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Solutions Manual to
R.P. KhareBirla Institute of Technology and Science
Pilani (Rajasthan)
Oxford University Press 2004. All rights reserved. Not for sale or further circulation.
Photocopying prohibited. For restricted use by faculty using Fiber Optics and Optoelectronics
as a text in the classroom.
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Preface
This solutions manual has been written to aid the instructors in teaching this
course. In the busy schedule it becomes difficult to find time for attempting newer
problems and solving them. The author, therefore, felt that a readymade solutions manual
should be made available to the instructors who are prescribing Fiber Optics and
Optoelectronicsas a textbook for their students. In fact a number of solved examples are
already given in the textbook and the key to multiple choice questions is also given at the
end of each set of such questions. However, the numerical review questions need
elaborate solutions. These are presented in this manual. I hope the reader would now
enjoy solving these problems.
The author will welcome suggestions from instructors for improvement in the
textbook or its presentation. They may mail to [email protected] .
R P Khare
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Chapter 2:Ray Propagation in Optical Fibers
2.3 The parameter that varies with the surrounding medium is acceptance angle mofthe fiber.
NA = nasin m
Since NA is a constant for a SI fiber, change in nawill change the value of
m. In the present case, NA = 0.244 (see Example 2.1), and na= 1.33, we have
nasin m= 1.33 Sin m= 0.244.
This gives m= 10.57o.
Other parameters do not depend on naand hence they will not be affected.
2.4 NA = nasin m = 1 sin 20o0.34
However, NA is also given by: NA = 2n1
Therefore, n1=03.02
34.0
= 1.388
Now nasin m= n1sin m= n1cos c= 0.34
This gives cos c388.1
34.0= 0.2449
and c76o
2.6 (a) NA = 0.17 = nasin m= 1.33 sin m
Therefore m= 7.34o
(b) Since NA = ( )21
2
2
2
1 nn = 0.17 and n2= 1.46
n1= ( ) ( )[ ] 47.146.117.0 2122 +
Now nasin m= n1cosc = 1.47cosc=0.17;which gives c= 83.35o
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Chapter 3: Wave propagation in planar waveguides
3.8: (a) V = ( ) ( )( )
( ) ( )[ ]21
222
12
2
2
1 455.146.1m30.1
m14nn
a2
=
V 4
Referring to Fig. 3.4 of the textbook, the arc of circle of radius V = 4,intersects three ua versus wa curves corresponding to m = 0, 1 and 2.
Therefore the guide supports these three modes. (The same information
can also be gathered by calculatingV2
and recalling that M is an integer
greater than
V2
.
(b)The abscissae of the intersecting points of ua versus wa curves (seeFig. 3.4 of text book) for m = 0, 1 and 2 with the quadrant of circle of radius
V = 4 give the values of uma and the corresponding ordinates give the value of
wm a. bm and m can also be calculated from these parameters. All theseparameters for different values of m are listed in the table below.
m uma (rad) um(m1
) wma wm(m1
)2
mm
V
awb
=
0 1.25235 1.7890105 3.79889 5.4269105 0.90197
1 2.47458 3.5351105 3.14269 4.4895105 0.61109
2 3.59531 5.1361105 1.75322 2.5046105 0.19211
The phase propagation constants m for different modes, m, can be calculatedusing the following relation:
( ){ }21
2
2
2
1m
2
2m b +=
where 1 = kn1= 1661 m100565.7
1030.1
46.12
n2 =
=
2= kn2= 1662 m100323.7
1030.1
455.12
n2 =
=
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3.9 The maximum thickness of the guide layer that can support M modes is given by
2a =
( )21
2
2
2
1 nn2
M
In the present case, M = 10, = 0.90 m, n1= 3.6 and n2= 3.58.
Therefore, 2a =( )
( ) ( )[ ]m875.11
58.36.32
m90.010
2
122
=
max= 1= kn1= 1561 m10513.2
1090.0
6.32
n2 =
=
min= 2= kn2=15
6
2
10499.21090.0
58.32
2
=
= mn
3.10 (a) V = ( ) ( )( )
( )02.025.1.m55.1
m102n
a21
=
V 6.
From Fig. 3.4, it can be easily seen that for V = 6, the guide will
support 4 modes: 2 symmetric modes corresponding to m = 0 and 2; and, 2
antisymmetric modes corresponding to m = 1 and 3.
(b) From Fig. 3.4 of textbook, the intersections of ua versus wa curves for
different values of m with the circle of radius V = 6 gives the values of
uma and aw m . The results are listed in the tabular form below.
m uma
(rad)um(m
1) wma(rad) wm(m
1)
2
m
mV
awb
=
0 1.34475 2.6895105 5.84736 1.169472106 0.94976
1 2.67878 5.35756105
5.36881 1.073762106
0.80067
2 3.98583 7.97166105 4.48477 8.96954105 0.558698
3 5.22596 1.045192106 2.94776 5.89552105 0.241369
( ){ }21
2
2
2
1m
2
2m b +=
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where 1 = kn1= 1661 m1008.6
1055.1
5.12
n2 =
=
2= kn2=16
6
2
m109588.51055.1
47.12
n2 =
=
This gives,
m= [1.459 bm+ 35.507] 106m1
3.11 For antisymmetric modes,
Ey(x) = B sin ux; | x | < a
= ( ) a|x|;|x|wexpD|x|
x
>
Power inside the guide layer,
( )
=a
a
2
0
in dxuxsinB2
1P
= ( ) a
0
2
0
dxux2cos1B2
=
a
0
2
0
ux2sinu2
1xB
2
=
ua2sinu2
1aB
2
2
0
(1)
Power outside the guide layer,
( ) ( )
+
=
a
a
2
y
2
y
0
out dxxEdxxE2
1P
=
a
2
y
0
dx)x(E22
1
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= ( )dxwx2expDa
2
0
=
a
wx22
0
e
w2
1D
=
wa22
0
ew2
1D (2)
The confinement factor
G =outin
in
PP
P
+(3)
Substituting the values of Pinand Poutfrom (1) and (2) in (3), we get
+
= wa22
0
2
0
2
0
e.w
1D.
