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36 2 Extremely Short-External-Cavity Laser Diode
where E 0 is the incident light amplitude and L ex is the external cavity length.
The total returned light E total is the sum of (E2.13) and is given as

∞ ∞
t 2 t k k
2

E total = E 0 r 2 + E k = E 0 r 2 + a (r 3 r exp [−j2β 2 L ex ]) .
2
r 2
k=1 k=1
(E2.14)
2
As t 2 t =1 − r ,and r = −r 2 the effective amplitude reflectivity r eff
2 2 2 2
(= E total /E 0 ) is given as
t 2 t ar 3 exp(−j2β 2 L ex )

eff
2
r = r 2 +
2
1 − ar 3 r exp(−j2β 2 L ex )
2
r 2 + ar 3 exp(−j2β 2 L ex ) eff
= ≡ R 2 exp(j∆ e ). (E2.15)
1+ ar 2 r 3 exp(−j2β 2 L ex )
Then the effective intensity reflectivity R eff and the phase ∆ ε are given as the
2
followingequations:
2
2 2
r + a r +2ar 2 r 3 cos(2β 2 L ex )
∗
3
2
eff eff
eff
R 2 = r r 2 = , (E2.16)
2
2 2 2
1+ a r r +2ar 2 r 3 cos(2β 2 L ex )
2 3
[r e ] −1 ar 3 (r − 1) sin(2β 2 L ex )
2
−1
2
∆ e = tan = tan 2 .
2
[r e ] r 2 + a r 2 r 3 + ar 3 (r + 1) cos(2β 2 L ex )
2
(E2.17)
Example 2.3. Derive the couplingcoefficient η which is defined by (2.8) as the
2
2
ratio of the feedback light power E to the original emitted light power E ,
0
z
2
where the Gaussian beam waists (1/e ) at the LD facet are w x0 and w y0 ,and
z =2L ex .
2
∞

E z E 0 dx dy
−∞
∞ 2 ∞ 2
η ≡
|E z | dx dy |E 0 | dx dy
−∞ −∞

2 2

4 1+ λz 1+ λz
πw 2 πw 2
x0 y0
=
. (2.8)
2
2

2+ λz 2+ λz
πw 2 πw 2
x0 y0
Solution 2.2. The light in the Gaussian form of (E2.18) travels in the z
direction in the form of (E2.19), where the beam waists at the LD facet are
w x0 and w y0 ,and C 0 and C z are the constants, respectively.

x y
2 2
E 0 = C 0 exp − 2 exp − 2 , (E2.18)
w x0 w y0
2 2
x y
E z = C z exp − 2 exp − 2 . (E2.19)
w (z) w (z)
x y