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Solutions Manual to Optoelectronics
and Photonics: Principles and Practices,
Second Edition
© 2013 Pearson Education
Safa Kasap
Revised: 11 December 2012

Check author's website for updates
http://optoelectronics.usask.ca
ISBN-10: 013308180X
ISBN-13: 9780133081800
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Solutions Manual (Preliminary) Chapter 2 2.2
11 December 2012
Preliminary Solutions to Problems and Question

Chapter 2
Note: Printing errors and corrections are indicated in dark red. Currently none reported.
2.1 Symmetric dielectric slab waveguide Consider two rays such as 1 and 2 interfering at point P
in Figure 2.4 Both are moving with the same incidence angle but have different m wavectors just
before point P. In addition, there is a phase difference between the two due to the different paths taken
to reach point P. We can represent the two waves as E1(y,z,t) = Eocos(tmymz + ) and E2(y,z,t) =
Eocos(tmymz) where the my terms have opposite signs indicating that the waves are traveling in
opposite directions.  has been used to indicate that the waves have a phase difference and travel
different optical paths to reach point P. We also know that m =k1cosm and m = k1sinm, and obviously
have the waveguide condition already incorporated into them through m. Show that the superposition of
E1 and E2 at P is given by
E( y, z, t)  2Eo cos( m y  12  ) cos(t   m z  12  )
What do the two cosine terms represent?
The planar waveguide is symmetric, which means that the intensity, E2, must be either maximum
(even m) or minimum (odd m) at the center of the guide. Choose suitable  values and plot the relative
magnitude of the electric field across the guide for m = 0, 1 and 2 for the following symmetric dielectric
planar guide : n1 = 1.4550, n2 = 1.4400, a = 10 m, = 1.5 m (free space), the first three modes have
1 = 88.84, 2 = 87.673 = 86.51. Scale the field values so that the maximum field is unity for m = 0
at the center of the guide. (Note: Alternatively, you can choose  so that intensity (E2) is the same at the
boundaries at y = a and y = a; it would give the same distribution.)
Solution
E( y)  Eo cos(t   m z   m z   )  Eo cos(t   m z   m z)
Use the appropriate trigonometric identity (see Appendix D) for cosA + cosB to express it as a product
of cosines 2cos[(A+B)/2]cos[(AB)/2],
E( y, z, t)  2Eo cos( m y  12  ) cos(t   m z  12  )
The first cosine term represents the field distribution along y and the second term is the propagation of
the field long the waveguide in the z-direction. Thus, the amplitude is
Amplitude = 2Eo cos( m y  21  )
The intensity is maximum or minimum at the center. We can choose  = 0 ( m = 0),  =  ( m = 1),  =
2 ( m = 2), which would result in maximum or minimum intensity at the center. (In fact,  = m). The
field distributions are shown in Figure 2Q1-1.
11 December 2012
Figure 2Q1-1 Amplitude of the electric field across the planar dielectric waveguide. Red, m = 0; blue, m = 1;
black, m = 2.
2.2 Standing waves inside the core of a symmetric slab waveguide Consider a symmetric planar
dielectric waveguide. Allowed upward and downward traveling waves inside the core of the planar
waveguide set-up a standing wave along y. The standing wave can only exist if the wave can be
replicated after it has traveled along the y-direction over one round trip. Put differently, a wave starting
at A in Figure 2.51 and traveling towards the upper face will travel along y, be reflected at B, travel
down, become reflected again at A, and then it would be traveling in the same direction as it started. At
this point, it must have an identical phase to its starting phase so that it can replicate itself and not
destroy itself. Given that the wavevector along y is m, derive the waveguide condition.
Figure 2.51 Upward and downward traveling waves along y set-up a standing wave. The condition for setting-up a
standing wave is that the wave must be identical, able to replicate itself, after one round trip along y.
Solution
From Figure 2.51 it can be seen that the optical path is
AB  BA  4a
With the ray under going a phase change  with each reflection the total phase change is
  4a m  2
11 December 2012
The wave will replicate itself, is the phase is same after the one round-trip, thus
  4a m  2  2m
2n1
and since  m  k1 cos  m  cos  m we get

2n1 (2a)
cos m  m  m

as required.
2.3 Dielectric slab waveguide

(a) Consider the two parallel rays 1 and 2 in Figure 2.52. Show that when they meet at C at a
distance y above the guide center, the phase difference is
   m = k12(a  y)cosmm
(b) Using the waveguide condition, show that
y
 m   m ( y)  m  (m  m )
a
(c) The two waves interfering at C can be most simply and conveniently represented as
E( y)  Acos(t)  Acos[t   m ( y)]
Hence find the amplitude of the field variation along y, across the guide. What is your conclusion?
Figure 2.52 Rays 1 and 2 are initially in phase as they belong to the same wavefront. Ray 1 experiences total internal
reflection at A. 1 and 2 interfere at C. There is a phase difference between the two waves.
Solution
(a) From the geometry we have the following:
(a y)/AC = cos
and  C/AC = cos(2)
The phase difference between the waves meeting at C is
 = kAC kAC = k1AC k1ACcos(2) 
11 December 2012
   = k1AC[1  cos(2)] k1AC[1 + cos(2)] 

   = k1(a y)/cos][ 1 + 2cos2  1]
   =k1(a y)/cos][2cos2]
   = k1(a y)cos
 2(2a)n1 
(b) Given, cos     m
 m m
 (mm ) mm
 cos  m  
2n1 (2a) k1 (2a)
mm
Then,  m  2k1 (a  y) cos  m  m  2k1 (a  y)  m
k1 (2a)
 y 
 m  1  m  m   m  m  (m  m )
y

 a  a
y y
 m  ( y)  m  (m  m )  m   m ( y)  m  (m  m )
a a
(c) The two waves interfering at C are out phase by ,
E( y)  A cos(t)  A cos[t   m ( y)]
where A is an arbitrary amplitude. Thus,
E  2A cos[t  12  m ( y)]cos1  m ( y)
2
or E  2A cos12  m ( y)cos(t  ) = Eocos(t + )
in which m/2, and cos(t +) is the time dependent part that represents the wave phenomenon,
and the curly brackets contain the effective amplitude. Thus, the amplitude Eo is
 m y 
E  2A cos  (m   )
o m
2 2a
To plot Eo as a function of y, we need to find m for m = 0, 1 , 2…The variation of the field is a
truncated) cosine function with its maximum at the center of the guide. See Figure 2Q1-1.
2.4 TE field pattern in slab waveguide Consider two parallel rays 1 and 2 interfering in the guide
as in Figure 2.52. Given the phase difference
y
 m   m ( y)  m  (m  m )
a
between the waves at C, distance y above the guide center, find the electric field pattern E (y) in the
guide. Recall that the field at C can be written as E( y)  Acos(t)  Acos[t   m ( y)] . Plot the field
11 December 2012
pattern for the first three modes taking a planar dielectric guide with a core thickness 20 m, n1 = 1.455
n2 = 1.440, light wavelength of 1.3 m.
11 December 2012
Figure 2.52 Rays 1 and 2 are initially in phase as they belong to the same wavefront. Ray 1 experiences total internal
reflection at A. 1 and 2 interfere at C. There is a phase difference between the two
Solution
The two waves interfering at C are out phase by ,
E( y)  A cos(t)  A cos[t   m ( y)]
where A is an arbitrary amplitude. Thus,
  1 
t   ( y) cos  ( y)
1
E  2A cos 
2 m 2 m

