Download Wave/ Fourier Optics

THE WAVE EQUATION
5.1. Solution to the wave equation in Cartesian coordinates
Recall the Helmholtz equation for a scalar field U in rectangular coordinates
 2U  r,     2 (r,  )U  r,    0,
(5.1)
Where  is the wavenumber, defined as
 2  r,     2   r,    i  r,  
 n 2  r,  
2
c2
(5.2)
 i  r,  
Assuming lossless medium (   0 ) and decoupling the vacuum contribution ( n  1 ) from   r,   , we re-write Eq. 2 to explicitly show the driving
term, namely
 2U  r,    k0 2U  r,    k0 2  n 2  r,    1 U  r,   ,
(5.3)
where k0   . Note that Eq. 3 preserves the generality of the Helmholtz equation. The Green’s function, h, (an impulse response) is obtained by
c
setting the driving term (the right hand side) to a delta function,
 2 h  r,    k0 2 h  r,     (3)  r 
(5.4)
   x    y    z 
In order to solve this equation, we take the Fourier transform with respect to r,
k 2 h  k ,    k0 2 h  k ,    1 ,
(5.5)
1
where k 2  k  k .
This equation breaks into three identical equations, for each spatial coordinate, as
h  k x ,   
1
k x  k0 2
2
(5.6)
1  1
1 




2 k 0 x  k x  k0 k x  k 0 
To calculate the Fourier transform of Eq. 6, we invoke the shift theorem and the Fourier transform of function 1 ,
k
f  k  a   eiax  f  x 
(5.7)
1
 i sign  x 
k
Thus we obtain as the final solution
h  x,   
1 ik0 x
e , x  0
ik0
(5.8)
The procedure applies to all three dimensions, such that the 3D solution reads
h  r,    eik0 k r ,

(5.9)
where k is the unit vector, k  k / k . Equation 9 describes the well known plane wave solution, which is characterized by the absence of amplitude
modulation upon propagation. This is an infinitely broad wavefront that propagate along direction k̂ (Figure 1).
2
y
r
k
x
Figure 5-1. Plane wave.
3
5.2. Solution of the wave equation in spherical coordinates
For light propagation with spherical symmetry, such as emission from a point source in free space, the problem becomes on dimensional, with the
radial coordinate as the only variable,
h  r,    U  r ,  
h  k ,    U  k ,  
1
2 r 2
(3)
 r   1
(3)
 r  
(1)
(5.10)
 r 
Recall that the Fourier transform pair is defined in this case as

h  r    h  k   sinc  kr  k 2 dk
0
(5.11)

h  k    h  r   sinc  kr  r dr
2
0
The Fourier properties of
(3)
(3)
  r  and  2 extend naturally to the spherically symmetric case as
 r  1
(5.12)
 2  k 2
Thus, by Fourier transforming Eq. 4, we obtain the frequency domain solution,
h  k ,   
1
k  k0 2
(5.13)
2
4
Not surprisingly, the frequency domain solutions for the Cartesian and spherical coordinates (Eqs. 6 and 13, respectively) look quite similar, except
that the former depends on one component of the wave vector and the latter on the modulus of the wave vector. The solution in the spatial domain
becomes
sin  kr  2
1

 k dk
k 2  k0 2
kr
0

h  r,    


1  1
1  eikr  e  ikr
dk



2r 0  k  k0 k  k0 
2i
(5.14)

eikr
1
dk .

