Page 119 - Fiber Optic Communications Fund

100 Fiber Optic Communications

The energy difference ΔE is
ΔE = ℏ,
15
=ΔE∕ℏ = 2.28 × 10 rad/s.
From Eq. (3.15), we have
2 3
A
B = ,
ℏ 3
8
where = c∕n is the velocity of light in the medium. With = 1.25 × 10 m/s, we obtain
0
8 3
2
8
5 × 10 × ×(1.25 × 10 )
B =
15 3
1.054 × 10 −34 ×(2.28 × 10 )
3
2
21
= 7.71 × 10 m ∕J ⋅ s .
Example 3.2
The energy levels of an atomic system are separated by 1.26 × 10 −19 J. The population density in the ground
−3
state is 10 19 cm . Calculate (a) the wavelength of light emitted, (b) the ratio of spontaneous emission rate to
stimulated emission rate, (c) the ratio of stimulated emission rate to absorption rate, and (d) the population
density of the excited level. Assume that the system is in thermal equilibrium at 300 K.
Solution:
(a) The energy difference

ΔE = ℏ = ℏ2f
= hf,
where h = ℏ2 = 6.626 × 10 −34 J ⋅ s and

ΔE 1.26 × 10 −19
f = = = 191 THz.
h 6.626 × 10 −34
The wavelength of the light emitted is given by

c 3 × 10 8
= = = 1.56 μm.
f 191 × 10 12
(b) The ratio of spontaneous emission to stimulated emission rate is given by Eq. (3.16),
R spont ℏ∕k B T ΔE∕k B T
= e − 1 = e − 1,
R
stim
k = 1.38 × 10 −23 J∕K,
B
T = 300 K,
R spont ( 1.26 × 10 −19 )
= exp − 1
R stim 1.38 × 10 −23 × 300
13
= 1.88 × 10 .