2
1ua2sin.
u2
1aB.
2
1
ua2sin.u2
1aB.
2
1
G
= 2wa2
B
De.
w
1ua2sin
u2
1a
ua2sin.2
1a
+
(4)
At x = a, B sin ua = D ewa
Therefore, uasin.eB
D wa= (5)
Substituting for
B
Dfrom (5) in (4) and manipulating, we get,
G =
+
uasin.w
1ua2sin
u2
1a
ua2sin.u2
1a
2
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This result can also be expressed in the following form:
G =( )
1
2
ua
uacos.uasin1wa
uasin1
+ (6)
(b) Given that 2a = 6.523 m, n1= 1.50, n2= 1.48 and = 1.0m
Therefore, V = ( )21
2
2
2
1 nna2
= ( ) ( )[ ]21
2248.15.1
1
523.6
5
So the guide supports 4 modes corresponding to m = 0, 1, 2 and 3.
m = 0, 2 are symmetric modes
m = 1, 3 are antisymmetric modes.
From Fig. 3.4, the values of ua and wa may be written as follows.
m ua(rad) wa(rad)
0 1.30644 4.8263
1 2.59574 4.27342
2 3.83747 3.20529
3 4.9063 0.963466
To calculate G-factors for symmetric modes, use formula (3.81) of textbook; and for antisymmetric modes use formula (6) derived in part (a) of thisquestion.
The respective calculations are given as follows:
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Gm = 0=
1
2
30644.1
)30644.1sin().30644.1cos(18263.4
)30644.1(cos1
+
+
= 0.98828
Gm = 1=
1
2
5974.2
)5974.2sin().5974.2cos(127342.4
)5974.2(sin1
+
= 0.94889
Gm = 2 =
1
2
83747.3
)83747.3sin().83747.3cos(120529.3
)83747.3(cos1
+
+
= 0.85992
Gm = 3 =
1
2
9063.4
)9063.4sin().9063.4cos(1963466.0
)9063.4(sin1
+
= 0.50960
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Chapter 4: Wave Propagation in Cylindrical Waveguides
4.2 (b) (i)2
1
2
2
2
1
n2
nn =
Therefore n2= n1(1 2)= 1.46 (1 2 0.01)= 1.445
(ii) 25)01.02(46.1)m(3.1
)m(502n
a2V 1
=
=
(iii) Ms=( )
3122
25
2
V22
==
4.3 (b) 16611max
m1093.10)m(1085.0
48.12n
2kn =
=
==
16
622minm1082.10
)m(1085.0
465.12n
2kn =
=
==
4.4 The cut off frequency for single mode operation is given by
Vc= 2.4052
1
21
+
This should be equal to V given by
V =
2na2
0
Therefore, for this case the following equality should be satisfied
2na2
0 = 2.4052
1
21
+
or 2a =
+
2n
21405.2
0
2
1
In the present problem, = 1, n0= 1.5 and
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Hence (i) for = 1.3 m,
( )( )
( )m1.7
013.025.1
21m3.1405.2a2
2
1
=
+=
and (ii) for = 1.55 m,
( )( )
( )m5.8
013.025.1
21m55.1405.2a2
2
1
=
+=
4.6 ( )
+
+
==3
2
cg
effg
kRn2
3
R
a2
2
21
2
1
M
)M((1)
In the present problem = 2, m,
nc= 1.48 and k =1)m(39.7
)m(85.0
22 =
=
.
Substituting the valves of different parameters in eq. (1),
we get
+=
3
2
R
1371.0
R
10010012
1 ;
where R is measured in m.
Solving this equation, we get
R 1.66 104m
or R 1.66 cm
4.7 The complete transverse electric field will be given by
Ez= A Jl
a
ur.e
il.ei(tz); r < a (1)
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and Ez= B Kl
a
wr.e
il.ei(tz); r > a (2)
where A and B are arbitrary constants.
The solutions for the transverse magnetic field are similar but the constants willbe different. Thus
Hz= C Jl
a
ur.e
il.ei(tz); r < a (3)
and Hz= D Kl
a
wr.e
il.ei(tz); r > a (4)
Let us derive the expression for Erfirst inside the core and then in the cladding. It
is given that
+
= zz2
r
r
H.
r
1
r
E
k
iE (5)
Substituting the values of Ezand Hzfrom (1) and (3) in (5), we get
+
= )zt(ilil)zt(iil
l2
r
r e.a
urCJ.e.
r
1ee.
a
urJ
rAB
k
iE
=
+
)zt(iill
)zt(iil
l2
r
e.ea
urJ.
r
iCee.
a
urJ.
a
uA
k
i
or )zt(iilll2r
r eea
urJ
r
iC
a
urJA
a
u
k
iE
+
= (6)
where ( ) [ ]2r0r02222r ..k ==
Inside the dielectric medium, such as glass fiber,
r= 1, r= n2and
00= 2C1
.
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Therefore
[ ]22222
222
r nkc
nk =
= (7)
where k ==
2
c= free space propagation constant.
