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©2000, D. L. Jaggard 1EE 511EE 511
EE 511: Introduction to EE 511: Introduction to Fourier Optics and Fourier Optics and
Image UnderstandingImage UnderstandingVolume 2
III. Scalar Diffraction Theory (Physical Optics)IV. Fresnel and Fraunhofer Diffraction
Dwight L. JaggardUniversity of Pennsylvania
308 Moore<[email protected]>
215.898.4411
©2000, D. L. Jaggard 2EE 511EE 511
Goal
Understand foundations of scalar diffraction, perform calculations and appreciate
results
©2000, D. L. Jaggard 3EE 511EE 511
Course OutlineCourse OutlineI. History and BackgroundII. Fourier Transforms and Linear Systems
III. Scalar Diffraction Theory (Physical Optics)
IV. Fresnel and Fraunhofer Approximations
V. Vector Diffraction TheoryVI. Geometrical and Ray Optics
VII. Properties of Lens
VIII. Coherent and Incoherent Imaging
IX. Partial Coherence TheoryX. Special Topics
©2000, D. L. Jaggard 4EE 511EE 511
Two Early Contributors to Wave Propagation and Interference
Chirstiaan Huygens (1629 -1695)Contributed to wave propagation &
concept of secondary sources
Thomas Young (1773 -1829)Discovered the principle of wave
interference (double -slit experiment)
©2000, D. L. Jaggard 5EE 511EE 511
III. Scalar Diffraction TheoryA. Maxwell’s EquationsB. Wave and Helmholtz EquationsC. Green’s FunctionD. Dirichlet SolutionE. Neumann SolutionF. Solution SummaryG. Babinet’s Principle
©2000, D. L. Jaggard 6EE 511EE 511
How do waves diffract?How do waves diffract?
Diffraction of light by saw tooth
©2000, D. L. Jaggard 7EE 511EE 511
How do waves diffract?How do waves diffract?
Diffraction of light by circular aperture
©2000, D. L. Jaggard 8EE 511EE 511
A. Maxwell’s Equationsl Time-Domain Equationsl Frequency-Domain Equations
James Clerk Maxwell (1831-1879)Showed light was electromagnetic
phenomena & discovered fundamentalequations for EM fields
©2000, D. L. Jaggard 9EE 511EE 511
Time-Domain Maxwell’s Equations
l Fields and sources are functions of time t and space x = (x,y,z) [Cartesian] or x = (r,θ,φ) [spherical]
l Current density j and charge density ρ are the sourcesl Electric and magnetic fields are connectedl Boldface denotes vectors
∇ × e(t) = −∂b(t)
∂t∇ •d(t ) = ρ(t )
∇ × h(t) = j(t ) +∂d(t)
∂t∇ • b(t) = 0
where
e,d,b,h, are EM fields
j,ρ are sources
©2000, D. L. Jaggard 10EE 511EE 511
Frequency-Domain Maxwell’s Equations
l The upper case fields are the F.T. of the lower case fieldsl Harmonic waves vary as exp( –jωt) [monochromatic]l Differentiation in the time domain becomes multiplication
by [–jωt] in the frequency domain
∇ ×E(x, ω) = + jωB( x,ω ) ∇ •D(x,ω ) = ρ (x, ω)
∇ ×H(x,ω ) = J(x,ω) − jωD(x,ω) ∇•B(x,ω ) = 0where
E(x,ω) = e(x, t)−∞
∞
∫ e+ jωtdt
e(x, t) =1
2πE(x,ω)
−∞
∞
∫ e− jωtdω
etc.
©2000, D. L. Jaggard 11EE 511EE 511
s• Operator (Divergence)
For vector A = Ax ˆ e x + Ayˆ e y + Az ˆ e z = Ar ˆ e r + Aθ ˆ e θ + Aφ ˆ e φ :
Cartesian Coordinates x = (x, y,z):
∇ • A =∂Ax
∂x+
∂Ay
∂y+
∂Az
∂z
Sperhical Coordinates x = (r,θ ,φ ):
∇ • A =
1r 2
∂∂r
r 2 Ar[ ]+1
r sinθ∂
∂θsinθAθ[ ]+
1r sinθ
∂Aφ
∂φ
©2000, D. L. Jaggard 12EE 511EE 511
sx Operator (Curl)
For vector A = Ax ˆ e x + Ayˆ e y + Az ˆ e z = Ar ˆ e r + Aθ ˆ e θ + Aφˆ e φ :
Cartesian Coordinates x = (x, y,z):
∇ × A =∂Az
∂y−
∂Ay
∂z
ˆ e x +∂Ax
∂z−
∂Az
∂x
ˆ e y +∂Ay
∂x−
∂Ax
∂y
ˆ e z
Sperhical Coordinates x = (r,θ ,φ ):
∇ × A =
1r sinθ
∂∂θ
sinθAφ( )−∂Aθ
∂φ
e r +
1r sinθ
∂Ar
∂φ
−
1r
∂∂r
rAφ( )
ˆ e θ +
1r
∂∂r
rAθ( )− ∂Ar
∂θ
ˆ e φ
©2000, D. L. Jaggard 13EE 511EE 511
B. Wave and Helmholtz Equations
l Find time-domain and frequency-
domain equations for EM field
l Scalarize result for optical fields
©2000, D. L. Jaggard 14EE 511EE 511
Wave and Helmholtz Equations
l Plane waves [e.g., exp( jkz)] and spherically waves [e.g., (1/r)exp( jkr)] are solutions to the Helmholtzequation for exp(–jωt) excitation
l These are vector equations and contain polarization information
Take curl of ∇ × e(x,t ) or ∇ ×h(x,t ) equations to
find wave equation for the source- free case:
∇2 − 1c2
∂2
∂t2
e(x,t )
h(x,t )
= 0
Take curl of ∇ ×E(x,ω) or ∇ ×H(x,ω) equations to
find Helmholtz equation for the source- free case:
∇2 + k 2[ ] E(x,ω)
H(x,ω)
= 0
Here v =1 / µε and k = ω / v
©2000, D. L. Jaggard 15EE 511EE 511
Scalarizing Helmholtz Equation
l One can neglect vector nature of EM fields in certain approximations:
l Very high frequency (e.g., large aperture and observer far from aperture
l Not interested in polarization effects
l Consider scalar form of Helmholtz equation where U is a scalar component (“optical field”) of one of the fields E or H
l Consider monochromatic case and suppress time -harmonic variation exp(–jωt)
©2000, D. L. Jaggard 16EE 511EE 511
