1. Gerd Keiser, Optical Fiber Communications, McGraw-Hill, 4th
ed., 2011 Problem Solutions for Chapter 2
2.1
E  1 00co s 2 1 08 t  3 0 ex  2 0co s 21 08t  5 0 ey
 4 0co s 21 08 t  2 10 ez
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 x1 and x2 yields
x1 + x2 = a1 [cos t cos 1 + sin t sin 1]
+ a2 [cos t cos 2 + sin t sin 2]
= [a1 cos 1 + a2 cos 2] cos t + [a1 sin 1 + a2 sin
2] sin t Since the a's and the 's are constants, we can set
a1 cos 1 + a2 cos 2 = A cos  (1)
a1 sin 1 + a2 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

2. A
2
(sin
2
 + cos
2
) = a
2 2
1  cos
2
1 1 sin
+
2 2
 cos
2
2 + 2a1a2 (sin 1 sin 2a2 sin 2
or
A
2
= 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 21 2
Thus we can write
x = x1 + x2 = A cos  cos t + A sin
2.4 First expand Eq. (2.3) as
Ey
= cos (t - kz) cos  - sin (t - kz) sin 
E0 y
Subtract from this the expression
E x
cos  = cos (t - kz) cos 
E0 x
to yield
Ey
-
Ex
cos  = - sin (t - kz) sin 
E
0 y
E
0x
+ cos 1 cos 2)
 sin t = A cos(t - )
(2.4-1)
(2.4-2)
Using the relation cos
2
 + sin
2
 = 1, we use Eq. (2.2) to write
sin
2
(t - kz) = [1 - cos
2
(t - kz)] =
 E 
1 
2
x

 
E 
 0x 
(2.4-3)
Squaring both sides of Eq. (2.4-2) and substituting it into Eq. (2.4-3) yields
2

3. E y
E
0 y
 E x
E0x
2
cos 

=
 E
1  x

E
 0x
2






sin
2

Expanding the left-hand side and rearranging terms yields
 2
 2
 E  E  E E  

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  = sin
2

(a) Apply Snell's law
n1 cos 1 = n2 cos 2
where n1 = 1, 1 = 33,and 2 = 90 - 33 = 57
 n2 =
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

4. 4

5. 2.8
Air r
Water

12 cm
Find c from Snell's law n1 sin 1 = n2 sin c = 1
When n2 = 1.33, then c = 48.75
Find r from tan c =
r
, which yields r = 13.7 cm.
12 cm
2.9
45 
Using Snell's law nglass sin c = nalcohol sin 90
where c = 45 we have
n
glass
= 1.45
= 2.05sin 45
2.10 critical= arcsin
n pure
= arcsin
n
doped
1.450
1.460
= 83.3
2.11 Need to show that n1 cos 2  n 2 cos 1  0 . Use Snell’s Law and the relationship
tan 
sin 
cos 
5

6. 2 2
2.12 (a) Use either NA = n1  n2
or
NA  n1 2= n1
2(n1  n2
n
1
1/ 2
= 0.242
)
= 0.243
(b) A = arcsin (NA/n) = arcsin 0.242
 1.0 
= 14
1  n 2  1 1.00
2.13 (a) From Eq. (2.21) the critical angle is c  sin    sin   41

n

1.50  1 

 (c) The number of angles (modes) gets larger as the wavelength decreases.
2 2 1/ 2 2 2
 (12.14 NA = n1  n2 = n1  n1
= n1
2 1 / 2
2 
Since  << 1, 
2
<< ; 
2.15 (a) Solve Eq. (2.34a) for jH:
)2 1/ 2
NA
 n1 
2
jH = j 

Er - 1 Hz
r 
Substituting into Eq. (2.33b) we have
j  Er +


E
rz
= 
Solve for Er and let
 Er  1 H j
z 
  r 
q
2
= 
2
 - 
2
to obtain Eq. (2.35a).
(b) Solve Eq. (2.34b) for jHr:
jHr = -j

E -
1 Hz
Substituting into Eq. (2.33a) we have r
6

7. 1 E  
j  E +
z
= - j E
r  
Solve for E and let q
2
= 
2

(c) Solve Eq. (2.34a) for jEr:
 1 H z 
 r
- 
2
to obtain Eq. (2.35b).
1 1  H
z
jEr =

 jrH Substituting into Eq. (2.33b) we haver  
 1  EH
z z
 jrH + = jH
 r  r
Solve for H and let q
2
= 
2
 - 
2
to obtain Eq. (2.35d).
(d) Solve Eq. (2.34b) for jE
jE= -
1 

H z 
jHr
r 
1 E  H - jHr 
z

z
r   r
Solve for Hr to obtain Eq. (2.35c).
Substituting into Eq. (2.33a) we have
(e) Substitute Eqs. (2.35c) and (2.35d) into Eq. (2.34c)
j 1  Hz Ez   Hz Ez

