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8/3/2019 Ch 9-Acoustooptic Effect
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ch 9.
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Region 1 Region 2
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Strain waves in a material produce a periodic structure thatdiffracts light. This can be used to modulate a beam.
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Photoelastic effect couples mechanical strain to a changein the optical index of refraction
where Skl is the strain tensor!! ! ! ! ! and Pijkl is calledthe strain-optic tensor. Diagonal terms of S are linearstrains (volume deformation), off-diagonal terms are shearstrains (shape deformation)
The index ellipsoid in the presence of a strain is
Using contracted notation we can write the effect of strain
as
ij = 1
n2ij
= PijklSkl
(ij + PijklSkl)xixj = 1
xk = Sklxl
3 1
n2
i
= PikSk, i, k = 1, 2 . . . , 6,
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The strain-optic tensor Pijkl can be written incontracted notation as Pik and has a form thatcan be found from symmetry considerations for
each crystal group. The form is identical to thatof the Kerr electro-optic tensor Sik.
For example in an isotropic material
Pik =
p11 p12 p12 0 0 0p12 p11 p12 0 0 0p12 p12 p11 0 0 0
0 0 0 12
(p11 p12) 0 00 0 0 0 1
2(p11 p12) 0
0 0 0 0 0 12
(p11 p12)
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Consider an acoustic plane wave in waterpropagating in the z-direction
producing a strain field, !! ! ! , of
Using the Pik tensor for an isotroic medium
S3 = KA sin (tKz) Ssin (tKz)
u (z, t) = Az cos(tKz)
Sij =ui
xj
PikSi =
p11 p12 p12 0 0 0p12 p11 p12 0 0 0p12 p12 p11 0 0 0
0 0 0 12
(p11 p12) 0 00 0 0 0 1
2(p11 p12) 0
0 0 0 0 0 12
(p11 p12)
00
Ssin(tKz)000
5warnin - this is not matrix multi lication
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i = PikSi =
p11 p12 p12 0 0 0p12 p11 p12 0 0 0p12 p12 p11 0 0 0
0 0 0 12
(p11 p12) 0 00 0 0 0 1
2(p11 p12) 0
0 0 0 0 0 12 (p11p12)
00
Ssin(tKz)00
0
11 = 1
n21
= p12Ssin (tKz) ,
22 =
1
n2
2
= p12Ssin (tKz) ,
33 =
1
n2
3
= p11Ssin (tKz) ,
ij = 0 for i = j
giving for
Resulting in an index ellipsoid of
6x2 + y2
1
n2
+ p12Ssin(tKz) + z2
1
n2
+ p11Ssin(tKz) = 1
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with new principle indices of refraction of
nx = ny = n1
2n3p12Ssin(tKz)
nz = n1
2n3p11Ssin(tKz)
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x2 + y2
1n2
+ p12Ssin(tKz)
+ z2
1
n2+ p11Ssin(tKz)
= 1
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Consider a transverse (shear) wave in GaAspolarized in the y direction, traveling in the z-direction
producing a strain wave of! ! ! which is
With the strain optic tensor for a cubic crystal
we get
u (z, t) = Ay cos(t kz)
Sij =ui
xj
Pij =
p11 p12 p120 0 0
p12 p11 p12 0 0 0
p12 p12 p11 0 0 0
0 0 0 p44 0 0
0 0 0 0 p44 0
0 0 0 0 0 p44
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S4 = KA sin (t kz) = Ssin (tKz)
23 = 32 = p44Ssin (t
Kz)
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The modified index ellipsoid is
We can choose a new set of principle coordinates
x, y and z that recast this equation into theform
From inspection x=x, so let z,y and z,y differ bya rotation of around the x-axis:
z=zcos-ysiny=zsin+ycos
x2
n2
x
+y
2
n2
y
+z
2
n2
z
= 1
9
1
n2
x2 + y2 + z2
+ 2yzp44Ssin(t kz) = 1
(z cos y sin )2
n2
+(z sin + y cos )2
n2
+x2
n2
+ 2(z cos y sin )(z sin + y cos )p44Ssin(tKz) = 1
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Evaluating
gives
cross term 2zycos(2)0 for =45 giving
or
z
z
y y
x2
n2+
1
n2 Ssin(tKz)
y2 +
1
n2+ Ssin(tKz)
z2 = 1
