Transcript 25.1-2x

Chapter 25
Waves and Particles
Wave Phenomena
• Interference
• Diffraction
• Reflection
Wave Description
 – wavelength: distance between crests (meters)
T – period: the time between crests passing fixed location (seconds)
v – speed: the distance one crest moves in a second (m/s)
f – frequency: the number of crests passing fixed location in one
second (1/s or Hz)
 – angular frequency: 2f: (rad/s)

v
T
1
f 
T
v  f
Wave: Variation in Time
 2 
E  E0 cos  t   E0 cos 
t
 T 
Wave: Variation in Space
 2x 
 2 
E  E0 cos
x
  E0 cos
  
  
Wave: Variation in Time and Space
 2 
E  E0 cos
t
 T 
2
 2
E  E0 cos
t

 T
 2 
E  E0 cos
x
  

x

‘-’ sign: the point on wave moves to the right
Wave: Phase Shift
2 
 2
E  E0 cos
t
x
 
 T
But E @ t=0 and x =0, may not equal E0
phase shift, =0…2
2
 2

E  E0 cos
t
x  

 T

 2

E  E0 cos
t     E0 cost   
 T

Two waves are ‘out of phase’
(Shown for x=0)
Wave: Amplitude and Intensity
E  E0 cost   
E0 is a parameter called amplitude (positive). Time dependence
is in cosine function
Often we detect ‘intensity’, or energy flux ~ E2.
For example: Vision – we don’t see individual oscillations
Intensity I (W/m2):
I  E02
Works also for other waves,
such as sound or water waves.
Interference
Superposition principle: The net electric field at any location is
vector sum of the electric fields contributed by all sources.
Can particle model explain the pattern?
Laser: source of radiation which has
the same frequency (monochromatic)
and phase (coherent) across the beam.
Two slits are sources of two waves
with the same phase and frequency.
Interference: Constructive
E1
Two emitters:
E2
Fields in crossing point
E1  E0 cost 
E2  E0 cost 
Superposition: E  E1  E2  2 E0 cost 
Amplitude increases twice: constructive interference
Interference: Energy
E1
Two emitters:
E2
E  E1  E2  2 E0 cost 
What about the intensity (energy flux)?
Energy flux increases 4 times while two emitters produce only
twice more energy
There must be an area in space where intensity is smaller than that
produced by one emitter
Interference: Destructive
E1  E0 cost 
E1
E2  E0 cost   
E2
E  E1  E2  E0 cost   cost     0
 cost 
Two waves are 1800 out of phase: destructive interference
Interference
Superposition principle: The net electric field at any location is
the vector sum of the electric fields contributed by all sources.
Constructive:
Destructive:
E1  E0 cost 
E1  E0 cost 
E2  E0 cost 
E  E1  E2  2 E0 cost 
Amplitude increases twice
E2  E0 cost   
E  E1  E2  E0 cost   cost   
Two waves are 1800 out of phase
Constructive: Energy flux increases 4 times while two emitters
produce only twice more energy
Interference
Intensity at each location depends on phase shift between two
waves, energy flux is redistributed.
Maxima with twice the amplitude occur when phase shift
between two waves is 0, 2, 4, 6 …
(Or path difference is 0, , 2 …)
Minima with zero amplitude occur when phase shift between
two waves is , 3, 5 …
(Or path difference is 0, /2, 3/2…)
Can we observe complete destructive interference if 1  2 ?
Predicting Pattern For Two Sources
C
Point C on screen is very far from sources
Need to know phase difference
Very far: angle ACB is very small
normal
Path AC and BC are equal
Path difference:
l  d sin(  )
If l = 0, , 2, 3, 4 … - maximum
If l = /2, 3/2, 5/2 … - minimum
Predicting Pattern For Two Sources
C
Path difference:
l  d sin(  )
If l = 0, , 2, 3, 4 … - maximum
If l = /2, 3/2, 5/2 … - minimum
What if d <  ?
normal
complete constructive interference
only at =00, 1800
What if d < /2 ?
no complete destructive
interference anywhere
Note: largest l for =/2
Intensity versus Angle
Path difference:
l  d sin(  )
If l = 0, , 2, 3, 4 … - maximum
If l = /2, 3/2, 5/2 … - minimum
Why is intensity maximum at
=0 and 1800 ?
Why is intensity zero at =90
and -900 ?
What is the phase difference at Max3?
d = 4.5 
Intensity versus Angle
Path difference:
l  d sin(  )
If l = 0, , 2, 3, 4 … - maximum
If l = /2, 3/2, 5/2 … - minimum
Two sources are /3.5 apart.
What will be the intensity pattern?
d = /3.5
Two-Slit Interference
Path difference: l  d sin(  )
If l = 0, , 2, 3, 4 … - maximum
If l = /2, 3/2, 5/2 … - minimum
L=2 m, d=0.5 mm, x=2.4 mm
What is the wavelength of this laser?
  d sin(  )
sin(  ) 