Note that for a SI fiber inside the core, n = n1and in the cladding n = n2.
Similarly, inside the cladding, (r > a).
)zt(iil
ll2
r
r e.ea
wrK
r
iD
a
wrK
a
wB
k
iE
+
= (8)
Expressions for E, Hrand Hcan be derived in a similar manner.
4.8 n(r) = n0 ar;a
r21
2
1
Substituting 2
0
2
c
2
0
n2
nn in the above equation and taking the square on both
sides, we get
=
a
r.
n2
nn21n)r(n
2
0
2
c
2
02
0
2 .
= ( )
a
rnnn 2c
2
0
2
0
Therefore the local numerical aperture, NA(r) would be given by
NA(r) = { }2
1
2c2 n)r(n
= ( )2
1
2
c
2
c
2
0
2
0 na
rnnn
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= ( )2
1
2
c
2
0a
r1nn
The mean square value of the numerical aperture taken over the core area may be
given by the following relation:
Mean square value of NA =
a
0
a
0
2
rdr2
rdr2).r(NA
=
( )
2
a
0
2
c
2
0
a
rdr2a
r1nn
=( )
a
0
22
2
2
c
2
0
)2(
r.
a
2
2
r2
a
nn
+
+
=( )
+
2
a2a
a
nn 222
2
c
2
0
= ( )
+
2
nn 2c2
0 .
Therefore, root mean square value of numerical aperture for an -profile wouldbe given by
(NA)rms= ( )2
1
2
c
2
0 nn.2
+
4.9 (a) (NA)rms= ( )2
1
2
c
2
0 nn.2
+
Given that = 2, n0= 1.460 and nc= 1.445
Therefore (NA)rms= ( ) ( ){ }2
1
22445.146.1
22
2
+
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= 0.1476
(b) 0= kn0= ( )1
30mm705646.1
mm103.1
2n
2 =
=
( ) ( )
( )0102.0
46.12
445.146.1
n2
nn2
22
2
0
2c
20 =
=
V = ( ) ( ) ( )[ ] 23.25445.146.13.1
50nn
a22
122
2
12
c
2
0 =
=
(c) Mg=( )
1592
23.25
22
2
2
V
2
22
=
+=
+
4.10 (a) Given that = 1, Mg= 500, n0= 1.46, 2a = 75 m and = 1.3 m.
Therefore Mg=6
V
2
V.
21
1
2
V.
2
222
=
+=
+
or V = 77.545006M6 g ==
As V can also be expressed as
V
2n
a2
0
, we have
021.046.175
77.543.1
2
1
n)a2(
V
2
122
0
=
=
(b) Vc= 2.405 165.41
21405.2
21
2
1
2
1
=
+=
+
If we assume that the core diameter for single mode operation is a2 , V
can be written as
V = 165.42na2
0 =
or ( )
m76.5021.0246.1
m3.1165.4a2 =
=
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4.11 (a) Given that = 2, 2a = 70m n0= 1.47, nc= 1.45 and = 1.3m. Then
0= kn0= 1660 m101048.7)m(103.147.12
n2
=
=
c= knc= 166c m101008.7)m(103.145.12
n2
=
=
( ) ( )
( )01351.0
47.12
45.147.1
n2
nn2
22
2
0
2
c
2
0 =
=
=
(b) V = 88.4001351.0247.13.1
702n
a20 =
=
Mg=( )
4184
88.40
4
V22
== .
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Chapter 5: Single Mode Fibers
5.2 (b) Given that
n1= 1.46, 2a = 8m,
In general,
V =
2na2
1
For single mode operation, in a SI fiber
V = Vc= 2.405,
Therefore,
( )
405.2
0052.0246.1m82n
V
a21
c
c
=
=
or c= 1.556m
5.4 V =
2na2
1
At
1= 1.31
m, V
1= 449.20036.024677.1
31.1
1.42
=
Therefore, using eq. (5.3) of the text book we get,
w =
( ) ( )
++
6
2
3449.2
879.2
449.2
619.165.01.4
= 4.1 [0.65 + 0.4224 + 0.01334]
or w = 4.452 mEmploying eq. (5.5), we get
wp= 4.452 4.1( )
+
7449.2
567.1016.0
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or wp= 4.374m
At 2= 1.55m, V2= 0705.20036.024682.155.1
1.42=
w = 4.1
( ) ( )
++
6
2
30705.2
879.2
0705.2
619.165.0
= 4.1 [0.65 + 0.5434 + 0.0365]
or w = 5.0427 m
wp= 5.0427 4.1
( )
+
7
0705.2
567.1016.0
or wp= 4.9377m
5.5 (a) 1= 1661 m833.410)m(103.12
n.2
=
=
and 2= 1562 m33.4810)m(103.12
n.2
=
=
Range of = 4.833 48.33 m1
(b) Lp1= m30.1)m(833.4
22
1
=
=
Lp2= m13.0)m(33.48
22
2
=
=
Lpvaries from 13 cm to 1.3 m
5.7 Given that n1= 1.48, = 1.32 and a = 4.4 m
n2n1(1
V = 2778.20027.0248.132.1
4.42=
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Using eq. (5.24) of the text book, we have
( )22
2
V834.2549.0080.0)bV(dV
Vd+
= 0.2498
From eq. (5.23), we get
( ) ( )2498.0
m32.1ms103
0027.0476.1D
18w
=
= 2.5138 1012s(m1) (km1)
= 2.5138 ps (nm1) (km1)
5.9 D = Dm+ Dw= 0
Since Dm= 7 ps nm1km1, Dw= 7 ps nm1km1
From eq. (5.23) and (5.24), we get
Dw= ( )[ ]22 V834.2549.0080.0c
n+
n1= 1.48, 2= n1(1 0.01)
