ece506 class 2014 1 to 40

ECE506. Optical interferometry and laser metrology Fall 2014. M 5.00-7.40 Eng B4

Mario C. Marconi

(970) 491-8299 (office)- (970) 481-4487 (cell phone)

[email protected]

Optical Interferometry. P. Hariharan “Optical Interferometry”

Other resources: D. Malacara “Interferogram analysis for Optical Testing”

Laser Fundamentals W. Silfvast.

Optics Hetch-Zajac.

Optical Physics. Lipson, Lipson & Lipson

Principles of optics. Born and Wolf

Topics:

Interferometry principles and applications

Laser devices

Sensors and metrology

Grading Midterm 20% Final 40% Homework 20% Presentation 20% The system +/- will be used in this class Exams dates to be defined

Misc Information Homeworks will be assigned after finishing each main topic

Brief review: Historic overview

Classical model of oscillating electron

Spontaneous decay. Classical model

Linewidth

Broadening processes

EM waves

Interference: a common phenomenon observed in soap bubbles, an oil slick on a wet road, in colored fringes seen between two glasses.

Historic Overview

Hooke postulated that these fringes were produced by interference and proposed a wave theory. Huyghens in 1690 put together the wave theory into its present form. But this theory made little progress because was opposed by Newton.

100 years….

1801-1803: Young stated the principles of interference. Summation of 2 beams could give rise to darkness

1809 Malus demonstrated that light can be polarized by reflection. This was in contradiction to Young’s and Huygens thought that light was propagated as longitudinal waves.

1818 Fresnel diffraction theory perfected the treatment of interference. Fresnel and Arago found that two orthogonally polarized beams could not interfere. This lead to the conclusion that light is a transverse wave. Fresnell postulated that light propagates in a medium called “luminiferous aether”.

Historic Overview

19th century. Most physicist supported the aether theory. Maxwell predicted in 1880 that the movement of the Earth though the aether should result in a change of the speed of light by a factor (v/c)2

1881 Michelson’s experiment to demonstrate the “aether drift”

1896. Michelson measured the bar of Pt-Ir (international prototype of the meter) in terms of the wavelenght of red Cd radiation. This experiment lead in 1960 the redefinition of the meter in terms of the wavelength of the orange radiation of Kr

Optical testing: Twyman in 1916 used Michelson interferometer to test optical components

Coherence: studies by Michelson also revealed the connection between the interference fringe visibility and the size and spectral purity of the light source. The phenomenon was observed in 1869 by Verdet, who showed that sun light going through two pinholes produced interference if the pinholes are less than 50 microns apart.

Historic Overview

The first quantitative concepts of coherence were formulated by Laue in 1907

Cittert (1934, 1939) and Zernike (1938) published the present coherence theory.

These concepts were developed in more extend later by Hopkins (1951, 1953) and Wolf (1954, 1955). Later Mandel and Wolf described higher order coherence effects.

Interference Spectroscopy: the origin of this technique can be tracked back to 1862, when Fizeau studied the effect of the separation of the plates of Newton’s rings formed with sodium light.

Changing the distance he observed that the rings almost disappeared at a separation corresponding to 490 fringes and reached a maximum again when 980 fringes had passed. Conclusion the Na line was a doublet.

Historic Overview

Michelson in 1891 plotted the visibility of the fringes as a function of the optical path difference for different spectral lines and observed that all of them except the red line of Cd exhibited a series of maxima and minima indicating multiple lines.

The laser: 1917 A. Einstein pointed out the phenomenon of stimulated emission. Schalow and Townes (1958) and Maiman (1960) demonstrated laser action in the visible region of the spectrum.

Heterodyne techniques: the observation of beats when the beams of two lasers operating at slightly different frequencies were mixed at a photodetector (Javan et al, 1962) led to the development of a range of heterodyne techniques to replace traditional methods of interpolation. Since measurements of the frequency or the phase of the beat can be done with high accuracy, the heterodyne techniques revolutionized length interferometry.

Fiber interferometers: Advantage of large optical paths in reduced space. Can be used to measure rotation (laser gyroscopes), temperature, strain, pressure, etc..

