255 39 Solutions Instructor Manual All Chapters

8/13/2019 255 39 Solutions Instructor Manual All Chapters

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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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1

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


3/53

2

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


4/53


5/53


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5

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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6

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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7

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


9/53

8

= ( )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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9

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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10

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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11

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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12

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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13

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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14

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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15

= ( )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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16

= 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 =

=


18/53

17

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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18

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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19

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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20

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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21

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


23/53

22

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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23

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


25/53

24

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.


26/53

25

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

+


27/53


28/53

27

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


29/53

28

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


30/53

29

(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=


31/53

30

(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.


32/53

31

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


33/53

32

= 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

=

=

=


34/53

33

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)


35/53

34

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


36/53

35

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,


37/53

36

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.


38/53

37

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


39/53

38

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


40/53

39

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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40

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

=

=

=


42/53

41

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=


43/53

42

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.


44/53

43

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

=


45/53

44

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.


46/53

45

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?


47/53

46

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.


48/53

47

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


49/53

48

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


50/53

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


51/53

50

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


52/53

51

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)


53/53

d =33.12

)s)(1053.0()ms(103

n2

tc 618

=

= 60 m

255 39 Solutions Instructor Manual All Chapters

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