# Lecture 7 Electro Optic Modulator ```Lecture 7
Electro Optic Modulator
The index of refraction is not constant but depends on the electric field
( )=
+
+
Pockels Effect
Kerr Effect
In uniaxial crystal le , ,
given by
+
+
be the principle axes of the index ellipsoid (indicatrix) so that ellipsoid is
=1
But in the ---- coordinate system , ,
1
+
1
+
1
1
+
the general indicatrix is
2
+
1
2
+
1
2
=1
So in general there are 6 components of the index of refraction
For the linear electro optic effect (Pockels effect) :
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
∆
∆
∆
∆
∆
∆
⎤
⎥
⎥
⎥
⎥=
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥∗
⎥
⎥
⎦
The Electro-Optic Tensor
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
∆
∆
∆
∆
∆
∆
⎤
⎥
⎥
⎥
⎥=
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎣
⎡
⎤ ⎢
⎥ ⎢
⎥∗⎢
⎥ ⎢2
⎥ ⎢
2
⎦
⎣2
⎤
⎥
⎥
⎥
⎥
⎥
⎦
Example: GaAs Isotropic (Cubic) Crystal
0
0
0
⎡
⎢
=⎢
⎢
⎢
⎣
0
0
0
0
0
0
0
0
0
0
0
⎤
⎥
⎥
⎥
⎥
⎦
0
( / ) ( = 0.9
( / ) ( = 1.15
= 3.6
= 1.1 ∗ 10
= 1.43 ∗ 10
)
)
Example: LiNbO3 Uniaxial Crystal
0
⎡ 0
⎢
= ⎢ 0
⎢ 0
⎢
⎣ - ̇
-
⎤
⎥
⎥
0 ⎥
0 ⎥
0 ⎦
0
0
= 9.6
= 6.8
= 30.9
= 32.6
1
=∆
1
+
1
= ∆
1
+
1
=∆
1
1
+
1
=∆
1
( / )
⃑ //
For case of GaAs, let the external field
∆
10
2
= 0, ∆
+2
1
=
(
)=0
So the spherical indicatrix is changed to ellipsoid with the O.A. in the direction of the applied electric
field rotating , coordinate 45 degree we get the principle axis.
=
=−
=
=
=
=
+
∅
+
∅
−
∅
−2
+2
−
+
+
−
+
[
−2
+
]+(
1
+(
1
) +2
[
1
)[
−
+2
+
−
+
]
]=1
Since
1
1
+
+
+
−2
1
2
+
1
+
−2
2
+
+2
(
−
1
)
=0
We should find the angle ∅ so that the coefficient of . In the case of isotropic material (GaAs)
) = 0 and ∅ = 45 .
=
= . So (
−
So the indicatrix becomes
1
1
=
+2
1
1
+
−2
+2
1
+
−→
=0
=
(1 +
=
1+
) =
Example of Phase Modulation
1. Transverse Modulation
2
( + ∆∅)
∅=
∆∅ =
1−
V
K
z
d
2
L
K
2. Longitudinal Modulation
2
∆∅ =
z
V
d
L
Example of Polarization Modulation
The polarization is orientation to make the angle is 45 with the
decomposed into components.
=
cos(
−
2
),
The phase difference between
=
and
∆∅ =
−
=
−
−
−
=
−
−
−
cos(
−
2
and
so the light is
)
is
,
If
∆∅ = 0 No change of polarization
∆∅ = Linear polarization
∆∅ = /2 Circular polarization
ℎ Elliptical Polarization
For Longitudinal Modulation
,
For Transverse Modulation
Optical Kerri Effect:
The refractive index changes in a function of the light intensity.
The part of the polarization vector that gives rise to this effect is:
( )(
( )=3∈
The total Polarization is
( )=
+
( )
Where
=
=∈
+3
( )|
− ) ( )
+
( )
( )+3∈
∗(
) ( )=3∈
( )|
( )|
( )(
( )=∈
; ,
,
) | ( )|
( )
( )
( )|
=1+
=
+
|&lt;
=
+
|&lt;
( ) &gt;| = 2 ( )
( ) &gt;|.
∗(
∈
1
) = 2| ( )|
2
By using equation 1 and equation 2
=1+
( )
,
=
( )
∈
```