32. Interference and Coherence - Brown University · 32. Interference and Coherence Interference...

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32. Interference and Coherence Interference Only parallel polarizations interfere Interference of a wave with itself The Michelson Interferometer Fringes in delay Measure of temporal coherence Interference of crossed beams Coherence Temporal coherence Spatial coherence Albert Michelson 1852-1931

Transcript of 32. Interference and Coherence - Brown University · 32. Interference and Coherence Interference...

Page 1: 32. Interference and Coherence - Brown University · 32. Interference and Coherence Interference • Only parallel polarizations interfere • Interference of a wave with itself The

32. Interference and CoherenceInterference• Only parallel polarizations interfere• Interference of a wave with itself

The Michelson Interferometer• Fringes in delay• Measure of temporal coherence• Interference of crossed beams

Coherence• Temporal coherence• Spatial coherence

Albert Michelson1852-1931

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Orthogonal polarizations don’t interfere.

The most general plane-wave electric field is:

where the amplitude is both complex and a vector:

0, Re exp ( )

E r t E j k r t

0 0 0 0, ,

x y zE E E E

don’t forget the complex conjugation!The irradiance is:

* * * *0 0 0 0 0 0 0 02 2

x x y y z zc cI E E E E E E E E

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Orthogonal polarizations don’t interfere (cont’d)Because the irradiance is given by:

combining two waves of different polarizations is different from combiningwaves of the same polarization.

* * * *0 0 0 0 0 0 0 02 2

x x y y z zc cI E E E E E E E E

Different polarizations (e.g., x and y):

* *1 1 2 2 1 22

x x y ycI E E E E I I

Same polarizations (e.g. both have x polarization):

**, , 1, 2, 1, 2,2 2

total x total x x x x xc cI E E E E E E

*1 21 2 Rec EI I I E Therefore: This is what is

called a “cross term.”

This cross term can give rise to very dramatic effects.

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The irradiance when combining a beam with a delayed replica of itself has “fringes.”

Suppose the two beams are E0 exp(jt) and E0 exp[jt)]. That is, a monochromatic wave and itself delayed by some time :

02 1 cos[ ] I I

Fringes (as a function of delay)

-

I

*1 1 2 2Re I I c E E I

*0 0 02 Re exp[ ] exp[ ( )] I I c E j t E j t

20 02 Re exp[ ] I c E j

20 02 cos[ ] I c E

How do we vary the delay of a beam?

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Varying the delay on purposeSimply moving a mirror can vary the delay of a beam by many wavelengths.

Since light travels 300 µm per ps in air, 300 µm of mirror displacement yields a delay of 2 ps. Delays of less than 0.1 fsec (10-16 sec) can be generated using this technique.

Moving a mirror backward by a distance L yields a delay of:

2 L /cNote the factor of 2.Light must travel the extra distance to the mirror—and back!

E(t–)Output beam

Translation stage

Input beam E(t)Mirror

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We can also vary the delay using a mirror pair or a corner cube.

Mirror pairs involve tworeflections and displace the return beam in space:But out-of-plane tilt yieldsa nonparallel return beam. Translation stage

Inputbeam

E(t)

E(t–)

MirrorsOutput beam

Corner cubes involve three reflections and also displace the return beam in space. Even better, they always yield a parallel return beam:

[EdmundScientific]

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output

The Michelson Interferometer

beam-splitter

inputbeam

delay

mirror

mirror

*1 2 0 1 0 2Re exp ( ) exp ( ) outI I I c E j t kz kL E j t kz kL

The Michelson Interferometer splits a beam into two and then recombines them at the same beam splitter.

Fringes (as a function of delay):

L = L2 – L1

Iout

Suppose the input beam is a plane wave. Then the intensity measured at the output is:

20 0 0 2 1 0 1 2 0 0 2 Re exp ( ) ( /2)since I I I jk L L I I I c E

0 2 1 cos( ) I k L

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The Michelson Interferometer - clarification

0 2 1 cos( ) outI I k LIf the path length difference is zero, this becomes: 0 4outI I

Are we getting more photons at the output than we put in? No!

But both beams that reach this point have passed through the 50/50 beam splitter twice, thus reducing their intensity (relative to the intensity of the input beam) by a factor of 4.

Thus: 0 4 inI I

In this expression, the symbol I0 refers to the intensity in either one of the two beams at this location.

beam-splitter

inputbeam

delay

mirror

mirror

output

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The Michelson-Morley experiment to measure the aether

Albert Michelson1852-1931

“The most famous failed experiment of all time”

The Aether Wind

Nobel Prize, 1907 (first American to win one of the science prizes)

Michelson’s laboratory, Case-Western University, 1887

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Interference of crossed beams

k

k

z

xˆˆcos sink k z k x

ˆˆcos sink k z k x

cos sink r k z k x

cos sink r k z k x

*0 0 02 Re exp[ ( )] exp[ ( )]

I I c E j t k r E j t k r

*0 0Re exp cos sin exp cos sin E j t kz kx E j t kz kx

Cross term is proportional to:

Fringes (as a function of x position)

x

Iout(x) Re exp 2 sin

cos(2 sin )

jkx

kx

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Big angle: small fringes.Small angle: big fringes.