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(z cos y sin )2
n2+
(z sin + y cos )2
n2+
x2
n2+ 2(z cos y sin )(z sin + y cos )p44Ssin(tKz) = 1
x2
n2+
y2
n2+
z2
n2
zy sin2
n2+
zy sin2
n2+ (z2 sin2 + 2zy cos2 y2 sin2)p44Ssin(tKz) = 1
z
n2+ y
n2+ z
n2+ (z2
y2)Ssin(t
Kz) = 1
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Equating
to
gives
x
n2
x
+y
n2
y
+z
n2
z
= 1
so
x2
n2+
1
n2 Ssin(tKz)
y2 +
1
n2+ Ssin(tKz)
z2 = 1
n2z = n2
11 + n2Ssin(tKz)
1
n2+ Ssin(tKz)
=
1
n2z
n2
z n
2(1 n2Ssin(tKz))
n
z n1
1
2
n2Ssin(tKz) n
y n1 +
1
2
n2Ssin(tKz)nx = n
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Light will diffract from the periodic mediumproduced by the acousto-optic effect. Since theacoutic wave is slow compared to the light wave
(vs/c10-5), the acousto-optic perturbation can betreated as a stationary volume grating
Strong coupling to the diffracted wave occurswhen the Bragg condition is met:
for typical parameters this occurs for 5
2k sin = K
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The Bragg condition is equivalentto! ! ! ! ! found fromgeometrical consideration
The effect of motion of theacoustic wave is to Dopplershift the frequency of thediffracted wave according to
where vs sin is the component of the acoustic wavevelocity in the direction of the incident beam. At the Braggcondition! ! ! ! ! ! for the diffracted beam
Figure 9.2
2 sin = /n
= 2vs sin
c/n
=2vs
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Alternatively consider the opticalwave as made up of photons ofenergy and momentum k
and the acoustic wave as made upphonons of energy and momentum
K. For one photon absorbing one phonon
Conservation of energy requires =+
Conservation of momentum requires k=k+K
k,k,
K,
k
kk
K2
2k sin = K
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Picture of momentum conservationfor photons and phonons involved indiffraction in isotropic material is an
isosceles triangle since kk=n/c
In anisotropic material kk since n isa function of propagation angle nn.We can use the intersection of the
normal shells with the plane ofdiffraction to reconsider momentumdiagram. The Bragg condition canbe generalized to
k
k
kK
2
Propagation in thex-z plane of a uniaxial crystal,
acoustic wave at 30 from z axisoptical wave polarized as e-wave
k
Kk
sin + ksin
= K
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k
Kk
incident wave polarized as o-wave,diffracted wave polarized as e-wave
kK
k
k
Kk
kK
k
incident wave polarized as e-wave,diffracted wave polarized as e-wave
incident wave polarized as e-wave,diffracted wave polarized as o-wave
incident wave polarized as o-wave,diffracted wave polarized as o-wave
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For a given direction of the acoustic wave, there isa maximum value for K that allows Braggdiffraction. The maximum value can easily be found
from the geometry of the normal shells diagram
diffraction in the x-z plane of auniaxial crystal with an acoustic
wave propagating in the z-directionand incident wave o-polarized,diffracted wave e-polarized
diffraction in the x-z plane of auniaxial crystal with an acoustic
wave propagating in the x-directionand incident wave o-polarized,diffracted wave e-polarized
17Kmax = k0(no + ne)Kmax
= 2k0
n0
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diffraction in the x-z plane of abiiaxial crystal with an acoustic wave
propagating in the z-direction,
incident wave polarized in xz plane,diffracted beam polarized along y
diffraction in the x-z plane of abiaxial crystal with an acoustic wave
propagating in the z-direction,incident wave polarized in xz plane,
diffracted beam polarized in xz plane