or n2= 1.4652
Dw= ( ) ( ) ( )[ ]2
18V834.2549.0080.0
m55.1ms103
01.04652.1+
= 31.509 (ps nm1km1) [0.080 + 0.549 (2.834 V)2]
Thus, in order to get D = 0, we should have
7 = 31.509 [0.080 + 0.549 (2.834 V)2]
This gives V = 2.325
Since V =
2na2
1 , we get
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Core radius, a =( )
01.0248.12
m55.1325.2
2n2
V
1
=
= 2.74 m
5.11
Fig. Q 5.11
Given that,
n1= 1.46, 2a = 8.2 m, 1= 25m
P(r) = P0exp
2
2
w
r2
At 1= 1.30 m,
V1= 241.2003.0246.13.1
2.82n
a21
1
=
=
and at 2= 1.55 m
8.2 m
25 m
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V2= 879.1003.0246.155.1
2.82n
a21
2
=
=
Therefore at 1= 1.3 m, (using eq. 5.3),
w 4.1( ) ( )
++
65.1241.2
879.2
241.2
619.165.0
= 4.1 [0.65 + 0.48259 + 0.0227]
w = 4.7368 m
and at 2= 1.55 m,
w = 4.1( ) ( )
++
65.1879.1
879.2
879.1
619.165.0
= 4.1 [6.65 + 0.6285 + 0.0654] = 5.51 m
At 1= 1.30 m, the fractional power at r = a1= 12.5 m
=
=
22
0 m7368.4
m5.122exp
w
r2exp
P
)r(P
7
0
109387.8)( =
P
rP(1)
and at 2= 1.55m, the fractional power at r = a1= 12.5m
5
2
0
103869.3m51.5
m5.122exp
P
)r(P =
= (2)
Eq. (1) & (2) give the fraction of optical power reaching the inner-outer claddinginterface and that may be lost by transmission into the outer parts of the cladding. In
practice a fraction of this power may be reflected and some may be refracted.
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Chapter 6: Optical fiber cables and connections
6.4 LF= log10(F) = 0.36 dB
This gives F= 0.92 = 4
2
)1k(
k16
+ (from eq. 6.3 of text book)
For air, n = 1, therefore k = 11 n
n
n=
Thus k22.17 k + 1 = 0
or 01n17.2n 12
1 =+
Solving this we get n1= 1.5. The second root of this equation is less than 1, which
is not possible.
6.6 Using eq. (6.11) of the text book,
Lang= 10 log10ang= 0.6 dB (1)
This is due to 4oangular misalignment.
From eq. (1), ang= 0.87
But from eq. (6.8),
87.0NA
n1ang =
In this problem, n = 1.46 and = 4o= 0.0698 rad and hence
NA = 25.013.0
0698.046.1=
6.8 Given that 12.0%12a2
y==
n1= 1.5 and n = 1.47
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Therefore, k = 02.147.1
5.1
n
n1 ==
F= 99957.0663.16
041.116
)102.1(
)02.1(16
)1k(
k164
2
4
2
=
=+
=+
And,
=
2
12
1
lata2
y1
a2
y
a2
ycos
2
or { }
= 2
12
lat )12.0(1)12.0(4505.12
= 0.8475
Hence the total coupling efficiency, Twould be
T= Flat= 0.99957 0.84757 = 0.8472
This gives a total loss at the joint,
LT= 10 log10T0.72 dB.
6.10 The Fresnel reflection coefficient, R is given by
R =
2
1
1
nn
nn
+
Its first derivative w.r.t. n1would be
( ) ( )( )
( )
+
+++
+
+
= nndn
d
nn
1nnnn
dn
d
nn
1
nn
nn2
dn
dR1
1
2
1
11
111
1
1
or( ) ( )( )
+
+
+= 1
2
1
1
111
1
1 dndn1
nnnn
dndn1
nn1
nnnn2
dndR
In this expression n is a constant as it is the refractive index of the
surrounding medium.
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Hence 0dn
dn
1
= . This gives us
( )
( )( ) ( )[ ]
( )
( )31
1
11
1
1
1 nn
nnn4nnnn
nn
nn2
dn
dR
+
=+
+
=
For small variation in n1, 1, will, therefore, produce a small variation given by:
( )
( )13
1
1 n.nn
nnn4
+
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Therefore inj =
+
23
213
10
10
1
1
02199.0
10035.11
1
= 0.9995
7.5 (a) The power radiated by the p-n diode as a function of photon energy,
P(Eph), may be written as
( ) ( ) ( ) 212ph dEEp.EnEP
with the constraint that E2 E1= Eph
or P(Eph) =( ) ( )
2
1vEE
EE
c2 dE
kT
EEexp.
kT
EEexp
phv
c2
+
=
where is a constant.
or P(Eph) =( ) ( )
212
EE
EE
vc dEkT
EEexp.
kT
EEexp
phv
c2
+
=
= exp +
phv
c
EE
E2
gphdE
kT
EE
= ( )
kT
EEexpEE
gph
gph
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Electron
EnergyE2
Ev+Eph
Ec n (E2)
Eg
Ev p(E1)
Ec-Eph
E1
Fig. Q 7.5
Fig. Q.7.5
(b)) ) )
0kT
EE-1
kT
EEexp
dE
EdP gphgph
ph
ph =
= , for maximum P.
