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Physics 116
Lecture 21
Wave optics
Nov 3, 2011
R. J. Wilkes
Email: ph116@u.washington.edu
11/3/11
1
Announcements
•! 3 clickers have quiz data logged, but no registration:
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If one of these is yours, see me pls
•! Exam 2 next Monday: same procedures as last time
•! Practice exam posted Thursday, in class Friday
•! YOU bring bubble sheet, pencil, calculator
•! Practice questions for Exam #2 are posted (in slides
directory), I will go over solutions in class tomorrow
•! Kyle will have special office hour: Monday, 11:30, B442
Lecture Schedule
(up to exam 2)
Today
11/3/11
3
Quiz: #8
Today, almost all large astronomical telescopes are
reflectors, because with a mirror instead of a lens…
A.! Since refraction is not involved, there is no color aberration
B.! There is only one surface to grind and polish, which is easier
and cheaper
C.!Both of the above
D.!None of the above
11/3/11
4
Physical (wave) optics
•! So far, we have been using simple ray-tracing methods to
understand behavior of light on a macroscopic scale
–! Analyze optical systems with diagrams and lens/mirror eqns.
–! Works as if light were a stream of particles (rays)
–! Accurate for system lengths from cm up to size of Universe
•! On a microscopic scale (few mm down to wavelength of light),
ray optics does not work: must employ the wave nature of light
–! Take into account the physical properties of waves:
superposition and interference effects
•! We’ve already discussed this: brief review follows
–! Wave optics applies to behavior of light in and around
•! Narrow apertures (mm or smaller)
•! Arrays of apertures (gratings, meshes, screens)
•! Edges of barriers (if we look closely enough)
•! Reflections in thin (few wavelengths thick) films
5
From mid-October:
Interference patterns
•! Imagine two kids paddling their hands in a pond in unison
•! Waves (ripples) propagate outward, and overlap
•! There will be lines of reinforcement, and lines of cancellation
(constructive and destructive interference)
lines of reinforcement, where peaks overlap
(constructive interference)
Halfway between are
lines of destructive
interference
11/3/11
6
From mid-October:
Interference = defining property of waves
•! If you see interference effects, you are looking at waves !
Case study: Isaac Newton thought light was a stream of particles
Newton’s “Opticks” (1687) explained all observations at the time
Thomas Young (120 years later) observed interference effects with light
Only waves could do that…
Wave theory of light replaced Newton’s
Important point to review:
•! Interference depends on phase relationship of overlapping waves
•! Phase relationship depends on distance from source
Recall: Phase at distance D from source = 2! (D/")
but sin/cos repeat every cycle, so all that matters is where we are
relative to start of latest cycle: fraction of a cycle
Phase at distance D from source = 2! [ fractional part of (D/") ]
Example: fractional part of 5.678 is 0.678
11/3/11
7
7
From mid-October:
Phase relationships
Example: source of sound has "=1.5m
Point A is 2.5 m away, B is 5.5 m away, C is 7 m away
Phase values are
To make this acoustic
example describe light
waves, just change
wavelength units from
meters to nanometers!
A
B
C
A, B, C are all 2/3
along a cycle, so
sound waves
have the same
phase at all these
points
D / " (distance, in units of wavelengths)
11/3/11
8
8
light
Example: 2 sound sources with same f
From mid-October:
Solid line: R = one wavelength,
dashed line: R= ! wavelength
1
B
C
A
2
These could be closely
spaced pinholes in an
opaque screen
Point A: sources 1 and 2 are in phase, so we get
constructive interference
Points B and C: 1 and 2 are “180 deg out of phase”, so
destructive interference
11/3/11
9
9
Another example reworked:
•! Instead of loudspeakers, let’s consider radio antennas
–! spaced 3 km apart, in phase: same electrical signal goes to both
Observer 4km away, parallel to antenna 1, finds constructive interference
(signal is double intensity of a single antenna’s)
3km
What f is being transmitted?
1
2
D1 = 4km
D2 =
( 3km )2 + ( 4km )2
D2 ! D1 = n"
for n = 1, " = D2 ! D1 = 1km
c 300, 000 km / s
f = =
= 300kHz
"
1km
= 5km
4km
5km
This is just the lowest possible frequency to give constructive interference.
Any integer multiple (harmonic) of this f will also produce constructive
interference at the observer’s location: then n > 1
Notice: this is just the stereo loudspeaker example, with meters changed to km
and speed of sound in air changed to speed of light
11/3/11
10
Coherence
•! One big difference between acoustics and optics: ordinary light sources
have billions of independent radiators
–! In acoustics, we only discussed simple cases of one source (or 2
sources “in phase”)
–! Light bulb = 1023 tungsten atoms, each emitting light on its own
schedule!
