fundamentals of opticsbucroccs.bu.ac.th/courses/documents/crcc1/opto_fourier.pdf · spherical lens...
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Light propagationLight propagation
� When distance z is large enough
x
f(x‘)
g(x)
r
Source plane Observation plane
� The field at the observation plane is
2
x'x
z
{ }z
x
zj
xfTFz
eig
λν
νπλ
λν == )'(.)(
2
Fourier transformFourier transform
� Fourier transformation of the source is observed at a distance z.
π4)'max( 2xk
z >>
� The frequency v is related to the location x
� The frequency v has units of 1/m and it is referred to as spatial-frequency
3
z
x
λν =
Spatial frequencySpatial frequency
x'x
αx
Source plane Observation plane
4
z
λα
ν
ααααλλ
ν
≈
≈==
tan then samll very is if tan1
z
x
Optical Fourier transformOptical Fourier transform
� The Fourier transform is then
� The phase term is
{ } ∫∫∞
∞−
−∞
∞−
− == ')'(')'()'(.'2
'2 dxexfdxexfxfTFxj
vxj λα
ππ
The phase term is
� Remember from the wave-vector definition
5
''2
xkx xx ααλπ
φ ==
'Hence,
),,(),,(
xk
kkkkk
x
zyxzyx
=
==
φ
αααr
Optical Fourier transformOptical Fourier transform
� The Fourier transform is then
� The inverse Fourier transform is
{ } ∫∞
∞−
−== ')'()'(..)('dxexfxfTFkFxjk
xx
The inverse Fourier transform is
� The exponent term represents a plane wave in a direction (kx,0,kz), where kz ≅ k
6
{ } ∫∞
∞−
− == x
xjk
xx dkekFkFTFxf x '1 )()(..)'(
PlanePlane--wave expansionwave expansion
� Any function f(x’) can be presented as an infinite summation of plane waves ( ) each propagates in a
( )zkxkj zxe+'
f(x’) F(k )
F(k1)
F(k2)
each propagates in a different direction ( ) and amplitude (F(k)).
7
kr f(x’) F(k0)
F(k-1)
F(k-2)
Fourier OpticsFourier Optics
� Basics of Fourier Optics
◦ Any field can be expanded into an infinite summation of plane waves.
◦ Manipulating each plane wave for specific task. This process is referred to as filtering.process is referred to as filtering.
◦ Recombining the resultant plane waves to reconstruct the output field.
� The question is how to manipulate the waves?
8
Plane wave expansion
F(k)
Filtering
H(k)
Plane waves recombinationF.T.-1{F(k)H(k)}
Inputf(x’)
Outputg(x)
Back to FresnelBack to Fresnel
� Fresnel propagation gives
∫
−
'
'2'2
2
')'()(2
2
x
xz
xjx
z
kj
xz
kj
dxeexfz
exg λ
π
λ2'x
kj
� One way to remove the phase is to have z very large (Fraunhofer).
� Another way is to introduce an element that has transmittance of
9
2'2xz
kj
e
2'2)'(xz
kj
ext−
=
Spherical lens and Fourier opticsSpherical lens and Fourier optics
� Consider a spherical lens of radius R and thickness do and refractive index n.
� The light at the output of
x’ dot(x)
� The light at the output of the lens gains phase φ(x’) due to the change of the delay with x’.
10
z
Input is a plane wave
)'()'( xjext φ=
Lens phaseLens phase
� Light passing by a point x’ experience a delay over distance d(x’) inside the lens with refractive index n and another delay over a
x’
R
dod(x’)
x’
t(x)
and another delay over a distance do-d(x) in air.