   
E  2A cos   m ( y)  cost    = Eocos(t + )
1
or
 2  
in which m/2, and cos(t +) is the time dependent part that represents the wave phenomenon,
and the curly brackets contain the effective amplitude. Thus, the amplitude Eo is
 m y 
E  2A cos  (m   )
o m
2 2a
To plot Eo as a function of y, we need to find m for m = 0, 1 and 2 , the first three modes. From
Example 2.1.1 in the textbook, the waveguide condition is
(2a)k1 cos  m  m  m
we can now substitute for m which has different forms for TE and TM waves to find,
2 1/ 2
   
 n
sin 2    2  
 m
  n
  1 
TE waves tan  ak1 cos  m  m   f TE ( m )
 2   
cos  m
11 December 2012
2 1/ 2
   
 n
sin 2    2  
 m
  n
  1 
TM waves tan ak1 cos  m  m   f TM ( m )
 2   
 n 
2
 2  cos  m
 n1 
The above two equations can be solved graphically as in Example 2.1.1 to find m for each choice of m.
Alternatively one can use a computer program for finding the roots of a function. The above equations
are functions of m only for each m. Using a = 10 m,  = 1.3 m, n1 = 1.455 n2 = 1.440, the results are:
TE Modes m=0 m=1 m=2
m (degrees) 88.84    
m (degrees) 163.75 147.02 129.69
TM Modes m=0 m=1 m=2
m (degrees) 88.84    
m (degrees) 164.08 147.66 130.60
There is no significant difference between the TE and TM modes (the reason is that n1 and n2 are very
close).
Figure 2Q4-1 Field distribution across the core of a planar dielectric waveguide
We can set A = 1 and plot Eo vs. y using
 m y 
E  2 cos  (m   )
o m
2 2a
with the m and m values in the table above. This is shown in Figure 2Q4-1.
11 December 2012
2.5 TE and TM Modes in dielectric slab waveguide Consider a planar dielectric guide with a core
thickness 20 m, n1 = 1.455 n2 = 1.440, light wavelength of 1.30 m. Given the waveguide condition,
and the expressions for phase changes  and  in TIR for the TE and TM modes respectively,
2 1/ 2 1/ 2
     n  
2
 n  
sin 2    2   sin 2    2  
m  n  m n 
       1 
1 

tan  21 m   and tan  12 m  
cos m  n2 
2
  cos  m
n
 1 
using a graphical solution find the angle  for the fundamental TE and TM modes and compare their
propagation constants along the guide.
Solution
The waveguide condition is
(2a)k1 cos  m  m  m
we can now substitute for m which has different forms for TE and TM waves to find,
2 1/ 2
   
 n
sin 2    2  
 m
  n
  1 
TE waves tan ak1 cos  m  m   f TE ( m )
 2   
 cos  m

 2 1/ 2
  
 n
sin 2    2  
 m
  n
  1 
TM waves tan ak1 cos  m  m   f TM ( m )
 2   
 n 
2
 2  cos  m
 n1 
The above two equations can be solved graphically as in Example 2.1.1 to find m for each choice of m.
Alternatively one can use a computer program for finding the roots of a function. The above equations
are functions of m only for each m. Using a = 10 m,  = 1.3 m, n1 = 1.455 n2 = 1.440, the results are:
TE Modes m=0
m (degrees) 88.8361
m = k1sinm  7,030,883 m-1  
11 December 2012
TM Modes m=0
m (degrees) 88.8340
m= k1sinm 7,030,878m-1
Note that 5.24 m-1 and the -difference is only 7.510-5 %.
The following intuitive calculation shows how the small difference between the TE and TM waves can
lead to dispersion that is time spread in the arrival times of the TE and TM optical signals.
11 December 2012
Suppose that  is the delay time between the TE and TM waves over a length L. Then,
 1 1    (5.24 m 1 )
   TE
 TM
 
L vTE vTM    (1.45 1015 rad/s)
= 3.610-15 s m-1 = 0.0036 ps m-1.

Over 1 km, the TE-TM wave dispersion is ~3.6 ps. One should be cautioned that we calculated
dispersion using the phase velocity whereas we should have used the group velocity.
2.6 Group velocity We can calculate the group velocity of a given mode as a function of frequency
 using a convenient math software package. It is assumed that the math-software package can carry
out symbolic algebra such as partial differentiation (the author used Livemath, , though others can also
be used). The propagation constant of a given mode is  = k1sin where  and  imply m and m. The
objective is to express  and  in terms of . Since k1 = n1 /c, the waveguide condition is
    sin 2
(n / n ) 2 
1/ 2
tan  a cos   m   2 1
 sin  2  cos 
so that 
tan 
arctansec sin   (n / n )  m( / 2)  F ( )
2 2
(1)
2 1 m
a
where Fm() and a function of  at a given m. The frequency  is given by
c c
  m ( ) (2)
F
n1 sin  n1 sin 
Both  and  are now a function of  in Eqs (1) and (2). Then the group velocity is found by
differentiating Eqs (1) and (2) with respect to  i.e.

 d   d  c  Fm( ) cos 
d
   1 
     ( )
v 
 d  n  sin  F     
g
d d sin 
2 m F ( )
  1    m 
c Fm ( ) 
i.e. vg  1  cot   Group velocity, planar waveguide (3)
n sin  F ( )
1  m 
where Fm = dFm/d is found by differentiating the second term of Eq. (1). For a given m value, Eqs (2)
and (3) can be plotted parametrically, that is, for each  value we can calculate  and vg and plot vg vs.
. Figure 2.11 shows an example for a guide with the characteristics in the figure caption. Using a
convenient math-software package, or by other means, obtain the same vg vs. behavior, discuss
intermodal dispersion, and whether the Equation (2.2.2) is appropriate.
11 December 2012
Solution
The results shown in Figure 2.11, and Figure 2Q6-1 were generated by the author using LiveMath based
on Eqs (1) and (3). Obviously other math software packages can also be used. The important conclusion
from Figure 2.11 is that although the maximum group velocity is c/n2, minimum group velocity is not
c/n1 and can be lower. Equation (2.2.2) in §2.2 is based on using vgmax = c/n2 and vgmin = c/n1, that is, taking
the group velocity as the phase velocity. Thus, it is only approximate.
Figure 2Q6-1 Group velocity vs. angular frequency  for three modes, TE0 (red), TE1 (blue) and TE4 (orange) in a planar
dielectric waveguide. The horizontal black lines mark the phase velocity in the core (bottom line, c/n1) and in the cladding
(top line, c/n1). (LiveMath used)
2.7 Dielectric slab waveguide Consider a dielectric slab waveguide that has a thin GaAs layer of
thickness 0.2 m between two AlGaAs layers. The refractive index of GaAs is 3.66 and that of the
AlGaAs layers is 3.40. What is the cut-off wavelength beyond which only a single mode can propagate
in the waveguide, assuming that the refractive index does not vary greatly with the wavelength? If a
11 December 2012
radiation of wavelength 870 nm (corresponding to bandgap radiation) is propagating in the GaAs layer,
what is the penetration of the evanescent wave into the AlGaAs layers? What is the mode field width
(MFW) of this radiation?
Solution
Given n1 = 3.66 (AlGaAs), n2 = 3.4 (AlGaAs), 2a = 210-7 m or a = 0.1 m, for only a single mode we
need
2a 
V  (n12  n22 )1/ 2  
 2
2 2 1/ 2 2 2 1/ 2
2a(n1 n2 ) 2(0.1 μm)(3.66 3.40 )
   = 0.542 m.
 
