Optics, Global Edition (5th Edition, 2017 – Eugene Hecht) | Complete Solutions Manual PDF

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SOLUTIONS MANUAL

,Chapter 2 Solutions 1


Chapter 2 Solutions

 2 1  2
2.1 
z 2  2 t 2

 2( z   t )
z
 2
2
z 2

 2 ( z   t )
t
 2
 2 2
t 2
It’s a twice differentiable function of ( z   t ), where  is in the negative z direction.

 2 1  2
2.2  2 2
y 2
 t
 ( y, t )  ( y  4t )2

 2( y  4t )
y
 2
2
y 2

 8( y  4t )
t
 2
 32
t 2
Thus,   4,  2  16, and,
 2 1  2
 2 
y 2 16 t 2
The velocity is   4 in the positive y direction.
2.3 Starting with:
A
 ( z, t ) 
( z   t )2  1
 2 1  2

z 2  2 t 2
 (z   t )
 2 A
z [( z   t )2  1]
 2  2( z   t )2 1 
 2 A   2 
z  [( z   t )  1] [( z   t )  1] 
2 2 3 2


 4( z   t )2 ( z   t )2  1 
 2 A   3 
 [( z   t )  1] [( z   t )  1] 
2 3 2


3( z   t )2  1
 2A
[( z   t )2  1]3

,2 Chapter 2 Solutions


 (z   t )
 2 A
t [( z   t )2  1]2
 2  (z   t ) 
 2 A  
t 2 t  [( z   t )2  1]2 
  4 ( z   t ) 
 2 A   (z   t )
 [( z   t ) 2
 1]2
[( z   t )2  1]3 
  [( z   t )2  1] 4 ( z   t )2 
 2 A   
 [( z   t )  1] [( z   t )2  1]3 
2 2


3( z   t )2  1
 2 A 2
[( z   t )2  1]3
Thus since
 2 1  2

z 2  2 t 2
The wave moves with velocity  in the positive z direction.
2.4 c  
c 3  10 8 m /s
   5.831  1014 Hz
 5.145  10 7 m
2.5 Starting with:
 ( y, t )  A exp[ a(by  ct )2 ]
 ( y, t )  A exp[ a(by  ct )2 ]  A exp[ a(by  ct )2 ]
 2 Aa c  c 
  2  y  t  exp[ a(by  ct )2 ]
t b b b 
 2 4 Aa 2 c 2 
2
c 
 4  y  b t  exp[ a(by  ct ) ]
2

t 2 b b2  
 2 Aa  c 
  2  y  t  exp[ a(by  ct )2 ]
y b  b 
 2 4 Aa 2
2
 c 
 4  y  b t  exp[ a(by  ct ) ]
2

y 2 b  
Thus  ( y, t )  A exp[ a(by  ct )2 ] is a solution of the wave equation with   c /b in the + y direction.

2.6 (0.003) (2.54  10 2 /580  10 9 )  number of waves  131, c   ,

  c /  3  10 8 /1010 ,   3 cm. Waves extend 3.9 m.

2.7   c /  3  10 8 /5  1014  6  10 7 m  0.6  m.

  3  10 8 /60  5  10 6 m  5  10 3 km.

2.8     5  10 7  6  10 8  300 m/s.

, Chapter 2 Solutions 3


2.9 The time between the crests is the period, so   s; hence
  1/  2.0 Hz. As for the speed   L /t  4.5 m/1.5 s  3.0 m/s. We
now know  ,  , and  and must determine . Thus,
  /  3.0 m/s/2.0 Hz  1.5 m.
2.10  =  = 3.5  103 m/s =  (4.3 m);  = 0.81 kHz.

2.11  =  = 1498 m/s = (440 HZ) ;  = 3.40 m.

2.12  = (10 m)/2.0 s) = 5.0 m/s;  = / = (5.0 m/s)/(0.50 m) = 10 Hz.
2.13     (/2 )  and so   (2 /).
2.14
q  /2  /4 0  /4 /2 3/4
sin q 1  2 /2 0 2 /2 1 2 /2
cos q 0 2 /2 1 2 /2 0  2 /2
sin(q  /4)  2 /2 1  2 /2 0 2 /2 1
sin(q   /2) 0  2 /2 1  2 /2 0 2 /2
sin(q 3 /4) 2 /2 0  2 /2 -1  2 /2 0
sin(q   /2) 0 2 /2 1 2 /2 0  2 /2

q  5/4 3/2 7/4 2
sin q 0  2 /2 1  2 /2 0
cos q 1  2 /2 0 2 /2 1
sin(q   /4) 2 /2 0  2 /2 1  2 /2
sin(q   /2) 1 2 2 0  2 /2 1
sin(q  3 /4) 2 /2 1 2 /2 0  2 /2
sin(q   /2) 1  2 /2 0 2 /2 1

sin q leads sin(q  p/2).

2.15
x /2 / 4 0 /4 /2 3/4 
2 x
kx   /2 0 /2  3/2 2

cos(kx   /4)  2 /2  2 /2 2 2 2 2  2 /2  2 /2 2 2
cos(kx  3 /4) 2 /2 2 /2  2 /2  2 /2 2 /2 2 /2  2 /2
2.16
t  /2  /4 0  /4  /2 3 /4 
 t  (2 / )t  /2 0 /2  3/2 
sin( t   /4)  2 /2  2 /2 2 2 2 2  2 /2  2 /2 2 /2
sin( /4   t )  2 /2 2 /2 2 /2  2 /2  2 /2 2 /2 2 /2