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SOLUTION MANUAL FOR OPTICAL FIBER COMMUNICATIONS 4TH EDITION BY GERD KEISER
Gerd Keiser,
Optical Fiber Communications
, McGrawHill, 4
th
ed., 2011 Problem Solutions for Chapter 2
2.1
E
1 00co s
2
1 0
8
t
3 0
e
x
2 0co s
2
1 0
8
t
5 0
e
y
4 0co s
2
1 0
8
t
2 10
e
z
2.2
The general form is:
y = (amplitude) cos(
t  kz) = A cos [2
(
t  z/
)]. Therefore (a) amplitude = 8
m (b) wavelength: 1/
= 0.8
m
1
so that
= 1.25
m (c)
= 2
(2) = 4
(d) At time t = 0 and position z = 4
m we have
y = 8 cos [2
(0.8
m
1
)(4
m)] = 8 cos [2
(3.2)] = 2.472
2.3
x1 = a1 cos (
t 
1) and x2 = a2 cos (
t 
2)
Adding x
1
and x
2
yields x
1
+ x
2
= a
1
[cos
t cos
1
+ sin
t sin
1
] + a
2
[cos
t cos
2
+ sin
t sin
2
] = [a
1
cos
1
+ a
2
cos
2
] cos
t + [a
1
sin
1
+ a
2
sin
2
] sin
t Since the a's and the
's are constants, we can set
a
1
cos
1
+ a
2
cos
2
= A cos
(1) a
1
sin
1
+ a
2
sin
2
= A sin
(2)
provided that constant values of A and
exist which satisfy these equations. To verify this, first we square both sides and add:
1
A
2
(sin
2
+ cos
2
) =
a
2 2
1
cos
2
1
1
sin
+
2 2
cos
2
2
+ 2a
1
a
2
(sin
1
sin
2
a
2
sin
2
or
A
=
a
2
a
2
+ 2a a
cos (

)
1 2
1 2 1 2
Dividing (2) by (1) gives
tan
=
a
1
sin
a
2
sin
1 2
a
cos
a
cos
1 2
1
2
Thus we can write x = x
1
+ x
2
= A cos
cos
t + A sin
2.4
First expand Eq. (2.3) as
E
y
= cos (
t  kz) cos
 sin (
t  kz) sin
E
0 y
Subtract from this the expression
E
x
cos
= cos (
t  kz) cos
E
0 x
to yield E
y

E
x
cos
=  sin (
t  kz) sin
E
0 y
E
0x
+ cos
1
cos
2
)
sin
t = A cos(
t 
)
(2.41) (2.42)
Using the relation cos2
+ sin2
= 1, we use Eq. (2.2) to write
sin
2
(
t  kz) = [1  cos
2
(
t  kz)] =
E
1
2
x
E
0x
(2.43) Squaring both sides of Eq. (2.42) and substituting it into Eq. (2.43) yields
2
E
y
E
0 y
E
x
E
0x
2
cos
=
E
x
E
0x
2
sin2
Expanding the lefthand side and rearranging terms yields
2
2
x
+
y
 2
x
y
E
0x
E
0y
E
0x
E
0y
2.5
Plot of Eq. (2.7).
2.6
Linearly polarized wave.
2.7 Air: n = 1.0 33
33
Glass 90
cos
= sin2
(a) Apply Snell's law n
1
cos
1
= n
2
cos
2
where n
1
= 1,
1
= 33
,
and
2 = 90
 33
= 57
n
2 =
cos
33
= 1.540
cos
57
(b) The critical angle is found from
n
glass
sin
glass
= n
air
sin
air
with
air = 90
and nair = 1.0
critical
= arcsin
1
= arcsin
1
= 40.5
n
g lass
1.540
3