-  r    
= jE
z2 
q r  r  r r 

r  
Upon differentiating and multiplying by jq
2
/ we obtain Eq. (2.36).
7

8. (f) Substitute Eqs. (2.35a) and (2.35b) into Eq. (2.33c)
j 1  Ez Hz  Ez Hz

  
= -jHz- 2   r   
q r  r   r r  
r  
 
Upon differentiating and multiplying by jq
2
/ we obtain Eq. (2.37).
2.16 For  = 0, from Eqs. (2.42) and (2.43) we have
Ez = AJ0(ur) e
j(t z ) and Hz = BJ0(ur) e
j(t z )
We want to find the coefficients A and B. From Eq. (2.47) and
(2.51), respectively, we have
C =
J
K


(ua)
(wa)
A and D =
J
K



(ua)
(wa)
B
Substitute these into Eq. (2.50) to find B in terms of A:
A
j 1

a
 2
u

1
w2
=
B 
J' (ua)  K' (wa)  
 (ua) wK(wa)uJ
  
For  = 0, the right-hand side must be zero. Also for  = 0, either Eq. (2.55a) or (2.56a)
holds. Suppose Eq. (2.56a) holds, so that the term in square brackets on the right-hand
side in the above equation is not zero. Then we must have that B = 0, which from Eq.
(2.43) means that Hz = 0. Thus Eq. (2.56) corresponds to TM0m modes.
For the other case, substitute Eqs. (2.47) and (2.51) into Eq. (2.52):
0 = 1 B jJ (ua)  A uJ' (ua)

u2
  a 1 
8

9. +
1  jJ (ua)  A wK' (wa)J (ua) B
 
  
w
2
a
2
K (wa) 
With k 2
1
= 
2
1 and k 2
2
= 
2
2 rewrite this as
B =
 
ja  1 


1 1

  w u
2 2 
(k1
2
J + k2
2
K) A
where J and K are defined in Eq. (2.54). If for  = 0 the term in square
brackets on the right-hand side is non-zero, that is, if Eq. (2.56a) does not hold,
then we must have that A = 0, which from Eq. (2.42) means that Ez = 0. Thus
Eq. (2.55) corresponds to TE0m modes.
2.17 From Eq. (2.23) we have
 =
n2
 n2
1
2n12
2
=
1 1

2
n 2
2
n 2
1


 << 1 implies n1  n2
Thus using Eq. (2.46), which states that n2k = k2  k1 = n1k, we have
2
k 2
 k 2 2
k 2
 k 2
2
n
2
 n
2 1 1
2.18 (a) From Eqs. (2.59) and (2.61) we have
M 
2
2
2
a2
n12  n22 
2 
2
2
a2
NA2
M 1/ 2  1000 1/ 2
0.85m
a  




 30.25m


NA 2

0.22
9

10. Therefore, D = 2a =60.5 m
(b)
2 2
M 
2 30.25m 2
 414  0.2
1.32m 2
(c) At 1550 nm, M = 300
2.19 From Eq. (2.58),
V =
2 (25 m)
0.82 m
(1.48)2 
2 1/
(1.46) 
2
= 46.5
Using Eq. (2.61)M  V
2
/2 =1081 at 820 nm.
Similarly, M = 417 at 1320 nm and M = 303 at 1550 nm. From Eq. (2.72)
Pclad
 P


total

4
3
M-1/2
=
4 100%
3 1080
= 4.1%
at 820 nm. Similarly, (Pclad/P)total = 6.6% at 1320 nm and 7.8% at 1550 nm.
2.20 (a) At 1320 nm we have from Eqs. (2.23) and (2.57) that V = 25 and M = 312.
(b) From Eq. (2.72) the power flow in the cladding is
7.5%. 2.21 (a) For single-mode operation, we need V  2.40.
Solving Eq. (2.58) for the core radius a
V 2 2 1/ 2 2.4 0(1.3 2m)
ma = n  n = = 6.55 
1 2  22
2 (1.4 80)  (1.4 78) 2 1/ 2
(b) From Eq. (2.23)
NA =

2 n
1
n 
2 1/ 2
2 =
 2
(1.480) 2
1/ 2
 (1.478) 
= 0.077
(c) From Eq. (2.23), NA = n sin A. When n = 1.0 then
10

11. A = arcsin NA =
 n 
arcsin
0.077
 1.0
 
= 4.4
2.22 n2 = 2
 NA
2
=
2 2
= 1.427n1 (1.458) (0.3)
a =
V
=
(1.30)(75)
= 52 m
2NA 2(0.3)
2.23 For small values of  we can write V 
2 a
n1 2

