www.bilkent.edu.tr/~ilday phys 415: optics review of interference and diffraction f. Ömer ilday...
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PHYS 415: OPTICS
Review of Interference and Diffraction
F. ÖMER ILDAY
Department of Physics,Bilkent University, Ankara, Turkey
I used the following resources in the preparation of almost all these lectures:Trebino’s Modern Optics lectures from Gatech (quite heavily used), andvarious textbooks by Pedrotti & Pedrotti, Hecht, Guenther, Verdeyen, Fowles and Das
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Interference vs. Diffraction
• Interference is when we add up multiple but a finite number of E&M waves
• Diffraction is when we add up a continuum of E&M waves.
• Fundamentally, there is no difference.
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Interference and Interferometers
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Varying the delay on purpose
Simply moving a mirror can vary the delay of a beam by many wavelengths.
Since light travels 300 µm per ps, 300 µm of mirror displacement yields a delay of 2 ps. Such delays can come about naturally, too.
Moving a mirror backward by a distance L yields a delay of:
2 L /cDo not forget the factor of 2!Light must travel the extra distance to the mirror—and back!
Translation stage
Input beam E(t)
E(t–)
Mirror
Output beam
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We can also vary the delay using a mirror pair or corner cube.
Mirror pairs involve tworeflections and displace the return beam in space:But out-of-plane tilt yieldsa nonparallel return beam.
Corner cubes involve three reflections and also displace the return beam in space. Even better, they always yield a parallel return beam:
“Hollow corner cubes” avoid propagation through glass.
Translation stage
InputbeamE(t)
E(t–)
MirrorsOutput beam
[EdmundScientific]
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The Michelson Interferometer
Beam-splitter
Inputbeam
Delay
Mirror
Mirror
Fringes (in delay):
*1 2 0 1 0 2
2
2 1 1 2 0 0
Re exp ( 2 ) exp ( 2 )
2 Re exp 2 ( ) ( / 2)
2 1 cos( )
outI I I c E i t kz kL E i t kz kL
I I I ik L L I I I c E
I k L
since
L = 2(L2 – L1)
The Michelson Interferometer splits a beam into two and then recombines them at the same beam splitter.
Suppose the input beam is a plane wave:
Iout
L1
where: L = 2(L2 – L1)
L2 Outputbeam
“Bright fringe”“Dark fringe”
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The Michelson Interferometer
Beam-splitter
Inputbeam
Delay
Mirror
Mirror
The most obvious application of the Michelson Interferometer is to measure the wavelength of monochromatic light.
L = 2(L2 – L1)
Iout
L1
L2 Outputbeam
2 1 cos( ) 2 1 cos(2 / )outI I k L I L
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Huge Michelson Interferometers may someday detect gravity waves.
Beam-splitter
Mirror
Mirror
L1
L2
Gravity waves (emitted by all massive objects) ever so slightly warp space-time. Relativity predicts them, but they’ve never been detected.
Supernovae and colliding black holes emit gravity waves that may be detectable.
Gravity waves are “quadrupole” waves, which stretch space in one direction and shrink it in another. They should cause one arm of a Michelson interferometer to stretch and the other to shrink.
Unfortunately, the relative distance (L1-L2 ~ 10-16 cm) is less than the width of a nucleus! So such measurements are very very difficult!
L1 and L2 = 4 km!
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The LIGO project
A small fraction of one arm of the CalTech LIGO interferometer…
The building containing an arm
The control center
CalTech LIGO
Hanford LIGO
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The LIGO folks think big…
The longer the interferometer arms, the better the sensitivity.
So put one in space, of course.
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Suppose the input beam is not monochromatic(but is perfectly spatially coherent):
Iout = 2I + c Re{E(t+2L1 /c) E*(t+2L2 /c)}
Now, Iout will vary rapidly in time, and most detectors will simply integrate over a relatively long time, T :
/ 2 / 2
1 2
/ 2 / 2
( ) 2 Re ( 2 / ) *( 2 / )
T T
out
T T
U I t dt U IT c E t L c E t L c dt
The Michelson Interferometer is a Fourier Transform Spectrometer
The Field Autocorrelation!
Beam-splitter
DelayMirror
L1
L2
2 Re ( ') *( ' 'U IT c E t E t dt
Recall that the Fourier Transform of the Field Autocorrelation is the spectrum!!
