Interference in One Dimension
We will assume that the waves have the same frequency, and travel to the
right along the x-axis.
E 1  E 01 sin  kx 1   t   10 
E 2  E 02 sin  kx 2   t   20 
source
1
E 1  E 01 sin  t   kx 1   10   E 01 sin  t   1 
E 2  E 02 sin  t   kx 2   20   E 02 sin  t   1 
,     kx   0 
E  E1  E 2
E  E 0 sin  t  
E0 
tan  

E 01  E 02  2 E 01 E 02 cos  
2
2
E 01 sin  1  E 02 sin  2
E 01 cos  1  E 02 cos  2
2
The Phase Difference
1  kx1   t  10
 2  kx 2   t   20
   1   2  ( kx1   t  10 )  ( kx 2   t   20 )
 k ( x1  x 2 )  (10   20 )  2 
x

  0
Path-length difference
Inherent phase difference
If the waves are initially in-phase
   
n  x  n  x1  x 2   
2

x 
2
0
n x  k0
The optical path difference (OPD)
If 0 is constant waves are said to be coherent.
3
4
if
E 01  E 02
,
 01   02 so   0  0
1
1
sin   sin   2 cos  2 (   )  sin  2 (   ) 
k x 

E   2 E 01 cos
sin

2 


x  

  t  k  x 
 
2 


If x << λ constructive interference
If x = λ/2 destructive interference
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The Phase Difference
The condition of being in phase, where crests are
aligned with crests and troughs with troughs, is that
 = 0, 2, 4, or any integer multiple of 2.
For constructive interference:
   2
x

   0  2 m  or

2

x


0
2
m
For identical sources, 0 = 0 rad , maximum constructive
interference occurs when x = m ,
Two identical sources produce maximum constructive
interference when the path-length difference is an
integer number of wavelengths.
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The Phase Difference
The condition of being out of phase, where crests are
aligned with troughs of other, that is,
 =, 3, 5 or any odd multiple of .
For destructive interference:
   2

2

x

x


   0  2( m  2 ) or
1
 0
2
m
1
2
For identical sources, 0 = 0 rad , maximum constructive
interference occurs when x = (m+ ½ ) ,
Two identical sources produce perfect destructive
interference when the path-length difference is
half-integer number of wavelengths.
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8
The Superposition of Many Waves
The superposition of any number of coherent harmonic waves having a
given frequency and traveling in the same direction leads to a harmonic
wave of that same frequency.
N
E 
E
0i
cos  i   t   E 0 cos    t 
i 1
N
E 
2
0
E
N
2
0i
i 1
 2
N
E
oi
E oj cos  i  
j

j  i i 1
N
tan  
E
oi
sin  i
E
oi
cos  i
i 1
N
i 1
9
The Mathematics of Standing
Waves
A sinusoidal wave traveling to the
right along the x-axis
D R  a sin kx   t 
An equivalent wave traveling to
the left is
sin      sin  cos   cos  sin 
D L  a sin kx   t 
D  x , t   D R  D L  a sin kx   t   a sin kx   t 
D  x , t   a (sin kx cos  t  cos kx sin  t )  a (sin kx cos  t  cos kx sin  t )
 ( 2a sin kx ) cos  t
E ( x , t )   2 E 01 sin kx  cos  t
D ( x , t )  A ( x ) cos  t
Where the amplitude function A(x) is defined as
A ( x )  2a sin kx
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Notes
A ( x )  2a sin kx
The amplitude reaches a maximum value
Amax = 2a at points where sin kx =1.
D ( x , t )  A ( x ) cos  t
The displacement is neither a function of (x-vt) or (x+vt) , hence it is not
a traveling wave.
The cos t , describes a medium in which each point oscillates in simple
harmonic motion with frequency f= /2.
The function A(x) =2a sin kx determines the amplitude of the oscillation
for a particle at position x.
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D ( x , t )  A ( x ) cos  t
The amplitude of oscillation,
given by A(x), varies from point
to point in the medium.
The nodes of the standing wave
are the points at which the
amplitude is zero. They are
located at positions x for which
A ( x )  2a sin kx  0
That is true if
kx m 
2 x m

 m
m  0 , 1, 2 , 3 , 
Thus the position xm of the mth node is
xm  m
Where m is an integer.

