Fundamentals of Opticsbucroccs.bu.ac.th/courses/documents/CRCC1/opto_fourier.pdf · Spherical lens...

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Fundamentals of Optics Fundamentals of Optics Dr. Waleed S. Mohammed 1

Transcript of Fundamentals of Opticsbucroccs.bu.ac.th/courses/documents/CRCC1/opto_fourier.pdf · Spherical lens...

Fundamentals of OpticsFundamentals of Optics

Dr. Waleed S. Mohammed

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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

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{ } ∫∞

∞−

− == 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?

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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’.

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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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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).

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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

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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

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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).

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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

Example: beam filteringExample: beam filtering

f(x‘)

H(k)“Pin hole”

g(x)

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f f f f

At the focal plane F(k) has low frequency component “Gaussian” and high frequency components “noise”

H(k) filters out the high frequency components “noise.” The output is

F(k)H(k)=exp[-2k2w2]