2 2
The cut-off wavelength is 542 nm.
When  = 870 nm,
2 (1 μm)(3.66 2  3.40 2 )1/ 2
V  = 0.979 < /2 
(0.870 μm)
Therefore,  = 870 nm is a single mode operation.

For a rectangular waveguide, the fundamental mode has a mode field width
V 1 0.979 1
2wo  MFW  2a  (0.2 μm) = 0.404 m.
V 0.979
The decay constant  of the evanescent wave is given by,
V 0.979
  =9.79 (m)-1 or 9.79106 m-1.
a 0.1 μm
The penetration depth 

   = 1/ = 1/ [9.79 (m)-1] = 0.102m.
The penetration depth is half the core thickness. The width between two e-1 points on the field decays in
the cladding is
Width = 2a + 2× = 0.2 m + 2(0.102) m = 0.404 m.
2.8 Dielectric slab waveguide Consider a slab dielectric waveguide that has a core thickness (2a)
of 20 m, n1 = 3.00, n2 = 1.50. Solution of the waveguide condition in Eq. (2.1.9) (in Example 2.1.1)
gives the mode angles 0 and 1for the TE0 and TE1 modes for selected wavelengths as summarized in
Table 2.7. For each wavelength calculate  and m and then plot  vs. m. On the same plot show the
lines with slopes c/n1 and c/n2. Compare your plot with the dispersion diagram in Figure 2.10
11 December 2012
Table 2.7 The solution of the waveguide condition for a = 10 m, n1 = 3.00, n2 = 1.50 gives the incidence angles 0 and 1
for modes 0 and 1 at the wavelengths shown.
m 15 20 25 30 40 45 50 70 100 150 200
0 77.8 74.52 71.5 68.7 63.9 61.7 59.74 53.2 46.4 39.9 36.45
1 65.2 58.15 51.6 45.5 35.5 32.02 30.17 ‐ ‐ ‐ ‐
Solution
Consider the case example for = 25 m = 25×10-6 m.
The free space propagation constant k = 2/ = 225×10-6 m = 2.513×105 m-1.
The propagation constant within the core is k1 = n1k = (3.00)( 2.513×105 m-1) = 7.540×105 m-1.
The angular frequency  = ck = (3×108 m s-1)( 2.513×105 m-1) = 7.54×1013 s-1.
Which is listed in Table 2Q8-1 in the second row under = 25 m.
The propagation constant along the guide, along z is given by Eq. (2.1.4) so that
m = k1sinm
or 0 = k1sin0 = (7.540×105 m-1)sin(71.5) = 7.540×105 m-1 = 7.15×105 m-1.
which is the value listed in bold in Table 2Q8-1 for the m = 0 mode at  = 25 m.
Similarly 1 = k1sin1 = (7.540×105 m-1)sin(51.6) = 7.540×105 m-1 = 5.91×105 m-1.
which is also listed in bold in Table 2Q8-1. We now have both 0 and  1 at  = 2.54×1013 s-1.
We can plot this 1 point for the m =0 mode at 0 = 7.15×105 m-1 along the x-axis, taken as the -axis,
and  = 2.54×1013 s-1 along the y-axis, taken as the -axis, as shown in Figure 2Q8-1. We can also plot
the 1 point we have for the m = 1 mode.
Propagation constants () at other wavelengths and hence frequencies () can be similarly calculated.
The results are listed in Table 2Q8-1 and plotted in Figure 2Q8-1. This is the dispersion diagram. For
comparison the dispersion  vs  for the core and the cladding are also shown. They are drawn so that
the slope is c/n1 for the core and c/n2 for the cladding.
Thus, the solutions of the waveguide condition as in Example 2.1.1 generates the data in Table 2Q8-1
for 2a = 10 m, n1 = 3; n2 = 1.5.
Table2Q8-1 Planar dielectric waveguide with a core thickness (2a) of 20 m, n1 = 3.00, n2 = 1.50.
m 15 20 25 30 40 45 50 70 100 150 200
 12.6 9.43 7.54 6.283 4.71 4.19 3.77 2.69 1.89 1.26 0.94
1013 s‐1
0 77.8 74.52 71.5 68.7 63.9 61.7 59.74 53.2 46.4 39.9 36.45
1 65.2 58.15 51.6 45.5 35.5 32.02 30.17 ‐ ‐ ‐ ‐
11 December 2012
0 12.3 9.08 7.15 5.85 4.23 3.69 3.26 2.16 1.37 0.81 0.56
105 1/m
1 11.4 8.01 5.91 4.48 2.74 2.22 1.89 ‐ ‐ ‐ ‐
105 1/m
Figure 2Q8-1 Dispersion diagram for a planar dielectric waveguide that has a core thickness (2a) of 20 m, n1 =
3.00, n2 = 1.50. Black, TE0 mode. Purple: TE1 mode. Blue: Propagation along the cladding. Red: Propagation
along the core.
Author's Note: Remember that the slope at a particular frequency  is the group velocity at that
frequency. As apparent, for the TE0 (m = 0) mode, this slope is initially (very long wavelengths) along
the blue curve at low frequencies but then along the red curve at high frequencies (very short
wavelengths). The group velocity changes from c/n2 to c/n1.
2.9 Dielectric slab waveguide Dielectric slab waveguide Consider a planar dielectric waveguide with
a core thickness 10 m, n1 = 1.4446, n2 = 1.4440. Calculate the V-number, the mode angle m for m = 0
(use a graphical solution, if necessary), penetration depth, and mode field width, MFW = 2a + 2, for
light wavelengths of 1.0 m and 1.5 m. What is your conclusion? Compare your MFW calculation
with 2wo = 2a(V+1)/V. The mode angle 0, is given as 0 = 88.85 for  = 1 m and 0 = 88.72 for  =
1.5 m for the fundamental mode m = 0.
Solution
 = 1 m, n1 = 1.4446, n2 = 1.4440, a = 5 m. Apply
2a
V 

n 2
1  n22 
1/ 2
to obtain V = 1.3079
11 December 2012
Solve the waveguide condition


  
1/ 2
 2
 n
sin 2    2  
 m
  n
  1 
tan ak1 cos  m  m   f ( m )
 2   
cos  m
graphically as in Example 2.1.1 to find: c = 88.35 and the mode angle (for m = 0) is o = 88.85.
Then use

1/ 2
 n  2
1  2 
2n2   sin   1
m
1  n2  
 m 
m 
to calculate the penetration depth:

