Transcript video slide

Chapter 35
Interference
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Copyright © 2012 Pearson Education Inc.
Goals for Chapter 35
• To consider interference of waves in space
• To analyze two-source interference of
light
• To calculate the intensity of interference
patterns
• To understand interference in thin films
• To use interference to measure extremely
small distances
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Introduction
• Why do soap bubbles
show vibrant color
patterns, even though
soapy water is colorless?
• What causes the
multicolored reflections
from DVDs?
• We will now look at
optical effects, such as
interference, that depend
on the wave nature of
light.
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Wave fronts from a disturbance
• Figure 35.1 at the right
shows a “snapshot” of
sinusoidal waves spreading
out in all directions from a
source.
• Superposition principle:
When two or more waves
overlap, the resultant
displacement at any instant
is the sum of the
displacements of each of
the individual waves.
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Constructive and destructive interference
• Figure 35.2 at the right
shows two coherent wave
sources.
• Constructive interference
occurs when the path
difference is an integral
number of wavelengths.
• Destructive interference
occurs when the path
difference is a half-integral
number of wavelengths.
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Interfering Sources
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Interfering Sources
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Two-source interference of light
• Figure 35.5 below-right shows Young’s double-slit experiment
with geometric analysis.
d sin   m (in phase, constructive)
d sin    m  12   (out of phase, destructive)
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Interference from two slits
•
Projection of two-slit interference on to a screen.
•
The linear dimension of the separation of fringes
obviously depends on the angle and the distance
from the screen.
ym

R
ym
m
 sin  
R
d
m
ym  R
d
tan  
•
Here, R is distance to screen, d is separation of
slits, and m is the “order” of the fringe.
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Two-slit interference
• Example 35.1: Given the measurements in the
figure, what is the wavelength of the light?
ym  R
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m
d
Broadcast pattern of a radio station
•
Example 35.2: Radiation pattern of two radio towers, 400 m apart, operating
at 1500 kHz, oscillating in phase. In what directions is the intensity greatest?
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Intensity in interference patterns
•
•
•
•
Consider two interfering waves with phase
different by phase angle f.
E1 (t )  E cos(t  f )
E2 (t )  E cos(t )
By superposition, we find the resultant wave by
simply adding.
EP cos t  E cos(t  f )  E cos(t )
We use a phasor diagram to show the vector
addition. Using the law of cosines
•
EP2  E 2  E 2  2E 2 cos(  f )
 E 2  E 2  2 E 2 cos f  2 E 2 (1  cos f )
But
f 
1  cos f  2 cos 2  
2
•
so
f 
E  4 E cos  
2
2
P
2
2
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Intensity is related to the
square of the electric field
through the Poynting vector
 0cEP2
f 
I
 2 0cE 2 cos2  
2
2
•
The maximum intensity
is
2
•
And in terms of the maximum
I 0  2 0 cE
f 
I  I 0 cos 2  
2
Intensity in interference patterns
•
What is the phase difference, f, at various angles  from the slits?
•
Think about the path difference r2 – r1. Whenever this path difference
increases by a wavelength, the phase difference increases by 2. Thus
r r
f
 2 1
2

•
But the path difference for a slit separation d is just
r2  r1  d sin  
•
•
•
f f

2 k
 f  kd sin 
f 
 kd sin  
Finally, then, the intensity pattern is I  I 0 cos 2    I 0 cos 2 

2
kdy

dy
y




2
Since sin   , I  I 0 cos 2 
  I 0 cos 

R
 2R 
 R 
Follow Example 35.3.
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

2
0
n

; k  nk0
Interfering Sources
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Interference in thin films
•
Fundamentally, the interference is due to
path-length differences for two coherent
sources. Any arrangement that causes such a
path-length difference will show interference
phenomena, as long as the
•
Figure 35.12 (below) shows interference of
an air wedge.
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Phase shifts during reflection
• Follow the text analysis of thin-film interference and
phase shifts during reflection. Use Figure 35.13 below.
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Wedge between two plates
• Read Problem-Solving Strategy 35.1.
• Follow Example 35.4, having air between the plates. Use
Figure 35.15 below.
• Follow Example 35.5, having water between the plates.
• Follow Example 35.6, another variation on the plates.
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Newton’s rings
• Figure 35.16 below illustrates the interference rings
(called Newton’s rings) resulting from an air film
under a lens.
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Using interference fringes to test a lens
• The lens to be
tested is placed
on top of the
master lens. If the
two surfaces do
not match,
Newton’s rings
will appear, as in
Figure 35.17 at
the right.
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Nonreflective coatings
• The purpose of the
nonreflecting film is
to cancel the
reflected light. (See
Figure 35.18 at the
right.)
• Follow Example
35.7.
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Michelson interferometer
• The Michelson interferometer is used to make precise
measurements of wavelengths and very small distances.
• Follow the text analysis, using Figure 35.19 below.
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