# Interference of light

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```Interference of light
• Interference of light waves similar to interference of
water waves
– two different waves arrive at the observation point
– the total influence is the sum of the two wave amplitudes at
each time and at each point in space
• High frequency of light has important consequences
– Cannot follow the fast cycling of the field
– Detectors measure the effect of many oscillations
– Only interference that persists over many periods is
observable
• Only two-beam interference is discussed in this module
– Multiple beam interference will be treated in next module
April 03
LASERS 51
What does an optical power meter
measure?
•
de
Amplitu
=A
Electron in
atom
Field wiggles electron causing it to escape
from atom. Becomes electrical current
• Electron is released due to wiggling
– Current from detector doesn’t go up and down with field
– Responds to the amplitude not phase
• Fast detectors can measure changes in 100 psec (10-10sec)
– This includes 60,000 cycles of the field
– Unlike radio waves where individual cycles can be measured
• The irradiance (power per unit area) is proportional to A2,
the square of the amplitude
April 03
LASERS 51
Power measurement with
interfering waves
• Detector measures square of amplitude of resultant
field (field obtained by adding waves at detector)
– Two 1000 Watt beams impinging on detector will give a zero
power reading if out of phase and 4000 Watts if in phase!!!
– Where does the power go? – somewhere else
• If the relative phase of the fields is not constant (e.g.
incoherent light) then interference effects go away
April 03
– This makes observation of interference difficult
LASERS 51
Two point sources create
interference pattern
• Along red lines
– Crests of two sources always
coincide
– Valleys of two sources always
coincide
– Net disturbance has twice the
amplitude
•Along green lines
–Crest of one wave always
occurs with valley of the other
–Water is undisturbed along
these lines
Interference of two spherical waves
source 1
crests of wave •
“Frozen” in time
– where two crests
coincide, amplitude
crests intersect is double that of a
single source
– where two troughs
lines of
constructive
coincide, amplitude
interference
is negative and
twice as deep as a
single source
troughs
intersect
source 2
April 03
Along indicated arrows waves have
twice amplitude of a single source
Constructive interference
LASERS 51
Interference of two spherical waves (cont)
source 1
• “Frozen” in time
crest wave 1
intersects
trough wave 2
lines of
destructive
interference
– If a crest of one
wave coincides
with a trough of
another wave,
there is no net
disturbance
crest wave 2
intersects trough
wave 1
Along indicated arrows waves have
zero amplitude!
source 2
April 03
Destructive interference
LASERS 51
Optical path difference (OPD)
source 1
source 2
d1
d2
OPD = d1-d2
observation point •
• Optical path length determines how long it takes light to travel
from the source to the observation point
– Phase at an instant of time depends on optical path length
• Optical path difference determines the phase difference at
observation point between light from two sources
• Understanding interference given the OPD is easy
• Finding the OPD in many cases is very complicated geometry
April 03
LASERS 51
amplitude_
(to tal)
amplitude
(wave 2)
amplitude
(wave 1)
OPD = 0, in-phase waves
1
Wave at P
due to
source 1
0
-1
1
0
-1
2
1
0
-1
-2
April 03
Wave at P
due to
source 2
source 1
d1
Observation
point, P
source 2
d2
d1=d2
•
• Optical path from either
source to observing point
is the same
• Resulting wave 2x
amplitude, same phase as
either component
• Intensity is four times that
of either source
LASERS 51
amplitude
(wave 2)
amplitude
(wave 1)
OPD = λ/2, 180&deg; out of phase
1
0
due to
source 1
-1
1
0
-1
2
2
amplitude (to tal)
Wave at P source 1
1
0
-1
April 03
Wave at P
due to
source 2
d1
Observation
Point, P
source 2
d2
•
d1=d2+λ/2
• Optical path difference
(OPD)=λ/2
– Phase difference=180&deg;
• Resultant amplitude zero
Conservation of energy works! Energy “missing”
due to destructive interference is redistributed to
regions of constructive interference
LASERS 51
amplitude
(wave 2)
amplitude
(wave 1)
Superposition of waves 90&deg;source
out 1of phase
source 2
1
0
-1
d1
d2
1
0
-1
2
amplitude (to tal)
d1=d2+λ/4
1
0
-1
-2
April 03
observation point •
• Path length from source 1
and 2 not the same
– Optical path difference
(OPD)=λ/4
– Phase difference=90&deg;
• Result: same frequency,
amplitude 1.4x, phase
different from either wave
• Intensity 2 times single
source intensity LASERS 51
More about superposition
• To find interference pattern
you need to know the OPD
Interference
observed on card
– Often difficult to calculate.
laser
Calculation not needed to
beam
understand interference.