  = 1/= 5.33 m.
 MFW = 2a + 2 = 20.65 m
We can also calculate MFW from
MFW = 2a(V+1)/V = 2(5 m)(1.3079+1)/(1.3079) = 17.6 m (Difference = 15%)
 = 1.5 m, V = 0.872, single mode. Solve waveguide condition graphically that the mode angle is o =
88.72.
  = 1/= 9.08 m.
 MFW = 2a + 2 = 28.15 m.
Compare with MFW = 2a(V+1)/V = 2(5 m)(0.872+1)/(0.872) = 21.5 m (Difference = 24%)
Notice that the MFW from 2a(V+1)/V gets worse as V decreases. The reason for using MFW =
2a(V+1)/V, is that this equation provides a single step calculation of MFW. The calculation of the
penetration depth  requires the calculation of the incidence angle  and .
Author's Note: Consider a more extreme case
 = 5 m, V = 0.262, single mode. Solve waveguide condition graphically to find that the mode angle is
o = 88.40.
  = 1/= 77.22 m.
 MFW = 2a + 2 = 164.4 m.
Compare with MFW = 2a(V+1)/V = 2(5 m)(0.262 + 1)/(0.262) = 48.2 m (Very large difference.)
11 December 2012
2.10 A multimode fiber Consider a multimode fiber with a core diameter of 100 m, core refractive
index of 1.4750, and a cladding refractive index of 1.4550 both at 850 nm. Consider operating this fiber
at  = 850 nm. (a) Calculate the V-number for the fiber and estimate the number of modes. (b) Calculate
11 December 2012
the wavelength beyond which the fiber becomes single mode. (c) Calculate the numerical aperture. (d)
Calculate the maximum acceptance angle. (e) Calculate the modal dispersion  and hence the bit rate 
distance product.
Solution
Given n1 = 1.475, n2 = 1.455, 2a = 10010-6 m or a = 50 m and = 0.850 m. The V-number is,
2a
V 

n 2
1  n22 
1/ 2

2π(50 μm)(1.475 2  1.455 2) 1/2
(0.850 μm)
= 89.47
Number of modes M,
V 2  89.47 2
M    4002
2 2
The fiber becomes monomode when,

2a
V

n 2
1  n22 
1/ 2
 2.405

2a n 2 n 2 
1/ 2
2(50 μm)(1.4752 1.4552 )1 / 2
or   1 2
 = 31.6 m
2.405 2.405
For wavelengths longer than 31.6 m, the fiber is a single mode waveguide.
The numerical aperture NA is
2 2 1/ 2 2 2 1/ 2
NA  (n1  n2 )  (1.475  1.455 ) = 0.242
If max is the maximum acceptance angle, then,

 NA 
 max  arcsin   arcsin(0.242 /1) = 14
 no 
Modal dispersion is given by
intermode n1 n 2 1.475 1.455

 
L c 3108 m s -1
= 66.7 ps m-1 or 67.6 ns per km

Given that   0.29, maximum bit-rate is
0.25L 0.25L 0.25
BL   
 total  intermode (0.29)(66.7 ns km -1 )
i.e. BL = 13 Mb s-1 km (only an estimate)

We neglected material dispersion at this wavelength which would further decrease BL. Material
dispersion and modal dispersion must be combined by
11 December 2012
 total
2
  intermode
2
  material
2
11 December 2012
For example, assuming an LED with a spectral rms deviation  of about 20 nm, and a Dm 200 ps
km-1 nm-1 (at about 850 nm)we would find the material dispersion as
material = (200 ps km-1 nm-1)(20 nm)(1 km)  4000 ps km-1 or 4 ns km-1,
which is substantially smaller than the intermode dispersion and can be neglected.
2.11 A water jet guiding light One of the early demonstrations of the way in which light can be
guided along a higher refractive index medium by total internal reflection involved illuminating the
starting point of a water jet as it comes out from a water tank. The refractive index of water is 1.330.
Consider a water jet of diameter 3 mm that is illuminated by green light of wavelength 560 nm. What is
the V-number, numerical aperture, total acceptance angle of the jet? How many modes are there? What
is the cut-off wavelength? The diameter of the jet increases (slowly) as the jet flows away from the
original spout. However, the light is still guided. Why?
Light guided along a thin water jet. A small hole is made in a plastic soda drink
bottle full of water to generate a thin water jet. When the hole is illuminated with a
laser beam (from a green laser pointer), the light is guided by total internal
reflections along the jet to the tray. Water with air bubbles (produced by shaking
the bottle) was used to increase the visibility of light. Air bubbles scatter light and
make the guided light visible. First such demonstration has been attributed to Jean-
Daniel Colladon, a Swiss scientist, who demonstrated a water jet guiding light in
1841.
Solution
V-number
2 1/2 -3 -9 2 2 1/2
V = (2a/)(n1 n2 ) 2
= (2×1.5×10 /550×10 )(1.330 1.000 ) = 15104
Numerical aperture
NA= (n12n22)1/2 = (1.33021.0002)1/2 = 0.8814

Total acceptance angle, assuming that the laser light is launched within the water medium
sinmax = NA/n0 = 0.113/1.33 or max = 41.4°.
Total acceptance 2o = 82.8
Modes = M = V2/2 = (15104)2/2 = 1.14×108 modes (~100 thousand modes)
The curoff wavelength corresponds to V = 2.405, that is V = (2a/)NA = 2.405
c = [2aNA]/2.405 = [(2)(4 m)(0.8814)]/2.405 = 3.5 mm
The large difference in refractive indices between the water and the air ensures that total internal
reflection occurs even as the width of the jet increases, which changes the angle of incidence.
2.12 Single mode fiber Consider a fiber with a 86.5%SiO2-13.5%GeO2 core of diameter of 8 m and
refractive index of 1.468 and a cladding refractive index of 1.464 both refractive indices at 1300 nm
11 December 2012
where the fiber is to be operated using a laser source with a half maximum width of 2 nm. (a) Calculate
the V-number for the fiber. Is this a single mode fiber? (b) Calculate the wavelength below which the
fiber becomes multimode. (c) Calculate the numerical aperture. (d) Calculate the maximum acceptance
angle. (e) Obtain the material dispersion and waveguide dispersion and hence estimate the bit rate
distance product (BL) of the fiber.
Solution
(a) Given n1 = 1.475, n2 = 1.455, 2a = 810-6 m or a = 4 m and =1.3 m. The V-number is,
2a 2(4 μm)(1.468 2  1.464 2) 1/ 2
V

n 2
1  n22 
1/ 2

(1.3 μm)
= 2.094
(b) Since V < 2.405, this is a single mode fiber. The fiber becomes multimode when
2a
V (n12  n22 )1/ 2  2.405


2a n 2 n 2 
1/ 2

2(4 μm) 1.4682 1.464 2 
1/ 2
or   1 2
 =1.13 m
2.405 2.405
For wavelengths shorter than 1.13 m, the fiber is a multi-mode waveguide.
(c) The numerical aperture NA is

NA  n
2
 n2

2 1/ 2 2
 (1.468 1.464 )
2 1/ 2
= 0.108
1
(d) If max is the maximum acceptance angle, then,

 NA 
 max  arcsin   arcsin(0.108 /1) = 6.2
 no 
so that the total acceptance angle is 12.4.

