Topic 3 Dielectric Waveguides and Optical Fibers 2-1 Symmetric ...

TIR at B & C
k1 (AB+BC) + phase change due to TIR = m(2π)
κ
E
λ
θ
θ θ
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
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k 1
A
β
B
C
n 2
Light
n 1
n 2
d = 2a
A light ray travelling in the guide must interfere constructively with itself to
propagate successfully. Otherwise destructive interference will destroy the
wave.
1
E
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2
θ
A
k 1
θ
A′
C
n 2
π−2θ
n 1
n 2
B′
2θ−π/2
Two arbitrary waves 1 and 2 that are initially in phase must remain in phase
after reflections. Otherwise the two will interfere destructively and cancel each
other.
?1999 S.O. Kasap, Optoelectronics (Prentice Hall)
B
θ
2a
y
x
1
y
x
z
z
5
7
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Waveguide Condition
k = kn =
1
1
2 1
π n / λ
For constructive interference, the phase difference between A
and C must be a multiple of 2π
Δφ(
AC ) = k ( AB + BC ) − 2φ
= m(
2π
)
1
1
d
BC = AB = BC cos( 2θ
)
cosθ
2
AB + BC = BC cos( 2θ
) + BC = BC[(
2cos
θ −1)
+ 1]
= 2d
cosθ
[ 2d
cosθ
] − 2φ
m(
2π
)
→ k =
Dividing (2) by 2 we obtain the waveguide condition
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⎡
⎢
⎣
πn
( 2a)
⎤
θm
− φm
= mπ
λ ⎥ cos
⎦
2 1
(1)
(2)
(3)
Resolve the wavevector k 1 into two propagation constants,
β and κ, along and perpendicular to the the guide axis z
m
⎛ 2πn1
⎞
βm = k1 sinθ
m = ⎜ ⎟sinθ m
⎝ λ ⎠
⎛ 2πn1
⎞
κ m = k1 cosθ
m = ⎜ ⎟cosθ m
⎝ λ ⎠
Φ = k φ ( a − y)
cosθ
−φ
( 1AC
− m)
− k1A'
C = 2k1
Φ
m
= Φ
m
y
( y) = mπ
− ( mπ
+ φm
)
a
m
m
6
8