m  0 , 1, 2 , 3 , 
2
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Example:
Cold Spots in a Microwave Oven
“Cold spots”, i.e. locations where objects are not adequately
heated in a microwave oven are found to be 1.25 cm apart.
What is the frequency of the microwaves?
d node  1.25 cm   / 2 so   2.50 cm
f 
c

(3.00  10 m /s)
8

 1.20  10
10
H z  12.0 G H z
(0.0250 m )
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Beats and Modulation
If you listen to two sounds with very different frequencies, you
hear two distinct tones.
But if the frequency difference is very small, just
one or two Hz, then you hear a single tone whose
intensity is modulated once or twice every second.
That is, the sound goes up and down in volume, loud,
soft, loud, soft, ……, making a distinctive sound
pattern called beats.
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When two light waves of different frequency
interfere, they also produce beats.
E tot ( x , t )  R e{ E 0 exp i ( k 1 x   1t )  E 0 exp i ( k 2 x   2 t )}
k ave 
Let
S im ilia rly,
 ave 
k1  k 2
and k 
k1  k 2
2
2
1   2
2
Take E0 to be real.
and  
1   2
2
So :
E tot ( x , t )  R e{ E 0 exp i ( k ave x   kx   ave t    t )  E 0 exp i ( k ave x   kx   ave t    t )}
 R e{ E 0 exp i ( k ave x   ave t )  exp i (  kx    t )  exp[  i (  kx    t )] }
 R e{2 E 0 exp i ( k ave x   ave t ) cos(  kx    t )}
 2 E 0 cos( k ave x   ave t ) cos(  kx    t )
For a nice demo of beats,
check out:
http://www.olympusmicro.com/
primer/java/interference/ 15
The group velocity is the velocity of the
envelope or irradiance: the math.
The carrier wave propagates at the phase velocity.
And the envelope propagates at the group velocity:
E ( t )  E 0 ( z  v g t ) exp[ ik ( z  v  t )]
Or, equivalently, the irradiance propagates at the group velocity:
E (t ) 
I ( z  v g t ) exp[ ik ( z  v  t )]
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Calculating the Group velocity
vg  d /dk
Now,  is the same in or out of the medium, but k = k0 n, where k0 is
the k-vector in vacuum, and n is what depends on the medium.
So it's easier to think of  as the independent variable:
v g   dk / d  
1
Using k =  n() / c0, calculate: dk /d = ( n +  dn/d ) / c0
vg  c0 /  n   dn/d) = (c0 /n) / (1 +  /n dn/dc )
Finally:
 dn 

v g  v / 1 

n
d



vg 
vg

n
kc dn
2
n dk
k dn 

 v 1 

n dk 

So the group velocity equals the phase velocity when dn/d = 0,
such as in vacuum. Otherwise, since n increases with , dn/d > 0,
and:
vg < v
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Calculating Group Velocity vs. Wavelength
We more often think of the refractive index in terms of wavelength ,so let's write
the group velocity in terms of the vacuum wavelength 0.
U se th e ch a in ru le :
N o w ,  0  2 c 0 /  , so :
R e ca llin g th a t :
w e have :
dn
d

dn d  0
d 0 d 
d 0
d

 2 c 0

2

 2 c 0
(2 c 0 /  0 )
 0
2
2

2 c 0
 dn 
c  
v g   0  / 1 
n d  
 n  

2 c 0
 c0 
v g    / 1 
n0
 n  

 dn

 d  0
   2   
0


 2 c  
0 


or :
c
vg   0
 n
 0 dn 
 

 / 1 
n
d

 
0 
vg 
c
n

kc dn
2
n dk
k dn 

v g  v 1 

18
n
dk


In regions of normal dispersion, dn/d is positive. So vg < c for these
frequencies.
vg 
vg
c
n

kc dn
2
n dk
k dn 

 v 1 

n dk 

19
The group velocity is less than the phase
velocity in non-absorbing regions.
vg = c0 / (n +  dn/d)
In regions of normal dispersion, dn/d is positive. So vg < c for these
frequencies.
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The group velocity can exceed c0 when
dispersion is anomalous.
vg = c0 / (n +  dn/d)
dn/d is negative in regions of anomalous dispersion, that is, near a
resonance. So vg can exceed c0 for these frequencies!
One problem is that absorption is strong in these regions. Also, dn/d is
only steep when the resonance is narrow, so only a narrow range of
frequencies has vg > c0. Frequencies outside this range have vg < c0.
Pulses of light (which are broadband) therefore break up into a mess.
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Home Work
• 7.5-7.6
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