Scalar Helmholtz Equation
l Need to find solution for optical field Ul Will use Green’s function approachl Note: λ is wavelength in the mediuml λ = λ0/nl For diffraction, usually n = 1
∇2 + k2[ ]U(x) = 0
Here the wavenumber is k = ω / v = 2π / λ
©2000, D. L. Jaggard 17EE 511EE 511
s2 Operator (Laplacian)
For scalar U (only):
Cartesian Coordinates x = (x, y,z):
∇2U (x, y, z) = ∂2
∂x2 + ∂ 2
∂y2 + ∂ 2
∂z2
U (x, y, z)
Sperhical Coordinates x = (r,θ ,φ ):
∇2U (r,θ,φ) =
1r2
∂∂r
r2 ∂∂r
+
1r2 sinθ
∂∂θ
sinθ ∂∂θ
+
1r2 sin 2 θ
∂2
∂φ 2
U(r,θ ,φ)
©2000, D. L. Jaggard 18EE 511EE 511
C. Green’s Functionl Green’s function G( x,x’) is similar
to impulse response (except Green’s function already has the shift included)
l Physically:
Green’s function [G( x,x’)] is the response of a system at x due to
a point source [ δ(x–x’)] at x’
©2000, D. L. Jaggard 19EE 511EE 511
More Green’s Functionl G(x,x’) allows fields to be written in
terms of integrals over sources f(x’) (if any) and values of fields on the surface
l For source-free fields, the optical field U(x) in a volume is uniquely found as a function of its value on the surface
l Reciprocity: G(x,x’) = G(x’,x)
©2000, D. L. Jaggard 20EE 511EE 511
Geometry – Two Views
x’ fx~x
y’
z
fy~y
U(fx,fy)U0(x’,y’)
S0
x’
y’ zU(x,y,z)
S0
S8
R
VdS’dS’
ez
λ
©2000, D. L. Jaggard 21EE 511EE 511
“Green Mathematics”
l Multiply the top equation by G( x,x’), the bottom by U(x’), and subtract
l Integrate result over volume V enclosed by surface S = S0 + S8 which encloses RHS
Let
′ ∇ 2 + k2[ ]U( ′ x ) = − f ( ′ x )
where f ( ′ x ) is a sourceThe associated Green' s function satisfies
′ ∇ 2 + k2[ ]G(x, ′ x ) = −δ(x − ′ x )
©2000, D. L. Jaggard 22EE 511EE 511
Expression for U(x)
l Volume integral vanishes in source-free regime f(x’) = 0
U(x) =G(x, ′ x ) f ( ′ x )d ′ x
V∫ +
G(x, ′ x ) ′ ∇ 2U( ′ x ) − U( ′ x ) ′ ∇ 2G(x, ′ x )[ ]d ′ x V∫
=G(x, ′ x ) f ( ′ x )d ′ x
V∫ +
G(x, ′ x ) ′ ∇ U ( ′ x )− U( ′ x ) ′ ∇ G(x, ′ x )[ ]• d ′ S S =S0 +S ∞
∫
=G(x, ′ x ) f ( ′ x )d ′ x
V∫ +
U ( ′ x ) ′ ∇ G(x, ′ x ) − G(x, ′ x ) ′ ∇ U( ′ x )[ ]• ˆ e zdS'S0
∫
©2000, D. L. Jaggard 23EE 511EE 511
U(x) in Source-Free Regime
lThis equation is exact for scalar Helmholtz equation
l Integration is over entire aperture plane S0 define by z = 0 [integration over S8 –> 0]
lU(x) is given anywhere in RHS z > 0 if U(x‘) and s‘U(x‘)l ez are known on the surface S 0, and if G(x,x’) is known
lUsual method is to assume U(x‘) and s‘U(x‘)l ez
U(x) = U ( ′ x ) ′ ∇ G(x, ′ x ) − G(x, ′ x ) ′ ∇ U( ′ x )[ ]• ˆ e zdS'
S0∫
©2000, D. L. Jaggard 24EE 511EE 511
Free-Space Green’s Function
l This solution is exact
l Positive (upper) sign is physically reasonable solution for causal system (it provides outgoingwaves)
lOnly valid if there are no boundaries
D.E.
∇2 + k2[ ]G(x, ′ x ) = −δ (x − ′ x )
B.C.
G(x, ′ x ) r→∞ → O1r
Solution
G(x, ′ x ) = e± jk x- ′ x
4π x- ′ x
©2000, D. L. Jaggard 25EE 511EE 511
Finding the Green’s Functionl Use the solution to the
homogenous equation (except at the singularity
l Find solution (to within a constant)
l Integrate over singularity to find the constant
l Use physical reasoning to keep outgoing wave only (i.e., discard acausal solution)
D.E. valid for x ≠ ′ x
∇ 2 + k2[ ]G(x, ′ x ) = 0
B.C.
G(x, ′ x ) r →∞ → O1r
Solution
G(x, ′ x ) = Ae+ jk x- ′ x
x - ′ x + B
e– jk x- ′ x
x - ′ x
Integrate over small volume at x = ′ x to find:
A + B = 1
4πChoose outgoing wave:
A = 1
4π and B = 0
©2000, D. L. Jaggard 26EE 511EE 511
Kirchhoff Assumptions
Reasonable physicalassumptions
l Kirchhoff B.C. approximation:
l U(x‘) and s‘U(x‘) l ez vanish on screen
l U(x‘) and s‘U(x‘)l ez are undisturbed in aperture
l Mathematical Problem:
l If U(x‘) and s‘U(x‘) l ez vanish anywhere on S, U(x) vanishes in V
©2000, D. L. Jaggard 27EE 511EE 511
Alternative Assumptionsl Dirichlet B.C.l GD(x,x’) = 0 for x’ on S0
l Estimate U(x‘) on S0
l U(x) = ?U(x‘) ?z’GD(x,x’) dS‘
l Neumann B.C.l ?z’GN(x,x’) = 0 for x’ on S0
l Estimate ?z’ U(x‘) on S0
l U(x) = −?GN(x,x’) ?z’ U(x‘) dS‘
©2000, D. L. Jaggard 28EE 511EE 511
Approximationsl This is scalar theory –– no polarizationl Usually fine for large apertures (high
frequencies)
l Usually fine if observer not too close to aperture
l Fields and derivatives are assumed undisturbed in the aperture (true only far from edges in high frequency limit)
l Solutions to scalar Helmholtz equation do not necessarily satisfy Maxwell’s equations
©2000, D. L. Jaggard 29EE 511EE 511
D. Dirichlet Solution
U (x) = U( ′ x ) ′ ∇ G(x, ′ x ) − G(x, ′ x ) ′ ∇ U ( ′ x )[ ]• ˆ e zd ′ S S0∫
UD(x) = U( ′ x ) ′ ∇ GD(x, ′ x )[ ]• ˆ e zd ′ S S0∫
= U( ′ x )∂ ′ z GD(x, ′ x )[ ]d ′ x S0 = Ap+ Screen∫ d ′ y
≈ U0( ′ x )∂ ′ z GD(x , ′ x )[ ]d ′ x Ap∫ d ′ y