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zδ
22 '
)'(
)'()1(
))'(()'()'(
xRR
dxd
kdxdnk
xddkxkndx
o
o
o
−−=
−=
+−=
−+=
δ
δ
φ
Lens phaseLens phase
� Using Fresnel approximation
� Hence
R
xRxR
2
''
222 −≈−
12
ooo
oo
kdR
xdnkkdxdnkx
R
xddxd
R
x
R
xRR
+
−−=+−=
−=−=
=−−≈
2
')1()'()1()'(
2
')'(
2
')
2
'(
2
2
22
φ
δ
δ
Lens phaseLens phase
� The lens phase is then
� Ignoring the constant phase term
okndR
xnkxx +−−==
2
')1()'()'(
2
φφ
Ignoring the constant phase term
� The lens transmittance t(x’) is
� Which is the needed term when z = f13
1 where,
2
')'()'(
2
−=−==n
Rf
f
xkxx φφ
2
'2
)'( f
xjk
ext−
=
Fourier transform lensFourier transform lens
x'x
f(x‘)
g(x)Source plane
Observation plane
f(x’)t(x’)
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x'
z
∫
−
='
'2'2
2
')'()'()(2
2
x
xz
xjx
z
kj
xz
kj
dxeextxfz
exg λ
π
λ
Fourier transform lensFourier transform lens
� The Fresnel propagation is then
∫
−−
='
'2'2
'2
2
')'()(
2
22
2
xkj
x
xz
xjx
z
kjx
f
kj
xz
kj
dxeeexfz
exg λ
π
λ
� If z = f
15
∫
−
−
='
'2'2
112
')'(
22
x
xz
xjx
k
fzj
xz
kj
dxeexfz
e λπ
λ
f
xv
xf
kj
x
xf
xj
xf
kj
xfTFf
edxexf
f
exg
λ
λπ
λλ =
−
== ∫ )}'(.{.')'()(
22
2
'
'22
Fourier transform lensFourier transform lens
� Lens cancels out the quadratic phase term in the Fresnel propagation at z = f.
� The lens brings Fraunhofer domain to the back focal plane.focal plane.
� The amplitude of the field is reduced by a factor of 1/f compared to 1/z in Fraunhofer case.
� The spatial frequency, v, is proportional to 1/f instead of 1/z (shrinks the space).
16
Optical Fourier transform Optical Fourier transform
�
f(x‘)g(x)=F(k)
� If the source is places at a distance f infront of the mirror the phase term outside the integration will cancel out
17
f f
f
xv
x
xf
xj
xfTFf
dxexff
xgλ
λπ
λλ =
−
== ∫ )}'(.{.1
')'(1
)('
'2
22--f systemf system
� This configuration is referred to as 2-f system.
f(x‘) g(x)=F(k)
� This is the building block for any Fourier optic system.
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f f
2-f � F.Tf(x‘) F(k)
44--f systemf system
� From Fourier transform properties
)'()}(.{.)}'(.{.)( xfkFTFxfTFkF −=→=
2-f � F.Tf(x‘) F(k)
2-f � F.Tf(-x‘)
� This configuration is referred to as 4-f system.19
2-f F.T 2-f F.T
f(x‘)
f f
F(k)
f f
f(-x)
Fourier opticsFourier optics
� In 2-f system, the output is an exact Fourier transform of the input
2-ff(x‘) F(k)
� In 4-f system the output is the same as the input (but flipped).
� The second 2-f system reconstructs the output from the Fourier transform
20
4-ff(x‘) f(-x)
Fourier opticsFourier optics
� The basics of Fourier optics
� The field at the back focal plane of the first lens
Plane wave expansion
2-f
FilteringH(k)??
Plane waves recombination
2-f
Inputf(x’)
Outputg(x)
The field at the back focal plane of the first lens is th F.T. of the source.
� Frequency v is proportional to space.
� The frequency components of the source can be manipulated by placing an element H(k).
21
πλ
πλα
λλν
22
1 xx kfx
k
f
x
f
x=→====
Fourier optics systemFourier optics system
f(x‘)
f f
H(k)
f f
g(x)
22
f f f f
{ })()(..)( kHkFTFxg =
Example: beam filteringExample: beam filtering
� In optics experiment, laser source usually produce noisy beam profile.
� A noisy Gaussian beam can be written asbe written as
� fn(x) is a rapidly varying noise function.
� Simple example is sin function with high frequency
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)()(2
2
2 xfexf nw
x
+=−
Example: beam filteringExample: beam filtering
� Noise corresponds to rapid variation and hence high frequency components.
� Gaussian beam corresponds to slow variation and hence low frequency components.
2
24
)()()}(.{.)(
)2cos()(
21
212
2
22
2
2
oo
wk
ow
x
vkvkexfTFkF
xvexf
++−+==
+=−
−
δδ
π
x k
F.T.
Noise
Gaussian