(e) At  =1.3 m, from D vs. , Figure 2.22, Dm  7.5 ps km-1 nm-1, Dw  5 ps km-1 nm-1.
 1/ 2 
 Dm  Dw 1/ 2
L
= |7.55 ps km-1 nm-1|(2 nm) = 15 ps km-1 + 10 ps km-1

= 0.025 ns km-1
Obviously material dispersion is 15 ps km-1 and waveguide dispersion is 10 ps km-1
The maximum bit-rate distance product is then
0.59L 0.59
BL   = 23.6 Gb s-1 km.
 1/ 2 0.025 ns km -1
11 December 2012
2.13 Single mode fiber Consider a step-index fiber with a core of diameter of 9 m and refractive
index of 1.4510 at 1550 nm and a normalized refractive index difference of 0.25% where the fiber is to
be operated using a laser source with a half-maximum width of 3 nm. At 1.55 m, the material and
11 December 2012
waveguide dispersion coefficients of this fiber are approximately given by Dm = 15 ps km-1 nm-1 and Dw
= 5 ps km-1 nm-1. (a) Calculate the V-number for the fiber. Is this a single mode fiber? (b) Calculate the
wavelength below which the fiber becomes multimode. (c) Calculate the numerical aperture. (d)
Calculate the maximum total acceptance angle. (e) Calculate the material, waveguide and chromatic
dispersion per kilometer of fiber. (f) Estimate the bit rate distance product (BL) of this fiber. (g) What
is the maximum allowed diameter that maintains operation in single mode? (h) What is the mode field
diameter?
Solution
(a) The normalized refractive index difference  and n1 are given.
Apply,  = (n1n2)/n1 = (1.451 n2)/1.451 = 0.0025, and solving for n2 we find n2 = 1.4474.
The V-number is given by
2a 2(4.5 μm)(1.4510 2  1.4474 2) 1/ 2
V  (n12  n22 )1/ 2  = 1.87; single mode fiber.
 (1.55 μm)
(b) For multimode operation we need

2a 2 2 1/ 2 2(4.5 μm)(1.4510 2 1.4474 2) 1/ 2
( ) 2.405
V n1  n2  
  
 < 1.205 m.
(c) The numerical aperture NA is
2 2 1/ 2 2 2 1/ 2
NA  (n1  n 2 )  (1.4510 1.4474 ) = 0.1025.
(d) If max is the maximum acceptance angle, then,

 NA 
 max  arcsin  = arcsin(0.1025/1) = 5.89
 no 
Total acceptance angle 2amax is 11.8 .

(e) Given, Dw = 5 ps km-1 nm-1 and Dm =  ps km-1 nm-1.
Laser diode spectral width (FWHM) 1/2 = 3 nm
Material dispersion 1/2/L = |Dm|1/2 = (15 ps km-1 nm-1)(3 nm)
= 45 ps km-1
Waveguide dispersion 1/2/L = |Dw|1/2 = (5 ps km-1 nm-1)(3 nm)
= 5 ps km-1
Chromatic dispersion, 1/2/L = |Dch|1/2 = (5 ps km-1 nm-1 + 15 ps km-1 nm-1)(3 nm)
= 30 ps km-1
(f) Maximum bit-rate would be
11 December 2012
0.59L 0.59 0.59

BL    = 20 Gb s-1 km
 1/ 2 (1/ 2 / L) (30 10 12 s km 1 )
i.e. BL  20 Mb s-1 km (only an estimate)

(g) To find the maximum diameter for SM operation solve,
2a
( n12  n22 )1/ 2  2(a μm)(1.4510  1.4474 )
2 2 1/ 2
V  2.405
 (1.55 μm)
 2a = 11.5 m.
(h) The mode filed diameter 2w is
2w  2a(0.65  1.619V 3/ 2  2.879V 6 ) = 12.2 m
2.14 Normalized propagation constant b Consider a weakly guiding step index fiber in which (n1 
n2) / n1 is very small. Show that
(  / k) 2  n 2 (  / k)  n
b 2
 2
n2  n2 n n
1 2 1 2
Note: Since  is very small, n2/ n1  1 can be assumed were convenient. The first equation can be
rearranged as
2 2 2 1/ 2 2 1/ 2 2 2 2
 / k  [n2  b(n1  n2 )]  n2 (1 x) ; x  b(n1  n2 ) / n2
where x is small. Taylor's expansion in x to the first linear term would then provide a linear relationship
between  and b.
Solution
1 1
  [n  b(n  n22 )] 2  n2 (1  x) 2
2 2
k 2 1
 n2 
where x  b 1 1
2
 n2 
Taylor expansion around x  0 and truncating the expression, keeping only the linear term yields,
 nx x   b  n12  b

  
k  n2  2  n2 1  2   n2 1  2 n 2  1   n2  n 2 n 2
    2    2 
2 1 
 n 2 
n1
then using the assumption  1 we get
n2
11 December 2012
  n  b(n  n )
2 1 2
k
and
11 December 2012
( / k ) n2
b
n1  n2
as required.
2.15 Group velocity of the fundamental mode Reconsider Example 2.3.4, which has a single mode
fiber with core and cladding indices of 1.4480 and 1.4400, core radius of 3 m, operating at 1.5 m. Use
the equation
( / k ) n2
b ;  = n k[1
2 + b]
n1  n2
to recalculate the propagation constant . Change the operating wavelength to  by a small amount, say
0.01%, and then recalculate the new propagation constant . Then determine the group velocity vg of
the fundamental mode at 1.5 m, and the group delay g over 1 km of fiber. How do your results
compare with the findings in Example 2.3.4?
Solution
From example 2.3.4, we have
2c
b  0.3860859 , k  4188790 m 1 ,    1.2566371015 s 1

 (1.4480 1.4400) 
  n k[1 b]  (1.4400)(4188790m 1 ) 1  (0.3860859)
2
1.4480
  6044795 m 1
   1.5μm(1  1.001)  1.5015μm , b  0.3854382 , k  4184606 m 1 ,   1.255382 1015 s1

 (1.4480 1.4400) 
   n k [1 b]  (1.4400)(418406m 1 ) 1  (0.3854382)
2
1.4800
   6038736m 1
Group Velocity
  (1.255382 1.256637) 10 15s 1

g   1  2.071310 m s
8 1
    (6.038736  6.044795)10 m
6
 g  4.83μs over 1 km.