•! Incoherent light source: jumble of wavelengths and phases
–! To have constant phase relationships, to observe interference
effects, we need to either
•! View a tiny part of an incoherent source (use slits, pinholes)
•! Have a source of coherent light (all atoms emit in synchrony)
–! Simple to do today: use a laser
–! This option did not exist until 1960s !
11/3/11
Broad spectrum
of wavelengths
mixed together
Monochromatic
(only 1 wavelength
is produced)
11
Constructive vs destructive interference
•! We see interference effects when waves overlap (superposition) after
travelling different paths from their source
•! For EM waves that are coherent (phase relationship at their source is
fixed, and does not vary with time)
–! Constructive interference if path difference is zero or an integer number of
wavelengths
–! Destructive interference if path difference is an odd integer number of
half-wavelengths (Even number of half-wavelengths = integer number of full
wavelengths)
Constructive if:
and
Destructive if:
! 2 ! !1 = m"
1
m = 0,1, 2…
1&
#
! 2 ! !1 = % m ! ( "
$
2'
#
# n&
alternative
statement:
!
!
!
=
%$ (' "
2
1
%$
2
m = 1, 2, 3…
&
n = 1, 3, 5…(
'
2
l1
l2
observer
(For phase relationships between these extremes, we get intermediate intensities)
11/3/11
12
Young’s 2-slit experiment
Some call it the most important experiment in physics
–! Fundamental impact on quantum theory – as we shall see…
•! Make a mask with 2 pinholes or narrow slits, closely spaced
–! What’s “narrow”? What’s “close”?
•! On the scale of visible light wavelength: fraction of mm
•! Illuminate with plane waves (optical equivalent of equally spaced
water waves at a beach, with all wave peaks parallel)
–! How do we get plane waves of visible light? Use a laser
–! How did Thomas Young manage in 1804?
•! Used a pinhole to view a tiny patch of lamp surface
•! “Partially coherent light” – pattern is partially
washed out
www.physics2000.com/
2-slit interference pattern
in laser light
•! The waves interfere to form bright and dark fringes
on a viewing screen some distance away
We can simulate this with water waves in a tank
plane waves
Slits = barrier with 2 holes
Huygens’ principle
Christiaan Huygens (English pronunciation: /"ha#$%nz/), Dutch, 1629 – 1695
Contemporary of Newton, advocate of wave theory of light
Proposed a mechanism for wave propagation:
•! Each point on a light wavefront acts like a point source of light
–! Produces expanding spherical wavelets
–! Superposition of wavelets from all points on original wavefront
produces the next wavefront
spherical wavelets
–! Repeat…
Picture shows wavelets from 3 points – imagine
every point adding in its contribution:
Today, Maxwell’s equations tell us
how EM waves propagate, but
Huygens’ Principle provides a
handy model to help us picture
interference phenomena.
plane wave
2-slit experiment in detail
•! Illuminate a pinhole or slit with light from a distant, coherent source
–! Nearby source would give spherical waves, distant source = plane waves
–! Coherent = from a laser*, or by using another pinhole to restrict area of light source
used
* Lasers supply plane waves at any distance
•! Place a mask with 2 slits in the beam of light
–! Slit spacing d should be “small” (not many wavelengths)
•! Place a screen far away (many wavelengths)
–! Rays* from each slit are nearly parallel when they arrive
* We’ll talk about ‘rays’ only to get path lengths –
then we’ll compare to wavelengths
•! We get a bright fringe on screen parallel to midpoint
–! Central bright fringe – equal path lengths from the 2 slits
–! Constructive interference
•! Where is the next bright fringe?
–! At what angle do we get next constructive interference?
–! Path difference from the 2 slits must be 1 or more wavelengths
!! = "
!! = 0
2-slit experiment in detail
•! Calculate the angle to the next bright fringe:
1
d
!
2
d sin !
l1
90
!
!
Slit mask
(angle to next bright fringe)
l2
Ray to next
bright fringe
Ray to slit
midpoint
Ray to next
bright fringe
Ray to slit
midpoint
On this scale, the screen is miles away…
! 2 ! !1 = d sin "
Constructive if d sin " = m#,
m = 0, ± 1, ± 2, ± 3…
!! = "
!! = 0
–! Central fringe for m=1, the next bright fringe on either side for +1, etc
–! Dark fringes = destructive interference: !! = #% m " 1 &( )
m = ±1, ± 2, ± 3…
$
2'
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