Non-linear interferometers: High intensity coherent laser beams allows the use of non linear effects. Some examples are second harmonic interferometry, phase conjugate interferometry, and high speed optical switching

Historic Overview

Wave equation for free space

Starting from Maxwell’s equations it is possible to derive the “wave equation”.

∇×H = J +ε0∂E∂t

+∂P∂t

∇×E = −µ0∂H∂t

+∂M∂t

We can eliminate H taking the curl of the first equation and replacing in the second

∇×∇×E = µ0∂∂t

∇×H( ) = −µ0ε0∂2E∂t2

Or

∇2E − 1c2

∂2E∂t2

= 0

Similarly we can derive an equation for H

∇2H −1c2

∂2H∂t2

= 0

c = 1µ0ε0

Wave equation for free space

E r, t( ) = Re E ω,k0( ) exp jω t − an ⋅ rc

#

$%

&

'(

)

*+

,

-.

/01

234

Any function of the form is a solution of the wave equation f t − an ⋅ rc

#

$%

&

'(

Or

E r, t( ) = Re E ω,k0( ) exp jωt( )exp − j k0 ⋅ r( ){ }

http://www.amanogawa.com/archive/wavesA.html

Harmonic waves and its superposition

Let us consider an elementary wave with wavenumber k and frequency ω

E x,t( ) = Re E ω,k( ) exp jωt( )exp − j k ⋅ x( ){ }= Re E0 exp j(ωt − k ⋅ x)!" #${ }

Now take an initial ( t = 0) superposition of such waves

g x,0( ) = Ei (x,0)i∑ = E0i exp − j(ki ⋅ x)$% &'

i∑

At time t each one of these elementary waves will evolve

g x,t( ) = E0i exp j(ωi t − ki ⋅ x)#$ %&i∑ = E0i exp jki (c t − x)#$ %&

i∑

Where we replaced ωi = ki cComparing the last equation with the definition of the superposition g (x,t), it is clear that

g x,t( ) = g (x − ct),0( )In words: the initial superposition g(x,0) has propagated with no change at velocity c. This simple result is a consequence of the substitution (*). This is the specific case of a non-dispersive medium. If we have a medium where different frequencies travel at different velocities (dispersive medium) the conclusion is different.

Attenuated waves and evanescent waves

Suppose the frequency ω is real but the dispersion relation leads to a complex value of k

k = k1 + j k2

Then the wave will be described by

E x,t( ) = E0 exp − k2 ⋅ x( )exp jωt( )exp − j k1 ⋅ x( )

Propagating wave Attenuation

In the case that the wavenumber is purely imaginary, there is no propagating component and we have an evanescent wave

E x,t( ) = E0 exp − k2 ⋅ x( )exp jωt( )

Uniform plane waves

Picture at a fixed position as time changes

( )0cosx mz

E E tω+

==

( )0

cosmy z

EH tωη

+

==

“linearly” polarized

Consider a E with two components

( ) ( )= cos cos+ +m1 m2E t z E t zω β ω β θ− + − +x yE a a

In the plane xy at z=0

( ) ( )z=0= cos cos+ +

m1 m2E t E tω ω θ+ +x yE a a

For the particular case 01 2 90m m mE E E θ+ + += = = −

( ) ( )0z=0= cos cos 90+ +

m mE t E tω ω+ −x yE a a

( ) ( )0cos sin 90+ +m mE t E tω ω= + −x ya a

“circularly” polarized

http://www.amanogawa.com/archive/wavesA.html

Wave equation: Snell’s Law

n1 n2

i

r

j

Without loss of generality we can treat two cases: when the E field is normal to the incidence plane (Ey ) and when E lies in the plane of incidence (E// ). Any other polarization state can be expressed as a combination of these two linear polarizations

z

x

y

E r, t( ) = Re E ω,k0( ) exp jωt( )exp − j k0 ⋅ r( ){ }

EyI = I exp − j k1 z cosi!+ k1 x sin i

!( )"#$

%&'

Incident beam

EyR = R exp − j k1 z cos j! + k1 x sin j

!( )"#$

%&'