4ir 0 k  k0
We recognize in Eq. 15 the Fourier transform of a shifted 1
k
function, which we encountered earlier (Eqs. 7). Thus, evaluating this Fourier
transform, we finally obtain Green’s function for propagation from a point source,
eikr
h  r,   
, r0
r
(5.15)
This solution defines a (outgoing) spherical wave.
5
z
cos 2 r 10
r
;r  x 2  y 2  z 2
y
x
Figure 5-2. Spherical wave
6
Chapter 6 – Fourier Optics
Gabriel Popescu
University of Illinois at Urbana‐Champaign
y
p g
Beckman Institute
Quantitative Light Imaging Laboratory
Quantitative
Light Imaging Laboratory
http://light.ece.uiuc.edu
Principles of Optical Imaging
Electrical and Computer Engineering, UIUC
ECE 460 – Optical Imaging
3.10 Lens as a phase transformer
3.10 Lens as a phase transformer
 E  Eo  ei becomes important
 How is the wavefront changed by a lens?
b(x,y)
(x,y)
b1
R1
x2  y 2
α
P
C1
R2
b2
b0
 ( x, y )  knb( x, y )  k[bo  b( x, y )]
glass
(3.32)
air
 kbo  k (n  1)b( x, y )
Chapter 3: Imaging
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ECE 460 – Optical Imaging
3.10 Lens as a phase transformer
3.10 Lens as a phase transformer
 Let’s calculate b(x,y); assume small angles
b1  R1  ( PC1 ) 
 R1  R12  ( R1 ) 2  R1 1  1   2 


 Taylor expansion:
1  x | x o  1 
   2 
2
 b1  R1 1  1  2    R1 2
 
 
   tan  
Small Angle Approx
(Gaussian)
(3.33)
x2  y 2
R1
2
2
x

y
 So: S
b1 ( x, y ) 
2 R1
Chapter 3: Imaging
x
2
(3.34)
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ECE 460 – Optical Imaging
3.10 Lens as a phase transformer
3.10 Lens as a phase transformer
 b( x, y )  bo  b1 ( x, y )  b2 ( x, y ) 
x2  y 2  1
1 
 bo 



2  R1 R2 
(3.35)
 This is the thickness approximation
 The phase φ becomes:
 ( x, y )  o  k (n  1)b( x, y ) 
1
x2  y 2
1
 o  k
(n  1)   
2
 R1 R2 
(3.36)
1
1
1 
 But we know:  (n  1)   
f
 R1 R2 
Chapter 3: Imaging
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ECE 460 – Optical Imaging
3.10 Lens as a phase transformer
3.10 Lens as a phase transformer
k
E(x y)
E(x,y)
 E '( x, y )  E ( x, y )  te ( x, y )
 The lens transformation is:
The lens transformation is:
te  ei 
e
Chapter 3: Imaging
iknbo
.e
i
k 2 2
(x y )
2f
E’(x
E
(x,y)
y)
(3.37)
(3.38)
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ECE 460 – Optical Imaging
3.10 Lens as a phase transformer
3.10 Lens as a phase transformer
 A lens transforms an incident plane wavefront into a parabolic p
p
shape
 Note: f > 0 covergent lens
f < 0 divergent
f
 So, if we know how to propagate through free space, then we can calculate field amplitude and phase through any imaging system
Chapter 3: Imaging
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ECE 460 – Optical Imaging
3.11 Huygens‐Fresnel
3.11 Huygens
Fresnel principle
principle
y
 Spherical waves:
p
eikR
 Wavelet: h
R
R
2
2
x

y
 R  x2  y 2  z 2  z 1 
z2
z
x
z
 We are interested close to OA, i.e. small angles
 1  x2  y 2 
 R  z 1  

2

 2 z
Chapter 3: Imaging
(3 39)
(3.39)
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ECE 460 – Optical Imaging
3.11 Huygens‐Fresnel
3.11 Huygens
Fresnel principle
principle
 For amplitude 1  1 is OK
R
z
e
ikR
Rz
 1  x2  y 2 
 For phase kR  kz 1  