Changing variables: t' = t + 2L1 /c and letting = 2(L2 - L1)/c and T
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Fourier Transform Spectrometer Interferogram
The Michelson interferometer output—the interferogram—Fourier transforms to the spectrum.
The spectral phase plays no role! (The temporal phase does, however.)
Inte
grat
ed ir
radi
ance
0 Delay
Michelson interferometer integrated irradiance
2/
1/
Frequency
Inte
nsity
Spectrum
A Fourier Transform Spectrometer's detected light energy vs. delay is called an interferogram.
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Fourier Transform Spectrometer Data
Interferogram
This interferogram is very narrow, so the spectrum is very broad.
Fourier Transform Spectrometers are most commonly used in the infrared where the fringes in delay are most easily generated. As a result, they are often called FTIR's.
Actual interferogram from a Fourier Transform Spectrometer
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Michelson-Morley experiment
19th-century physicists thought that light was a vibration of a medium, like sound. So they postulated the existence of a medium whose vibrations were light: aether.
Michelson and Morley realized that the earth could not always be stationary with respect to the aether. And light would have a different path length and phase shift depending on whether it propagated parallel and anti-parallel or perpendicular to the aether.
Mirror
Supposed velocity of earth through the aether
Parallel and anti-parallel propagation
Perpendicular propagation
Beam-splitter
Mirror
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Michelson-Morley Experiment: Results
Michelson and Morley's results from A. A. Michelson, Studies in Optics
Interference fringes showed no change as the interferometer was rotated.
The Michelson interferometer was (and may still be) the most sensitive measure of distance (or time) ever invented and should’ve revealed a fringe shift as it was rotated with respect to the aether velocity.
Their apparatus
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The “Unbalanced” Michelson Interferometer
Now, suppose an object isplaced in one arm. In additionto the usual spatial factor, one beam will have a spatiallyvarying phase, exp[2if(x,y)].
Now the cross term becomes:
Re{ exp[2if(x,y)] exp[-2ikx sinq] }
Distorted fringes (in position)
Place anobject inthis path
Misalign mirrors, so beams cross at an angle.
z
x
Beam-splitter
Inputbeam
Mirror
Mirror
q
exp[if(x,y)]
Iout(x)
x
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The "Unbalanced" Michelson Interferometercan sensitively measure phase vs. position.
Phase variations of a small fraction of a wavelength can be measured.
Placing an object in one arm of a misaligned Michelson
interferometer will distort the spatial fringes.
Beam-splitter
Inputbeam
Mirror
Mirror
q
Spatial fringes distorted by a soldering iron tip in
one path
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The Mach-Zehnder Interferometer
The Mach-Zehnder interferometer is usually operated “misaligned” and with something of interest in one arm.
Beam-splitter
Inputbeam
Mirror
Mirror
Beam-splitter
Output beam
Object
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Mach-Zehnder Interferogram
Nothing in either path Plasma in one path
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The Sagnac Interferometer
The two beams take the same path around the interferometer and the output light can either exit or return to the source.
Beam-splitter
Inputbeam
Mirror
Mirror
Mirror
Beam-splitter
Inputbeam
Mirror
Mirror
It turns out that no light exits! It all returns to the source!
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Why all the light returns to the source in a Sagnac interferometer
For the exit beam:
Clockwise path has phase shifts of p, p, p, and 0.
Counterclockwise path has phase shifts of 0, p, p, and 0.
Perfect cancellation!!
For the return beam: Clockwise path has phase shifts of p, p,p, and 0. Counterclockwise path has phase shifts of 0, p, p, and p. Constructive interference!
Beam-splitterInput
beam
Mirror
Mirror
Reflection off a front-surface mirror yields a phase shift of p (180 degrees).
Reflective surface
Reflection off a back-surface mirror yields no phase shift.
Exit beam
Returnbeam
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The Sagnac Interferometer senses rotation
Suppose that the beam splitter moves by a distance, d, in the time, T, it takes light to circumnavigate the Sagnac interferometer.
As a result, one beam will travel more, and the other less distance.
If R = the interferometer radius, and W = its angular velocity:
Thus, the Sagnac Interferometer's sensitivity to rotation depends on its area. And it need not be round!