Kmax = 2nxKmax = nx + ny
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Geometrical arguments give a constraint on theBragg angle for acousto-optic modulation, butdetermining the efficiency of conversion requires
we analyze the coupling of the undiffracted anddiffracted modes
Consider a total electric field
where A1 and A2 are coupled by the effect ofthe acoustic wave
E= A1E1e
i(1tk1.r) +A2E2e
i(2tk2.r)
ij = 0
ij= pijklSklcos (tKz)
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The perturbation to the impermeability tensor
is equivalent to a perturbation to the dielectrictensor
where 1 is a tensor of the first Fouriercoefficients of the tensor (z) and is given by
This perturbation will couple the fieldamplitudes A1 and A2
ij = 0
ij= pijklSklcos (tKz)
(z, t) = 21 cos(tKz) cos(tKz)
1 = (pijklSkl)
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The wave equation in a perturbed system can bewritten as
using the solutions in the unperturbed material
along with k2=2 and assuming A is slowly
varying (dA/dz
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Consider the cases of smallangle diffraction, shown in figure
9.7a of the textbookThe interaction region is thewidth of the acoustic beam andthe mode amplitudes are
functions of only x giving
k1
k2
2
1
12
22
2i1dA1
dxE1e
i(1t1x1z) 2i2
dA2
dxE2e
i(2t2x2z)
= 2
1e
i(tKz)+ 1e
i(tKz)
A1
E1ei(1t1x1z) +A2
E2ei(2t2x2z)
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k1
k2
2
1
12
The term is related to the asymmetry ofthe incident and diffracted waves relative tothe acoustic wave.
The conversion from A1 to A2 occurs when A1and A2 are in phase, but when they drift outof phase due to the acoustic wave converts field A2into field A1
When the Bragg condition is satisfied 1=2 so =0 and
the conversion efficiency is maximized
dA1
dx= i12A2e
ix,
dA2
dx= i
12A1e
ix,
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k1
k2
2
1
12
At the Bragg condition
which has solutions
where!! ! ! ! If A2(0)=0 this gives
dA1
dx= i12A2,
dA2dx
=i12A1
b = sin1
K
2k
= sin
1
2
A1 (x) = A1 (0) cos x i12
A2 (0) sin x,
A2 (x) = A2 (0) cos x i
12
A1 (0) sinx,
= |12|.
A1 (x) = A1 (0) cosx
A2 (x) = i
12
A1 (0) sinx 25
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The fraction of the power transfered to the diffractedbeam after an interaction distance x is
with full power transfer occurring at x=/2. Where can be expressed in terms of the original acousto-strainpijklSkl matrix
which depends on the intensity of the acoustic wave (viathe strain tensor Skl), the material properties, and thegeometry.
Idiffracted
Iincident =| A2 (x) |
2
| A1 (0) |2= sin
2
x,
=n3
4c cosB
|p1. (pijklSkl) p2|.
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ch 9.
Consider the cases of largeangle diffraction, shown in figure9.7b of the textbook
The mode amplitudes only varyalong z giving
giving k1
k2
2
1
1
2
E=A1 (z) E1e
i(1t1z) +A2 (z) E2ei(2t2z)
eix
dA1
dz= i
1
|2|12A2e
iz
dA2
dz= i
2
|2|12A1e
iz
=
1
2 Kwith where12 =
2
|12|
p
1 1p2 27
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For the case where 12>0 (i.e. forward diffraction)the solutions to
subject to the boundary conditions A2(0)=0 is
dA1
dz
= i12A2eiz
dA2
dz= i
12A1e
iz
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A1 (z) = A1 (0) ei1
2z
cos szi
2ssin sz
A2 (z) = iA1 (0) ei
1
2z
12
ssin sz
with s2 = |12|2 +
1
2
2
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The transfer of energy from A1 to A2 after interactionthrough a length L is given by
which is maximized when =0 (no phase mismatch ofincident and diffracted beams) and requires =0 ,L=/212 for 100% conversion.