Reflected beam

EyT =T exp − j k2 z cosr! + k2 x sin r

!( )"#$

%&'

Transmitted beam

The boundary condition at (z=0, and x=0) imposes I R T+ =

For the plane z=0 the oscillatory parts must be also be identical thus k1 sin i = k1 sin j

= k2 sin r

From which we can derive the Snell’s Law j = π − i

sin i = k2k1sin r = n2

n1sin r

If we define the reflection and transmission coefficients R = R / I and T = T / I , the coefficients for this polarization are:

R ⊥=n1 cosi

− n2 cosr

n1 cosi+ n2 cosr

T

⊥=2n1 cosi

n1 cosi+ n2 cosr

i

jr

Wave equation: Snell’s Law

For the other polarization the formula can be worked out similarly to obtain

R =n1 cosr

− n 2 cosi

n1 cosr + n 2 cosi

T =2n1 cosi

n1 cosr + n 2 cosi

Plots of the reflected intensity for the parallel polarization (blue) and the normal polarization (green). The reflection for the parallel component cancels at the Brewster angle

R =n1 cosr

− n 2 cosi

n1 cosr + n 2 cosi

= 0 ⇒ n1 cosr − n 2 cosi

= 0

n1 cosr − n 2 cosi

= 0 ⇒cosr

cosi=n 2n 1

=sin i

sin r

That can be written as tan i = n2n1

Total internal reflection

sin i = n2n1sin r

If n1>n2 and the refraction angle is 90, Snell Law imposes a limit to the incidence angle. This is the critical angle or total internal reflection angle defined as

ic = sin−1n2n1

Wave equation: Snell’s Law, Brewster angle SUMMARY

Snell’s Law: describes the direction of propagation of the waves produced when a plane wave impinges in an interface between two media

1 1sinncω

φ θ=

This relationship imposes that the incident and reflected waves make the same angle with respect to the normal to the surface. For the transmitted wave

1 1 2 2sin sinn nc cω ω

θ θ= 1 1 2 2sin sinn nθ θ=

Brewster’s angle: When the incidence angle is θB there is no TM reflected component

1 2 1 1 2 2 2 1 2 1sin sin sin cos2 2

n n n nπ πθ θ θ θ θ θ⎛ ⎞+ = → = = − =⎜ ⎟

⎝ ⎠Therefore

2

1

tan Bnn

θ =

Classical model of oscillating electron

•  Atoms composed by a massive nucleus and an electronic cloud •  The atom has well defined “resonances” •  These resonances are simple (no harmonics at 2ω, 3ω., etc.) •  They are dipole electric transitions

When an electric field E(t) is applied, the electronic cloud is displaced x(t), and the nucleus tends to recover the original position with a force - k x(t)

)()()(2

2

teEtxkdttxdm −=+

k = m ω2 where ω = √(k/m)

Assume: !

1212

EEmk −

=== ωω

* Experimentally was observed with molecular monolayer of dyes adsorbed on a metallic surface → the radiative dipole is modified according the classical calculation

Spontaneous decay. Classical model

The motion equation for an oscillating dipole adding the dumping term is

0)(4)()( 20

202

2

=++ txmdttdxm

dttxdm νπγ

Solving for γ0<<ω0 tit eexx 00 2/

0ωγ −−=

An oscillating dipole according classical EM will emit. Hence the emitted field E(t) will be:

( )

0 0/ 20( ) ( ) 00 0

t i tE t E t e e tE t t

γ ω− −⎧ = ≥⎪⎨

= <⎪⎩

)()()(2

2

teEtxkdttxdm −=+

k = m ω2 where ω = √(k/m) =2πυ0 Assume: !