l Dirichlet B.C.l GD(x,x’) = 0 for x’ on S0
l Estimate U(x‘) on S0 [replace by U 0(x‘) the undisturbed field]
l U(x) ∼ ?U ο(x‘) ?z’GD(x,x’) dS‘
©2000, D. L. Jaggard 30EE 511EE 511
Dirichlet Green’s Function
l |x-x’| and |x-x’’ | define “image points” with respect to the z = 0 plane
GD (x, ′ x ) = e+ jk x- ′ x
4π x- ′ x − e+ jk x- ′ ′ x
4π x- ′ ′ x
where
x - ′ x = (x − ′ x )2 + (y − ′ y )2 + (z − ′ z )2
x - ′ ′ x = (x − ′ x )2 + (y − ′ y )2 + (z + ′ z )2
©2000, D. L. Jaggard 31EE 511EE 511
Explicit Form for ?z’GD(x,x’)
∂′ z GD(x, ′ x )
′ z =0= e+ jk x- ′ x
2π x - ′ x z
x- ′ x
− jk + 1
x - ′ x
≈ − jke+ jk x- ′ x
2π x - ′ x z
x - ′ x
l For the observer 10 wavelengths or more away from the aperture:
|–jk| >> |x-x’|–1
©2000, D. L. Jaggard 32EE 511EE 511
E. Neumann Solution
U(x) = U ( ′ x ) ′ ∇ G(x, ′ x ) − G(x, ′ x ) ′ ∇ U( ′ x )[ ]• ˆ e zd ′ S S0
∫UN (x) = − GN (x, ′ x ) ′ ∇ U( ′ x )[ ]• ˆ e zd ′ S
S0∫
= − GN (x, ′ x )∂ ′ z U( ′ x )[ ]d ′ x S0 = Ap+ Screen∫ d ′ y
≈ − GN (x, ′ x )∂ ′ z U0 ( ′ x )[ ]d ′ x Ap∫ d ′ y
l Neumann B.C.l ?z’GN(x,x’) = 0 for x’ on S0
l Estimate ?z’ U(x‘) on S0 [replace by ?z’ U 0(x‘) the undisturbed field derivative]
l U(x) ∼ −?GN(x,x’) ?z’ U(x‘) dS‘
©2000, D. L. Jaggard 33EE 511EE 511
F. Solution Summary
U(x) ≈ kj2π
U 0( ′ x )′ z =0
e +jk x- ′ x
x- ′ x
Θ(θ, ′ θ )[ ]d ′ S
Ap∫where
Θ(θ, ′ θ ) =
12
cosθ + cos ′ θ [ ] Kirchhoff Approximation
cosθ Dirichlet Approximationcos ′ θ Neumann Approximation
l All forms are approximations l Kirchhoff form has mathematical problems but is not
restricted to planar screens/aperturesl All agree with each other and with experiment and exact
calculations in far -zone and near axis in high frequency regime
©2000, D. L. Jaggard 34EE 511EE 511
Our Usual Solution
U(x) ≈
kcosθj2π
U0 ( ′ x ) ′ z = 0
e+ jk x- ′ x
x - ′ x
d ′ S Ap∫
l This is the starting point for Fresnel and Fraunhofer diffraction
l This is a mathematical description of Huygen’s principle of secondary sources
Diffracted field is sum ofspherical waves weighted by
value in aperture [i.e, Huygen’s principle]
Constants conserve energyand provide correct units
while “j” provides phase shiftassociated with diffraction
©2000, D. L. Jaggard 35EE 511EE 511
G. Babinet’s Principlel Consider a diffracting screen S A and its
complement SB [two screens are complementaryif the opening in one corresponds to the opaque region of the other and vice versa]
l Let S0= SA+ SB [S0 could be the z = 0 plane]l Fields (from scalar diffraction theory):
l With no screen, the diffracted field is Ul With screen SA the field is U Α (integrate over opening)l With screen SB the field is U Β (integrate over opening)
l Result:
U = U Α+ U Β
Babinet’s Principle relatesdiffraction by a screen andits complement using scalar
diffraction theory sinceEach field is found by
integration over the aperture
©2000, D. L. Jaggard 36EE 511EE 511
Babinet Geometry
|UA|2
SA
SB
|UΒ|2
ΙΑ = |UΑ|2 = |UΒ|2 = ΙΒ except near z axiswhere light would be focused by lens
z
z
©2000, D. L. Jaggard 37EE 511EE 511
Two Babinet Conclusionsl If UΑ = 0 –> UΒ = Ul At points in which the intensity is zero in
the presence of one of the screens, the intensity in the presence of the other is the same as if no screen was present
l If U = 0 –> UΑ = –UΒ
l At points where U = 0, the magnitudes of UΑ and UΒ are equal and their phases are opposite –> IA = IB (e.g., case of lens in previous slide)
©2000, D. L. Jaggard 38EE 511EE 511
Babinet’s Statement
Two complementary screens produce the same illumination
(intensity) at all points in space not illuminated in their
absence
©2000, D. L. Jaggard 39EE 511EE 511
Course OutlineCourse OutlineI. History and BackgroundII. Fourier Transforms and Linear Systems
III. Scalar Diffraction Theory (Physical Optics)
IV. Fresnel and Fraunhofer Approximations
V. Vector Diffraction TheoryVI. Geometrical and Ray Optics
VII. Properties of Lens
VIII. Coherent and Incoherent Imaging
IX. Partial Coherence TheoryX. Special Topics
©2000, D. L. Jaggard 40EE 511EE 511
IV. Fresnel and FraunhoferDiffraction
A. Overview
B. Fraunhofer Approximation
C. Fresnel Approximation
D. Summary of Approximations
E. Fractal Example
F. Talbot Images
G. Scalar Diffraction Summary
©2000, D. L. Jaggard 41EE 511EE 511
Key Players in Diffraction Theory
Josef von Fraunhofer (1787 -1826)Developed diffraction gratings and
increased understanding of diffraction
Augustin Jean Fresnel (1788 -1827)Made contributions to transverse nature
of light and diffraction theory
©2000, D. L. Jaggard 42EE 511EE 511
Related Contributor
Jean-Baptiste Joseph Fourier (1768 -1830)Discovered harmonic analysis
(the basic math for Fourier Optics)
©2000, D. L. Jaggard 43EE 511EE 511
Overview
