Page 101 - Fiber Optic Communications Fund

82 Fiber Optic Communications

(NA) 2
Δ= , (2.243)
2n 2
1
NA = 0.2, n = 1.45.
1
From Eq. (2.243), we find
Δ= 0.0095.
When the difference between the core index and the cladding index is small, n ≈ n and Eq. (2.17) can be
1 2
approximated as
n LΔ
1
ΔT ≈= , (2.244)
c
8
n = 1.45, L = 2km, Δ= 0.0095, and c = 3 × 10 m/s.
1
Substituting these values in Eq. (2.244), the delay between the shortest and longest paths is
ΔT = 91.95 ns.

Example 2.13
6
The propagation constant at the wavelength = 1550 nm is 6 × 10 rad/m. Calculate the propagation con-
0
2
stant at = 1551 nm. Assume = 0.5 × 10 −8 s/m and =−10 ps /km. Ignore , n > 2.
1 1 2 n
Solution:
From Eq. (2.102), we find
2
( )= + ( − )+ ( − ) ∕2. (2.245)
1 0 1 1 0 2 1 0
Using c = f,wehave
2c 15
= 2f = = 1.2161 × 10 rad/s,
0 0

0
2c 15
= 2f = = 1.2153 × 10 rad/s,
1 1

1
11
− =−8.168 × 10 rad/s,
1 0
16
= 6 × 10 rad/m.
0
Substituting these values in Eq. (2.245), we find
6
( )= 5.9959 × 10 rad/s.
1
Note that the change in propagation constant is very small.