1212

EEmk −

=== ωω

•  Atoms composed by a massive nucleus and an electronic cloud •  The atom has well defined “resonances” •  These resonances are simple (no harmonics at 2ω, 3ω., etc.) •  They are dipole electric transitions

Equation of motion for a e in a harmonic potential

Collisional decay

Collisions may: Induce decay → reduces the relaxation time. Interrupt the phase

Tcolisu

1=γ

The non-radiative processes are included in tuu

ueNN γ−= 0

colisu

raduu γγγ += ∑=

iui

radu Aγ

colisraduu γγγτ

+==

11

Ttot

111+=

ττBroadening mechanisms * Collisions with the discharge tube walls * Vibrations of the crystal (thermal). Levels change the energy by interaction with crystal lattice ------> phonon broadening * The field in one dipole acts of the other dipole: The fields are coupled and change the phase Collision: two atoms get close each other, modify its energy level structure (≈ 0.1 ps). The interaction changes the phase in the emission

collisions

Linewidth

Nu

NL

hυ

ων !==Δ hE

The emission is not infinite (in time) → Δυ ≠ 0

tit eetEtE 00 2/0 )()( ωγ −−=

We take the FT:

01( ) ( )2

i tE E t e dtωωπ

∞

−∞

= ∫

Solving: ( )

( )0 0 / 20 0

0 00

1( )/ 22 2

i i tE EE e dti i

ω ω γωω ω γπ π

∞− +⎡ ⎤⎣ ⎦= = −

− +⎡ ⎤⎣ ⎦∫

( )[ ]4/2/)()( 2

02

0

00

2

γωωπγ

ωω+−

== IEI

Lorenzian

0 50 100 150 200 250 300

0,0

0,2

0,4

0,6

0,8

1,0

Y Ax

is Tit

le

X Axis Title

B

Classic linewidth

( )[ ]4/2/)()( 2

02

0

00

2

γωωπγ

ωω+−

== IEI

Lorenzian

The characteristic parameter is FWHM (Full Width at Half Maximum)

0

0)(πγ

ωII =

Solving for (ω-ω0)

4)(1

)4/()()2/()(

21

)4/()()2/( 2

0202

02

0

20

0

020

20

00 γωω

γωωγ

πγω

γωωπγ

=−⇒=+−

→==+−

III max

This is valid for N atoms -------> homogeneous broadening (all atoms have the same linewidth)

Example: Crystals. Nd:YAG (Yttrium Aluminum Garnet)

2/001 γωω −= 2/002 γωω += 01 2 clasicω γ π ντ

Δ = = = Δ

Linewidth

0

0

2Iπγ

0ω

Doppler broadening

v

hν

detector

x

vx 0v(1 )xc

ν ν= ±

Doppler broadening is important in gaseous lasers

At a given temperature T, the velocity mean value for an atom is 8v kTMπ

=

Considering a Maxwell distribution

3/ 22 2 2(v , v , v ) exp (v v v ) v v v

2 2x y z x y z x y zM MP d d dkT kTπ

⎛ ⎞ ⎡ ⎤= − + +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

For an atom moving towards the observe the frequency shifts according

0v(1 )xc

ν ν= +

The probability G(ν) dν that the frequency is between ν and ν+dν equals to the probability that the velocity vx falls between

