Solutions Manual to Optoelectronics and Photonics: Principles and Practices, 2nd Edition © 2013 Pearson Education Safa Kasap A+ Latest

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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S.O. Kasap, Optoelectronics and Photonics: Principles and Practices, Second Edition, © 2013 Pearson Education
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright
and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a
retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or
likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education,
Inc., Upper Saddle River, NJ 07458.




© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

, Solutions Manual (Preliminary) Chapter 1 1.2
11 December 2012


Preliminary Solutions to Problems and Questions
Chapter 1

Note: Printing errors and corrections are indicated in dark red. See Question 1.47. These are
correct in the e-version of the textbook
1.1 Maxwell's wave equation and plane waves
(a) Consider a traveling sinusoidal wave of the form Ex = Eo cos(tkz + o). The latter can also be
written as Ex = Eo cos[k(vtz) + o], where v = /k is the velocity. Show that this wave satisfies
Maxwell's wave equation, and show that v = (oor)1/2.
(b) Consider a traveling function of any shape, even a very short delta pulse, of the form Ex =
f[k(vtz)], where f is any function, which can be written is Ex = f(),  = k(vtz). Show that this
traveling function satisfies Maxwell's wave equation. What is its velocity? What determines the form
of the function f?
Solution
(a)
Ex = Eo cos(tkz + o)
 2 Ex
 0
x 2
 2 Ex
and 0
y 2
 2 Ex
and   k 2 E0 cos(t  kz  0 )
z 2


 2 Ex
   2 E0 cos(t  kz  0 )
t 2


2E 2E 2E 2E
Substitute these into the wave equation        0 to find
x 2 y 2 z 2 t 2
o r o


k 2 E cos(t  kz  0 )   o r o   2 E0 cos(t  kz  0 )  0
2 1
   
k 2
 o r o
 1
    ( o r o ) 2
k
1
   v  ( o r o ) 2




(b) Let
Ex  f [k (v t  z )]  f ( )
Take first and second derivatives with respect to x, y, z and t.




© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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, Solutions Manual (Preliminary) Chapter 1 1.3
11 December 2012

 2 Ex
0
x 2
 2 Ex
0
y 2
Ex df
 k
z d
 2 Ex 2
2 d f
 k
z 2 d 2
Ex df
 kv
t d
 2 Ex 2
2 2 d f
k v
t 2 d 2

2E 2E 2E 2E
Substitute these into the wave equation        0 to find
x 2 y 2 z 2 t 2
o r o



d2 f 2
2 2 d f
k2     k v 0
d 2 d 2
o r o



1
  v2 
 o r  o
1
 v  ( o r o ) 2




1.2 Propagation in a medium of finite small conductivity An electromagnetic wave in an
isotropic medium with a dielectric constant r and a finite conductivity  and traveling along z obeys
the following equation for the variation of the electric field E perpendicular to z,
d 2E 2E E
      o  
t t
2 o r o 2
dz
Show that one possible solution is a plane wave whose amplitude decays exponentially with
propagation along z, that is E = Eoexp(z)exp[j(t – kz)]. Here exp(z) causes the envelope of the
amplitude to decay with z (attenuation) and exp[j(t – kz)] is the traveling wave portion. Show that in
a medium in whichis small, the wave velocity and the attenuation coefficient are given by
 1 
v  and 
k o o r 2 o cn

where n is the refractive index (n = r1/2). (Metals with high conductivities are excluded.)

Solution




© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

, Solutions Manual (Preliminary) Chapter 1 1.4
11 December 2012

We can write E = Eoexp(z)exp[j(t – kz)] as E = Eoexp[jt – j(k – j)z]. Substitute this into the
wave resonance condition
[– j(k – j)]2Eoexp[jt – j(k – j)z]  (joroEoexp[jt – j(k – j)z] =
      jo Eoexp[jt – j(k – j)z]
 (k – j)2  oro = jo
 k2 jk – 2 + oro = jo
Rearrange into real and imaginary parts and then equating the real parts and imaginary parts
  k2 – 2 + oro jk = jo
Real parts
k2 – 2 + oro = 0
Imaginary parts
k = o
      c 
Thus,  o   o  o 
2k k 2 2n 2 o n
where we have assumed /k = velocity = c/n (see below).
From the imaginary part
k 2   2  o o r   2
Consider the small  case (otherwise the wave is totally attenuated with very little propagation). Then
k 2   2  o  o r
and the velocity is
 1
v 
k  o o r

1.3 Point light source What is the irradiance measured at a distance of 1 m and 2 m from a 1 W
light point source?

Solution
Then the irradiance I at a distance r from O is
P 1W
I  o2  = 8.0 W cm-2
4r 4 (1 m) 2


which drops by a factor of 4 at r = 2 m to become 2.0 W cm-2

1.4 Gaussian beam A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a
Gaussian beam, what is the divergence of the beam? What are its Rayleigh range and beam width at 10
m?
Solution
Using Eq. (1.1.7), we find,
4 4(633  109 m)
2   = 1.0110-3 rad = 0.058
 (2wo )  (0.8 103 m)
The Rayliegh range is
 wo2  [ 12 (0.8 103 m)]2
zo   = 0.79 m
 (633  109 m)



© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.