Page 442 - Fiber Optic Communications Fund

Nonlinear Effects in Fibers 423

flux density D instead of E in Maxwell’s equations for free space. Substituting Eq. (10.21) in Eq. (10.24),
0
we obtain
[ (1) ]
D = 1 + E. (10.26)
0
In Section 1.2, we defined the electric flux density as
D = E. (10.27)
0 r
Therefore, the relative permittivity is
(1)
= 1 + . (10.28)
r
2
Since the relative permittivity and refractive index are related by n = , we obtain the result
r
Nq 2 e
2
(1)
n = 1 + = 1 + . (10.29)
2
m ( − )
2
0 0
From Eq. (10.29), we see that the refractive index is dependent on the frequency of the incident electromag-
netic signal. This is known as chromatic dispersion. In free space, n = 1 for all frequencies and the medium
is not dispersive. In a dispersive medium, suppose < .As increases (with < ), the denominator
0
0
of Eq. (10.29) decreases and the refractive index increases with frequency. This explains why a prism bends
light more at the violet end than the red end of the visible spectrum.
10.2.1 Absorption and Amplification
From Eq. (10.9), when = , we see that the displacement becomes infinite, which is unphysical. This
0
is because we ignored the loss effects. The situation is similar to that of an oscillating simple pendulum in
which the oscillations decrease with time due to frictional forces that carry away the energy. In the case of
an electron cloud, its vibration leads to the generation of electromagnetic waves that carry away the energy
and as a result, the vibration is damped. There are also other reasons for the dissipation of energy, such as
collision between atoms. Mathematical modeling of the vibrating electron cloud should include a damping
force of the form
dx
F =−r , (10.30)
damp
dt
to account for energy dissipation. Here, r is the damping coefficient. When this force is included, the equation
of motion (Eq. (10.5)) is modified as

2
d x dx
m + r + Kx = q E , (10.31)
e x
dt 2 dt
Assuming that the applied electric field intensity is of the form given by Eq. (10.6) and proceeding as before,
the expression for the displacement is
q E
e x
x(t)= . (10.32)
2
2
m( − − ir∕m)
0
Using Eq. (10.32), the polarization becomes
2
Nq E
e
P = 2 (10.33)
2
m( − − ir∕m)
0