Solutions Manual for Optics, Global Edition by Eugene Hecht – Step-by-Step Exercise Solutions

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SOLUTIONS MANUAL




F I F T H E D I T I O N

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SOLUTIONS MANUAL FOR OPTICS, GLOBAL EDITION
Chapter Solutions

2  1  
2
2.1
z2  2 t 2

 2(z   t)
z
2
2
z2

 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
y2  t2
2


 (y, t)  (y  4t)2

 2(y  4t)
y
2 
2
y2

 8(y  4t)
t
2 
32
t 2
Thus,   4,  2  16, and,
2 1 2

2
y2 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
z2  2 t 2
 (z t)
 2A
z [(z t)2  1]

2  2(z t)2 1 
 2A 
 
z2 [(z t)  1] [(z t)  1] 
2 3 2 2

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

3(z t)2 1
 2A
[(z t)2  1]3

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 (z t)
 2A
t [(z t)2  1]2
2
 (z t) 
 2A  
t 2 t  [(z t)  1] 
2 2

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


z2  2 t 2
The wave moves with velocity  in the positive z direction.
2.4 c  
c
   3  108 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 ]
 2Aa c  c 
  2  y  t  exp[a(by  ct)2 ]
t b b b
 
2
  4 Aa c  y c t  exp[a(by  ct)2 ]
2 2 2
 
t
2
b4 b2  c b 
2Aa  
  2  y  t  exp[a(by  ct)2 ]
y b b
 
2
2 4 Aa  y c t  exp[a(by  ct)2 ]
2
 4  
y2 b  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  108 /1010 ,   3 cm. Waves extend 3.9 m.

2.7   c /  3  108 /5  1014  6  107 m  0.6 m.

  3  108 /60  5  106 m  5  103 km.

2.8     5  107  6  108  300 m/s.

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2.9 The time between the crests is the period, so   1/ 2 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