Diverging
lens
– Changing OPD by λ, 2λ, 3λ, etc.
Glass plate
doesn’t change interference
– Constructive interference: OPD=0, λ, 2λ, 3λ, etc. (integral number
of wavelengths)
– Destructive interference: OPD=λ/2, 3λ/2, 5λ/2, etc. (half-integral
number of wavelengths)
• Interference between waves with unequal amplitudes
– OPD=half-integral number of waves
• incomplete cancellation
• Dark fringes are not completely dark
– When OPD=integral no. of waves
April 03 • intensity less than four times that of a single wave
LASERS 51
amplitude (wave 1)
Coherence
1
Wave due to
source 1 has
random
phase jumps
0.5
0
-0.5
-1
-1.5
amplitude (wave 2)
1.5
Wave due to
source 1 has
different
random
phase jumps
1
0.5
0
-0.5
-1
amplitude (to tal)
1.5
Observation
Point, P
d2
•
– On a femtosecond timescale
interference still occurs but it
is not observable
1
0.5
0
-0.5
-1
-1.5
April 03
d1
source 2
• No interference observed
-1 5
2
-2
source 1
Superposition has some regions
of constructive other regions of
destructive interference
• If phase jumps in each
source are the same
interference returns
LASERS 51
Young’s double-slit experiment
• First demonstration of wave nature of light
– Thomas Young, 1802
– not accepted until Fresnel’s work 12 years later
point
source
• Division of wavefront
opaque screen
with two slits
April 03
– single wavefront from point source strikes both
slits simultaneously
– source is nearly monochromatic
– point source must be nearly same distance
from each pinhole
LASERS 51
Young’s experiment—fringe spacing
Essentially two
sources, Huygen’s
principle
D
d
δ=OPD
L
slits
OPD calculation
easy if:
Observation
point
screen
• Assume L is very
large compared to
D or d
• By (approximately)
similar triangles
Photo of pattern
on screen
D δ
=
L d
• Constructive interference (bright fringes) if δ is 0 or an
integral number of wavelengths
L
D = nλ where n = 0,1,2, L
d
Spacing between bright fringes is λL/d
April 03
LASERS 51
Two slit interference pattern
• Interference of two slits
gives a sinusoidal
variation in intensity
• This pattern can be
modulated by an overall
intensity pattern
Intensity
– This pattern is charcteristic
of all two beam
interference
– Due to diffraction as in
next module, or uniformity
of illumination in other
cases
April 03
Position on screen
If intensity at low points
don’t go to zero (unequal
illumination of slits or
partial coherence) fringes
are indistinct or fuzzed out
(low visibility)
LASERS 51
Interference using nonmonochromatic sources
• Each wavelength produces
interference fringes
screen
– spacings are different
• At center n=0, all wavelengths
have a bright fringe
• For larger n, fringes become
colored, red on outside blue on
slits
inside
• Finally for larger n fringes
become completely washed out
White-light interference can only
be observed for OPD &lt;&lt; λ
April 03
Fringes in
red light
Fringes
in blue
light
LASERS 51
Young’s fringes - large source
• For a large source each
point on the source
produces a set of fringes
point 2
• Fringes are shifted
relative to each other
• Waves from each point
interfere add
point 1
“incoherently”
– Interfere with themselves
only
• A light field incident on
the set of slits is said to
have spatial coherence if
an interference pattern is
produced on the screen
April 03
screen
slits
Fringes
from point 1
Fringes
from
point 2
LASERS 51
Why do fringes from two different points
wash out?