l Typical Diffraction Experiment
l Regions of Validity
l Intensity
l General Approximations
©2000, D. L. Jaggard 44EE 511EE 511
Typical Diffraction Experiment
BeamExpander
DiffractingScreen
ObservationScreen
CoherentSource
©2000, D. L. Jaggard 45EE 511EE 511
Regions of Validity
l Diffraction becomes easier to calculate with distance
DiffractingScreen
z~1 mm
~10 m
~10 cm
Fraunhofer Regime(far-zone)
Easier Calculations
Fresnel RegimeDifficult Sol’n
Near ZoneVery Hard
B.V.Sol’n
VeryNearZone
~1 cm
©2000, D. L. Jaggard 46EE 511EE 511
Note On Intensity
l For EM (vector) fields:
Poynting' s Vector (= "power density" ~ "intensity")
S ≡12
Re E × H•{ } [watts/ m 2 ]
For spherical waves in the far- zone
S ≡1
2ηE
2 ˆ e r [watts/ m 2] where η =µε
[ohms]
P = S •∫ ˆ e r dS [watts]
©2000, D. L. Jaggard 47EE 511EE 511
Scalar Wave Intensity
l For scalar optical fields:
2
2
Optical field intensity
for monochromatic waves
for quasi-monochromatic waves
[total "power"]
I U
I U
P IdS
≡
≡
= ∫
©2000, D. L. Jaggard 48EE 511EE 511
General Approximationsl Start with exact Dirichlet formulation:
UD(x) ≈ kj2π
U( ′ x ) e+ jk x- ′ x
x - ′ x
z
x - ′ x
1+ jk x - ′ x
d ′ S Ap∫
′ z =0
l Invoke approximations:l High frequencyl Observer in far-zone
©2000, D. L. Jaggard 49EE 511EE 511
High-Frequency Approximation
l kR >> 1 where R = |x – x’|
1 + jkx - ′ x
→ 1
U( ′ x )′ z =0
→ U0 ( ′ x )′ z =0
©2000, D. L. Jaggard 50EE 511EE 511
Observer in Far-Zone
0
cos cosx - x
1 1 1 1x - x x
z
z
R r
β θ′=
= → ′
= → =′
U(x,y,z)
z
R=|x–x’|
r=|x ||x’|
θ ∼ ββ
l R >> D = |x’|max and R >> λ
Bring termoutside integral
©2000, D. L. Jaggard 51EE 511EE 511
Apply Approximationsl Approximations yield:
UD(x) ≈
kj2πr
cosθ U0 ( ′ x )e+ jk x- ′ x d ′ S Ap∫
′ z =0
l Investigate two approximations for the exponential (this is more sensitive to approximation than {|x–x’|–1} term)
lDirichlet, Neumann and Kirchhoff B.C.s all reduce to same expression for cosθ –> 1 and cosθ’ –> 1
©2000, D. L. Jaggard 52EE 511EE 511
Fraunhofer Approximationl Fraunhofer Phase Term
l Examples for Simple Aperturesl Rectangular
l Circular
l Gaussian
l Examples for Multiple Apertures
l Other Illumination
©2000, D. L. Jaggard 53EE 511EE 511
C. FraunhoferApproximation
l Fraunhofer Assumption
l Examples for Simple Apertures
l Rectangular
l Circular
l Examples for Multiple Apertures
©2000, D. L. Jaggard 54EE 511EE 511
Fraunhofer Phase Term
l Restriction is on distance (far-zone)l Few restrictions on angle θl Can use lens to cancel residual phase
R ≡ x - ′ x ′ z = 0
= x − ′ x ( )2 + y − ′ y ( )2 + z( )2
= x2 + y2 + z 2
r21 2 4 3 4 − 2 x ′ x + y ′ y ( )+ ′ x 2 + ′ y 2( )
= L
R ≈ r −x ′ x + y ′ y
r where r = x = x( )2 + y( )2 + z( )2
valid for
k( ′ x 2 + ′ y 2 )2r
<< 2π
©2000, D. L. Jaggard 55EE 511EE 511
Fraunhofer Expression
U(x) ≈k
j2πrcosθ U0( ′ x )e + jk x- ′ x d ′ S
Ap∫′ z =0
~k
j2πcosθ
e+ jkr
rU0( ′ x )e
− jkr
x ′ x + y ′ y [ ]d ′ S
Ap∫′ z = 0
U(x, y, z) ≈ kj2π
cosθ e+ jkr
rt( ′ x , ′ y )e − j2π fx ′ x + f y ′ y [ ]d ′ x d ′ y
S0∫ fx = x /λr
fy =y / λr
U(x, y, z) ≈CF t( ′ x , ′ y ){ } fx =x /λr
f y = y /λrwhere
C =k
j2πcosθ
e+ jkr
rt( ′ x , ′ y ) =U0( ′ x , ′ y ,0) in the aperture and zero elsewhere
l Fraunhofer diffraction is proportional to the Fourier transform of the undisturbed aperture field
©2000, D. L. Jaggard 56EE 511EE 511
Two-Dimensional Functions(Reminder)
circ(r / a) ≡1 r / a < 10 r / a > 1
jinc(ρ) ≡ 2J1(2πρ)
2πρ
Gaus(ax ,by) ≡ exp[−π (a 2x2 + b 2y2 )]
Gaus(ar) ≡ exp[−πa2r2 ] = exp[−πa 2(x2 + y2)]
δ (x, y) ≡ δ (x)δ (y) = δ (r)πr
comb(ax)comb(by) ≡ δ (ax − n,by − m)m = −∞
∞
∑n= −∞
∞
∑
©2000, D. L. Jaggard 57EE 511EE 511
Two-Dimensional F.T. Pairs
rect (x / a)rect (y / b) ⇔ absinc(afx )sinc(bfy )
circ(r / a) ⇔ π a 2 jinc(aρ)
δ (r)πr
⇔1
δ (ax,by ) ⇔1ab
1r
⇔1ρ
1 ⇔δ ( fx, fy )
cos(πr2 ) ⇔ sin(πρ2 )
exp(± jπr2 ) ⇔ ± j exp(mjπρ 2 )
Gaus(ax,by ) ⇔1ab
Gaus( fx / a, f y / b )
Gaus(ar) ⇔ 1a 2 Gaus( fρ / a)
comb(x / a)comb(y / b) ⇔ abcomb( afx )comb(bfy)
©2000, D. L. Jaggard 58EE 511EE 511
Calculation for Rectangular Aperture
l Nulls for ax/λr = m or bλ/λr = m(m = ±1, ±2, ±3 . . )
l Maximum sidelobe level ~ –13 dB
t( ′ x , ′ y ) = rect( ′ x / a)rect( ′ y / b)
F t( ′ x , ′ y ){ } ⇔ absinc(afx )sinc(bf y)
U(x, y, z) = CF t( ′ x , ′ y ){ } fx = x / λrf y = y / λr
U(x, y, z) = Cab sinc ax / λr[ ]sinc ay / λr[ ]
U(x, y, z) =k
j2πcosθ
e+ jkr
rab
sin(πax / λr )(πax / λr )
sin(πby / λr)(πby / λr)
©2000, D. L. Jaggard 59EE 511EE 511
Calculation for Circular Aperture
l Nulls at zeros of first -order Bessel function (non-periodic)
l Maximum sidelobe level ~ –17 dB
t( ′ x , ′ y ) = circ ′ x 2 + ′ y 2 / a[ ]= circ[ ′ r / a]
F t( ′ x , ′ y ){ } ⇔ a 2 jinc a fx2 + fy
2[ ]= a 2 j inc aρ[ ]