Comparing to Example 2.3.4
2.0713  2.0706
%diff  100%  0.03%
2.0706
2.16 A single mode fiber design The Sellmeier dispersion equation provides n vs.  for pure SiO2
and SiO2-13.5 mol.%GeO2 in Table1.2 in Ch. 1. The refractive index increases linearly with the addition
11 December 2012
of GeO2 to SiO2 from 0 to 13.5 mol.%. A single mode step index fiber is required to have the following
properties: NA = 0.10, core diameter of 9 m, and a cladding of pure silica, and operate at 1.3 m. What
should the core composition be?
Solution
The Sellmeier equation is
A 2 A 2 A 2
n 1
2 1
 2
 3
2  12 2  22 2  32
From Table1.2 in Ch.1. Sellmeier coefficients as as follows

Sellmeier A1 A2 A3 1 2 3
m m m
SiO2 (fused silica) 0.696749 0.408218 0.890815 0.0690660 0.115662 9.900559
86.5%SiO2‐13.5%GeO2 0.711040 0.451885 0.704048 0.0642700 0.129408 9.425478
Therefore, for  = 1.3 m pure silica has n(0) = 1.4473 and SiO2-13.5 mol.%GeO2 has n(13.5)= 1.4682.
Confirming that for NA=0.10 we have a single mode fiber
2a 2(4.5 μm)
n2  NA  (0.1) = 2.175
 (1.3 μm)
Apply NA  n 2  n 2 
1/ 2
to obtain n  NA2  n 2  
1/ 2
=(0.12+1.44732)1/2 = 1.4508
1 2 1 2
The refractive index n(x) of SiO2-x mol.%GeO2, assuming a linear relationship, can be written as
 x  x
n(x)  n(0)1    n(13.5)
 13.5  13.5
Substituting n(x) = n1 = 1.4508 gives x = 2.26.
2.17 Material dispersion If Ng1 is the group refractive index of the core material of a step fiber, then
the propagation time (group delay time) of the fundamental mode is
  L / v g  LN g1 / c
Since Ng will depend on the wavelength, show that the material dispersion coefficient Dm is given
approximately by
d  d 2 n
Dm  
Ld c d2
Using the Sellmeier equation and the constants in Table 1.2 in Ch. 1, evaluate the material dispersion
at= 1.55 m for pure silica (SiO2) and SiO2-13.5%GeO2 glass.
Solution
From Ch. 1 we know that
11 December 2012
dn
Ng  n  
d
Differentiate  with respect to wavelength  using the above relationship between Ng and n.
L LN g1
 
vg c
d L dN g1 L dn d 2 n dn   L d 2 n
       
d c d c  d d2 d  c d2
d  d 2n
Thus, Dm   (1)
Ld c d2
From Ch. 1 we know that the Sellmeier equation is

A 2 A 2 A 2
n 1 
2 1
 2
 3
3
(2)
 
2 2
1
 
2 2
 
2 2
The Sellmeier coefficients for SiO2-13.5%GeO2.

The 1, 2, 3 are in m.
A1 A2 A3 1 2 3
SiO2‐13.5%GeO2 0.711040 0.451885 0.704048 0.0642700 0.129408 9.425478
We can use the Sellmeier coefficient in Table1.2 in Ch.1 to find n vs. , dn/d and d2n/d, and, from
Eq. (1), Dm vs as in Figure 2Q17-1. At  = 1.55 m, Dm =14 ps km-1 nm-1
Figure 2Q17-1 Materials dispersion Dm vs. wavelength (LiveMath used). (Other math programs such as
Matlab can also be used.)
11 December 2012
2.18 Waveguide dispersion Waveguide dispersion arises as a result of the dependence of the
propagation constant on the V-number, which depends on the wavelength. It is present even when the
refractive index is constant; no material dispersion. Let us suppose that n1 and n2 are wavelength (or k)
independent. Suppose that  is the propagation constant of mode lm and k = 2π/in which is the free
space wavelength. Then the normalized propagation constant b and propagation constant are related by
 = n2k[1 + b] (1)
The group velocity is defined and given by
d dk
vg  c
d d
Show that the propagation time, or the group delay time,  of the mode is
L Ln2 Ln2  d (kb)
   (2)
vg c c dk
Given the definition of V,
V  ka[n12  n22 ]1/ 2  kan2 (2)1/ 2 (3)

and
d (Vb)

d
bkan (2)   an (2)
1/ 2 1/ 2 d
(bk) (4)
2 2
dV dV dV
Show that
d Ln2  d 2 (Vb)
 V (5)
d c dV 2
and that the waveguide dispersion coefficient is

d n2 d 2 (Vb)
D   V (6)
Ld c
w 2
dV
Figure 2.53 shows the dependence of V[d2(Vb)/dV2] on the V-number. In the range 1.5 < V < 2.4,
d 2 (Vb) 1.984
V  2
dV 2 V
Show that,
n2  1.984 (n1 n2 ) 1.984

D   (7)
c c
w
V2 V2
which simplifies to
11 December 2012

1.984
Dw   (8)
c(2a) 2 2n
2
11 December 2012
83.76
i.e. D (ps nm 1 km 1 )   Waveguide dispersion coefficient (9)
 (μm)
w
[a(μm)] 2 n
2
Consider a fiber with a core of diameter of 8 m and refractive index of 1.468 and a cladding refractive
index of 1.464, both refractive indices at 1300 nm. Suppose that a 1.3 m laser diode with a spectral
linewidth of 2 nm is used to provide the input light pulses. Estimate the waveguide dispersion per
kilometer of fiber using Eqs. (6) and (8).
1.5
V[d2(Vb)/dV2]
0.5
0
0 1 2 3
V-number
Figure 2.53 d2(Vb)/dV2 vs V-number for a step index fiber. (Data extracted from W. A. Gambling et al. The Radio and
Electronics Engineer, 51, 313, 1981.)
Solution
Waveguide dispersion arises as a result of the dependence of the propagation constant on the V-number
which depends on the wavelength. It is present even when the refractive index is constant; no material
dispersion. Let us suppose that n1 and n2 are wavelength (or k) independent. Suppose that  is the
propagation constant of mode lm and k = 2/where is the free space wavelength. Then the
normalized propagation constant b is defined as,
(  / k) 2  n22
b (1)
n2  n2
1 2
Show that for small normalized index difference  = (n1  n2)/n1, Eq. (1) approximates to
( / k ) n2
b (2)
n1  n2
which gives  as,

 = n2k[1 + b] (3)
The group velocity is defined and given by
d dk
vg  c
d d
Thus, the propagation time  of the mode is
11 December 2012
L d
L Ln Ln  d (kb)
  2  2
c dk
    (4)
vg
c  dk  c
where we assumed   constant (does not depend on the wavelength). Given the definition of V,
V  ka[n 2  n 2 ]1/ 2  ka[(n  n )(n  n )]1/ 2
1 2 1 2 1 2
1/ 2
  n  n 

 ka (n1  n2 )n1  1 2
  (5)

  n1 
 ka[2n2 n1]1/ 2  kan2 (2)1/ 2
From Eq. (5),
d (Vb)

d
bkan (2)   an (2)
2
1/ 2
2
1/ 2 d
(bk)
dV dV dV
This means that  depends on V as,

Ln2 Ln2  d (Vb)
  (6)
c c dV
Dispersion, that is, spread  in due to a spread  can be found by differentiating Eq. (6) to obtain,
d Ln  dV d d (Vb) Ln   V  d 2 (Vb)
 2
 2
 
d c d dV dV c    dV
2
(7)
Ln2  d (Vb) 2
 V
c dV 2
The waveguide dispersion coefficient is defined as

d n2  d 2 (Vb)
D   V (8)
Ld c
w 2
dV
Figure 2.53 shows the dependence of V[d2(Vb)/dV2] on the V-number.

