Physics 214 Lecture 2

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Lecture 2 Interference
l
S1
S2

• Overview
• Coherent vs. incoherent sources
• Interference of Sound waves
• Two-Slit Interference of
• Phasors (See text secs 35.3, 36.3, 36.4)
• Multiple-Slit Interference

See Text. Young and Freeman Chapters 35, 36
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Overview

• Review from Lecture 1
• Superposition, Amplitude and Intensity
• Coherent and Incoherent sources
• Interference of coherent waves
• Interference is a key phenomenon that shows the
  nature of waves
• Examples of sound and light waves

See Text. Young and Freeman Chapter 35, Secs.
1-3
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Review from last lectureSuperposition of waves
Amplitude and Intensity

• Example Suppose we have two waves with the same
  amplitude A1 and angular frequency ?. Then their
  wave numbers k are also the same. Suppose that
  one starts at phase ? after the other

y1 A1 cos(k x - ? t) and y2 A1
cos(k x - ? t ?)
Spatial dependence of 2 waves at t
0 Resultant wave

• Intensity I amplitude2 y2
• Interference of waves depends on the relative
  phase ?
• More today

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Coherent and Incoherent Waves (I)

• Two waves are incoherent if their phase relation
  is random
• Waves from two unrelated sources
• Intensities add average of constructive and
  destructive interference is no interfernce!
• Two waves are coherent if they have a definite
  phase relation. Coherent sources

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Coherent and Incoherent Waves (IA)

• Added slide NOT in printed notes

• Familiar examples of incoherent waves

Earth
Light from two stars (also lightfrom two points
on the same startor light from two different
partsof the same incandescent light bulb,
Sound from two independentsources

• The waves add -- superposition
• But the interference averages out sometimes
  constructive sometimes destructive no net
  effect of interference

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Coherent and Incoherent Waves (II)

• Light from independent sources is incoherent
• The sun or a light bulb light from two points
  on the sun or two points on the hot filament in
  the bulb
• Intensities add average of constructive and
  destructive interference is no interfernce!
• Coherent light sources

A laser produces light that iscoherent and very
monochromatic A purely quantum effect!(More
later!)
Laser
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Interference of Waves

• When two waves are present at the same point in
  space and time they lead to interference. Add
  amplitudes (e.g., pressures or electric fields).
• What we observe however is Intensity (absorbed
  power).
• I A2

In this lecture we confine ourselves to waves
with the same wavelengths.
? 0 waves add in phase (constructive) ? I
2 A12 4A12 4I1 ? p waves add out of
phase (destructive) ? I 2 A102 0
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Act 1 Changing phase of the Source

• Each speaker alone produces an intensity of I1
  1 W/m2 at the listener

I I1 1 W/m2
Drive the speakers in phase. What is the
intensity I at the listener?
Now shift phase of one speaker by 90o.What is the
intensity I at the listener?
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Act 1 Solution

• Each speaker alone produces an intensity of I1
  1 W/m2 at the listener

I A12 I1 1 W/m2
Drive the speakers in phase. What is the
intensity I at the listener?
I (2A1)2 4I1 4 W/m2
Now shift phase of one speaker by 90o.What is the
intensity I at the listener?
I 4 I1cos2(450) 2.0 I1 2.0 W/m2
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Spatial Interference
The relative phase of (two or more) waves also
depends on the relative distances to the sources
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Spatial Interference
The relative phase of (two or more) waves also
depends on the relative distances to the sources
Path difference
Phase difference
Here we use equal intensities.
d f I 0
4I1 l/4 2I1 l/2
0 l 4I1
A 2A1cos(f/2)
A 2A1cos(f/2)
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Amplitude vs. Intensity (for 2 interfering waves)
cos(f/2) cos2(f/2)
Plot here as a function of f.
For equal intensities.
What is the spatial average intensity?
Iave 4I10.5 2I1
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Sound wave example

• Each speaker alone produces intensity I1 1 W/m2
  at the listener, and f 900 Hz. Drive the
  speakers in phase. Compute the intensity I at
  the listener

For sound use sound velocity v 330 m/s
Procedure 1) Compute path-length difference
d 2) Compute wavelength l 3) Compute
phase difference (in degrees) f 4) Write a
formula for the resultant amplitude A 5)
Compute the resultant intensity, I
Nice demo on web www.falstad.com/interference
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Sound wave example

• Each speaker alone produces intensity I1 1 W/m2
  at the listener, and f 900 Hz. Drive the
  speakers in phase. Compute the intensity I at
  the listener