Therefore (Eph)peak= Eg+ kT;
and P (at peak Eph) = .e
kT
7.7 (a) Using eq. (7.61) of text book
( )
( )0821.0
7.32
50.1
n2
nF
2
2
2
s
2
a ===
Eph
Free ElectronDensity
Hole Density
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(b) Given that as= 10% = 0.1 t = 1 and inj= 0.40
ext= inj(1 as) (1) F.
= 0.40 (1 0.1) (1) 0.0821
= 0.0295
(c) The fraction of incident radiation collected and propagated by the fiber isgiven by (see eq. 7.113)
( ) ( )
( )01137.0
50.1
16.0
n
NA2
2
2
a
2
m
===
(d) T= extm
= 0.0295 0.01137 = 3.35 104
(e) If the LED is emitting in air,
( ) ( )
( )0256.0
1
16.0
n
NA2
2
2
a
2
m
===
7.8 (a) 4545.0
100
1201
1
1
1
nr
rr
int =+
=
+
=
(b) phintint Ee
I.
=
= 0.4545 ( ) ( )[ ]Je1.42Ae
10100 3
= 0.0645 W = 64.5 mW
Power supplied to LED = VI = 1.5 100 103= 150 103W
= 150 mW.
Therefore internal power efficiency = .43.0150
5.64=
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(c) If the diode is emitting in air, the external quantum efficiency would be
given by
( )( )2ss
sintext1nn
2a1
+=
=( )
( )217.37.3
21.014545.0
+
= 0.010
( )e42.1e
10100010.0E
e
I 3
phextext
=
=
or ext= 1.42 103W = 1.42 mW
Therefore, external power efficiency = 0095.0150
42.1=
(d) T= ext.( ) ( )
2
a
2
1ext2
a
2
n
2n
n
NA =
Given that na= 1.5, NA = 2n1 = 1.46 02.02 =0.292
Using eq. (7.65), we can calculate extfor this part as follows
ext= int(1 as)( )2sas
3
a
nnn
n2
+
=2
3
)7.35.1(7.3
)5.1(2)1.01(4545.0
+
= 0.0275
Therefore, T= 0.0275 ( )
( )3
2
2
10042.15.1
292.0
=
and T= T ( )e42.1e
1010010042.1E
e
I3
3
ph
=
or T= 1.4796 104W = 0.14796 mW.
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Thus overall source-fiber power coupling efficiency
= 41086.9mW150
mW14796.0 =
and optical loss = 10 log10(9.86 104) = 30 dB.
7.9 In the presence of back enrission, the fraction, F, of the total optical power that
can be collected at the semiconductor-air surface is given by
F =2
s
2
a
n2
n(see eq. 7.61 of the text book)
In the absence of back enrission, this factor will be doubled and the new value of
F will be
2
s
2
a
n
nF =
Further, R and t remain the same but as= 0 and hence T = 1.
Therefore ext= int tTF
=
( )
2
sa
sa
2
s
2
a
int
nn
nn4
n
n
+
1
=( )2sas
3
a
intnnn
n4
+
Given that na= 1, ns= 3.7, int= 0.60
( )
029.07.317.3
1460.02ext
=+
=
(a) Optical power emitted within the LED
= int= int phE.e
I
= 0.60 ( )eeV43.1e
A10120 3
= 102.96 103W
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= 102.96 mW
The total power consumed by the device
= 120 103
(A) 1.5 (V)
= 180 103 W
= 180 mW.
Therefore internal power efficiency = 572.0180
96.102=
(b) Optical power emitted in air by the LED
= ext= ext phE.eI
= 0.029 ( )eeV43.1e
A10120 3
= 4.976 103W
= 4.976 mW
Therefore external power efficiency = 0276.0180
976.4 =
7.11 Using eq. (7.91) of text book, we can calculate the cavity length, L as follows
L =( )
( )m1062.1
Hz102507.32
ms103
n2
c 49
18
=
=
or L = 162 m
The number of longitudinal modes can be calculated using eq. (7.90) as follows:
( )
( )m85.0
m1627.32nL2
1nL2.
cm
=
=
=
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or m = 1410
7.12 In terms of wavelength m can be expressed as follows:
( )( )m85.0m5007.32nL2m
=== 4353
m =
nL2.1
2
Therefore mode separation in terms of wavelength will be given by (putting
m = 1),
( )m1095.1
5007.32
85.0
nL2
m 422
=
=
=
or = 0.195 nm
In terms of frequency, m is given by eq. (7.90), i.e.
m =
c2nL; and m =
c
nL2. Therefore the mode separation in terms of
frequency will be given by (with m = 1)
10
6
18
101.8)m(105007.32
ms103
nL2
cm =
==
Hz = 81 GHz.
7.13 The threshold gain coefficient for an ILD is given by (see eq. 7.92 of text book).
gth=
+
21eff
RR
1ln
L2
11
Given that gthJth. Therefore
Jth=
+
21eff
RR
1ln
L2
11(1)
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In the present problem, = 0.02 cm A1, eff= 12 cm1, L = 300 m = 0.03 cm,width = W = 100 m = 0.01 cm
n = 3.7 and hence R1= 33.017.3
17.3
1n
1n22
=
+
=
+
and R2= 1.
