Electro-Optics - Applied Optics Fundamentals and Device Applications - page 86

or

Δ n x = 0

n

3

0

Δ ʹ = − n r E

y x

41

2

n

3

0

Δ ʹ = + n r E

x x

41

2

The results of this analysis show that one must consider both the waveguide
alignment and the polarization of the optical field launched in the wave-
guide. This shows that an index change is encountered only by field compo-
nents lying in the plane of the wafer. Furthermore, the index change depends
on the crystal direction of the optical propagation.
The i index change will vary from positive along the directions <011>
or <

011 to negative along the directions <01 - > and <0 - 1>; along four of
the crystal axes, <001>, <00 - >, <010>, and <0 - 0>, the positive and negative
changes cancel and there is no net index change.
To determine the ideal waveguide direction, two points must be consid-
ered. First, arbitrary polarization of the launched E-field will generally result
in an induced elliptical polarization. Only by launching the field parallel to
the wafer will the linear polarization be maintained. The second consider-
ation is that the waveguiding of any optical field is accomplished by making
the guiding region a higher index of refraction than the cladding; therefore,
it is usually desirable to align the guide such that the index change is positive
and leads to greater confinement.
With these points in mind, we use the waveguide alignment shown in
Figure 2.64. The optical TE mode is launched in the <011> direction and the
E-field is parallel to the <0 - 1> or Z ′ axis. The important effect for the opera-
tion of the double-Y device example is that the induced index change can be
related to a phase change in the launched optical field; therefore, the field
splitting to the guide with the applied dc field will undergo a phase advance
relative to the field that travels through the neutral guide.
The phase change for a given length l of the guide is

>

Δ∅ = 2π

λ Δ nl

(2.15)

The applied field in the x direction can be rewritten in terms of the applied
voltage E x = V/t , where V is the applied voltage and t is the guide thickness.
Then

r n V
t

Δ ʹ = n

3

(2.16)

41 0

2