2
0 0
20
exp( ) exp( )
sin ( )
outI E ikd E ikd
I kd
2
20
(2 / ) 2 ( ) / 2 /
sin (2 / )out
d R T R R c R c c
I I k c
Area
Area
qR
qWT
d
d = Rq
Sagnac Interfer- ometer
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Newton's Rings
Get constructive interference when an integral number of half wavelengths occur between the two surfaces (that is, when an integral number of full wavelengths occur between the path of the transmitted beam and the twice reflected beam).
This effect also causes the colors in bubbles and oil films on puddles.
You see the colorl when: 2L = ml
L
You only see bold colors when m = 1 or 2. Otherwise the variation withl is too fast for the eye to resolve.
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Newton's Rings
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Multiple-beam interference: The Fabry-Perot Interferometer or Etalon
A Fabry-Perot interferometer is a pair of parallel surfaces that reflect beams back and forth. An etalon is a type of Fabry-Perot interferometer, and is a piece of glass with parallel sides.
The transmitted wave is an infinite series of multiply reflected beams.
Transmitted wave:
Incident wave: E0
Reflected wave: E0r
d = round-trip phase delay inside medium = 2kL
2 2 2 2 2 2 2 2 30 0 0 0 0( ) ( ) ...i i i
tE t E t r e E t r e E t r e E
Transmitted wave: E0t
r, t = reflection, transmission coefficients from glass to air2
0t E2 2
0it r e E
2 2 20( )it r e E
2 2 30( )it r e En nair = 1nair = 1
L
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The Etalon (cont'd)
The transmitted wave field is:
2 2 2 2 2 2 2 2 30 0 0 0 0
2 2 2 20
( ) ( ) ...
1 ( ) ( ) ...
i i it
i i
E t E t r e E t r e E t r e E
t E r e r e
2 20 0 / 1 i
tE t E r e
2
1
1 sin / 2T
F
2 2
2
2 2
1 1
r rF
r R
The transmittance is:
2 22 40
2 2 20 1 (1 )(1 )t
i i i
E t tT
E r e r e r e
4 2 2 2 2
4 4 2 2 2 4 2 2
(1 ) (1 )
{1 2cos( )} {1 2 [1 2sin ( / 2)]} {1 2 4 sin ( / 2)]}
t r r
r r r r r r
where:
Dividing numerator and denominator by 2 2(1 )r
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Etalon Transmittance vs. Thickness, Wavelength, or Angle
The transmittance varies significantly with thickness or wavelength.We can also vary the incidence angle, which also affects d (via L).
As the reflectance of each surface (r2) approaches 1 (F increases), the widths of the high-transmission regions become very narrow.
Transmission maxima occur when d/ 2 = mp:
2pL/l = mp
or:
/ 2L m
2
1
1 sin / 2T
F
2 4 /kL L
Tra
nsm
ittan
ce
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Does this look familiar?
Recall that a finite train of identical pulses can be written:
where g(t) is a Gaussian envelope over the pulse train.( ) {III( / ) ( )} ( )E t t T g t f t
( ) {III( / 2 ) ( )} ( )E T G F
g(t) = exp(-t/t)The light field trans-mitted by the etalon!
The peaks become Lorentzians.
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The Etalon Free Spectral Range
lFSR =Free Spectral
Range
The Free Spectral Range is the wavelength range between transmission maxima.
4 4 4 42 2
[1 / ]FSR FSR
L L L L
24 [1 / ] 1 /4 1 12
2 2FSR FSR
FSR
LL
L L
lFSR
2 4 /kL L
Tra
nsm
ittan
ce
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Etalon Linewidth
The Linewidth dLW is a transmittance peak's full-width-half-max (FWHM).
Setting d equal to dLW/2 should yield T = 1/2:
2
1
1 sin / 2T
F
2 21 sin / 2 / 2 2 or sin / 4 1/LW LWF F
2/ 4 1/ 4 /LW LWF F
4 /L
2 1
2LW
R
L r
22
1
rF
R 2
4 12LW
L R
r
2
4LW LW
L
Substituting and we have:
Or:
For d << 1, we can make the small argument approx: l
lLW
Tra
nsm
ittan
ce
The linewidth is the etalon’s wavelength-measurement accuracy.