00

v ( )xc

ν νν

= − 00

v v ( )x xcd dν ν νν

+ = + −and Independent of vy and vz

At room temp the typical velocities are 100-1000 m/s

To obtain G(ν). We replace in P(vx,vy,vz) 00

v ( )xc

ν νν

= −0

vxcd dνν

=

Integrating P(vx,vy,vz) in vz and vy

3/ 2 22 2 2

020 0

( ) exp (v v ) v v exp ( )2 2 2y z y zM M M c cG d d d dkT kT kT

ν ν ν ν νπ ν ν

∞∞

−∞−∞

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

∫ ∫

0 50 100 150 200 250 300

0,0

0,2

0,4

0,6

0,8

1,0

Y A

xis

Title

X Axis Title

B

0 50 100 150 200 250 300

0,0

0,2

0,4

0,6

0,8

1,0

Y A

xis

Title

X Axis Title

B

Normalizing ∫∞

=0

1)( νν dG and

Using ∫∞

=0

0)( IdI νν Is obtained

⎟⎟⎠

⎞⎜⎜⎝

⎛−−= 2

020

2

00

))()(2(exp

2)( νν

νπνν

ckTMI

kTMcI

Gauss Lorenz

Doppler broadening

Each integral in vy and vz equals to MkTπ2 Hence

22

020 0

( ) exp ( )2 2

c M M cG d dkT kT

ν ν ν ν νν π ν

⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

ν0

ΔνD

We define “Doppler broadening” as the FWHM of this function

0 50 100 150 200 250 300

0,0

0,2

0,4

0,6

0,8

1,0

Intensidad

frecuencia

)(2 0ννν −=Δ D

N

D

MT

MckT

07

20 1016.72ln22 ννν −==Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−= 2

020

2

00

))()(2(exp

2)( νν

νπνν

ckTMI

kTMcI

20

0 2

4 ln 2 ( )2 ln 2( ) exp( )DD

I I υ υυ

υυ π

⎡ ⎤⎛ ⎞−= −⎢ ⎥⎜ ⎟ΔΔ ⎝ ⎠⎣ ⎦

Doppler broadening

20

0 2

4 ln 2 ( )2 ln 2( ) exp( )DD

I I υ υυ

υυ π

⎡ ⎤⎛ ⎞−= −⎢ ⎥⎜ ⎟ΔΔ ⎝ ⎠⎣ ⎦

λ

( )I λ

0 (1 )xvc

υ υ= ±

Detector (measuring frequencies)

xULυΔ

laser λ [nm] ΔνN [Hz] ΔνD [Hz]HeNe (Ne) 632.8 5.4 105 1.5 109

Ar 488 1.2 107 2.7 109

HeCd (Cd) 441.6 2.2 106 1.1 109

Cu 510.5 2.2 107 2.3 109

Doppler broadening

Laser sources

Uncertainty relations

Concept of spatial and temporal coherence

How is it constructed? 1- Stimulated emission 2- Optical feedback

Amplifier medium

mirror 100% Output coupler

Resonant cavity

Output (useful beam)

Something to remember:

ν : frequency Hertz (1/s) λ : wavelength nm, µm, Ǻ

hν : energy Joules, eV [1.602 10-19 C 1 eV = 1.602 10-19 Joules]

* if hυ = 1 eV ⇒ h c / λ = 1.602 10-19 J → λ = 1.23 µm

pump

Sources: Lasers

30 µm 10 µm 3 µm 1 µm 0.3 µm 0.1 µm 30 nm 10 nm

FIR IR visible UV RX

EM spectrum

Far IR: 10 µm a 1000 µm Medium IR: 1 µm a 10 µm Near IR: 0.7 µm a 1 µm Visible : 0.4 µm a 0.8 µm → 3.1 eV a 1.54 eV Ultraviolet: 0.2 µm a 0.4 µm Vacuum UV (VUV) : 0.1 µm a 0.2 µm Soft XR : 1 nm a 50 nm Hard XR: 0.1 nm a 1 nm

FIR

CO HF

diodes dyes

HeNe HeNe Nd3+

Ruby

RX

KrF ArF

XeCl

TiSa Ar+

N2

HeCd

Au

Cu

Pb HBr

H-like

Li-like

Be-like

Ne-like

Co-like

Ni-like Free e

Alexandrite

CrLiSAF CrLiCaF

Er

CO2

Uncertainty relationships

In communication theory it is clear that there is a minimum requirement in the bandwidth Δω of a signal in order to pass a pulse with a Δt risetime. This relationship holds for any Fourier conjugated variables

Δω Δt ≥ 12

This equation can be easily transformed into the Heisenberg uncertainty principle multiplying in both sides of the equation by h/2π

ΔE Δt ≥ 4π

For the spatial coordinates the uncertainty relationship holds for each coordinate; the table below summarizes the relationships between conjugate variables

Δkx Δx ≥12

; Δky Δy ≥12

; Δkz Δz ≥12

Variable Conjugate variable

E t kx Propag. along x x ky Propag. along y y kz Propag. along z z ω Angular frequency t px Momentum along x x py Momentum along y y pz Momentum along z z

ΔE Δt ≥ 4πω

Δkz Δz ≥1 2

Δω Δt ≥1 2

Δpz Δz ≥ h 4π

Δpx Δx ≥ h 4π

Δpy Δy ≥ h 4π

Δky Δy ≥1 2

Δkx Δx ≥1 2

Uncertainty relationships

Example: Particle in a box: a quantum particle is confined in a one dimension potential of length L. The eigenfunctions in the space and momentum space are