Substituting the values of given parameters in (1), for a strong carrier confinement
( )1= , we get
+
=
33.0
1ln
03.02
112
102.0
1J th
1524 A cm2
The threshold current, Ithwill be given by
Ith= JthLW = 1524 (A cm2
) (0.03 cm) (0.01 cm)
= 0.4572 A
7.14 m185.05.32
)m(3.11
2n
mLor.
nL2m =
=
=
=
Chapter 8: Optoelectronic Detectors
8.3 Given that = 50% = 0.5 at = 0.90 m.
(a))ms(103)Js(10626.6
)m(1090.0)C(106.15.0hce
1834
619
==
= 0.362 AW1
(b) Ip= 106
A
Using eq. (8.4) of the text book, we get
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Pin= ( )==
1
6p
AW362.0
A10I2.76 106W = 2.76 W
(c) The rate of photons received will be given by
rp= ( )183466
inin
ms103)Js(10626.6
)m(1090.0)W(1076.2
hc
P
h
P
=
=
= 1.25 1013s1
8.4 Given that = 0.50 at = 1.3 m. Its responsivity would be
523.010310626.6
103.1106.150.0
hc
e834
619
=
=
=
Therefore Ip= . Pin= 0.523 0.4 106= 0.2092 106A
or Ip= 0.2092 A
The output photocurrent after avalanche gain is
I = 8 A.
Thus the multiplication factor M = 38
2092.0
8
I
I
p
= .
8.5 For an ideal p-n photodiode, = 1.
(a) At = 0.85 m
1
834
619
AW684.010310626.6
1085.0106.11
=
=
(b) At = 1.30 m,
1
834
619
AW046.110310626.6
103.1106.11
=
=
(c) At = 1.55 m,
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1
834
619
AW248.110310626.6
1055.1106.11
=
=
8.6 Ip= Pin = 0.40 (AW1
) 100 106
( )2
2 mm2mm
W
or Ip= 80 106A = 80 A.
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Chapter 9: Optoelectronic Modulators
9.2( ) ( )
m10856.0486.1658.14
)m(103.589
nn4x 6
9
e0
=
=
=
= 0.856 m
9.3( ) ( )
m1088.32738544.1553.12
)m(103.589
nn2x 9
9
0e
=
=
=
= 0.0327 mm
9.5 From Table 9.1 of the text book, for lithium niobate (LiNbO3),
n0= 2.29, ne= 2.20, and r33= 30.8 1012m/V.
It is given that
l = 5 mm, = 550 nm, and V = 100 volts
(a) Ez= m/V102105
100
l
V 43
=
=
4123
z33
3
0 102108.30)29.2(2
1E.rn
2
1n ==
or 106
= 369
10510698.310550
2ln
2
=
= 0.067
Therefore
= 2= 0.1344
(b) V= 123
9
33
3
0 108.30)29.2(2
10550
rn2
=
or V= 743.5 V
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9.6 (a) (due to intrinsic birefringence) = ( )e0 nnl2
Substituting the values of constants, we get
).20.229.2(m10550
l2
2 9
=
Therefore, .m528.1m105277.1)09.0(4
m10550l 6
9
=
=
(b) (due to external field) = .V.d
lnr 3033
( ) 20
d
10528.129.2108.30
m105502
6312
9
=
d = .m041.0m101.4120)528.1()29.2(550
m108.302 939
==
(c) V= V528.1
041.0
108.30)29.2(
10550
l
d
rn2 123
9
63
3
0
=
= 40 V
9.8 Sin m= m
=
m1037533.1
m10633
1n 6
9
0
= 1.269 10 3
( )
=
==
m10375s104
ms1500
Hzfrequency
waveaccousticofVelocity 616
1
m = 1
m= 0.0727o
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Chapter 10 : Optical Amplifiers
10.3 (a) Psat=
=
cgcg
hcAAh
From the given parameters, the cross-sectional area of the active region,
A = 5 106(m) 0.5 106(m) = 2.5 1012m2
= 0.4, g= 3 1020m2, c= 1 ns = 1 109s and
= 1.3 m = 1.3 106m.
Therefore, Psat=)m(103.1)s(101)m(1034.0
m105.2)ms(103)Js(10626.669220
2121834
= 3.185 102W
= 31.85 mW.
(b) Zero (or small) signal gain coefficient,
g0= g
tr
c NeV
I
= 0.4 3 1020(m2)
)m(100.1)m(105)C(106.1
)s(101)A(1.0 32431619
19
where the volume of the active region has been calculated as follows:
V = 5 106(m) 0.5 106(m) 200106(m) = 5 1016(m3)
Thus g0= 3000m1.
(c) Zero (or small) signal net gain over the length L of the active region will
be given by
G0= exp (g0L) = exp (3000 2 104) = 1.82.
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10.4 (i) Using eq. (10.15) of the text book, for R1= R2= 0.01%, we get
2
44
442
21s
21s
1010101
1010101
RRG1
RRG1
+=
+
= 004.1101
1012
3
3
=
+
(ii) For R1= R2= 1%
22
9.0
1.1
01.001.0101
01.001.0101
=
+
= 1.4938
10.6 If Ps, inand Ps, outare signal powers at the input and output ends of an EDFA at the
signal wavelength, s; and Pp, in is the input pump power at a wavelength of p;using the principle of conservation of energy, the following inequality should hold
true:
Ps, outPs, in+s
p
Pp, in
Therefore PCE sin,p
in,sout,s p
P
PP
As p< s, PCE < 1.
The maximum value that PCE can take iss
p
.