• Atoms excited by some
energy source
• Excited atoms decay at
random times
– average lifetime (~10-8 sec)
– each emission results in very
short burst (wavepacket)
– no phase relationship
Atoms
between different packets
in gas
• Emission between different
points, or even same point
at different times are
incoherent
April
03
Wave packets emitted
by each atom at
random times in
random directions
LASERS 51
Interference between incoherent sources
a m plitude (wa ve 1)
1.5
1
0.5
0
-0.5
-1
a m plitude (wa ve 2)
-1.5
1.5
1
0.5
0
-0.5
-1
-1.5
2
1.5
a m plitude (tota l)
1
0.5
0
-0.5
-1
-1.5
-2
April 03
• Shown is superposition of
a randomly restarted wave
with a perfect sine wave
• sudden jumps represent
termination of light from
one atom and start of light
from another
• Superposition sometimes
in phase sometimes out of
phase
– average over many cycles
shows no enhancement by
interference
LASERS 51
a m plitude (wa ve 1)
1.5
1
0.5
0
-0.5
-1
a m plitude (wa ve 2)
-1.5
1.5
1
0.5
0
-0.5
-1
-1.5
2
1.5
a m plitude (tota l)
1
Coherence length of a laser
• Because of feedback,
laser light has longer
“memory” of its phase
• Nevertheless, the phase
drifts and eventually goes
out of phase
0.5
0
-0.5
-1
-1.5
-2
April 03
LASERS 51
amplitude (wave 2)
amplitude (wave 1)
1.5
1
Superposition of waves with different
frequencies
• May be from two different
lasers or two different
atoms with different
Doppler shifts
• Resultant varies
periodically in amplitude as
the two waves go in and
out of phase
• Hetrodyne detection of fm
signal
0.5
0
-0.5
-1
-1.5
1.5
1
0.5
0
-0.5
-1
-1.5
2
1.5
amplitude (total)
1
0.5
0
-0.5
-1
-1.5
-2
“A photon can only interfere with itself.”-P. Dirac
April 03
LASERS 51
Fringe visibility
• Perfectly coherent sources of
High
equal intensity give maxima
visibility
of intensity 4x intensity of one
source and zero intensity minima
• Coherent source of unequal
Low
intensities give fringes with
visibility
maxima of I1+I2+2 √(I1I2) and
minima of I1+I2-2 √(I1I2)
• Partially coherent sources give
fringes with lower maxima and higher minima
depending on the degree of coherence
• Completely incoherent sources give no fringes (intensity
everywhere = 2x intensity of one source)
I max − I min
Visibility ≡
I max + I min
LASERS 51
April 03
Lloyd’s mirror
point source
image of
point source
mirror
Interference fringes
in overlap region
• Light reflected from mirror interferes with light
directly from point source
• Considered as interference between the point
source and its image, this is almost identical to
Young’s fringes
– one significant difference is that center fringe is dark
due to phase change on reflection
April 03
LASERS 51
Division of amplitude
Michelson interferometer
• Source may also be a non- expanded
laser point source or even laser beam
an extended source
– compensator plate required in
this case
• Tilting one mirror produces
straight-line fringes
• Often used for optical
testing
April 03
mirror 2
beamsplitter
mirror 1
observation
screen
LASERS 51
Interference from wedged plates
input wave
• Waves from two surfaces
interfere at observation point
– OPD determined only by path
between reflecting surfaces –
other parts of path are common
• Interference from other
d
surfaces occurs with laser
sources
wave reflected
α
– With incoherent sources OPD
for other surfaces is greater than
wave reflected
OPD = 2d + λ / 2
from second surface
coherence length
phase change on reflection
– Small gap, nearly
d = αx for small angles
monochromatic source needed
OPD = 2αx + λ / 2
to observe interference with
incoherent sources
Bright fringes when OPD=integral number
λ 5λ 9λ
of waves
– Spatial coherence not needed
x=
,
,
, etc
4α 4α 4α
LASERS 51
April 03
x
from first surface
Newton’s rings
monochromatic,
extended light source
eye
part under test
OPD, optical
path difference
High quality
optical flat
• Common optical testing technique
– optical flat may be replaced by a curved test plate
• Optical path difference varies with shape of tested part
– radius of part, as well as size of defects can be measured
– dark fringe in center due to phase change on reflection
April 03
LASERS 51
Antireflection (AR) coatings
Air,
index na
Glass, index ng
Coating,
index nc
Reflection coefficient
of uncoated glass
R=
(ng − na )
2
(ng + na ) 2
• Reflection from air/coating interface interferes with
reflection from coating/glass interface
– If nc=√(ng*na) and the coating is λ/(4nc) thick the two reflections
cancel completely for one wavelength
– For ng=1.5 and na=1.0 this requires nc=1.225 (unknownium)
– a 1/4-wave coating of a relatively low index material MgF(n=1.38)
can be applied to obtain a reflectance of 1.4% (at one wavelength)
– to get lower reflectance or more than one wavelength multiple
LASERS 51
April 03layers must be used
Shear plate interferometer
shear
Front surface
• For a plane input beam,
reflection
Back surface fringes are straight lines
reflection
perpendicular to tilt
direction
– Spacing of fringes depends
on tilt angle and
wavelength
input beam
Top view
Tilt
Side view
April 03
• Working out the OPD for
any other given wavefront
is complicated
• Shear can be measured
using shadow of object in
the input beam
LASERS 51
Shear-plate interferometer fringe
patterns
• Horizontal lines indicates a plane
wave
– Infinite radius of curvature
• Tilted straight lines indicates a
spherical wavefront
– Tilts one direction for converging wave,
opposite direction for diverging wave
• Deviation from straight lines means
wavefront is not plane or spherical
– i.e. aberrations
April 03
LASERS 51
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