In the range 2 < V < 2.4,
d 2 (Vb) 1.984
V 
dV 2 V2
so that Eq. (8) becomes,
n2  1.984 (n1 n2 ) 1.984

D   (9)
c c
w
V2 V2
11 December 2012
We can simplify this further by using


1/ 2
n2  1.984 1.984n2    
D  
 2an 2 (2)1/ 2 
c c
w
V2 
11 December 2012

1.984
 Dw   (10)
c(2a) 2 2n
2
Equation (6) should really have Ng2 instead of n2 in which case Eq. (10) would be
1.984 N g 2
Dw    (11)
c(2a) 2 2n22
Consider a fiber with a core of diameter of 8 m and refractive index of 1.468 and a cladding
refractive index of 1.464 both refractive indices at 1300 nm. Suppose that a1.3 m laser diode with a
spectral linewidth of 2 nm is used to provide the input light pulses. Estimate the waveguide dispersion
per kilometer of fiber using Eqs. (8) and (11).
2a 2(4 μm)(1.468 2  1.464 2) 1/ 2
V

n 2
1  n22 
1/ 2

(1.3 μm)
= 2.094
and  = (n1  n2)/n1 = (1.4681.464)/1.468 = 0.00273.

From the graph, Vd2(Vb)/dV2 = 0.45,
n2  d 2 (Vb) (1.464)(2.73 10 3 )
Dw   V   (0.45)
c dV 2 (3108 m s -1 )(1300 10 9 m)
 Dw  4.610-6 s m-2 or 4.6 ps km-1 nm-1

Using Eq. (10)
1.984 1.984(1300 10 9 m)
Dw     
c(2a) 2 2n 2 (3 108 m s -1 )[2  4 10 6 m]2 2(1.464)]
 Dw  4.610-6 s m-2 or 4.6 ps km-1 nm-1

For 1/2 = 2 nm we have,
 1/2 = |Dw|L1/2 = (4.6 ps km-1 nm-1)(2 nm) = 9.2 ps/km
2.19 Profile dispersion Total dispersion in a single mode, step index fiber is primarily due to
material dispersion and waveguide dispersion. However, there is an additional dispersion mechanism
called profile dispersion that arises from the propagation constant of the fundamental mode also
depending on the refractive index difference . Consider a light source with a range of wavelengths 
coupled into a step index fiber. We can view this as a change in the input wavelength. Suppose that
n1, n 2, hence depends on the wavelength . The propagation time, or the group delay time, g per unit
length is
 g  1/ v g  d / d  (1/ c)(d / dk) (1)
where k is the free space propagation constant (2/), and we used dcdk. Since  depends on n1, 
and V, consider g as a function of n1,  (thus n2), and V. A change in will change each of these
quantities. Using the partial differential chain rule,
11 December 2012
g 
g n1 g V  g 
   (2)
 n1  V   
The mathematics turns out to be complicated but the statement in Eq. (2) is equivalent to
Total dispersion = Material dispersion (due to ∂n1/∂)
+ Waveguide dispersion (due to ∂V/∂)
+ Profile dispersion (due to ∂/∂)
in which the last term is due to  depending on; although small, this is not zero. Even the statement in
Eq. (2) above is over simplified but nonetheless provides an insight into the problem. The total
intramode (chromatic) dispersion coefficient Dch is then given by
Dch = Dm + Dw + Dp (3)
in which Dm, Dw, Dp are material, waveguide, and profile dispersion coefficients respectively. The
waveguide dispersion is given by Eq. (8) and (9) in Question 2.18, and the profile dispersion coefficient
is (very) approximately1,
N g1  d 2 (Vb)  d 
Dp   V 2
  (4)
c dV d
  
in which b is the normalized propagation constant and Vd2(Vb)/dV2 vs. V is shown in Figure 2.53,we can
also use Vd2(Vb)/dV2  1.984/V2.
Consider a fiber with a core of diameter of 8 m. The refractive and group indices of the core
and cladding at  = 1.55 m are n1 = 1.4500, n 2 = 1.4444, Ng1 = 1.4680, Ng 2 = 1.4628, and d/d = 232
m-1. Estimate the waveguide and profile dispersion per km of fiber per nm of input light linewidth at this
wavelength. (Note: The values given are approximate and for a fiber with silica cladding and 3.6%
germania-doped core.)
Solution
Total dispersion in a single mode step index fiber is primarily due to material dispersion and waveguide
dispersion. However, there is an additional dispersion mechanism called profile dispersion that arises
from the propagation constant of the fundamental mode also depending on the refractive index
difference . Consider a light source with a range of wavelengths  coupled into a step index fiber. We
can view this as a change in the input wavelength. Suppose that n1, n 2, hence depends on the
wavelength . The propagation time, or the group delay time, g per unit length is
1 1  d 
 g     (1)
vg c  dk 
Since  depends on n1,  and V, let us consider g as a function of n1,  (thus n2) and V. A change
in will change each of these quantities. Using the partial differential chain rule,
1
J. Gowar, Optical Communication Systems, 2nd Edition (Prentice Hall, 1993). Ch. 8 has the derivation of this equation..
11 December 2012
g 
g n1 gV g 
   (2)
 n1  V   
The mathematics turns out to be complicated but the statement in Eq. (2) is equivalent to
Total dispersion = Materials dispersion (due to n1/)
+ Waveguide dispersion (due to V/)
+ Profile dispersion (due to /)
where the last term is due  depending on; although small this is not zero. Even the above statement in
Eq. (2) is over simplified but nonetheless provides an sight into the problem. The total intramode
(chromatic) dispersion coefficient Dch is then given by
Dch = Dm + Dw + Dp (3)
where Dm, Dw, Dp are material, waveguide and profile dispersion coefficients respectively. The
waveguide dispersion is given by Eq. (8) in Question 2.6 and the profile dispersion coefficient away is
(very) approximately,
N g1  d2 (Vb)  d 
Dp   V 2
  (4)
c dV d
   
where b is the normalized propagation constant and Vd2(Vb)/dV2 vs. V is shown in Figure 2.53. The
term Vd2(Vb)/dV2  1.984/V2.
Consider a fiber with a core of diameter of 8 m. The refractive and group indexes of the core and
cladding at  = 1.55 m are n1 = 1.4504, n 2 = 1.4450, Ng1 = 1.4676, Ng 2 = 1.4625. d/d = 161 m-1.
2a 2(4 μm)(1.4504 2  1.4450 2) 1/ 2
V

n 2
1  n22 
1/ 2

(1.55 μm)
= 2.03
and  = (n1  n2)/n1 = (1.4504-1.4450)/1.4504 = 0.00372

From the graph in Figure 2.53, when V = 2.03, Vd2(Vb)/dV2  0.50,
Profile dispersion:

N  d 2 Vb  d  
Dp    g1
V
( ) 
   1.4676
0.50 161 m 1 
c  dV 2  d  3 108 m s 1
 Dp = 3.8 10-7 s m-1 m-1 or 0.38 ps km-1 nm-1

Waveguide dispersion:
1.984 1.984(1500 10 9 m)
Dw     
c(2a) 2 2n 2 (3 10 8 m s -1 )[2  4 10 6 m]2 2(1.4450)]
 Dw   5.6 ps km-1 nm-1