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The Interferometer or Etalon Finesse
The Finesse is the number of wavelengths the interferometer can resolve.
2
22
1 12
rLR R
L r
F
/ [1 ]R FTaking 1r
The Finesse, F , is the ratio of the Free Spectral Range and the Linewidth:
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How to use an interferometer to measure wavelength
1. Measure the wavelength to within one Free Spectral Range using a grating or prism spectrometer to avoid the interferometer’s inherent ambiguities.
2. Scan the spacing of the two mirrors and record the spacing when a transmission maximum occurs.
3. If greater accuracy is required, use another (longer) interferometer with a FSR ~ the above accuracy (line-width) and with an even smaller line-width (i.e., better accuracy).
Interferometers are the most accurate measures of wavelength available.
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Anti-reflection Coating
Notice that the center of the round glass plate looks like it’s missing.It’s not! There’s an “anti-reflection coating” there (on both the front and back of the glass).
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Anti-reflection Coating Math
Consider a beam incident on a piece of glass (n = ns) with a layer of material (n = nl) if thickness, h, on its surface.
It can be shown that the Reflectance is (for such thin media, we need to go back to Maxwell’s equations):
2 2 2 2 2 20 0
2 2 2 2 2 20 0
( ) cos ( ) ( ) sin ( )
( ) cos ( ) ( ) sin ( )l s s l
l s s l
n n n kh n n n khR
n n n kh n n n kh
Notice that R = 0 if:2
0l sn n n
/ 2 / 4kh h At normal incidence, and if (i.e., )2 2
02 2
0
( )
( )s l
s l
n n nR
n n n
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Multilayer coatings
Typical laser mirrors and camera
lenses use many layers.
The reflectance and transmittance
can be tailored to taste!
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Stellar interferometry
Taken from von der Luhe, of Kiepenheuer-Institut fur Sonnenphysik, Freiburg, Germany.
Stars are too small to resolve using normal telescopes, but interferometry can see them.
Stellar interferometers operate in the radio (when the signals are combined electronically) and visible (where the beams are combined optically).
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“Photonic crystals” use interference to guide light—sometimes around corners!
Interference controls the path of light. Constructive interference occurs along the desired path.
Augustin, et al., Opt. Expr., 11, 3284, 2003.
Yellow indicates peak field regions.
Borel, et al., Opt. Expr. 12, 1996 (2004)
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Convolution
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The Convolution
The convolution allows one function to smear or broaden another.
( ) ( ) ( ) ( )
( – ) ( )
f t g t f x g t x dx
f t x g x dx
changing variables:
x t - x
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The convolution can be performed visually.
Here, rect(x) * rect(x) =
(x)
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Convolution with a delta function
Convolution with a delta function simply centers the function on the delta-function.
This convolution does not smear out f(t). Since a device’s performance can usually be described as a convolution of the quantity it’s trying to measure and some instrument response, a perfect device has a delta-function instrument response.
( ) ) ( ) ( – )
( )
f t t a f t u u a du
f t a
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The Convolution Theorem
The Convolution Theorem turns a convolution into the inverse FT of the product of the Fourier Transforms:
Proof:
{ ( ) ( )} = ( ) ( )f t g t F G wF
{ ( ) ( )} ( ) ( – ) exp( )
( ) ( ) exp(– )
( ){ ( exp(– )}
( ) exp(– ) ( ( (
f t g t f x g t x dx i t dt
f x g t x i t dt dx
f x G i x dx
f x i x dx G F G
F
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The Convolution Theorem in action
2
{ ( )}
sinc ( / 2)
x
k
F{rect( )}
sinc( / 2)
x
k
F
rect( ) rect( ) ( )x x x
2sinc( / 2) sinc( / 2) sinc ( / 2)k k k
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The symbol III is pronounced shah after the Cyrillic character III, which is
said to have been modeled on the Hebrew letter (shin) which, in turn,
may derive from the Egyptian a hieroglyph depicting papyrus plants
along the Nile.
The Shah Function
The Shah function, III(t), is an infinitely long train of equally spaced delta-functions.
t
III( ) ( )m
t t m
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The Fourier Transform of the Shah Function
If = 2n, where n is an integer, the sum diverges; otherwise, cancellation occurs. So:
{III( )} III(t F
) exp( )
)exp( )
exp( )
m
m
m
t m i t dt
t m i t dt
i m
t
III(t)
F {III(t)}
2
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Fraunhofer Diffraction
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Diffraction
Light does not always travel in a straight line.