U(x, y, z) = CF t( ′ x , ′ y ){ } fx = x/ λrf y = y/ λr
U(x, y, z) = Cπ a 2 jinc aρ[ ]ρ = x
2+y
2/ λr
U(x, y, z) = kj2π
cosθ e+ jkr
rπ a 2
J1 2πa x 2 + y2 / λr[ ]aπ x 2 + y2 / λr[ ]
jinc a x 2 + y2 / λr[ ]1 2 4 4 4 3 4 4 4
©2000, D. L. Jaggard 60EE 511EE 511
Intensity Results
I( x, y, z) = U(x, y, z) 2
For rectangular aperture:
I( x, y, z) =k cosθ
2πr
2
ab 2sin2(πax / λr)(πax / λr) 2
sin2(πby / λr)(πby / λr)2
sinc 2 ax/ λr[ ]sinc 2 by / λr[ ]1 2 4 4 4 4 4 3 4 4 4 4 4
For circular aperture:
I( x, y, z) = k cosθ2πr
2
π 2 a 4J1
2 2πa x 2 + y2 / λr[ ]π 2a2( x2 + y 2) / ( λr)2
jinc2 a x2 +y 2 /λr[ ]1 2 4 4 4 3 4 4 4
©2000, D. L. Jaggard 61EE 511EE 511
sinc and jinc Fuctionsjinc(r/2)
jinc2(r/2)sinc2(x)
sinc(x)
Note similarities and differences between sinc and jinc functions
©2000, D. L. Jaggard 62EE 511EE 511
Square Aperture Diffraction
©2000, D. L. Jaggard 63EE 511EE 511
Circular Aperture Diffraction
©2000, D. L. Jaggard 64EE 511EE 511
Experimental Results for Rectangular and Circular
Aperture Diffraction
©2000, D. L. Jaggard 65EE 511EE 511
Gaussian “Aperture”and Apodization
l Aperture edges create high spatial frequencies [remember F.T. properties]
l To reduce sidelobes , apertures are apodized or tapered
l Consider Gaussian transmission function or “aperture” with unit area and its Fraunhofer diffraction
©2000, D. L. Jaggard 66EE 511EE 511
Calculation for Gaussian “Aperture”
t( ′ x , ′ y ) =1
2πσexp − ′ r 2 / 2σ 2[ ]= Gaus(a ′ r ) with a =
12πσ
F t( ′ x , ′ y ){ } ⇔ exp −2π 2σ 2( fx2 + f y
2 )[ ]U(x,y, z) = CF t( ′ x , ′ y ){ } fx = x /λr
f y = y /λr
U(x,y, z) = C exp −2π 2σ 2 (x 2 + y2 ) / λ2r2[ ]I (x,y, z) = kcosθ
2πr
2
exp −4π 2σ 2 (x 2 + y2 ) / λ2r2[ ]
l No nulls – need to use half -power points to determine width
l Low (non-existent) sidelobes obtained at expense of increased size of main beam
©2000, D. L. Jaggard 67EE 511EE 511
Equivalent Widths
l If two functions have equal height and the same area, this defines their “equivalent width”
l There are other similar “widths”
∆xy ≡g(x , y)dxdy
−∞
∞
∫−∞
∞
∫g(0,0 )
and ∆ f x f y≡
G( fx , fy)dfxdfy
−∞
∞
∫−∞
∞
∫G(0,0)
I.C.B.S.T. ∆ xy∆ f x f y=1 (equivalent width in x - y × equivalent
bandwidth in fx − fx = 1)Also:
∆rmsxy ≡
x 2y2 g(x, y) 2 dxdy−∞
∞
∫−∞
∞
∫
g(x, y) 2 dxdy−∞
∞
∫−∞
∞
∫
1/2
and ∆rmsfx f y ≡
fx2 fy
2 G( fx , fx ) 2 dfxdfx−∞
∞
∫−∞
∞
∫
G( fx , fx ) 2 dfxdfx−∞
∞
∫−∞
∞
∫
1/2
I.C.B.S.T. ∆ rmsxy∆rms
f x f y ≥14π
(uncertainty relation)Assumes functionscentered on their
centroids
©2000, D. L. Jaggard 68EE 511EE 511
Equivalent WidthDiffraction Example
l For same equivalent widths, Guassian aperture is broader
l For same equivalent widths, Gaussian aperture diffraction is narrower
l Tapering aperture can reduce sidelobes at the expense of increased aperture diameter or main beam size
ˆ Ψ ( ′ r )Circular
= circ( π ′ r )
ˆ Ψ ( ′ r )Gaussian
= exp(−π ′ r 2 )
these have same equiv. widths
Ψ(x , y,z)Circular
=k cosθ
2πr
1π
J1 2π x2 + y2( ) / π λr[ ]x2 + y2( )/ π λr[ ]
Ψ(x, y,z )Gaussian
=k cosθ
2πr
exp −π x2 + y2( )/ λ2 r2[ ]
diffracted fields
©2000, D. L. Jaggard 69EE 511EE 511
Two Aperture Diffraction
l Consider diffraction from two circular apertures
x’
y’
2d
b
©2000, D. L. Jaggard 70EE 511EE 511
Two Aperture Intensity
I( x, y, z)I(0,0, z)
=J1
2 2πb fx2 + fy
2( )π 2b2 ( f x
2 + fy2 )
cos2(2π fy d)12[1 +cos( 4 πf y d )
1 2 4 3 4 fx = x /λr
fy = y/ λr
l Result yields fringes along y axis modulated by jinc function
l First zero of J1 when argument = 3.832l First zero of cosine when argument = π/2
©2000, D. L. Jaggard 71EE 511EE 511
Diffraction Gratings and Multiple Apertures
l For N multiple identical apertures, use geometric progression and sum
l Result is similar to antenna array factorl Can be used for diffraction grating calculations
Let g(x) ⇔ G( f x )
F g(x − na)-
N -12
N -12
∑
= G( f x)
Usually slowlyvarying due tosinge aperture
1 2 3 sin(Nπ fx a)sin(πfx a)
Usually sharply peakedfrom multiple apertures
1 2 4 3 4
©2000, D. L. Jaggard 72EE 511EE 511
Scaling and Shiftingfor Circular Apertures
©2000, D. L. Jaggard 73EE 511EE 511
Multiple Circular Apertures
©2000, D. L. Jaggard 74EE 511EE 511
Amplitude Diffraction Grating
l Periodic grating gives characteristic 0 th and 1 st
order diffractionl Phase grating is more efficient
t( ′ x , ′ y ) = 12
+ m2
cos 2πf0 ′ x ( )
rect ′ x
a
rect ′ y
b
U( x, y, z) = CF t( ′ x , ′ y ){ } f x =x / λrf y =y / λr
= C ab sinc bf y( ) sinc afx( ) + m2
sinc a f x − f0( )[ ]+ m2
sinc a fx + f0( )[ ]
f x =x / λr
f y =y / λr
I( x,y, z) ~ C2 ab 2 sinc 2 bf y( ) sinc 2 af x( ) + m 2
4sinc 2 a f x − f0( )[ ]+ m 2
4sinc 2 a fx + f0( )[ ]
f x = x / λr
f y= y / λr
Diffraction Efficiencies:
η0 =1 / 4
η+1 = η−1 = m2 / 16
What happens forsquare-wave grating?
