s  2  ( i x i   i y i )   ( i ,  i )   Ao S  

(65)

i



B o m ( i ,  i ) Ao

S

(66)

re

M

-p

The modulation of cosine fringe on the x i y i -plane can be obtained from the Eq. (65):

In addition, according to the Eq. (60), the modulation of cosine fringe on the x o y o plane

o



Bo Ao

(67)

na

M

lP

can be obtained:

ur

The modulation transfer function ( M T F ) of the lens obtained from Eq. (66) and Eq. (67) is: M i ( i , i )

M o ( o , o )



m ( i , i )

(68)

S

Jo

M T F ( i , i ) 

Since the background brightness of the object plane does not affect the phase of cosine

fringe of the image plane, according to Eq. (55) and Eq. (67), the following equation can be obtained: O T F ( i , i ) = M T F ( i , i )P T F ( i , i ) 

1 S

Substitute the Eq. (53) into the Eq. (69), and get:

m (  i ,  i ) e x p  i (  i ,  i ) 

(69)

O T F ( i , i ) 

1 S

 P ( x   d  i



i

, y   d i i ) P ( x , y ) d x d y

(70)

The results of Eq. (70) and Eq. (59) are the same. They are OTF of aberration system based on the method of non-Fourier transform, namely OTF of optical system is equal to the normalized autocorrelation function of pupil function. This result is consistent with OTF theory based on mathematical analysis. [1] The results of Eq. (70) and Eq. (59) are the same, indicating that the method of non-Fourier transform OTF can objectively represent the performance of the lens, independent of the object surface brightness function. OTF analysis of complex optical system

ro of

8.

According to the method of non-Fourier transform OTF and the transmission process of light beam, OTF is directly calculated according to the structure of optical system. For the

-p

system with complex structure, OTF of the system can also be calculated by the method of

re

non-Fourier transform OTF, which is the advantage of this method. Pupil function of the imaging system P ( x , y ) determines the OTF of the system. For a single thin lens, the pupil

lP

function is simple. The actual optical system is generally consisting of multiple lenses, so the pupil function must be obtained by combining the system structure and using conditions. For

na

example, the optical system in the Fig. 5 contains two lenses L1 and L 2 . If there is aberration at each surface of the lens, W 1 is the aberration of the front surface of L1 , W 2 is

ur

the aberration of the back surface of L1 , W 3 is the aberration of the front surface of L 2 , and

Jo

W 4 is the aberration of the back surface of L 2 .

S1'

xoyo

S2'

xiyi

s'

s'

S3'

o

sʺ

do

S1ʺ

S2ʺ L1

sʺ

S3ʺ

S4' i

S4ʺ

L2 di

Fig.5 Propagation path of two beams of interference pair in the optical system The cosine fringe of object surface in the Fig. 5 emits interference pairs, s  and s  are two beams of an interference pair. The beam s  goes through S 1 , S 2 , S 3 and S 4 to x i y i -plane, and the beam s  through S 1 , S 2 , S 3 and S 4 to x i y i -plane. The total aberration of beam s  in the optical system is: W  = W 1 ( S 1 )  W 2 ( S 2 )  W 3 ( S 3 ) + W 4 ( S 4 )

and the total aberration of beam s  in the optical system is: W  = W 1 ( S 1)  W 2 ( S 2 )  W 3 ( S 3) + W 4 ( S 4 )

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(71)

(72)

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The generalized pupil function in Fig.5 is:

(73)

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P ( x 4 , y 4 )  P ( x 4 , y 4 ) e x p ( ik W ）

system in Fig.5 can be: 1 S

 P



( x 4 , y 4 ) P ( x 4   d i  i , y 4   d i i ) d x 4 d y 4

(74)

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O T F ( i , i ) 

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Where the exit pupil coordinates is ( x 4 , y 4 ) .According to the Eq. (70), the OTF of optical

In the actual process, in order to simplify the OTF calculation of the system, a certain number of

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beams can be selected according to the system construction, the total aberration of each beam can be calculated, the generalized pupil function of the optical system can be fitted out, and then

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the generalized pupil function can be normalized and autocorrelated, at last, the OTF of optical system can be obtained. When calculating the generalized pupil function, the more the amount of beams is selected, the more detailed the generalized pupil function is obtained, and the more accurate the OTF is calculated. This paper takes the imaging of the center area of the object plane as an example to study the effect of aberrations on OTF, this method is also suitable for the whole object plane.

Conclusions The essence of the non-Fourier transform OTF method is to study images in the frequency domain, this method combines the intuitiveness of geometric optics and the accuracy of wave optics. The non-Fourier transform OTF method starts with the interference pairs，according to the propagation of the interference pair in the imaging process, the OTF of the lens can be obtained. It is found that the aperture and the aberration of the lens affect the utilization efficiency and utilization quality of the interference pair respectively. The non-Fourier

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transform OTF method replaces geometric optical rays with light waves, this method can also be extended to reflective optical systems. Under the condition of incoherent imaging, all the interference pairs are equally probable. when partial coherent light imaging conditions, can

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also be studied by using the non-Fourier transform OTF method. In this case, each interference

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pair is no longer the same probability.

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Declaration of interests

The authors declare that they have no known competing financial interests or

paper.

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personal relationships that could have appeared to influence the work reported in this

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References

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[1] J.W. Goodman. Introduction to Fourier Optics.[M]. Englewood: Roberts & Company Publishers.2005. [2] P.M. Duffieux. LʼIntėgral de Fourier et ses Applications à lʼOptique. Rennes.Societé Anonyme des Imprimeries Oberthur.1946. [3] H.H. Hopkins. Wave Theory of Aberrations.[M]. Oxford: Oxford University Press.1950. [4] Charles S. Williams, Orville A. Becklund. Introduction to the Optical Transfer Function. [M]. New York: Wiley.1989. [5] Tom L Williams. Optics and Optoelectronics Series. [M]. Bristol: Institute of

Physics.1999. [6] Adolf W. Lohmann Edited by Stefan.Optical Information Processing. [M]. Sinzinger Universitätsverlag Ilmenau. 2006. [7] Zhenming Zhao. Investigation of Optical Transfer Function Without Using Fourier Transformation. [EB/OL]. Beijing: Sciencepaper Online [2017-12-26] http://www.paper.edu.cn/releasepaper/content/201712-319. [8] Fangjie Li, Yu Bai, Zhenming Zhao. Optical transfer function without using fourier transformation. Optik. Volume176, January 2019, Pages 410–418

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[9] Jiaqi Chen, Yu Bai, Xiangyan Zhou, Zhenming Zhao,Study on eigenstate of OTF theory,Optik,Volume 200,2020,163363