Interference
See Chapter 9 of Hecht
A single point
creates
waves with
concentric
circles of light
and dark
bands.
This picture does not
show an interference
pattern. It is simply
the concentric waves
of two points sources
drawn in the same
plane. Contrast
this with the image
to the right
This is the
interference pattern of
two waves.The two
waves add or
subtract to form the
light and dark regions
of the interference
pattern
Waves are not simple two dimensional objects.
When they interfere with each other, peaks and
valleys are formed. Many interference patterns
look like two-dimensional systems of light and dark
bands because they are being viewed from above.
In this picture the system tilted so it can be viewed
from the side.
Consider two waves intersecting at some location
E1
E2



 


E1  E01 cos k1  r  1t  1
 


E2  E02 cos k 2  r   2t   2

At the point where the waves intersect the total
electric field will be:
 
 



ET  E01 cos k1  r  1t  1  E02 cos k2  r  2t  2




Almost always the intensity of the radiation is detected
2
I   0c E
TA = Time Average
TA
For convenience, neglect constants and say:
2
I E
TA
2
I E
TA
2
 E1
TA
2
 E2
TA
 
 2 E1  E2
TA
 I1  I 2  I12
Interference may, or may not, arise depending in the nature of I12
Getting interference is not easy.

 

 
 
 
I12  2 E01  E02 cos k1  r  1t  1 cos k 2  r   2t   2
TA
 
 
  cos k1  r  1 cos1t  sin k1  r  1 sin 1t 
 2 E01  E02
 
 
cos k 2  r   2 cos 2t  sin k 2  r   2 sin  2t
 
 






 
I12  0 if E01  E02  0
The two beams must have some common polarization


TA
Temporal Coherence
Waves should have the same frequency for interference. We will
need:
cos1t cos 2t
TA
1 T
  dt cos1t cos 2t
T 0
0
1
sin T
dt cost cos    

T0
2T
T
This term decays away in a time called the coherence time:
Tcoh  2 
Coherence Length:
Lcoh  cTcoh  2c   2c f
This is the length for which wavetrains stay in phase.
It is extremely difficult to maintain coherence for two beams
UNLESS they come from the same laser.
Consider two different, but similar lasers. At l=850 nm one can
“lock” laser to an atomic Cs transition.
Possible to have two lasers with:
f  100kHz f  3.531014 Hz
Lcoh  c f  3 108 m / s 100kHz  3km
Tcoh  1 / f  10s
Long, but not infinite.
What is worse – “Mode Hops”
Every s or so a laser will randomly shift its phase. Two lasers
will do this independently and interference shifts.
Back to Interference Criteria
Assume two light beams from the same source.
 
 
 
I12  2 E01  E02 cos k1  r   t  1 cos k 2  r   t   2
TA
 
 
  cos k1  r  1 cost  sin k1  r  1 sin t 
 2 E01  E02
 
 
cos k 2  r   2 cost  sin k 2  r   2 sin t TA
 
 
  1  cos k1  r  1 cos k 2  r   2  

 2 E01  E02 
 
 
2  sin k1  r  1 sin k 2  r   2 
 
 
 
 E01  E02 cos k1  r  1  k 2  r   2
 
 E01  E02 cos
Depends on the phase difference of 2 waves
 
 






 
 
 










Note that:
 
I12  E01  E02 cos
 2
2
I1  E1
 E01 2
TA
2
I 2  E2
TA
 2
 E02 2
 I  I1  I 2  2 I 2 I 2 cos
Constructive interference:
I  I1  I 2  2 I 2 I 2 for   0,2 ,4 ,...
Destructive interference:
I  I1  I 2  2 I 2 I 2 for    ,3 ,5 ,...
Spatial Coherence
Source 1
Observation
point
Source 2
Rays coming from extended source will have different
phases. Hard to get interference between sources 1 and
2. Phase of light all mixed up.
Source 1
Source 2
For point sources, arriving light has well defined phase.
Observation
point
2 Pinhole Sources
Grimaldi’s experiment of 1665
Observation
screen – no
interference
observed. No
“fringes”
Sun
2 pinholes
or slits
Sun is an extended source – no spatial coherence. No interference
Young
Thomas Young, 1805. Used the sun, but an additional pinhole
creates a (small) source with spatial coherence
Sun
1 pinhole
2 pinholes
Observation
screen
Difference in path lengths is:
r2
r1
r1  r2  d sin   d
But   y / L
Constructive interference when
r1  r2  ml
where m is an integer
 lL 

 4 I 0 cos2 
 dy 


 2 lL 
2

I  4 I 0 cos2  1 2   4 I 0 cos2 
 k and
 2screen.
r troughs
r   on observing
yd 
Peaks
I1  I 2  I 0
predicts
location
peaks
0 2
0 cosof
1  r2of troughs,
 Correctly
I

2
I
I
k
r

But not envelope (next week).
Waves versus particles (or so we think)
Cool demonstration of double slit on the web. See:
http://micro.magnet.fsu.edu/primer/java/doubleslit/
You can change wavelength of laser and the distance
between the two slits.
Light & Matter, Waves &
Particles – de Broglie
Wave and interference effects can be seen with matter too!
Quantum wave properties:
E  hf ,
p  hf / c  h / l for photons
De Broglie said, why not matter too?!?
l  h/ p
Interference properties seen with electrons, neutrons,
atoms, and now even molecules like C60 and C70!
Interferometers or the Double Slit: Interference seen even
when only one particle is in system. Particle (be it
electron, photon, atom, etc) goes through both slits at
once.
Observed interference of
60
C
See results of Prof. Anton Zeilinger and his group
http://www.quantum.univie.ac.at/research/
and
70
C
Standing Waves
Consider two counter-propagating waves from a single laser.
Say too that they have equal amplitudes.
Standing Waves
Consider two counter-propagating waves from a single laser.
Say too that they have equal amplitudes.
Etotal