Maximum efficiencyof ~6%
©2000, D. L. Jaggard 75EE 511EE 511
Other Illuminationsl Oblique Incidencel Center of diffracted beam is moved
(think “Modulation Theorem”)l Oblique incidence produces the
expected result
l Spherical Wave Sourcel Quadratic phase approximation
©2000, D. L. Jaggard 76EE 511EE 511
Oblique Incidence
l Both Fraunhofer and Fresnel diffraction patterns are shifted as expected
zα
Diffraction patternis centered here
r r sin α
If t( ′ x , ′ y ) = e jk xsinα + z cosα[ ] and
U(fx , fy) = CF t( ′ x , ′ y ){ }= T( fx , fy ) for α = 0
Then U(x ,y ,z) = CΨ fx −ksinα
2π, fy
for α ≠ 0
©2000, D. L. Jaggard 77EE 511EE 511
Spherical Wave Incidence
For sperhical source e jkRs
Rs
at (xs , ys ,zs) with
Rs = x − xs( )2+ y − ys( )2
+ z − zs( )2
The diffraction integral is evaluated at
x = ′ x , y = ′ y ,z = ′ z = 0
Rs ≈ zs 1+′ x − xs( )2
2 zs2 +
′ y − ys( )2
2zs2
z
Source at (xs, ys, zs)
Rs
©2000, D. L. Jaggard 78EE 511EE 511
D. Fresnel Approximations
l Fresnel Assumptions
l Example for Slit Diffraction
l Example for Edge Diffraction
l Examples for Circular Aperture
©2000, D. L. Jaggard 79EE 511EE 511
Fresnel Assumptionsl Fraunhofer result breaks down for
knife edge diffraction
l Need better approximation for the phase term exp{j2πk|x–x’|}
©2000, D. L. Jaggard 80EE 511EE 511
Knife Edge Example
U0 ( ′ x , ′ y ) =1 + sgn( ′ x )
2
F U0( ′ x , ′ y ){ }=12
δ ( fx )+1
jπf x
fy
U(x,y,z) =C2
δ( fx ) +1
jπf x
f y
f x =x / λr
fy =y / λr
Ψ(x,z)
z
x’ x
Does this Fraunhofer
resultmake sense?
©2000, D. L. Jaggard 81EE 511EE 511
Fresnel Phase Term
l Restriction on z (not R)l Observer must be close to axisl Valid in “near zone” as well as far zone
R ≡ x - ′ x ′ z = 0
= x − ′ x ( )2 + y − ′ y ( )2 + z( )2
= z 1+x − ′ x ( )2 + y − ′ y ( )2
z 2
= L
R ≈ z +(x - ′ x )2 + (y − ′ y 2 )[ ]
2z
valid for
k (x - ′ x )2 + (y − ′ y 2 )[ ]8z 2 << 2π
©2000, D. L. Jaggard 82EE 511EE 511
Fresnel Expression
U(x) ≈ kj2πr
cosθ U0( ′ x )e+ jk x- ′ x d ′ S Ap∫
′ z =0
≈k
j2πe+ jkz
zU0( ′ x )e
+jk
2 z( x− ′ x )2 +( y − ′ y ) 2[ ]
d ′ S Ap∫
′ z = 0
U(x, y, z) ≈ kj2π
e+ jkz
zt( ′ x )e
+jk2z
(x − ′ x )2 + (y − ′ y )2[ ]d ′ x d ′ y S0
∫
l Fresnel diffraction has quadratic phase in integrandl The phase expansion is valid near the z axis
©2000, D. L. Jaggard 83EE 511EE 511
Fresnel Diffraction by Rectangular Aperture -I
U(x, y, z) ≈k
j2πe+ jkz
zC
1 2 4 3 4 t ( ′ x )e
+jk
2 z(x − ′ x ) 2 + ( y− ′ y )2[ ]
d ′ x d ′ y S0
∫
= C e+
jk2z
( x− ′ x )2 + (y − ′ y ) 2[ ]d ′ x d ′ y where C = ke jkz
j2πz− Lx / 2
Lx /2
∫− Ly /2
L y / 2
∫
Let ξ ≡ kπz
( ′ x − x) and dξ = kπz
d ′ x
Let η ≡ kπz
( ′ y − y) and dη = kπz
d ′ y
Then ξ1
2
= m kπz
( Lx
2± x) and η1
2
= m kπz
(Ly
2± y)
U(x, y, z) = Cπzk
e+
jπ2
ξ2
dξ πzk
e+
jπ2
η2
dηη1
η2
∫ξ 1
ξ 2
∫
©2000, D. L. Jaggard 84EE 511EE 511
Fresnel Diffraction by Rectangular Aperture -II
U(x, y, z) = C πzk
e+
jπ2
ξ2
dξ πzk
e+
jπ2
η2
dηη1
η2
∫ξ1
ξ2
∫
where ξ ≡ kπz
( ′ x − x) and η ≡ kπz
( ′ y − y)
Note: close to aperture ξ and η are large
far from aperture ξ and η are small
l Calculations given in terms of tabulated Fresnel Integrals
©2000, D. L. Jaggard 85EE 511EE 511
Fresnel Integral Definitions
I(ξ ) ≡ e+ jπ
2t 2
dt 0
ξ
∫ [Complex Fresnel Integral I]
I(ξ ) = C(ξ )+ jS(ξ ) [Fresnel Integrals C and S]
C(ξ ) = cosπ2
t2
dt
0
ξ
∫
S(ξ ) = sin π2
t2
dt
0
ξ
∫
©2000, D. L. Jaggard 86EE 511EE 511
Fresnel Integrals & Cornu Spiral
©2000, D. L. Jaggard 87EE 511EE 511
Plot of Fresnel Integrals
©2000, D. L. Jaggard 88EE 511EE 511
Cornu Spiral
©2000, D. L. Jaggard 89EE 511EE 511
Fresnel Approximations
For ξ → 0
e+
jπ2
t2
→1 + jπ2
ξ2 −12!
π2
2
ξ 4 − j13!
π2
3
ξ6 L
∴ I(ξ ) = e+
jπ2
t2
dt 0
ξ
∫ → ξ + j11!
