Transcript PPT - LSU Physics & Astronomy

Physics 2102
Jonathan Dowling
Lecture 26: TUE 27 APR 2010
Ch. 36: Diffraction
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EXAM III
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AVG = 60
STD = 15
A=100–85
B=84–75
C=74–45
D=44–40
F=39–0
Michelson Interferometers:
As we saw in the previous example, interference is a spectacular way
of measuring small distances (like the thickness of a soap bubble), since
we are able to resolve distances of the order of the wavelength of the
light (for instance, for yellow light, we are talking about 0.5 of a
millionth of a meter, 500nm). This has therefore technological
applications.
In the Michelson interferometer, light from a
source (at the left, in the picture) hits a semiplated mirror. Half of it goes through to the right
and half goes upwards. The two halves are
bounced back towards the half plated mirror,
interfere, and the interference can be seen by the
observer at the bottom. The observer will see
light if the two distances travelled d1 and d2 are
equal, and will see darkness if they differ by half
a wavelength.
Michelson-Morley Experiment
Michelson won the Nobel prize in 1907, "for his
optical precision instruments and the spectroscopic
and metrological investigations carried out with
their aid"
"The interpretation of these results is that there is no displacement of the
interference bands. ... The result of the hypothesis of a stationary ether is thus
shown to be incorrect." (A. A. Michelson, Am. J. Sci, 122, 120 (1881))
The largest Michelson interferometer in the world is in Livingston, LA,
in LSU owned land (it is operated by a project funded by the National
Science Foundation run by Caltech and MIT, and LSU collaborates in
the project).
http://www.ligo-la.caltech.edu
Mirrors are suspended
with wires and will move
detecting ripples in
the gravitational field due
to astronomical events.
Gravitational Waves Interferometry:
an International Dream
GEO600 (British-German)
Hannover, Germany
LIGO (USA)
TAMA (Japan)
Mitaka
Hanford, WA and Livingston, LA
AIGO (Australia),
Wallingup Plain, 85km north of Perth
VIRGO (French-Italian)
Cascina, Italy
Things You Should Learn from
This Lecture
1. When light passes through a small slit, is spreads out and produces a
diffraction pattern, showing a principal peak with subsidiary maxima
and minima of decreasing intensity. The primary diffraction maximum
is twice as wide as the secondary maxima.
2. We can use Huygens’ Principle to find the positions of the diffraction
minima by subdividing the aperture, giving qmin = ±p l/a, p = 1, 2, 3, ... .
3. Calculating the complete diffraction pattern takes more algebra, and
gives Iq=I0[sin(a)/a]2, where a = p a sin(q)/l.
4. To predict the interference pattern of a multi-slit system, we must
combine interference and diffraction effects.
Single Slit Diffraction
When light goes through a
narrow slit, it spreads out
to form a diffraction
pattern.
Analyzing Single Slit Diffraction
For an open slit of width a, subdivide the opening into segments and
imagine a Hyugen wavelet originating from the center of each segment. The
wavelets going forward (q=0) all travel the same distance to the screen and
interfere constructively to produce the central maximum.
Now consider the wavelets going at an angle such that l = a sin q @ a q.
The wavelet pair (1, 2) has a path length difference Dr12 = l/2, and
therefore will cancel. The same is true of wavelet pairs (3,4), (5,6), etc.
Moreover, if the aperture is divided into p sub-parts, this procedure can be
applied to each sub-part. This procedure locates all of the dark fringes.
p
l
a
= sin q p @ q p ;
p = 1, 2, 3,
(angle of the p th dark fringe)
Conditions for Diffraction
Minima
p
l
= sin q p @ q p ; p = 1, 2, 3,
a
(angle of the p th dark fringe)
Pairing and Interference
Can the same technique be used to find the maxima, by choosing pairs
of wavelets with path lengths that differ by l?
No. Pair-wise destructive interference works, but pair-wise
constructive interference does not necessarily lead to maximum
constructive interference. Below is an example demonstrating this.
Calculating the
Diffraction Pattern
We can represent the light through the
aperture as a chain of phasors that “bends” and
“curls” as the phase Db between adjacent
phasors increases. b is the angle between the
first and the last phasor.
Calculating the
Diffraction Pattern (2)
Eq = 2r sin  b / 2
b = Emax / r; r = Emax / b
Eq =
Emax
sin a
sin  b / 2  = Emax
b /2
a
a
b
2
=
I = CE 2
pa
sin q
l
 sin a 
Iq = I max 

 a 
2
Minima : a = mp or a sin q = ml
Diffraction Patterns
1
l = 633 nm
0.8
a = 0.25 mm
0.6
0.5 mm
0.4
1 mm
0.2
2 mm
0.03
pa
a=
sin q
l
0.02
0.01
 sin a 
Iq = I max 

 a 
0
q (radians)
2
Blowup
0.01
0.02
0.03
The wider the slit opening a, or the
smaller the wavelength l, the narrower
the diffraction pattern.
Radar: The Smaller The Wavelength the Better The Targeting Resolution
X-band:
l=10cm
Ka-band:
l=1cm
K-band:
l=2cm
Laser:
l=1 m
Angles of the Secondary Maxima
The diffraction
minima are precisely at
the angles where
sin q = p l/a and a = pp
(so that sin a=0).
However, the
diffraction maxima are
not quite at the angles
where sin q = (p+½) l/a
and a = (p+½)p
(so that |sin a|=1).
1
l = 633 nm
a = 0.2 mm
 sin a 
Iq = I max 

 a 
2
3
2
4
p
(p+½) l/a
qMax
1
0.00475
0.00453
2
0.00791
0.00778
3
0.01108
0.01099
4
0.01424
0.01417
5
0.01741
0.01735
5
q (radians)
To find the maxima, one must look near sin q = (p+½) l/a, for places
where the slope of the diffraction pattern goes to zero, i.e., where
d[(sin a/a)2]/dq = 0. This is a transcendental equation that must be solved
numerically. The table gives the qMax solutions. Note that qMax < (p+½) l/a.