Page 181 - Introduction to Information Optics

166 3. Communication with Optics

and SnelPs law,

n 0 sin B ic = n i sin B r. (3,4)
Substituting Eqs. (3.2) and (3.3) into Eq. (3.4), we have

n 0 sin B ic = y nf ~~ n\ (3.5)
Note that « 0 sin 9 ic is called the numerical aperture (NA) of the fiber and B ic is
called the acceptance angle of the fiber. When the incident angle is B i < B ic, the
light can propagate in the optical fiber without severe attenuation since the
total internal reflection happens at the intersection surface between the core
and the cladding. However, when 0,- > 9 ic, the leakage happens at the intersec-
tion surface between the core and the cladding, which results in severe
attenuation. This is the basic principle of how light can propagate in optical
fiber from a geometric optics point of view. Since the fiber length is very long,
there are many incidences of reflection. To ensure low attenuation, we need
100% perfect reflection. A little bit of loss in each reflection can result in huge
attenuation after many reflections, illustrated in the following example.

Example 3.2. Consider a step index fiber with parameters n t — 1.475, n 2 =
1.460, n 0 = 1.000, and core radius a = 25 ^m.
(a) Calculate the maximum incident angle and numerical aperture (NA) of
the fiber.
(b) Under the maximum incident angle, how many total reflections happen
for a 1-km-long fiber?
(c) If the power loss is 0.01% for each reflection, what is the total loss of
this 1-km-long fiber (in dB)?

Solve:
2 2
(a) NA = n 0 sin 0, = Jn\ - n\ = VI.475 - 1.460 =0.21.
(b) From the relationship n Qs'm 9, ; = n l smB r, as shown in Fig. 3.2, the
internal refractive angle, 0 r, can be calculated as

Then, the propagation length for each reflection is 2a/tan 8 r. Thus, the total
number of reflections is

3
L 10 m 6
N = — - _—__—- tan 8.19° = 2.88 • 10 times,
6
2a/tan0 r 2-25-10~ m
where L is the length of the fiber.