11 December 2012
 than
Profile dispersion is more 10    than
times smaller waveguide
 dispersion.
11 December 2012
2.20 Dispersion at zero  dispersion

  coefficient
   the spread in the group delay  along a fiber
Since
depends on the , the linewidth of the sourcewe can expand  as a Taylor series in . Consider the
expansion at  = 0 where Dch = 0. The first term with  would have d /d as a coefficientthat is
Dch, and at 0 this will be zero; but not the second term with ( that has a differential, d2/d or
dDch/d. Thus, the dispersion at 0 would be controlled by the slope S0 of Dch vs.  curve at 0. Show
that the chromatic dispersion at 0 is
L
  S 0 ( )
2
2
A single mode fiber has a zero-dispersion at 0 = 1310 nm, dispersion slope S0 = 0.090 ps nm2 km. What
is the dispersion for a laser with  = 1.5 nm? What would control the dispersion?
Solution
Consider the Taylor expansion for , a function of wavelength, about its center around, say at 0, when
we change the wavelength by For convenience we can the absolute value of  at 0 as zero since we
are only interested in the spread . Then, Taylor's expansion gives,
d 1 d 2 
  f ( )  ( )  ( ) 2  
d 2! d2

d 2 d  d  d
1   
 D
1 1
    0  ( ) 2  0   ( ) 
2
( ) 2
2! d2 2! dt  d 
ch
2! dt


L
  S 0 ( ) 
1 km 2 -2 -1

0.090 ps nm km 2 nm = 1.01 ps
2

2 2
This can be further reduced by using a narrower laser line width since depends on (
2.21 Polarization mode dispersion (PMD) A fiber manufacturer specifies a maximum value of 0.05
ps km-1/2 for the polarization mode dispersion (PMD) in its single mode fiber. What would be the
dispersion, maximum bit rate and the optical bandwidth for this fiber over an optical link that is 200 km
long if the only dispersion mechanism was PMD?
Solution
Dispersion
  DPMD L1/ 2  0.05ps km 1/ 2 (200km)1/ 2  0.707 ps
Bit rate
0.59 0.59
B   8.35 Gb s -1
 0.707ps
Optical bandwidth
f op  0.75B  (0.75)(8.35 Gb s 1 )  6.26 GHz
11 December 2012
2.22 Polarization mode  dispersion

    aparticular
Consider   single mode fiber (ITU-T G.652
compliant) that has a chromatic dispersion of 15 ps nm-1 km-1. The chromatic dispersion is zero at 1315
nm, and the dispersion slope is 0.092 ps nm-2 km-1. The PMD coefficient is 0.05 ps km-1/2. Calculate the
total dispersion over 100 km if the fiber is operated at 1315 nm and the source is a laser diode with a
linewidth (FWHM)  = 1 nm. What should be the linewidth of the laser source so that over 100 km,
the chromatic dispersion is the same as PMD?
Solution
Polarization mode dispersion for L = 100 km is PMD  DPMD L1/ 2 = 0.05  100 ps = 0.5 ps
We need the chromatic dispersion at 0, where the chromatic dispersion Dch = 0. For L = 100 km, the
chromatic dispersion is
L
  S ( )2 = 1000.092(1) /2 = 4.60 ps
2
ch 0
2
The rms dispersion is
2
2
 rms   PMD   ch = 4.63 ps
The condition for  PMD   ch is

2DPMD
  = 0.33 nm
S0 L1 / 2
2.23 Dispersion compensation Calculate the total dispersion and the overall net dispersion
coefficient when a 900 km transmission fiber with Dch = +15 ps nm-1 km-1 is spliced to a compensating
fiber that is 100 km long and has Dch = 110 ps nm-1 km-1. What is the overall effective dispersion
coefficient of this combined fiber system? Assume that the input light spectral width is 1 nm.
Solution
Using Eq. (2.6.1) with  = 1 nm, we can find the total dispersion
 = (D1L1 + D2L2)
= [(+15 ps nm-1 km-1)(900 km) + (110 ps nm-1 km-1)(100 km)](1 nm)
= 2,500 ps nm-1 for 1000 km.
The net or effective dispersion coefficient can be found from  = DnetL,
Dnet =  /(L = (2,500 ps)/[(1000 km)(1 nm)] = 2.5 ps nm-1 km-1
2.24 Cladding diameter A comparison of two step index fibers, one SMF and the other MMF shows
that the SMF has a core diameter of 9 m but a cladding diameter of 125 m, while the MMF has a core
diameter of 100 m but a cladding diameter that is the same 125 m. Discuss why the manufacturer has
chosen those values.
Solution
For the single mode fiber, the small core diameter is to ensure that the V-number is below the cutoff value
for singe mode operation for the commonly used wavelengths 1.1 m and 1.5 m. The larger total
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Preliminary Solutions to Problems and Question

2.3 Dielectric slab waveguide

   = k1AC[1  cos(2)] k1AC[1 + cos(2)] 

or E  2A cos12  m ( y)cos(t  ) = Eocos(t + )

= 3.610-15 s m-1 = 0.0036 ps m-1.

Therefore,  = 870 nm is a single mode operation.

The penetration depth 

1 65.2 58.15 51.6 45.5 35.5 32.02 30.17 ‐ ‐ ‐ ‐

1 65.2 58.15 51.6 45.5 35.5 32.02 30.17 ‐ ‐ ‐ ‐

Solve the waveguide condition

to calculate the penetration depth:

The fiber becomes monomode when,

If max is the maximum acceptance angle, then,

intermode n1 n 2 1.475 1.455

= 66.7 ps m-1 or 67.6 ns per km

i.e. BL = 13 Mb s-1 km (only an estimate)

NA= (n12n22)1/2 = (1.33021.0002)1/2 = 0.8814

(d) If max is the maximum acceptance angle, then,

so that the total acceptance angle is 12.4.

= |7.55 ps km-1 nm-1|(2 nm) = 15 ps km-1 + 10 ps km-1

(b) For multimode operation we need

(d) If max is the maximum acceptance angle, then,

Total acceptance angle 2amax is 11.8 .

0.59L 0.59 0.59

i.e. BL  20 Mb s-1 km (only an estimate)

 nx x   b  n12  b

   1.5μm(1  1.001)  1.5015μm , b  0.3854382 , k  4184606 m 1 ,   1.255382 1015 s1

  (1.255382 1.256637) 10 15s 1

 g  4.83μs over 1 km.

From Table1.2 in Ch.1. Sellmeier coefficients as as follows

Substituting n(x) = n1 = 1.4508 gives x = 2.26.

d c d c  d d2 d  c d2

From Ch. 1 we know that the Sellmeier equation is

The Sellmeier coefficients for SiO2-13.5%GeO2.

V  ka[n12  n22 ]1/ 2  kan2 (2)1/ 2 (3)

and that the waveguide dispersion coefficient is

n2  1.984 (n1 n2 ) 1.984

which gives  as,

This means that  depends on V as,

The waveguide dispersion coefficient is defined as

Figure 2.53 shows the dependence of V[d2(Vb)/dV2] on the V-number.

n2  1.984 (n1 n2 ) 1.984

We can simplify this further by using

and  = (n1  n2)/n1 = (1.4681.464)/1.468 = 0.00273.

 Dw  4.610-6 s m-2 or 4.6 ps km-1 nm-1