π2
ξ3
3−
12!
π2
2 ξ5
5− j
13!
π2
3 ξ7
7L
C(ξ) = cosπ2
t2
dt
0
ξ
∫ → ξ 1 −1
2!5π2
ξ2
2
+1
4!9π2
ξ2
4
L
S(ξ ) = sin π2
t2
dt
0
ξ
∫ → ξ 11!3
π2
ξ2
− 1
3!7π2
ξ 2
3
+ 15!11
π2
ξ 2
5
L
For ξ → ∞
C(ξ) → ±12
+sin πξ2 / 2( )
πξL for z
>
<0
S(ξ )→ ±12
−cos πξ2 / 2( )
πξL for z
><
0
©2000, D. L. Jaggard 90EE 511EE 511
Fresnel Intensity byRectangular Aperture
U(x, y, z) = Cπzk
[C (ξ2 ) − C(ξ1 )] + j[ S(ξ2 ) − S(ξ1 )]{ }Ae jψ
1 2 4 4 4 4 4 4 3 4 4 4 4 4 4 o
[C(η2 ) − C(η1)] + j[S(η2 ) − S(η1)]{ }Be jχ
1 2 4 4 4 4 4 4 3 4 4 4 4 4 4
I (x, y, z) =14
A2 B2 (for I0 = 1)
l Can use Cornu spiral to find characteristics– A is the length on the Cornu spiral from ξ1 to ξ2
– B is the length on the Cornu spiral from η1 to η2
©2000, D. L. Jaggard 91EE 511EE 511
Fresnel Cornu Calculations
l Let ∆v = ξ2 – ξ1 = Lx(k/ πz)1/2 = constant for given slitlAnd ξ0 = (ξ2 – ξ1)/2 = x(k/ πz)1/2 – moves as x (the center
of the aperture) moves
Ae j ψ(ξ0= 0)
with
∆v = 2.0
ψ
A
©2000, D. L. Jaggard 92EE 511EE 511
Case: ∆v = 1.5
lHere ∆v =15 = ξ2 – ξ1 = Lx(k/πz)1/2
lAnd ξ0 = (ξ2 – ξ1)/2 = x(k/ πz)1/2
∆v = 1.5
Α(ξ0=0) ~ 1.45Α(ξ0=0.25) ~ 1.37Α(ξ0=1.00) ~ 0.50Α(ξ0= 2.00) ~ 0.36
A(ξ 0 = 2.00)
A(ξ 0 = 1.00)
A(ξ 0 = 0)
A(ξ0 = 0.25)
lFor ξ0 =0 (aperture centered), draw line ∆v =15 from u = –0.75 to u = +0.75lFind A (optic
field) and A 2
(intensity)lNext move the
aperture ξ0, repeat and plot
©2000, D. L. Jaggard 93EE 511EE 511
Cases: ∆v = 0.6 to ∆v = 10.1A2 vs ξ 0
ξ0 = (ξ2 – ξ1)/2 = x(k/πz)1/2 –>
∆v = 0.6 = ξ2 – ξ1= L(k/πz)1/2
∆v = 1.5
∆v = 2.4
∆v = 3.7
∆v = 5.5
∆v = 6.2
∆v = 10.1
∆v = 4.7
©2000, D. L. Jaggard 94EE 511EE 511
Notes on Fresnel Diffractionl For ∆v > 10, pattern approaches geometrical
optics patternlRipples always appearlRipples become finer with increasing ∆v
l For ∆v < 1, pattern approaches Fraunhofer diffraction pattern
l For ∆v << 1, pattern spreads greatly in accordance with far-zone beam divergence
l For 1 < ∆v < 10, pattern undergoes pronounced change
l Here ∆v = [8NF]1/2, where NF (= Lx2/4λz) is
the Fresnel number used by Goodman
©2000, D. L. Jaggard 95EE 511EE 511
Fresnel Slit
Diffraction Evolution
∆v = 0.283
∆v = 2.83
Fourier transform-likeresult near far-zone
∆v = 5.65
∆v =8.94
Sloughing off of diffraction“shoulders” with distance
Geometric “image” ofslit near aperture
©2000, D. L. Jaggard 96EE 511EE 511
Infinite ApertureLet x dimensions → ±∞
Ae jψ = [C(ξ1) − C(ξ2 )]+ j[S(η1 ) − S(η2 )]x→ ±∞
= [C(∞) − C (−∞)] + j[S(∞) − S(−∞ )]
= 1+ j
= 2e jπ / 4
If y dimensions also → ±∞
Be jχ = 2e jπ / 4
Then
U(x, y, z ) =e jkz
2 j2ejπ /4 2e jπ /4 Ψ0
U(x, y, z ) = e jkzΨ0
This validates the constants infront of the diffraction integral
©2000, D. L. Jaggard 97EE 511EE 511
Fresnel Knife Edge Diffraction
DiffractingKnife Edge
z
ξ0 = 0
ObservationScreen
k=2π/λ
x’
l Here:l ξ2 –> –8l η2 –> –8 and η1 –> +8
©2000, D. L. Jaggard 98EE 511EE 511
Fresnel Knife Edge Diffraction
DiffractingKnife Edge
z
ξ0 < 0
ObservationScreen
k=2π/λ
x’
x
Shift coordinateupward by x
l Now, move coordinate system upward and find A
©2000, D. L. Jaggard 99EE 511EE 511
Knife Edge CalculationsLet y dimensions → ±∞
Be jχ = 2e jπ / 4
Then
U (x, y,z ) =e jkz
2 jAe jψ 2e jπ / 4U0
I (x, y,z ) =12
A 2 (for I0 = U0
2=1)
Here Ae jψ = [C (ξ1) − C(∞)]+ j[S(ξ1) − S(∞)]
= [C (∞) − C(ξ1 )]+ j[S(∞) − S(ξ1 )]
if ξ1 taken negative Use Cornu spiral to see
the physics of diffraction
©2000, D. L. Jaggard 100EE 511EE 511
Case: Knife EdgelLet ξ2 –> –8
lVary ξ1 (< 0)
lFind A (optical field) and A 2
(intensity)
lCalculate I = A2/2
A(ξ 0 = –0.5)
Max. V
alue
A(ξ 1 = 0)
A(ξ0= –1.22)
©2000, D. L. Jaggard 101EE 511EE 511
Fresnel Knife Edge Diffraction
©2000, D. L. Jaggard 102EE 511EE 511
Fresnel Strip Diffraction
l Think Babinet!