2
z
 z


R
 The wavelet becomes:
h ( x, y ) 
e
ikz
z
k ( x2  y 2 )
i
e 2z
(3 40a)
(3.40a)
z
 Remember, for the lens we found:
io
te ( x , y )  e e
i
f ( x, y )  x 2  y 2
f ( x)  x 2
k 2 2
(x y )
2f
(3.40b)
Negative Lens
 Free
Free space acts on the wavefront like a divergent
space acts on the wavefront like a divergent lens lens
(note “+” sign in phase)
Chapter 3: Imaging
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ECE 460 – Optical Imaging
3.11 Huygens‐Fresnel
3.11 Huygens
Fresnel principle
principle
 At a given plane, a field is made of point sources
y
x
E ( x, y )    E ( x ', y ') ( x  x ') ( y  y ')dx ' dy '
 Eq 3.40 a‐b represent the impulse response of the system (free space or lens)
space or lens)
 Recall linear systems (Chapter 2, page 12, Eq 2.16)
p
( p )
p
 Final response (output) is the convolution of the input with the impulse response (or Greeen’s function)
 Nice! Space or time signals work the same!
 f ( x ') ( x0  x ')dx ' 
Chapter 3: Imaging
f ( x0 )
9
ECE 460 – Optical Imaging
3.11 Huygens‐Fresnel
3.11 Huygens
Fresnel principle
principle

y
θ

U ( , )
z
U ( x, y )
x
U ( x, y )    U ( , )h( x   , y   )d d
U ( x, y )    U ( ,
Remember:
Chapter 3: Imaging
ik
( x  ) 2  ( y  ) 2 
)e 2 z
d
 d
(3.41)
V
10
ECE 460 – Optical Imaging
3.11 Huygens‐Fresnel
3.11 Huygens
Fresnel principle
principle
 Fresnel diffraction equation = convolution
 Fresnel diffraction equation is an approximation
of Huygens principle (17th century)
1
eikR ( , )
U ( x, y )    U ( , )
cos ( , )d d
i
R( , )

 x2  y 2  
 R  z 1 
2 

2
z



(3.42)
 ! Fresnel is good enough for our pourpose
 Note: we don
Note: we don’tt care about constants A (no x
care about constants A (no x‐yy dependence)
dependence)
U ( x, y )    U ( ,
Chapter 3: Imaging
ik
( x  )2  ( y  ) 2 
)e 2 z
d
 d
11
ECE 460 – Optical Imaging
3.12 Fraunhofer Approximation
3.12 Fraunhofer Approximation
Z2
Z1
Z3
 One more approximation (far field)
x
 The phase factor in Fresnel is:
Near Field
k
 ( x, y )  ( x   ) 2  ( y   ) 2  
Fresnel
2z
2z
k
Fraunhofer
 ( x 2  y 2 )  ( 2   2 )  2( x  y ) 
2z
(3.43)
≈ 0
 If z  k ( 2   2 ) , we obtain the Fraunhofer equation:

U ( x, y )  A  U ( , )e

i 2
( x  y )
z
d
 d
(3.44)
 Thus eq. 3.44 defines a Fourier transform
 Useful to calculate diffraction patterns !
Chapter 3: Imaging
12
ECE 460 – Optical Imaging
3.12 Fraunhofer Approximation
3.12 Fraunhofer Approximation
x
z
y
fy 
z
 Let’s define: f x 

 U ( f x , f y )    U ( , ).e
i 2 ( f x  f y )
d d
(3.45)
 Example: diffraction on a slit

x
k
z
a
Chapter 3: Imaging
13
ECE 460 – Optical Imaging
3.12 Fraunhofer Approximation
3.12 Fraunhofer Approximation
 One dimensional:
x
U ( x)     
a
a, |x| < a/2
0 rest
0, rest
 The far‐field is given by Fraunhofer eq:

U ( f x )   U ( x)ei 2 xff dx 
x

  x 
      
  a 

a
2
a
2

[ ( x)]  sin c( f x )
 Similarity Theorem + :
 U ( f x )  a sin c(af x ) 
sin(af x )
a
(3.46)
af x
x
 fx 
z
Chapter 3: Imaging
14
ECE 460 – Optical Imaging
3.12 Fraunhofer Approximation
3.12 Fraunhofer Approximation
 Always measure intensity  the diffraction pattern is:
2  sin( af x ) 
2
I ( fx )  U ( fx )  a 