For sound use sound velocity v 330 m/s
Procedure 1) Compute path-length difference
d r2 - r1 1 m 2) Compute wavelength l
v/f (330 m/s)/(900 Hz) 0.367 m 3) Compute
phase difference (in degrees) f 360? (d/l)
360?(1/0.367) 981? 4) Write a formula for the
resultant amplitude A 2A1cos(f/2), A1
?I1 5) Compute the resultant intensity, I 4
I1cos2(f/2) 4 (1 W/m2 ) (0.655)2
1.72W/m2
Nice demo on web www.falstad.com/interference
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Act 2 Speaker interference
What happens to the intensity at the listener if
we decrease the frequency f by a small amount?
(Recall the phase shift was 981?.) a. decrease
b. stay the same c. increase
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Act 2 Speaker interference - Solution
What happens to the intensity at the listener if
we decrease the frequency f by a small amount?
(Recall the phase shift was 981?.) a. decrease
b. stay the same c. increase
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More examples of interference

• Loud Speaker with and without a wood baffle
  Interference of sound from front and back of
  speakerWhy your speakers are enclosed!
• Noise canceling headphonesElectronic circuit
  sends sound out of phase withexternal noise to
  cancelthe noise

baffle
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SummaryInterference of coherent waves

• Resultant intensity of two equal-intensity waves
  of the same wavelength at the same point in
  space
• I 4 I1cos2(f/2)
• For nonequal intensities, the maximum and minimum
  intensities are
• Imax A1 A22 Imin A1 - A22
• The phase difference between the two waves may be
  due to a difference in their source phases or a
  difference in the path lengths to the observer,
  or both. The difference due to path lengths is

f 2p(d/l)
with d r2 r1
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http//webphysics.ph.msstate.edu/jc/library/24-2/h
uygens.htm
Huygens principle (1678) All points on
wavefront are point sources for spherical
secondary wavelets with speed, frequency
equal to initial wave.

• What happens when a plane wave meets a small
  aperture?
• Answer The result depends on the ratio of the
  wavelength l to the size of the aperture a

Similar to a wave from a point source.
l gtgt a
Diffraction Interference of waves from objects
or apertures
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Historical Note Light - Particle or Wave?

• Diffraction of light played an important
  historical role.
• 1818 French Academy held a science competition
• Fresnel proposed the diffraction of light.
• One judge, Poisson, knew light was made of
  particles, and thought Fresnels ideas
  ridiculous he argued that if Fresnel ideas were
  correct, one would see a bright spot in the
  middle of the shadow of a disk.
• Another judge, Arago, decided to actually do the
  experiment

• Conclusion Light must be a wave, since
  particles dont diffract!

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Transmission of light through narrow slits
Monochromatic light source at a great distance,
or a laser.
Observation screen
Slit pattern
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Double-slit interference

• Light (wavelength l) is incident on a two-slit
  (two narrow, rectangular openings)
  apparatus

I1

• If either one of the slits is closed, a diffuse
  image of the other slit will appear on the
  screen. (The image will be diffuse due to
  diffraction. We will discuss this effect in more
  detail later.)

Diffraction profile
screen
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Its just like sound waves

• In both cases, I 4 I1cos2(f/2) with f
  2p(d/l), d r2 - r1

However, for light observer distance L is
generally much greaterthan the wavelength l or
the slit spacing d L gtgt l, L gtgt d.
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Simple formula for the path difference, d , when
the observer is far from sources.

• Assume 2 sources radiating in phase

r
When observer distance gtgt slit spacing
(r gtgt d)
d dsinq
f 2p(d/l) 2p(d sinq/l)
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Two-Slit Interference
Basic result
Constructive Interference
d dsinq ml
m 0, 1, 2,...
Destructive Interference
d dsinq (m 1/2)l
Usually we care about the linear (as opposed to
angular) displacement y of the pattern (because
our screens are often flat)
y L tanq
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Two-Slit Interference, small angles
The slit-spacing d is often large compared to l,
so that q is small. Then we can use the small
angle approximation to simplify our results
For small angles (q ltlt 1 radian)
y L tan q ? L q (in radians)
sin q ? q ? tan q (only in radians!)
Constructive Interference
q ? m(l/d)
y ? m(l/d)L
m 0, 1, 2,...
Destructive Interference
q ? (m 1/2)(l/d)
y ? (m 1/2)(l/d)L
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Act 3 2-slit interference
A laser of wavelength 633 nm is incident on two
slits separated by 0.125 mm.
Dy
1. What is the spacing Dy between fringe maxima
on a screen 2m away? a. 1 mm b. 1 mm c. 1 cm
2. If we increase the spacing between the
slits, what will happen to Dy? a. decrease b.
stay the same c. increase 3. If we instead use
a green laser (smaller l), Dy will? a. decrease
b. stay the same c. increase
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Act 3 2-slit interference - solution
A laser of wavelength 633 nm is incident on two
slits separated by 0.125 mm.
1. What is the spacing Dy between fringe maxima
on a screen 2m away? a. 1 mm b. 1 mm c. 1 cm
2. If we increase the spacing between the
slits, what will happen to Dy? a. decrease b.
stay the same c. increase 3. If we instead use a
green laser (smaller l), Dy will? a. decrease
b. stay the same c. increase
First question can we use the small angle
approximation? d 125 mm l 0.633 mm ? d
gtgt l ? q is small d sinqi mi l d qi ?
qi ? mi (l/d) Dy ? L(q2 q1) ? L(2 1)
(l/d) Ll/d (2 m)(0.663 mm)/125 mm 0.01 m
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Act 3 2-slit interference - solution
A laser of wavelength 633 nm is incident on two
slits separated by 0.125 mm.
Dy
1. What is the spacing Dy between fringe maxima
on a screen 2m away? a. 1 mm b. 1 mm c. 1 cm
2. If we increase the spacing between the
slits, what will happen to Dy? a. decrease b.
stay the same c. increase 3. If we instead use
a green laser (smaller l), Dy will? a. decrease
b. stay the same c. increase
Since Dy 1/d, the spacing decreases. Note
This is a general phenomenon the far-field
interference pattern varies inversely with slit
dimensions.
Since Dy l, the spacing decreases.
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Phasors
See text, Secs 35.3, 36.3, 36.4 and appendix to
lecture 3Phasors make it easier to solve other
problems later on

