coupled wave theory by daniel marks september 10, 2009 ece 299 holography and coherence imaging...
DESCRIPTION
Volume holography Two plane waves interfere inside a medium. Intensity of two superimposed plane waves Spatial frequency k 1 /2 Spatial frequency k 2 /2 Spatial frequency ( k 2 - k 1 )/2 Wave #1 Wave #2 Interference pattern The hologram records a periodic pattern which has a spatial frequency given by the difference between the spatial frequencies of the interacting waves.TRANSCRIPT
Coupled wave theory
By Daniel Marks
September 10, 2009
ECE 299 Holography and Coherence Imaging Lecture 7
Duke University
What is coupled wave theory?
Interactingmedium
(e.g. holographic emulsion)
Interactingmedium
(e.g. holographic emulsion)
Two or more waves interact in a medium (e.g. holographic emulsion) altering each other.
In holography, waves are coupled by a pattern recorded in the emulsion.
Volume holography
Two plane waves interfere inside a medium.
22010 )exp()exp()( xikExikExI
Intensity of two superimposed plane waves
Spatial frequency k1/2
Spatial frequency k2/2
Spatial frequency (k2- k1)/2
Wave #1
Wave #2
Interference pattern
The hologram records a periodic pattern which has a spatial frequency given by the difference between the spatial frequencies of the interacting waves.
The pattern recorded by two plane waves (consider x direction only)
22010 )exp()exp()( xikExikExI
)exp()exp(Re22)( 212
02
0 xikxikEExI
xkkiEExI )(expRe22)( 212
02
0
xkkEExI )(cos22)( 212
02
0
This is the periodic pattern recorded in the emulsion.
The emulsion electric permittivity changes in proportion to the intensity dose on the film.
xkkxIx m )(cos)()( 21
There are many mechanisms for this photosensitivity (photochemical change, trapped charge, etc.)
The wave equation in a periodic hologram
0)(22 UxU Wave equation in
inhomogeneous medium (scalar approximation).
0)(cos 2122 UxkkU
Wave equation in periodically modulated
permittivity medium
We use coupled wave theory to approximately solve this equation for two incoming plane waves.
Assumption of this derivation: the incoming plane waves vary spatially on a length scale much bigger than a wavelength (slowly varying envelope approximation).
Coupled wave theory.
)exp()()exp()(),( zikxikzSzikxikzRzxU szsxrzrx
Express the field U(x,z) as a sum of two slowly varying plane waves R & S.
)exp( zikxik rzrx
)(zR )(zSandSlowly varying amplitudes in z direction of R & S waves.
)exp( zikxik szsx
X component of spatial frequency of plane waves
X component of spatial frequency of plane waves
Coupled wave theory (contd).
Insert U(x,z) into the wave equation…
…expand out all of the derivatives…
…omit of the terms proportional to and2
2
dzRd
2
2
dzSd
because R(z) & S(z) are slowly varying.
Some terms are proportional to exp(ikrzz) and some are proportional to exp(ikszz). We separate these into two equations because the spatial oscillations at these frequencies are “out of phase” and interact very little.
Yet more coupled wave theory
We also remove a common propagation phase exp[ik(krz+ksz)z/2] and we get the two coupled differential equations:
0)(exp)(2
2
zkkizSkdzdRik rzszrz
0)(exp)(2
2
zkkizRkdzdSik szrzsz
Note krx-ksx=K to make the x plane wavecomponents cancel the hologram phase.
22 k 222 )( kKkk rzsz
222 )( kKkk rzsz
Phase matching condition
zkxk szsx ˆˆ
zkzk rzsz ˆˆ zkxk rzrx ˆˆ
xKxkxk rxsx ˆˆˆ
How to solve these equations
Define zkkizSzS rzsz )(exp)()(~ and substitute…
0)(~2
2
zSkdzdRikrz
0)(exp)(2
~)(~ 2
zkkizRkSkkkdzSdik szrzszrzszrz
0)0( RR
0)0(~S
0z dz Initial conditions, no reference, a signal wave
?)(zR
?)(~ zSHologram
How to solve these equations (contd.)
Use guess of sum of complex exponentials and solve the indicial equation. Back substitute and you find
Insert boundary conditions, solve for constants of integration, and you get…
)sin(2
)cos(2
exp)( 0 zkizzkiRzR zz
)sin(2
exp2
)(2
0 zzkik
kiRzS z
sz
szrz kkk )/(21 2242 szrzz kkkk
So what does this solution mean?
S(z) is proportional to -1
2/12242 )/()(~ szrzz kkkkzS
The larger kz, the more phase mismatch and the less power exchanged from the reference to the signal beam. )(~ zS
zk
1
2/1
)/(2szrzkkk
Diffraction efficiency
For the ideal case kz=0 with no phase mismatch, we find
)cos()( 0 zRzR )sin()( 0 ziRzS
Maximum of power to the signal occurswhen sin(z)=0 or z=/2
Efficiency at power transfer (diffraction efficiency)
2
0RS
Diffraction efficiency is 1 when
k
kk
k szrz
And now for the simulations….
I performed a simulation of coupled wave theory.
Simulated forming a hologram, reconstructinga hologram, Bragg diffraction.
Instead of these equations which are approximatebut analytically tractable, I used the FiniteDifference Time Domain (FDTD) method.