Page 182 - Introduction to Information Optics

3.2. Light Propagation in Optical Fibers 3 67

(c) The total loss (in dB) is

4 2 88 106
Loss = 10-log, 0 ^ = 10-logioO - io- ) - ' = -1238dB.
* in
Again, this example tells us that we need 100% reflectivity for each reflection.
A reflectivity of 99.99% is absolutely not acceptable.

So far, we have explained how light can propagate in optical fiber from a
geometric optics point of view. The next question we need to answer is how
high the bandwidth can be, which can be estimated as follows: Assume that the
fiber length is L. The minimum time, t min , that light can pass through this fiber
corresponds to the case of incident angle $ f = 0°. Mathematically, t min is given
by

(16)
- '

The maximum time, t max, corresponds to the critical incident angle case. In this
case, the traveling distance becomes

-ma x ~~ ~ i ~ •
sin 4> c n 2/n 1 n 2
Thus, the maximum traveling time is

nc
The traveling time difference, At, is

,

Note that this traveling time difference results in a basic limitation on the
maximum bandwidth that can be used for transmission. To avoid confusion
among differing time information, the maximum bandwidth, B, is

(3.10)
At n lL/c(n l/n 2 — 1)

For a quantitative feeling about Eq. (3.10), let us look at the following example