2 /(2 sin )/(2sin )

k

The fringe spacing, :

As the angle decreases to zero, the fringes become larger and larger, until finally, at = 0, the intensity pattern becomes constant.

posi

tion

Small angle:

Large angle:

posi

tion

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You can't see the spatial fringes unlessthe beam angle is very small!

sin /(2 )

/(2sin )

= 0.1 mm is about the minimum fringe spacing you can see with the naked eye:

0.5 / 200 1/ 400 rad 0.15

m m

The fringe spacing is:

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The MichelsonInterferometer(Misaligned)

Suppose we misalign the mirrors, so the beams cross at an angle when they recombine at the beam splitter. And we won't scan the delay, so the lengths are equal.

Beam-splitter

Inputbeam

Mirror

Mirror

z

x

Fringes (in position)

x

Iout(x)

*0 0Re exp cos sin exp cos sin E j t kz kx E j t kz kx

If the input beam is a plane wave, the cross term becomes:

Re exp 2 sin

cos(2 sin )

jkx

kx

Crossing beams maps delay onto position If the path length difference changes, the fringes shift.

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The Michelson Interferometer: A question

beam-splitter

inputbeam

delay

mirror

mirror

Let’s go back, for now, to the well-aligned Michelson interferometer.

If we move the moveable mirror further and further back, do we continue to see fringes forever?

If not, then how far can we go before they disappear?

output

L = L2 – L1= large

02 1 cos( ) outI I k L

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The Temporal Coherence Time and the Spatial Coherence LengthThe temporal coherence time is the time over which the beam wave-fronts remain equally spaced. Or, equivalently, over which the field remains sinusoidal with a given wavelength:

The spatial coherence length is the distance over which the beam wave-fronts remain flat:

Since there are two transverse dimensions, we could talk about two different coherence lengths. Instead, we define a coherence area.

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Spatial and Temporal Coherence

Beams can be coherent or only partially coherent (or, even incoherent) in both space and in time.

Spatial and Temporal Coherence:

Temporal Coherence; Spatial Incoherence

Spatial Coherence; Temporal Incoherence

Spatial and Temporal Incoherence

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A nearly monochromatic light source has a large coherence time:

If we know there is a maximum here...

...and we wait a time equal to an integer number of periods...

...then we know that we will find another maximum.

time

E-field ampitude

The coherence time of monochromatic light

In the real world: highly stabilized lasers can have coherence times on the order of a few seconds. That’s amazing! More than 1015 cycles!

Thus, an ideal monochromatic light source has an infinite coherence time.

For a perfect cosine, the integer could be as large as you want (up to infinity), and this would still be true.

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A polychromatic light source has a smaller coherence time.

Here’s an example: an E-field composed of a superposition of several monochromatic waves, each with a slightly different frequency.

Start at this maximum.

Wait N periods, where N = 1, 2, 3, ...

What is the value of the E-field at each successive value of N? Is it still a local maximum?

1 2 3 45

6 7 8 9101112131415

1617

181920 2122

2324

25

2627

The coherence time of polychromatic light

The coherence time for a given waveform is the average amount of time one has to wait from an arbitrary starting point before coherence is lost.

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What determines the coherence time?

More spectral components = more rapid loss of coherence

sum of 3 sine waves

S()

E fi

eld

time

sum of 5 sine waves

S()

E fi

eld

time

sum of 10 sine waves

S()

E fi

eld

time

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The coherence time is the reciprocal of the bandwidth.

The coherence time is given by:

where is the light bandwidth (the width of the spectrum).

1/c v

Short optical pulses also have small coherence times, roughly equal to their duration.

12

354

6

Sunlight is temporally very incoherent because its bandwidth is very large (the entire visible spectrum).

“Coherence length” = (c0/n)c

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Why are we interested in coherence time?Because the notion is relevant to measurements that we often do. Let us suppose that we take a wave and interfere it with a copy of itself:

a wave

a time-delayed replica of the same wave

If the time delay is zero ( = 0): perfect constructive interference at every point. The net irradiance is large.

If the time delay is half the period ( = ): nearly perfect destructive interference at every point. The net irradiance is zero.

If the time delay is the period ( = 2): constructive interference at every point. The net irradiance is large.

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a wave

a time-delayed replica of the same wave

If the time delay is large ( > c): There is no correlation between the peaks of the wave and the peaks of the time-delayed version. Interference is sometimes constructive, sometimes destructive. The net irradiance no longer depends on the delay .

time delay

Interferogram: the pattern formed by the net irradiance as a function of delay.

at this time, constructive interference

destructive interference here

Why is this interesting? One reason: because interferograms are easy to measure using a Michelson interferometer.

What if the delay time is large?