 E0 coskz  t     E0 cos kz  t 
I  I1  I 2  2 I 2 I 2 cos2kz    
 2 I 0  2 I 0 cos2kz     4 I 0 cos2 kz   / 2
I  4I 0 cos kz   / 2
2
Two traveling waves produces a standing wave.
Microwave Ovens
Standing microwaves
Peaks and troughs => Hot spots and cold spots
=> Nodes and anti-nodes
Spinning dish hopefully brings all parts of food into
contact with nodes
Demo with marshmallows and a microwave
Beamsplitters
In order to create multiple beams from a single laser one
needs to use a “beamsplitter”
R*I0
Laser
r*E0
E0
t*E0
I0
T*I0
Usually (but not always) us 50-50 beamsplitter; half the light
transmitted, half reflected. Example- half silvered mirror.
Polarizing Beamsplitters
Some beamsplitters separated light according to polarization

E//
 

E  E  E//
Laser

E
Mach-Zender Interferometer
Mach-Zender Interferometer

E0

t * E0
Say that the length for the top
path is L1 and L2 for the bottom.

r * E0
At detectors
D1 and D2

 ikL1
 ikL2
E1  rtE0e  rtE0e

 ikL1
 ikL2
2
2
E2  t E0e  r E0e
If other factors would change the
acquired phase for the two paths it
would affect counts at D1 and D2.
 k L1  L2  
I1  I 0 cos 

2


2
 k L1  L2  
I1  I 0 sin 

2


2
L1
Mirror 2
BS2
L2
L2
Mirror 1
BS1
L1
Atoms or neutrons: Say interferometer is in a gravitational field.
The arms of length L1 are parallel to the ground, while when
particles are in the arms of length L2 they climb up against gravity.
Interferometer pivoted about bottom arm by angle .
Changed phase between two interferometer paths by insertion of Al in
one path, or rotating interferometer in Earth’s gravitational field.
Atom Interferometry –
Overlapping Na atoms from
Bose-Einstein Condensate
Atom Laser: Results from Wolfgang Ketterle’s group, MIT.
Ketterle shares 2001 Noble Prize in Physics
Michelson Interferometer

E0
L2
L1
I0
I out  I 0 sin 2 k L1  L2 
For equal arm lengths => L1 = L2
No light out of that beamsplitter port for all wavelengths –
White Light Fringe
Michelson-Morley’s Search for
the Aether
Homework 7, problem 1
Mirror 2
L
Mirror 1
Incident light
L
Beamsplitter
Calculate time for
light to traverse each
arm, according to
ether theory. Take v
to be Earth’s orbital
velocity.
Ether wind of speed v
Gravitational Radiation Detection
Laser Interferometric Gravitational Wave Observatory
LIGO
HANFORD
Washington
MIT
Boston
CALTECH
Pasadena
LIVINGSTON
Louisiana
Hanford Observatory
4 km
2 km
Livingston Observatory
4 km
LIGO Interferometers
Power Recycled
Michelson
Interferometer
With Fabry-Perot
Cavities
end test mass
Light bounces back
and forth along arms
about 30 times
Light is “recycled”
about 50 times
input test mass
Laser
signal
4 km Fabry-Perot arm cavity
beam splitter
Vibration Isolation Systems
Core Optics
Core Optics Suspension and
Control
Core Optics Installation and
Alignment
Washington 2k Pre-stabilized
Laser
Custom-built
10 W Nd:YAG
Laser
Stabilization cavities
for frequency
and beam shape
Fabry-Perot Interferometer
Er
E0
Mirror 1
Mirror 2
Et
L
Cavity of length L, incident light of amplitude E0 and
wavelength l, k=2/l
Et  E0t1t2eikL  E0t1r2 r1t2eik 3 L  E0t1r2 r1r2 r1t2eik 5 L  ...
Er  r1E0  E0t1r2t1eik 2 L  E0t1r2 r1r2t1eik 4 L  ...
Real Mirrors have losses
Er
E0
Ir
Mirror
Et
I0
It
R T  A 1
A is the loss coefficient. For a very good mirror A~10-4 to 10-3
Free Spectral Range
When 2kL changes by 2 we get another resonance
2
2L
l
 2m
4Lf
 2m
c
Where m is some integer
f 
c
m
2L
f 
c
2L
Free Spectral Range (FSR) = c/2L
L=2cm => FSR=7.5GHz
Cavity resonances function as “fence posts” or references
Fabry-Perot Interferometer as an
Optical Spectrum Analyzer
What’s going on with a laser???
c
f laser 
n
2L
where n is an int eger
An integral number of half-wavelengths fit into a
laser cavity
gain
frequency=(c/2L)n, n an integrer
The laser medium will have some gain profile,
as a function of frequency.
Overlap of gain profile and possible longitudinal modes
Resulting observed laser modes at various frequencies.
Fabry-Perot mirror mounted to a piezo-electric crystal
Vary cavity length, and scan frequencies that resonate
Mirror 1
Mirror 2
AC Voltage
PZT
L
The Speckle Effect