d
x
tan(  ) 
L
Small angle limit: sin() tan() 
 x

d L
xd

 6  107 m  600 nm
L
Application: Interferometry
Using interference effect we can measure distances with submicron
precision
laser
Detector
Multi-Source Interference: X-ray Diffraction
Coherent beam of X-rays can be used to reveal the structure of
a crystal.
Why X-rays?
- they can penetrate deep into matter
- the wavelength is comparable to interatomic distance
Diffraction = multi-source interference
Multi-Source Interference
Diffraction = multi-source interference
X-ray
lattice
Electrons in atoms will oscillate causing secondary radiation.
Secondary radiation from atoms will interfere.
Picture is complex: we have 3-D grid of sources
We will consider only simple cases
Generating X-Rays
Accelerated
electrons
Copper
Electrons knock out inner
electrons in Cu. When these
electrons fall back X-ray
X-rays is emitted.
(Medical equipment)
Synchrotron radiation:
Electrons circle around accelerator.
Constant acceleration leads to radiation
X-Ray: Constructive Interference
Simple crystal: 3D cubic grid
first layer
Simple case: ‘reflection’
incident angle = reflected angle
phase shift = 0
X-Ray: Constructive Interference
Reflection from the second layer
will not necessarily be in phase
Path difference:
l  2d sin  
Each layer re-radiates. The total
intensity of reflected beam depends
on phase difference between waves
‘reflected’ from different layers
Condition for intense X-ray reflection:
2d sin    n where n is an integer
Simple X-Ray Experiment
x-ray
diffracted
turn crystal
2d sin    n
crystal
May need to observe several maxima to find n and deduce d
X-ray of Tungsten
Using Crystal as Monochromator
Suppose you have a source of X-rays which has a continuum
spectrum of wavelengths.
How can one make it monochromatic?
incident broadband X-ray
2d sin    n
crystal
reflected single-wavelength X-ray
X-Ray of Powdered Crystals
Powder contains crystals in all possible orientations
Note:
Incident angle doesn't have to be equal to
scattering angle.
Crystal may have more than one kind of
atoms.
Crystal may have many ‘lattices’ with
different d
polycrystalline LiF
X-Ray of Complex Crystals
(Myoglobin)
1960, Perutz & Kendrew
Reflection of Visible Light
Why do we see reflection of light
from any smooth surface?
Condition for intense X-ray reflection:
2d sin    n where n is an integer
Visible light:
 ~ 6000 Å >> interatomic spacing
Reflection from many layers is almost
in-phase
Reflection of Visible Light
Constructive interference:
The only possible difference
in path length is zero.
There will be maxima only
when incident angle is equal
to scattering angle.
Thin-Film Interference
Thin films such as soap bubbles are often colored: interference
Consider thin /2-thick film
There are ~3000 atomic layers
Layer 1 and (N/2+1):
destructive interference
For each layer i=1…N/2 there is a layer i+N/2
which re-radiates with 1800 phase shift
resulting in zero intensity – there will be no
reflection of light for this particular wavelength
Thin-Film Interference
Destructive interference: for film thickness n/2.
Constructive interference: for film thickness /4, 3/4, 5/4…
Why are soap bubbles so colorful?
Why after a while soap bubbles lose their color?
Why there is no such effect for thick glass plates?
Other examples or thin-film interference:
oil or gasoline on water
butterfly wings (in some cases)
bird feathers (in some cases)
Index of Refraction
Wavelength of light in dense materials is shorter than in vacuum.
Atoms get polarized due to the E induced
by EM wave and due to the field created
by other polarized atoms.
The crest-to-crest distance in the net electric field is reduced
v=f
since  is reduced, the speed v is slower
Index of refraction: n=c/v, or v=c/n
Index of Refraction
Index of refraction: n=c/v, or v=c/n
Water: n=1.33
Glass: n~1.5
Frequency of light: not affected
v=f
Wavelength: ’ = /n
X-rays: very high frequency, barely
polarize atoms, speed almost not affected
1 = /n1
2 = /n2
1 n1= 2 n2