Spring 2016
PHYS 306: Final Exam Review
Exam date – May 2nd 9:45-11:45
Two final practical exams – one due 26th in class
and the other for yourself
Last lecture – April 26th on chapter 8: nonlinear harmonic oscillator
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Phasor Presentation
Rotating Arrow + Phase Angle
t
  t
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Representation of a complex number in
terms of real and imaginary components
Additions of Two Parallel SHOs in 1D
1=5, 2 =6
1=11, 2 =12
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Damped SHOs
b2
  0
4
2
0
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x(t )  bx (t )   x(t )  0
2
0
0

A2  F0 (m 02 )
 0 2
1
(  )
2
Q
0 
 
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20
Q
tan  
1/ Q
 0


0 
Variation of Pav with 
 
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0
Q
Coupled oscillators and normal modes
x
t
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Coupled oscillators and normal modes
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Wave Motion as Coupled Oscillations
 Oscillation of infinite no. of coupled particles (lattice/medium):
2

y
2 y
2
c
2
t
x 2
y ( x, t )  f ( x  ct )
yn  An cos(nt  kx)
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Coupled Oscillations of a Loaded String
y
T
a
x
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Reflection
Wave eqations
and atConditions
an interface
Wave Equations
& Boundary
y A  yB
y A y B

x
x
y1 t  x v A  y3 t  x vB  y5 t  x vC 
y2 t  x v A  y4 t  x vB 
y
A
B
x
combine forward and
reflected waves to give total
fields for each region
apply continuity conditions
for separate components
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hence derive fractional11
C
transmission and reflection
 2E
 2E
  2  0
2
x
t
y
2
M2
y  x, t 
M1
1
T
δx
x
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T
x+δx
2
2 y T 2 y
2  y

c
2
2
t
 x
x 2
x
Light is an Electromagnetic Wave


 E
2
 E   2  0
t
2

B
2
 B   2  0
dt
2

~ i ( kx t )
E y (r , t )  E y e

~ i ( kx t )
Bz ( r , t )  Bz e
1. The electric field, the magnetic field, and the k-vector are all perpendicular:
  
EB  k
2. The electric and magnetic fields are in phase.
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c
1
 0 0
 3 *108 m / s
The Fresnel’s problem

ki
Er
Ei
Bi

kr
ni
y
Br
i r
Interface
z
t
Et
Bt

kt
I r Ar
R  Reflected Power / Incident Power 
I i Ai
I t At
T  Transmitted Power / Incident Power 
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I i Ai
nt
x
Note that
R+T =1
R  Reflected Power / Incident Power 
wi
i r
ni
nt
I r Ar
I i Ai
wi
R  r2
I t At
T  Transmitted Power / Incident Power 
I i Ai
wi
ni
nt
i
t
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wt
  nt cos t    2
T 
t
  ni cos i   
r  E0 r / E0i   ni cos( i )  nt cos( t )  /  ni cos( i )  nt cos( t )
t  E0t / E0i  2ni cos( i ) /  ni cos( i )  nt cos( t ) 

ki
Er
Ei
Bi
Interface
Br
i r
t
Et
Bt
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
kr

kt
ni
nt
r||  E0 r / E0i   ni cos(t )  nt cos(i ) /  ni cos(t )  nt cos(i ) 
t||  E0t / E0i  2ni cos(i ) /  ni cos(t )  nt cos(i ) 

ki
Bi
Ei
Br
i r

kr
ni
×
Er
Interface
Beam geometry
for light with its
electric field
parallel to the plane of incidence
(i.e., in the page)
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t
Et
Bt

kt
nt
The Tangential Field Components are Continuous
Parallel polarization
Perpendicular polarization
1.0
1.0
T
T
.5
.5
R
R
0
0
0°
30°
60°
Incidence angle, i
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90°
0°
30°
60°
Incidence angle, i
90°
Calculate R and T for normal incidence
from an air-glass interface.
 nt  ni 
R  

 nt  ni 
2
T

4 nt ni
 nt  ni 
For an air-glass interface (ni = 1 and nt = 1.5),
R = 4% and T = 96%
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2
Energy and energy transport in waves
U  UE UB   E2
 2E
 2E
  2  0
2
x
t
E(x,t) = A cos[( t- k x + ]
y
2
M2
y  x, t 
M1
1
T
δx
x
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T
x+δx
2
2 y T 2 y
2  y

c
2
2
t
 x
x 2
x
Light is an Electromagnetic Wave
1. The electric field, the magnetic field, and the k-vector are all perpendicular:
  
EB  k
2. The electric and magnetic fields are in phase.
3. The light density an EM field:
1
Bz ( x, t )  E y ( x, t )
c
So the electrical and magnetic energy densities in light are equal.
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The Energy Transport in Waves
EM waves: U B 
11 2
1
E     E 2  U E

2
2
U  UE UB   E2
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S  c 2 E  B  k
Group Velocity
Light-wave beats
Etot(x,t) = 2E0 cos(kavex–avet) cos(kx–t)
This is a rapidly oscillating wave: [cos(kavex–avet)]
with a slowly varying amplitude [2E0 cos(kx–t)]
The phase velocity comes from the rapidly varying part: v = ave / kave
What about the other velocity—the velocity of the amplitude?
Define the "group velocity:" vg   /k
In general, we define the group velocity as:
vg  d /dk
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vg  dE /dp
Group vs. Phase Velocity
V   
Vg   
c
n 
c
dn
n   
d
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 < vphase.
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Group and phase velocity
Quiz: vg and vph directions?
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Fourier transform
We desire a measure of the frequencies present in a wave. This will
lead to a definition of the term, the "spectrum.“
Plane waves have only
one frequency, .

F ( ) 
 f (t ) exp(it ) dt
FourierTransform

f (t )
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
1
2

 F ( ) exp(i t )d 

Inverse Fourier Transform
Long vs. Short Pulses
Long pulse
Short pulse
Time-bandwidth product (The uncertainty principle)
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CB ~ 0.5 in most cases
  t  2cB
Example: the Fourier Transform of a
rectangle function: rect(t)
1/ 2
1
[exp(  i t )]1/1/2 2
F ( )   exp(  i t ) dt 
 i
 1/ 2
1
[exp(  i / 2)  exp(i 
 i
 exp(i / 2)  exp(  i  sin( 


 
 
2i

F (   sinc(
Imaginary
Component = 0
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The Fourier Transform of the Triangle
function, (t), is sinc2(/2)

Convolution
f (t )  g (t ) 

f ( x) g (t  x) dx


g ( x=
) dxF ( w
{f (t)  f (gt –(xt))}
) G ( )

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
rect(x) * rect(x) = (x)
We can perform
convolution visually.
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Interference and diffraction
Division of amplitude
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Division of wavefront
Matter of scale
Superposition of Waves
The irradiance is given by:


I  I1  c Re E1  E2  I 2
 
*
I  2 I 0  2 I 0 cos(2kx)
Fringes (in delay)
I
I  2 I 0  2 I 0 cos[ ]
‐ 
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Spatial Fringes
The fringe spacing, :
Large angle:
  2 /(2k sin  )
  /(2sin  )
  sin    /(2)
   0.5 m / 200  m
Small angle:
 1/ 400 rad  0.15
0.1 mm is about the minimum fringe spacing seen by eye
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Diffraction and Fourier Transform
Fraunhofer Diffraction from
a slit is simply the Fourier
Transform of a rect function,
which is a sinc function.
The irradiance is then sinc2
.
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