Course Notes for EE1232 Introduction to Lasers and Optical Electronics
Chapter 7: waveguides
1
NOTES for Waveguide
This lecture covers Chapter 7.3 and 7.4
1. One-dimensional waveguide
2. Waveguide mode, transcendental equation, number of
modes, cut-off wavelength
3. Field distribution, mode confinement, evanescent
wave
4. Group velocities
5. Optical fiber
Planar Dielectric Waveguides
We know that for internal refraction, (n1>n2 for the figure showing
below). If the incident angle is less than the critical angle, we will
have total internal reflection (TIR).
cθ
θ <
Attention: we follow the angle assignment in the figure below, in
Snell’s law, the critical angle is the angle respect to the normal
direction (i.e. the critical angle in the snell’s law θc:
c
o
c
θ
θ
−
= 90
Now, if n1 is sandwiched by n2, TIR will take place on both
boundaries, which will “trap” the light in the layer of n1, and guide the
light in this layer. We call this structure waveguide. In this case (the
figure below), we have 1-dimensional (y-axis) confinement, we call 1-
D waveguide as planar (slab) waveguide.
Now, is it true that as long as
cθ
θ <
, we will have guided light
propagated in the waveguides? The answer is NO. Why????
For light travels in waveguide, the wave vector has two components
z
y
k
k
k
r
r
r
+
=
Course Notes for EE1232 Introduction to Lasers and Optical Electronics
Chapter 7: waveguides
2
Along y-axis, it is an optical resonator, we know that an optical
resonator only allows discreet # of modes, which is for a round trip,
we need has a phase change of 2mπ!!!
π
ϕ
m
d
k
r
y
2
2
2
=
−
ϕr is the phase change on the dielectric boundary. Let’s rewrite the
above formula:
π
ϕ
θ
λ
π
m
d
n
r
2
2
sin
4
0
1
=
−
Question: why
r
yd
k
ϕ
2
2
−
rather than
r
yd
k
ϕ
2
2
+
We can prove that for the TE mode, we have the phase change on
the dielectric boundary to be (your assignment):
2
/
1
2
2
1
sin
sin
2
tan
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=
θ
θ