Page 153 - Fiber Optic Communications Fund

134 Fiber Optic Communications

Solution:
(a) The frequency separation Δf is given by Eq. (3.45),
c 3 × 10 8
Δf = f − f = = = 142.8 GHz.
n+1 n −6
2nL 2 × 3.5 × 300 × 10
(b) Since
c
f = ,

we have
−c
df = d
2
or −6 2 11
2 (1.3 × 10 ) ×(1.428 × 10 )
|Δ| = Δf = m = 0.8 μm.
c 3 × 10 8

Example 3.10
In a direct band-gap material, an electron in the valence band having a crystal momentum of 9 × 10 −26 kg ⋅ m/s
14
makes a transition to the conduction band absorbing a light wave of frequency 3.94 × 10 Hz. The band gap
is 1.18 eV and the effective mass of an electron in the conduction band is 0.07m, where m is the electron rest
mass. Calculate the effective mass of the electron in the valence band.
Solution:
From Eq. (3.103), we have
2 2
ℏ k 1
hf = E + , (3.144)
g
2m r
where
1 1 1
= + (3.145)
m r m eff,1 m eff,2
and
E = 1.18 eV = 1.18 × 1.602 × 10 −19 J = 1.89 × 10 −19 J.
g
For a direct band-gap material, k ≈ k . The crystal momentum ℏk = 9 × 10 −26 kg ⋅ m/s and the energy of
1 2 1
the photon is
14
hf = 6.626 × 10 −34 × 3.94 × 10 J = 2.61 × 10 −19 J.
From Eq. (3.144), we have
2 2
ℏ k
1 −19 −19 −20
= hf − E =(2.61 × 10 − 1.89 × 10 ) J = 7.2 × 10 J,
g
2m
r
(9 × 10 −26 2 −32
)
m = = 5.62 × 10 kg.
r
14.4 × 10 −20