For a = 5 m we have  0.002, so that at 0.82 m
V 
2 (5
0.82
m)
m
1.45 2(0.002) = 3.514
Thus the fiber is no longer single-mode. From Figs. 2.18 and 2.19 we see that
the LP01 and the LP11 modes exist in the fiber at 0.82 m.
2.25 From Eq. (2.77), Lp = 2

=

n
y
 n
x
For Lp = 10 cm
For Lp = 2 m
Thus
ny - nx =
ny - nx =
1.3 10 6
101
m
1.3 106
2 m
m
m
= 1.310
-5
= 6.510
-7
6.510
-7
ny - nx 1.310
-5
2.26
We want to plot n(r) from n2 to n1. From Eq. (2.78)
n(r) = n1 1 2(r / a)
1 / 2
= 1.48 1 0.02(r / 25)
1 / 2
11

12. n2 is found from Eq. (2.79): n2 = n1(1 - ) =
1.465 2.27 From Eq. (2.81)
 2 2 2
 2an 2
M    1 
a k n 2  2   
1
where
 =
n 1  n 2
n
1
= 0.0135
At  = 820 nm, M = 543 and at  = 1300 nm, M = 216.
For a step index fiber we can use Eq. (2.61)
M
step 
V2
2
=
1 2a 2
 
n2  n2 2   
1 2
At  = 820 nm, Mstep = 1078 and at  = 1300 nm, Mstep = 429.
Alternatively, we can let  in Eq. (2-81):
M
step
=
2an1 2
 = 
1086 at 820 nm

432 at 1300 nm
2.28 Using Eq. (2.23) we have
(a) NA =

2
n
1

n 
2 1/ 2
2 =
 2
(1.60) 
2 1/
(1.49) 
2
= 0.58

2 2

1/ 2
= 0.39(b) NA = (1.458)  (1.405)
2.29 (a) From the Principle of the Conservation of Mass, the volume of a
preform rod section of length Lpreform and cross-sectional area A must equal the
volume of the fiber drawn from this section. The preform section of length
Lpreform is drawn into a fiber of length Lfiber in a time t. If S is the preform feed
speed, then Lpreform = St. Similarly, if s is the fiber drawing speed, then Lfiber = st.
Thus, if D and d are the preform and fiber diameters, respectively, then
Preform volume = Lpreform(D/2)
2
= St (D/2)
2
12

13. and
Equating these yields
D
St 2
Fiber volume = Lfiber (d/2)
2
= st (d/2)
2
2 d 2 D 2
    
 = st   or s = S 
d

2
(b) S = s
d 2
D 
= 1.2 m/s
0.125 mm
 9 mm
2

= 1.39 cm/min
2.30 Consider the following geometries of the preform and its corresponding fiber:
25 m
R
4 mm
62.5 m
3 mm
FIBER
PREFORM
We want to find the thickness of the deposited layer (3 mm - R). This can be
done by comparing the ratios of the preform core-to-cladding cross-sectional
areas and the fiber core-to-cladding cross-sectional areas:
A
A
preform core
preform clad
=
A
A
fiber core
fiber clad
or
2 2
)(3  R
2 2
)(4  3
=
 (25)2
 (62.5)2  (25)2 
from which we have
R =


7(25)2 1/ 2
= 2.77 mm9
2 2 
(62.5)  (25) 
13

14. Thus, the thickness = 3 mm - 2.77 mm = 0.23 mm.
2.31 (a) The volume of a 1-km-long 50-m diameter fiber core is
V = r
2
L =  (2.510
-3
cm)
2
(10
5
cm) = 1.96 cm
3
The mass M equals the density  times the volume V:
M = V = (2.6 gm/cm
3
)(1.96 cm
3
) = 5.1 gm
(b) If R is the deposition rate, then the deposition time t is
t =
M
=
5.1 gm
= 10.2 min
R 0.5 gm / min
2.32 Solving Eq. (2.82) for  yields
 K 2
 =  where Y =  for surface flaws.Y
Thus
 =
(20 N / mm 3/ 2
)2
(70 MN/ m
2 2
)
= 2.6010
-4
mm = 0.26 m
2.33 (a) To find the time to failure, we substitute Eq. (2.82) into Eq. (2.86)
and integrate (assuming that  is independent of time):
 f
  b / 2 d
i
t
= AY
b

b 
0
dt
which yields
1
1 
or
t =
b1
f
 b / 2
1
i
 b/ 2
 = AYbbt
2
2 f b i
 b)/ 2 2 b)/ 2(2 (
(b  2)A(Y)
(b) Rewriting the above expression in terms of K instead of yields
14

15. 2  2 b
Kf
2 b 
t =
Ki 


(b  2)A(Y)
b     
Y Y 
2K
2 b
b 2 b 2

i
if K orb i << K f
(b  2)A(Y)
K 2 b
 K 2 b
i f
15