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 Orthogonal polarizations don’t interfere.

The most general plane-wave electric field is:  E rt,Reexp( E jk r t )    0  

where the amplitude is both complex and a vector:    EEEE0000  xyz,,

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

cc   IEEEEEEEE****  2200 0xx 0 0 yy 0 0 zz 0 Orthogonal polarizations don’t interfere (cont’d) Because the irradiance is given by: cc   IEEEEEEEE****  2200 0xx 0 0 yy 0 0 zz 0 combining two waves of different polarizations is different from combining waves of the same polarization. Different polarizations (e.g., x and y): c I EE** E E I I 2 11xx 2 y 2 y 12 Same polarizations (e.g. both have x polarization):

cc *  * IEE EEEE 22total, x total , x 1, x 2, x 1, x 2, x  * This is what is Therefore: III 12cE Re 12 E  called a “cross term.” This cross term can give rise to very dramatic effects. The irradiance when combining a beam with a delayed replica of itself has “fringes.”

* I  Ic1122 Re EE I

Suppose the two beams are E0 exp(jt) and E0 exp[jt)]. That is, a monochromatic wave and itself delayed by some time : * IIc200 Re E exp[ jtE ]  0 exp[  jt (  )]

2 2Ic00 Re E exp[ j ] 2 Fringes (as a function of delay) 2cos[]IcE00  I

II21cos[]0    -  How do we vary the delay of a beam? Varying the delay on purpose

Simply moving a mirror can vary the delay of a beam by many wavelengths. Input Mirror beam E(t) Output beam E(t–) Translation stage

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

Note the factor of 2.   2 L /c Light must travel the extra distance to the mirror—and back!

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

Mirror pairs involve two E(t) Input beam reflections and displace Mirrors the return beam in space: Output But out-of-plane tilt yields E(t–) beam a nonparallel return beam. Translation stage

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

[Edmund Scientific] The Michelson Interferometer input beam The Michelson Interferometer splits a beam into two and then recombines mirror output them at the same beam splitter. beam- splitter delay Suppose the input beam is a plane wave. mirror Then the intensity measured at the output is:

* IIIcEout 12 Re 0 exp jtkzkLE (  1 )  0 exp  jtkzkL (  2 ) 2 I00II 2 0 Re exp jkLL ( 21  ) since IIIc 012  ( 0 / 2) E 0

 2I0  1 cos(kL ) Iout

Fringes (as a function of delay):

L = L2 –L1 The Michelson Interferometer - clarification

IIout  20  1 cos( kL )

If the path length difference is zero, this becomes: IIout  4 0

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

In this expression, the symbol I0 refers input to the intensity in either one of the two beam beams at this location. mirror But both beams that reach this point output beam- have passed through the 50/50 beam splitter splitter twice, thus reducing their intensity (relative to the intensity of the input delay mirror beam) by a factor of 4.

Thus: II0  in 4 The Michelson-Morley experiment to measure the aether “The most famous failed experiment of all time”

The Aether Wind

Albert Michelson 1852-1931

Michelson’s laboratory, Case-Western University, 1887

Nobel Prize, 1907 (first American to win one of the science prizes) Interference of crossed beams x   k kkcos zkˆ sin xˆ   kkcos zkˆ sin xˆ   z   krkcos zk  sin x    krk cos zk  sin x k    *  IIc200 Re E exp[j (tkrE  )] 0  exp[ j (tkr  )]

Cross term is proportional to: * ReE00 exp j t kz cos kx sin E exp   j t kz cos kx sin 

Re exp 2jkx sin  Fringes (as a function of x position) I (x)  cos(2kx sin ) out

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

Large angle: The fringe spacing, :

2/(2sin)k  position  /(2sin )

As the angle decreases to Small angle: zero, the fringes become larger and larger, until finally, at

 = 0, the intensity pattern position becomes constant. You can't see the spatial fringes unless the beam angle is very small!

The fringe spacing is:

   /(2sin )

 = 0.1 mm is about the minimum fringe spacing you can see with the naked eye:   sin /(2 )  0.5mm / 200  1/ 400 rad 0.15 The Michelson Input beam x Interferometer  z (Misaligned) Mirror Beam- splitter Suppose we misalign the mirrors, so the beams cross at an angle when they recombine at the beam Mirror splitter. And we won't scan the delay, so the lengths are equal. If the input beam is a plane wave, the cross term becomes: * ReEjtkzkxE00 exp cos sin exp   jtkzkx cos sin 

Re exp 2jkx sin  Fringes (in position) Iout(x)  cos(2kx sin )

x Crossing beams maps delay onto position If the path length difference changes, the fringes shift. The Michelson Interferometer: input A question beam

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

L = L2 –L1 If we move the moveable mirror = large further and further back, do we continue to see fringes forever?

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

IIout 21cos()0  kL The Temporal Coherence Time and the Spatial Coherence Length The 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. Spatial Spatial and Temporal and Coherence: Temporal Coherence Temporal Coherence; Spatial Incoherence Beams can be coherent or only partially coherent (or, even incoherent) Spatial in both space and in Coherence; Temporal time. Incoherence

Spatial and Temporal Incoherence The coherence time of monochromatic light

A nearly monochromatic light source has a large coherence time: E-field ampitude

time

If we know there is ...and we wait a time equal to ...then we know that we a maximum here... an integer number of periods... will find another maximum.

For a perfect cosine, the integer could be as large as you want (up to infinity), and this would still be true. Thus, an ideal monochromatic light source has an infinite coherence time. 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! The coherence time of polychromatic light 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. 1 2 3 4 5 6 11 7 910 12 27 8 1314 15 26

16 25 17 24 18 23 19 22 20 21

What is the value of the E-field Start at this Wait N periods, at each successive value of N? maximum. where N = 1, 2, 3, ... Is it still a local maximum?

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. What determines the coherence time?

S() sum of 3 sine waves E field

time  S() sum of 5 sine waves E field

time  S() sum of 10

sine waves E field

time  More spectral components = more rapid loss of coherence The coherence time is the reciprocal of the bandwidth.

The coherence time is given by:

 c 1/v

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

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

1 Short optical pulses also have 2 3 small coherence times, roughly 4 equal to their duration. 5 6

“Coherence length” = (c0/n)c 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. What if the delay time  is large? at this time, constructive interference destructive interference here

a wave

a time-delayed replica of the time delay  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 . Interferogram: the pattern formed by the net irradiance as a function of delay.

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