It tends to bend around objects. This tendency is called diffraction.
Any wave will do this, including matter waves and acoustic waves.
Shadow of a hand
illuminated by a
Helium-Neon laser
Shadow of a zinc oxide
crystal illuminated
by aelectrons
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Why it’s hard to see diffraction
Diffraction tends to cause ripples at edges. But poor source temporal or spatial coherence masks them.
Example: a large spatially incoherent source (like the sun) casts blurry shadows, masking the diffraction ripples.
Untilted rays yield a perfect shadow of the hole, but off-axis rays blur the shadow.
Screenwith hole
A point source is required.
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Diffraction of a wave by a slit
Whether waves in water or electromagnetic radiation in air, passage through a slit yields a diffraction pattern that will appear more dramatic as the size of the slit approaches the wavelength of the wave.
slit size
slit size
slit size
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Diffraction of ocean water waves
Ocean waves passing through slits in Tel Aviv, Israel
Diffraction occurs for all waves, whatever the phenomenon.
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Diffraction Geometry
We wish to find the light electric field after a screen with a hole in it.
This is a very general problem with far-reaching applications.
What is E(x1,y1) at a distance z from the plane of the aperture?
Incident wave This region is assumed to be
much smaller than this one.
x1x0
P10
A(x0,y0) y1y0
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Diffraction Solution
The field in the observation plane, E(x1,y1), at a distance z from the aperture plane is given by a convolution:
0 0
1 1 1 0 1 0 0 0 0 0
( , )
011 0 1 0
01
2 2201 0 1 0 1
( , ) ( , ) ( , )
exp( )1 ( , )
A x y
E x y h x x y y E x y dx dy
ikrh x x y y
i r
r z x x y y
where :
and :
A very complicated result! And we cannot approximate r01 in the exp by z because it gets multiplied by k, which is big, so relatively small changes in r01 can make a big difference!
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Fraunhofer Diffraction: The Far Field
0 0
2 21 1
1 1 0 1 0 1 0 0 0 0
( , )
exp( ), exp exp ,
2A x y
x yikz ikE x y ik x x y y E x y dx dy
i z z z
Let D be the size of the aperture: D 2 = x02 + y0
2.
When kD2/2z << 1, the quadratic terms << 1, so we can neglect them:
0 0
2 22 20 1 0 1 0 01 1
1 1 0 0 0 0
( , )
( 2 2 ) ( )exp( ), exp exp ,
2 2 2A x y
x x y y x yx yikzE x y ik ik E x y dx dy
i z z z z
Recall the Fresnel diffraction result:
This condition corresponds to going far away: z >> kD2/2 = D2/If D = 100 microns and = 1 micron, then z >> 30 meters, which is far!
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Fraunhofer Diffraction Conventions
As in Fresnel diffraction, we’ll neglect the phase factors, and we’ll explicitly write the aperture function in the integral:
1 1 0 1 0 1 0 0 0 0 0 0, exp ( , ) ( , )ik
E x y x x y y A x y E x y dx dyz
This is just a Fourier Transform!
Interestingly, it’s a Fourier Transform from position, x0, to another position variable, x1 (in another plane). Usually, the Fourier “conjugate variables” have reciprocal units (e.g., t & , or x & k). The conjugate variables here are really x0 and kx = kx1/z, which have reciprocal units.
So the far-field light field is the Fourier Transform of the apertured field!
E(x0,y0) = constant if a plane wave
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The Fraunhofer Diffraction formula
where we’ve dropped the subscripts, 0 and 1,
, exp ( , ) ( , )x y x yE k k i k x k y A x y E x y dx dy
E(x,y) = const if a plane wave
Aperture function
We can write this result in terms of the off-axis k-vector components:
kx = kx1/z and ky = ky1/z
, ( , ) ( , )x yE k k A x y E x y F
kz
ky
kx
x = kx /k = x1/z and y = ky /k = y1/z
and:
or:
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Fraunhofer Diffraction from a slit
Fraunhofer Diffraction from a slit is simply the Fourier Transform of a rect function, which is a sinc function. The irradiance is then sinc2 .