=
|A2 (L) |
|A1 (0) |2 =
|12|
|12|2 +1
2
2 sin
2
sL.
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For the case where 12
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The transfer of energy from A1 to A2 after interactionthrough a length L is given by
which is identical to that of a Bragg reflector. Thisconfiguration can be thought of as a periodic Braggreflector, since the speed of the acoustic wave is somuch smaller than the speed of light that it can betreated as equivalent to a static perturbation of thepermittivity tensor.
R =
A2 (0)
A1 (0)
2
=
|12|2 sinh2 sL
s2 cosh2 sL+1
2
2sinh2 sL
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ch 9.
thus! ! ! ! ! giving
from which the index ellipsoid becomes
i = pi6S6
1
n2o
x2 +
1
n2o
y2 +
1
n2e
z2 2
ip41Ku0e
i(tKy)xz
i (p11 p1
1
n2o
x2 +
1
n2o
y2 +
1
n2e
z2 2
ip41Ku0e
i(tKy)xz
i (p11 p12)Ku0e
i(tKy)xy = 1
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sn.
Our expressions for the optical field diffracted fromthe acoustic wave assume there is only one diffractedorder (m=1). Is it possible to have|m|>1, corresponding to more than 1 phonon absorbed?
Conservation of energy requires k and k be almost thesame length (except for the relatively small increase ofk due to absorbed energy of phonon). This isntpossible for more than one order at a time, with a
unique K vector.
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k
kk
K2k
k1
K2
k2
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sn.
If the acoustic wave is a finite beam with width X,Heisenbergs uncertainty relation xp/2 tells us it
will have a range of K-vectors K=1/(2X).
If the range of K-vectors is large compared to theBragg angle, multiple phonos to be absorbed withoutchan in len th of k relative to k.
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k
kk
K2k
k1
K2
k2
k
K2
k1k
2
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sn.
Acoustic beam spread is K=K
Thus the Bragg angle (b=/(2n) )can beexpressed in terms of the acoustic beam spread
Q>1 is called the Bragg regime and correspondsto single order diffraction
Q
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sn.
Consider a narrow acoustic beam introducing achange in the index of refraction in a material of
an optical wave expressed by
is incident at x=0, upon exiting the acoutic beamat x=X it can be written as
with
where is the direciton of the optical beam
relative to the x-axis37
E= E0ei(tkr)
n(0 < x < X) = n0 sin(t K r)
E= E0ei(tkr)
=
X
0
1
cos
cndx
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sn.
IfX is small (i.e. in the Rama-Nath regime), index n(x,t) seen by optical beam can be considered constantacross beam width
giving
where
is called the modulation index. This can be expandedusing a form of the Jacobi-Anger identity
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=X
cos cn =
X
cos cn0 sin(t
K
r)
E= E0ei(tkr sin(t Kr))
X
cos
cn0
ei sinx =
m=
Jm()eim
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sn.
Expanding
into Bessel functions gives
and the diffraction efficiency for the mth orderdiffracted beam is
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E= E0ei(tkr sin(t Kr))
E= E0
m=
Jm()ei((m)t(km K)r)
m = J
2
m() = J2
m
X
cos
cn0
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sn.40
m = J2
m() = J2
m
X
cos
cn0
1 2 3 4 5 6 7 8 9 1
-
-0.5
0.5
J0()
J1() J2()
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sn.
Calculate the acoustic beam width X tomaximize diffraction into the m=1 order forn0=0.0001, =00 and =633 nm. What fraction
of the power gets diffracted into the m=1 beam?J1() is max at 1.85, solving for X we getX=1.9 mmEvaluating =J12(1.85)0.33
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Yariv & Yeh Optical Waves in Crystals chapter 9
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