DiffractingSlit
z
Case: ∆v = L(k/πz)1/2 = 0.6
L
ObservationScreen
k=2π/λ
©2000, D. L. Jaggard 103EE 511EE 511
Fresnel Strip Calculations
©2000, D. L. Jaggard 104EE 511EE 511
Fresnel Strip Diffraction
l Note: this is different than diffraction by slit with ∆v = 0.6l Babinet’s Principle is relation for amplitudes
©2000, D. L. Jaggard 105EE 511EE 511
Fresnel Diffraction by Circular Apertures
©2000, D. L. Jaggard 106EE 511EE 511
Fresnel Zone
l Radius of first Fresnel zone is a = (λr)1/2
l Keeping aperture such that a << (λr)1/2 is equivalent to Fresnel approximation
Aperture
zr0
k=2π/λ
x’
r0 + λ/2a
©2000, D. L. Jaggard 107EE 511EE 511
Fresnel Zone Platel Consider radial transmission plate t(r’)
with absorption corresponding to the Fresnel zones
l Such plates can focus light due to constructive interference
t( ′ r ) =12
1 + cos(α ′ r 2 )[ ]circ( ′ r / a) [sinusoidal variation]
or
t( ′ r ) = 12
1 + sgn{cos(α ′ r 2)}[ ]circ( ′ r / a) [square- wave variation]
©2000, D. L. Jaggard 108EE 511EE 511
Fresnel Zone Plate & Lens
z
r0
x’
k=2π/λ
r0 +λ
r0+2λ
Focused lightfrom constructive
interference
l Fresnel zone plate has openings so that constructive interference occurs at distance r 0
l Similar idea can be applied to a lens
r0 +nλ
an
Radii of openings:an
2 =[r0+nλ]2 – r02
©2000, D. L. Jaggard 109EE 511EE 511
Fresnel On-Axis Diffraction
l Intensity along z axis can be found for circular aperturelCalculations can be done for
complementary disk
Aperture
z
k=2π/λ
x’
a
©2000, D. L. Jaggard 110EE 511EE 511
E. Fractal Resultsl Fractals are self-similar objects:l Are often “grown” through iterationl Have roughness on all scalesl Look similar at various
magnifications
l Fractal dimension D is measure of roughness or “area filling” nature
l For curves:l D –> 1 means smooth curvel D –>2 means very rough curve
©2000, D. L. Jaggard 111EE 511EE 511
FractallySerrated
Aperturesl Analytic and
experimental results available
l Fractal dimension Dis measure of perimeter “roughness”
l For lines: 1 < D < 2
D = 1.01
D = 1.50
D = 1.99
©2000, D. L. Jaggard 112EE 511EE 511E
xam
ple
s o
f Fra
ctal
A
per
ture
s
©2000, D. L. Jaggard 113EE 511EE 511
Stage 0
Stage 1 Stage 2
CantorSquare
Diffraction
l Stage 0 result is for square aperture
l Diffraction for higher stages can be found by iteration
l Scaling and complexity of aperture is reflected in diffraction pattern
©2000, D. L. Jaggard 114EE 511EE 511
Selected Other Fractal Results Using Scalar &
Vector Theoriesl Antenna array analysis/synthesis
l Aperture array analysis/synthesis
l Antenna element design
l Rough surface scattering
l Beam propagation & diffraction in turbulence
©2000, D. L. Jaggard 115EE 511EE 511
F. Talbot Images
x’ fx~x
y’
z
fy~y
U(fx,fy)t(x’,y’)
L
l Consider diffraction by infinite periodic grating
l Use Fresnel appoximation
t( ′ x , ′ y ) =12
+m2
cos 2π ′ x / L( )
©2000, D. L. Jaggard 116EE 511EE 511
Alternative Fresnel Method
U(x, y, z) =k
j2πe+ jkz
zC
1 2 4 3 4 t( ′ x , ′ y )e
+ jk2 z
( x− ′ x )2 +( y− ′ y )2[ ]d ′ x d ′ y
S0
∫
Or in the frequency domain,
we write, the transfer function
H( fx , fy ) = Fke jkz
j2πzexp j
k2 z
x2 + y2( )
= e jkz exp −jπλz fx2 + fy
2( )[ ]→
ignore phase term{ exp − jπλz fx
2 + fy2( )[ ]
Now write transmission functionin frequency domain and multiply
by transfer function H( fx,fy ) –then take inverse transform
Easiest methodfor periodic
input
©2000, D. L. Jaggard 117EE 511EE 511
Talbot Calculation - I
F t( ′ x , ′ y ){ }= F 12
1+ m cos 2π ′ x / L( )[ ]
= 12
δ ( fx , fy ) + m4
δ ( fx − 1L
, fy ) + m4
δ ( fx + 1L
, fy )
Note:
H ± 1L
, 0
= exp − j πλz
L2
After propagation over distance z
F t( ′ x , ′ y ){ }=12
δ ( fx , fy ) +m4
e− j
π λz
L 2 δ ( fx −1L
, fy ) +m4
e− j
π λz
L2 δ ( fx +1L
, fy )
Now take the inverse transformto find the field
©2000, D. L. Jaggard 118EE 511EE 511
Talbot Calculation - II
U(x, y, z) =12
+m4
e− j
πλz
L2e
+ j 2πxL +
m4
e− j
πλz
L2e
− j 2πxL
=12
1+ me− j
πλz
L2 cos2πx
L
Or:
I(x, y, z) =14
1+ 2mcosπλzL2
cos2πxL
+ m 2 cos2 2πxL
This result produces perfect“Talbot mages” or “self -images”
of the original grating
©2000, D. L. Jaggard 119EE 511EE 511
Talbot Calculation - IIICase I:
πλzL2 = 2nπ (n is an integer)
I(x ,y,z) =14
1 + m cos2πxL
2
Case II: πλzL2 = (2n + 1)π (n is an integer)
I(x ,y,z) =14
1 − m cos2πxL
2
Case III: πλzL2 = (2n − 1)π / 2 (n is an integer)
I(x ,y,z) =14
1 + m2 cos2 2πxL
=14
1 +m2
2
+
m2
2cos
4πxL
Produces perfect“Talbot images” or “self -images”
Produces contrast reversed“Talbot images” or “self -images”
Produces lower contrast“Talbot sub-images” with
twice the spatial frequency
©2000, D. L. Jaggard 120EE 511EE 511
G. Scalar Diffraction Summary
©2000, D. L. Jaggard 121EE 511EE 511
Course OutlineCourse OutlineI. History and BackgroundII. Fourier Transforms and Linear Systems
III. Scalar Diffraction Theory (Physical Optics)
IV. Fresnel and Fraunhofer Approximations
V. Vector Diffraction TheoryVI. Geometrical and Ray Optics
VII. Properties of Lens
VIII. Coherent and Incoherent Imaging
IX. Partial Coherence TheoryX. Special Topics
![Reminder Fourier Basis: t [0,1] nZnZ Fourier Series: Fourier Coefficient:](https://img.pdfslide.us/doc/110x75/56649d395503460f94a13929/reminder-fourier-basis-t-01-nznz-fourier-series-fourier-coefficient.jpg)