Therefore ( ) ( ) .1.PCE.QCEs
p
p
s
maxp
s
max
=
=
=
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10.7 (a) The rate of absorption per unit volume from Er3+
level E1to pump level E3due to the pump at p(assuming N2= 0)
t
p
pppa
1
pp
ppaN
hca
PN
ha
P
= ; [N1Ntas N20]
)ms(103)Js(10626.6)m(105.8
)m(105)m(1098.0)Js(1030)m(1017.21834212
324613225
=
= 1.8879 1028m3s1
(b) The rate of absorption per unit volume from Er3+
level E1to the metastable
level E2due to the signal at s(assuming N2N12
N t )
2
N
hca
PN
ha
P t
s
sssa
1
ss
ssa =
=
)m(103)Js(10626.6)m(105.8
)m(105.2)m(1055.1)Js(10200)m(1057.21834212
324616225
=
= 1.1788 1026m3s1
The rate of stimulated emission per unit volume from level E2to level E1
due to the signal at s(assuming again N2 N1Nt/2)
2
N
hca
PN
ha
P t
s
ssse
2
ss
sse =
=
)m(103)Js(10626.6)m(105.8
)m(105.2)m(1055.1)Js(10200)m(1041.31834212
324616225
=
= 1.564 1026m3s1
10.8 (a) In the present problem p= 0.98 m, s= 1.55 m
Ps, in= 0 dBm = 1 mW
Ps, out= 20 dBm = 100 mW
Therefore, the gain of the amplifier, G = 100mW1
mW100=
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G(dB) = 10 log10100 = 20 dB.
(b) Assuming there is no spontaneous emission
Gin,ss
in,pp
in,s
out,s
PP1
PP
+=
In order to achieve a specific maximum gain, G, the input pump power
should be
( ) in,sp
s
in,p P.1GP
The limiting value of
Pp, in= ( ) W1058.15610198.0
55.11100 33 =
or Pp, in156.6 mW.
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Chapter 11: Wavelength Division Multiplexing
11.2 (b) Using eq (11.2) and (11.3) of the text book
P1= P0cos2(z)
and P2= P0sin2(z)
If the splitting ratio is 10 : 90,the required interaction length L will begiven by
2
2
1 cot9
1
P
P== (L)
or tan (L) = 3
which gives L 1.25/
11.5 (b) For demultiplexer with 32 channels spaced at 50 GHz, the FSR should
be at least 32 50 = 1600 GHz.
Since the centre wavelength is 1.55 m, corresponding frequency is
Hz10935.11055.1
103C 146
8
c
c =
=
=
Using eq. (11.15), we get
m =( )
121101600
10935.19
14
FSR
c
=
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Chapter 12:Fiber Optic Communication Systems
12.2 Given that Ptx = 13 dBm, Prx = 40 dBm, f = 0.5 dB/km, splice = 0.1dB/splice, con = 0.5 dB/connector, Ms = 7 dB and the continuous length of asingle piece of fiber = 5 km.
Assuming the repeaterless distance to be L, the number of splices would be
5
L1
5
L
; and the related splice loss would be ( )dB1.0
5
L .
The attenuation within the fiber would be f.L = 0.5L (dB)
Using eq. (12.1) and (12.2) of the text book, we have
Ptx= Prx+ fL + con+ splice+ Ms
or 13 dBm = 40 dBm + 0.5 L (dB) + 0.5 (dB) 1 + 0.1 (dB) 5
L+ 7 dB
Solving this, we get L 37.5 km.
12.3 Link Power Budget
Power supplied by transmitter, Ptx= 0 dBm
Channel loss, CL= (0.5 dB/km) L + 1 dB 2 (connectors)
where L is link length
or CL= 0.5L + 2 (dB)
Safety margin, Ms= 6 dB
Receiver sensitivity = Prx= 35 dBm
Therefore we should have Ptx= Prx+ CL+ Ms
or 0 = 35 dBm + (0.5 L + 2) + 6.This gives L = 54 km.
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12.4 Given that L = 12 km, B = 100 Mb/s
ttx= 10 ns, trx= 12 ns
The source (LED) has = 30 nm, mean wavelength = 0.85m = 850 nm.
Fiber has core index n1= 1.46, = 0.01 and
Dm= 80 ps km1nm
1.
Therefore the pulse broadening due to material dispersion would be given by
tmat= DmL = (80 ps km1nm1) (12 km) (30 nm).
= 28.8 ns.
tintermodal= ( ) ;Lc
nnnn 21
2
1
( ) 98.046.121nn 12 == = 1.445
Thus, tintermodal= )m(1012ms103
445.146.1
445.1
46.1 318
.
= 6.06 107s
= 606 ns
Using eq. (12.8) of the text book, we get
tsys= [ ]21
2
rx
2
f
2
tx ttt ++
= ( ) ( ) ( ){ } ( )[ ]21
2222ns12ns8.28ns606ns10 +++
= 607 ns = 0.607 s
For NRZ format, tsys ns7s100.710100
70.0
B
70.0 96
==
=
The total rise time of the system is much greater than 0.70/B, and hence thesystem can not operate with NRZ format.
What bit rate is possible?
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B = 66
1015.1s10607.0
70.0
bits/s = 1.15 Mb/s.
12.6 With reference to Fig. 12.6 (b) of the text book; if the optical loss within the Bus
(the optical fiber) is not neglected, the power available at the Nth tap will be givenby
PN= PTC [(1 C) (1 ) (1 )]N1 (1)
where PTis the transmitted power, C is the fraction of optical power coupled out
at each tap, is the fractional loss (assumed to be same) at each tap, is theaverage fractional loss within the fiber between two taps (assumed to be same at
equal distances of 50 m) and N is the number of subscribers.
In the present case, PT= 1 mW, PN= 40 dBm = 104mW, and C = 0.05.
The insertion loss at each tap = Ltap= 0.3 dB.
Now Ltap= 10 log10(1 ) = 0.3 dB
This gives = 0.06.
The minimum distance between two taps is 50 m and the optical fiber is
exhibiting a loss of 0.01 dB/m. Therefore the optical loss, Lf, within the fiber
between two taps is 0.50 dB.