• Finding the resultant amplitudes of two or more
  waves with different apmplitudes, using phasors

Represent a wave by a vector with magnitude (A1)
and direction (f). One wave has f 0.
To get the intensity, we simply square this
amplitude
where I1 A12 is the intensity when only one
slit is open
This is identical to our previous result !
More generally, if the phasors have different
amplitudes A and B, C2 A2 B2 2AB cos f
Here f is the external angle.
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Phasors for 2-Slits

• Plot the phasor diagram for different f

f
2p
-2p
0
4I1
I
0
q
l/d
-l/d
y
(l/d)L
-(l/d)L
Slits Demo
(Small-angle approx. assumed here)
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Multi-Slit Interference

• What changes if we increase the number of slits
  (e.g., N 3, 4, 1000?)
• First look at the principle maxima
• If slit 1 and 2 are in phase with each other,
  than slit 3 will also be in phase.

Conclusion Position of principle interference
maxima are the samefor any number of
slits! (i.e., d sinq m l)
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Act 4 Multiple Slits

• 1. In 2-slit interference, the first minimum
  corresponds to f p. For 3-slits, we have a
  secondary maximum at f p (see diagram). What is
  the intensity of this secondary maximum? (Hint
  Use phasors.)

2. What value of f corresponds to the first zero
of the 3-slit interference pattern?
3. What value of f corresponds to the first zero
of the 4-slit interference pattern?
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Act 4 Multiple Slits - Solution

• 1. In 2-slit interference, the first minimum
  corresponds to f p. For 3-slits, we have a
  secondary maximum at f p (see diagram). What is
  the intensity of this secondary maximum? (Hint
  Use phasors.)

2. What value of f corresponds to the first zero
of the 3-slit interference pattern?
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Act 4 Multiple Slits - Solution

• 1. For 3-slits, the intensity of this secondary
  maximum is

2. What value of f corresponds to the first zero
of the 3-slit interference pattern?
(a) fp/2
(c) f3p/4
(b) f2p/3
3. What value of f corresponds to the first zero
of the 4-slit interference pattern?
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Act 4 Multiple Slits - Solution

• 1. For 3-slits, the intensity of this secondary
  maximum is

2. What value of f corresponds to the first zero
of the 3-slit interference pattern?
(a) fp/2
(c) f3p/4
(b) f2p/3
3. What value of f corresponds to the first zero
of the 4-slit interference pattern?
(b) f2p/3
(c) f3p/4
(a) fp/2
For N slits the first zero is at 2p/N.
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General properties of N-Slit Interference

• The positions of the principal maxima of the
  intensity patterns always occur at f 0, 2p,
  4p, ... f is the phase between adjacent slits
  (i.e., dsinq ml, m 0, 1, 2,).
• The principal maxima become taller and narrower
  as N increases.
• The intensity of a principal maximum is equal to
  N2 times the maximum intensity from one slit.
  The width of a principal maximum goes as 1/N.
• The of zeroes between adjacent principal
  maxima is equal to N-1. The of secondary
  maxima between adjacent principal maxima is N-2.

Slits Demo Laser Demo
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Summary

• Review from Lecture 1
• Superposition, Amplitude and Intensity
• Coherent and Incoherent sources
• Waves from independent sources are incoherent
• Coherent sources of waves sound waves from same
  source,waves emanating from two slits created by
  a single incident wave, light from a laser,
  incident and reflected waves, . . .
• Interference of coherent waves
• Changing phases of the sources and path length
  differences Examples of sound waves
• Interference of waves from narrow slits small
  angle approximation Two-slit and multiple slit
  interference Examples with light waves
• General solutions for many slits using phasors
• Conclusions
• Classical wave behavior is demonstrated clearly
  by interference
• Careful experiments show conclusively that light
  is a wave Remember this when we start quantum
  mechanics!

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Next time

• Diffraction and Spectroscopy
• Text Ch. 36 added material

• Diffraction gratings

• Spectroscopy an important scientific tool