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Fraunhofer Diffraction from a Square Aperture
The diffracted field is a sinc function in both x1 and y1 because the Fourier transform of a rect function is sinc.
Diffractedirradiance
Diffractedfield
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Diffraction from a Circular ApertureA circular aperture
yields a diffracted
"Airy Pattern,"
which involves a
Bessel function.
Diffracted field
Diffracted Irradiance
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Diffraction from small and large circular apertures
Recall the Fourier scaling!
Far-field intensity pattern
from a small aperture
Far-field intensity
pattern from a large aperture
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Fraunhofer diffraction from two slits
A(x0) = rect[(x0+a)/w] + rect[(x0-a)/w]
0 x0a-a
1 0( ) { ( )}E x A x F
1 1
1 1
sinc[ ( / ) / 2]exp[ ( / )]
sinc[ ( / ) / 2]exp[ ( / )]
w kx z ia kx z
w kx z ia kx z
1 1 1( ) sinc( / 2 ) cos( / )E x wkx z akx z kx1/z
w w
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Diffraction from one- and two-slit screens
Fraunhofer diffraction patterns
One slit
Two slits
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Diffraction from multiple slits
Slit DiffractionPattern Pattern
Infinitely many equally spaced slits yields a far-field pattern that’s the Fourier transform
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Young’s Two Slit Experiment and Quantum Mechanics
Imagine using a beam so weak that only one photon passes through the screen at a time. In this case, the photon would seem to pass through only one slit at a time, yielding a one-slit pattern.
Which pattern occurs?
Possible Fraunhofer diffraction patterns
Each photon passes
through only one slit
Each photon passes
through both slits
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Fresnel Diffraction
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Fresnel Diffraction: Approximations
2 22 20 1 0 10 1 0 11 1
12 2 2 2
x x y yx x y yz z
z z z z
2 2
2 22 0 1 0 101 0 1 0 1 1
x x y yr z x x y y z
z z
But, in the denominator, we can approximate r01 by z.
And, in the numerator, we can write:
0 0
2 2
0 1 0 11 1 0 0 0 0
( , )
1( , ) exp ( , )
2 2A x y
x x y yE x y ik z E x y dx dy
i z z z
This yields:
1, 1 1 / 2 But if
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Fresnel Diffraction: Approximations
Multiplying out the squares:
0 0
2 2 2 20 0 1 1 0 0 1 1
1 1 0 0 0 0
( , )
( 2 ) ( 2 )1, exp ( , )
2 2A x y
x x x x y y y yE x y ik z E x y dx dy
i z z z
0 0
2 22 20 1 0 1 0 01 1
1 1 0 0 0 0
( , )
( 2 2 ) ( )exp( ), exp exp ,
2 2 2A x y
x x y y x yx yikzE x y ik ik E x y dx dy
i z z z z
This is the Fresnel integral. It's complicated!
It yields the light wave field at the distance z from the screen.
Factoring out the quantities independent of x0 and y0:
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Diffraction Conventions
We’ll typically assume that a plane wave is incident on the aperture.
0 0( , )E x y const
And we’ll explicitly write the aperture function in the integral:
And we’ll usually ignore the various factors in front:
2 22 2
0 1 0 1 0 01 11 1 0 0 0 0
( 2 2 ) ( )exp( ), exp exp ( , )
2 2 2
x x y y x yx yikzE x y ik ik A x y dx dy
i z z z z
2 2
0 1 0 1 0 01 1 0 0 0 0
( 2 2 ) ( ), exp ( , )
2 2
x x y y x yE x y ik A x y dx dy
z z
It still has an exp[i(t – k z)], but it’s constant with respect to x0 and y0.
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Fresnel Diffraction: Example
Fresnel Diffraction from a single slit:
Far from
the slit
zClose to the slit
Incident plane wave
Slit
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Fresnel Diffraction from a Slit
This irradiance vs. position emerges from a slit illuminated by a laser.
x1
Irra
dia
nce
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Diffraction Approximated
The approximate intensity vs. position from an edge:
Such effects can be modeled by measuring the distance on a “Cornu Spiral”
But most useful diffraction effects do not occur in the Fresnel diffraction regime because it’s too complex.
For a cool Java applet that computes Fresnel diffraction patterns, try http://falstad.com/diffraction/