 Also Babinet’s Principle
af x
2


(3.48)
I(fx)
f ( x)  F [ ]
1  f ( x)   ( )  F [ ]
-2π
Chapter 3: Imaging
-π
π
2π
afx
15
ECE 460 – Optical Imaging
3.12 Fraunhofer Approximation
3.12 Fraunhofer Approximation
 
 fx 
 Note: sin(af x )  sin  
1
 
 a 
 narrow slit: a1

 wide slit:

a2
1
  width of diffraction pattern
a
 Similarity Theorem ↔ uncertainty principle
Chapter 3: Imaging
16
Quiz:
What is the diffraction pattern from 2 slits of size a separated by d?
a
d
ECE 460 – Optical Imaging
3.13
Fourier Properties of lenses
3.13 Fourier Properties of lenses
U(x2,y2)
U(x1,y1)
U(x
( 3,
3 y3)
U(x
( 4,
4 y4)
OA
F
F’
F
d1
 Propagation:
U(x1,y1)
U(x2,y2)
U(x3,y3)
Chapter 3: Imaging
z
d2
Fresnel
Lens
Fresnel
U(x2,y2)
U(x3,y3)
U(x4,y4)
18
ECE 460 – Optical Imaging
3.13 Fourier Properties of lenses
3.13 Fourier Properties of lenses
 U ( x2 , y2 )  A12  U ( x1, y1 )e
 U ( x3 , y3 )  A23U ( x3 , y3 )e
i
ik 
2
2

x

x

y

y




2
1
2
1

2 d1 
dx1dy1
k
 x3 2  y3 2 
2f
 U ( x4 , y4 )  A34  U ( x3 , y3 )e
ik 
2
2




x
x
y
y




4
3
4
3

2d2 
dx3dy3
(3.49)
 Combining Eqs (3.49) is a little messy, but there is a special Combining Eqs (3.49) is a little messy, but there is a special
case when eqs simplify  very useful
Chapter 3: Imaging
19
ECE 460 – Optical Imaging
3.13 Fourier Properties of lenses
3.13 Fourier Properties of lenses
 If d1 = d2 = f
U1
F
U ( x4 , y4 )  A41 
fx 



U1 ( x1, y1 )e
U4
f
f
F’
i 2 ( x1 f x  y1 f y )
x4
y
; fy  4
f
f
dx1dy1
(3 50)
(3.50)
What is [U ]
2
[ U ] 
2
 Same
Same eq as (3.45); now z 
eq as (3 45); now z  f
 Lenses work as Fourier transformers
 Useful for spatial filtering
Chapter 3: Imaging
[U  U *] 
U  U *
Autocorrelation
20
ECE 460 – Optical Imaging
3.13 Fourier Properties of lenses
3.13 Fourier Properties of lenses
 Exercise: Use Matlab to FFT images (look up “fft2” in help)
a)
Granting (amplitude)
b)
iFFT
FFT
Final
High‐pass
FFT
iFFT
Final
Band‐pass
 Note the relationship between the frequencies passed and the details / contrast in the final image
Chapter 3: Imaging
21
Fourier Optics
Uinput
F
Uoutput
f
f
FF’
2D FT pairs diffraction patterns
(Using ImageJ)
FT is in Log(Abs) scale!! Why?
Time‐domain:
Time
domain: sound
sound
• load
load your mp3
your mp3
• plot time‐series
• plot frequency amplitude, phase, power l f
li d
h
spectrum, linear/ log
• show frequency bands, i.e. “equalizer”
• adjust and play in real time
j
p y
Equalizer from mp3 player
Space‐domain:
Space
domain: image
image
• load
load your image
your image
• display image
• show 2D frequency amplitude, phase, power h 2 f
li d
h
spectrum, linear/ log
• show rings of equal freq., “image equalizer”
• adjust and display in real time‐
j
p y
example on p
next slide
Fourier Filtering (ImageJ)