Since, Lf= 10 log10(1 ) = 0.50 dB
where is the fractional loss; we have
= 0.10
Substituting the values of relevant parameters in eq. (1) above, we get
104= 1 0.05 [(1 0.05) (1 0.06) (1 0.10)]N1
This gives N = 29
12.7 (a) Fiber attenuation between A and B = (0.1 dB/km) ((25 km) = 2.5 dB
Attenuation between B and C = (0.12 dB/km) (40 km) = 4.8 dB
Attenuation between C and D = (0.1 dB/km) (35 km) = 3.5 dB.
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Total fiber attenuation between A and D = 10.8 dB.
Losses at 2 couplers (one each at A & D) = 0.5 dB 2 = 1 dB.
In the forward direction (when the signal is traversing from A to D). There will be
loss at the joint B due to mismatch of NA and profile parameter .
This loss can be calculated as follows:
The total coupling efficiency at B when the signal is traversing in the forward
direction.
T= NA
=
+
+
2
1
2
1
2
/21
/21
NA
NA
=
+
+
9.1
21
0.2
21
20.0
17.02
= (0.7225) (0.9743)
= 0.70
Therefore loss at joint B = 10 log10T= 1.52 dB
There will be no loss at joint C in the forward direction.
Similarly, in the backward direction (when the signal is transversing from D to
A), there will be 1.52 dB loss at joint C and no loss at B. Therefore, in either
direction the total channel loss CL= 13.32 dB.
The desired safety margin, Ms= 7 dB.
The receiver sensitivity Prx= 0.5 w = 33 dBm.Therefore,
Ptx= Pry+ CL+ Ms= 33 dBm + 13.32 dB + 7 dB
= 12.68 dBm
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or Ptx= 54 W (Both sides)
(b) The pulse broadening due to material dispersion would be given by
tmat= [Dm1(Length of segments AB + CD) + Dm2(Length of
segment BC)) or, tmat= [(70 ps km
1(nm1) (25 + 35) (km)
+ (80 ps km1
(nm1
) (40 km)] (20 nm)
or tmat= 0.148 s
It is given that intermodal dispersion is negligible. Therefore total risetime of the system,
tsys= [ ] s1488.0ttt 212
rm
2
mat
2
tx =++
Maximum bit rate for NRZ format = s/bits107.4101488.0
70.0 66
=
= 4.7 Mb/s
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49
Chapter 13 :Fiber Optic Sensors
13.3 (a) Sang(normalized) =
NA
n1
=( )
)17.0(
180/100.11
= 0.673
NA = ( ) ( )[ ]21
22 45.146.1 = 0.17
Range of Sang= (1 to 0.673)
(b) ang= ( )
+ )17.0(
180/100.11)46.11()46.1(16
4
2
= 0.627
Lang= 10 log10ang= 2.027 dB
Range of loss = 0 to 2.027 dB
(c) PT= Loss + PR
PT(dBm) = (Lang) dB 30 dBm = 27.973 dBm PT= 0.0016 mw = 1.6 w
13.7 We know that
=
C
LD2
It is given that
L = 400m, D = 0.12 m, = 5 104rad/s
= 0.633 106m, C = 3 108ms1
Therefore
= rad10633.0103
10512.0400268
4
= 0.066 rad
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Chapter 14 :Laser Based Systems
14.2 The pump energy that must fall on each cm2of Ruby crystal surface to achieve
threshold inversion is (see example 14.1 of the text book) equal to
)().(hN2
Given that = 5.1 1014Hz, () = 2 cm1, )( = 0.95 and N0= 1.5 1019
atoms of Cr3+
per cm3.
Taking N22
N0 = 7.5 1018, we get
the required pump energy =95.02
101.510626.6105.7 143418
= 1.328 J cm2
14.3
If we assume that the loss per unit length within the cavity is , length of thecavity is L, R1and R2 are the reflectivities of the two end mirrors and P0 is the
incident power, then in one round trip the wave will return with power P2Lgiven
by
P2L= P0R1R2exp (2 L).
R1PL
R1R2P2L
L
P0
R2 R1
Fig. Q. 14.3
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Therefore fractional loss in power per round trip is
0
2100
0
L20
P
)L2exp(RRPP
P
PP =
= 1 R1R2exp (2 L).
If n is the refractive index of the active medium and C speed of light in free space,
the time duration of this fractional loss is .C
Ln2 This time corresponds to an
exponential decay time constant tc(of the power) approximately given by (Yariv1997)*
tc( ))R1C
nL
, assuming = 0 and R1R2R 1.
and the population inversion density at threshold
Nth= (N2N1)th=c
3
spont
23
tc
tn8
Given that R1= 1, R2= R = 0.99, L = 10 cm = 0.1 m, n = 1, = 109Hz,= c/= 4.74 1014Hz and tspont= 0.1 s. Substituting values of relevant parameters, weget
( )( )s1033.3
99.01ms103
)m(1.01
)R1(c
nlt 8
18c
=
=
=
and( )
)s(10)s(1033.3ms103
)s(10)s1074.4()1(8)NN(N 19
8318
721143
th12th
==
= 6.28 1014m3
= 6.28 108cm3
14.8 The time difference between the two pulses is given by
t = ,c
dn2where the d is the depth of the object.
Given that s and n = 1.33. Therefore,
* A. Yariv : Optical Electronics in Modern Communications, (5
thEd. ), Oxford University Press, N.Y.,
p. 188 (1997)
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d =33.12
)s)(1053.0()ms(103
n2
tc 618
=
= 60 m
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