( ) ( )sin,

0

ni tn

nA k x ex t

ω

ψ−⎧

= ⎨⎩

Inside the box Outside the box

( )( )( )2 2 2 2

1 1,

nn i tikL

n

e eLp tn k L

ωπ

φπ

−− −=

−h

With standard deviations given by 2

2 22 2

6112x pL n

n Lπ

σ σπ

⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

h

If we calculate the smallest value (for n = 1) 2

2 0.5682 3 2x p

πσ σ = − ≈ >

h hh

Do you know what was your speed???

Hummm.. Not really…. But I can tell you

exactly where I am!

Uncertainty relationship

Consider a finite beam propagating in the z direction with an uneven energy distribution across the y direction

z

y

y

x

( )2

20

0

y

E y E e ω−

=

2 ω0

y

z

kz ≈ ω / c

( )2

20

0

y

E y E e ω−

=

E ky( ) = π ω0E0 e−ky2 ω0

2

4

We assume a beam with a smooth Gaussean profile in space (spot). The finite extend of the intensity in the y direction produces also a spread in the direction of propagation of the beam given by the uncertainty principle

If we have E(y) we can calculate E(ky) with the Fourier relationship

y

E(y)

Δy = ω0

E(ky)

ky

Δk = 2 / ω0

The vector k is then

kz ≈ ω/c = 2 π / λ Δky = 2/ω0

k θ/2

θ/2 = Δky / kz = λ / π ω0 ⇒ θ = 2 λ / π ω0

A beam perfectly collimated (θ=0) must have infinite extension (ω0 = ∞)

Few numbers:

λ = 632.8 nm - HeNe laser

2 ω0 = 0.1 cm

θ = 2 * 632.8 10-9 m π * 0.001 m

= 4 10-4

(diverges 0.4 mm per meter)

Uncertainty relationship

E ky( ) = π ω0E0 e−ky2 ω0

2

4( )2

20

0

y

E y E e ω−

=

Gaussean profile transforms into a Gaussean, but the principle holds for any arbitrary profile

Temporal (longitudinal) coherence

Perfectly coherent field = the phase can be determined in time and space

Suppose a field that is separated into two branches by an interferometer (any). The two fields (beams) are delayed by a distance (L1-L2) and re-combined with a certain small angle Θ to produce interference fringes

(L1-L2) detector

Δx = λ2 sinθ

≅λ2θ

E1 =E02exp − j k cosθ

2z+ k sinθ

2x

"

#$

%

&'

(

)*

+

,-exp − j2kL1( )exp − jΔφ( )


2z− k sinθ

2x

"

#$

%

&'

(

)*

+

,-exp − j2kL2( )

Depends only of t, phase jump during the time that the signal is delayed


( ) ( )01 1exp cos sin exp 2 exp

2 22EE j k z k x j kL jθ θ

ϕ⎡ ⎤⎛ ⎞= − + − − Δ⎜ ⎟⎢ ⎥

⎝ ⎠⎣ ⎦


2z− k sinθ

2x

"

#$

%

&'

(

)*

+

,-exp − j2kL2( )

The intensity in the detector plane is obtained combining these two fields

I x, y( ) = E1 +E2( )• E1 +E2( )∗= E02

η0

"

#$

%

&'cos2 k L2 − L1( )+ k θ

2x + Δφ

2"

#$

%

&'

*

+,

-

./

Depends on the alignment

Random phase

The random phase will make the position of the black and bright fringes wander. The effect is that the fringes will be blurred (because the typical time for the phase change is much more smaller than the typical integration time of the detector). The “blurring” of the fringes is evaluated with the “visibility” function defined as

V =Imax − IminImax + Imin

The visibility varies from 0 to 1. It measures how deep the field is modulated in the interference pattern


The fringes will blur completely when the phase change due to the random term is π (the bright fringes will fall in the dark fringes positions and vice versa) For example if we assume that the change in the phase is

dφdt

≅10−5 ω → Δφ =dφdt

Δt = π ⇒ 10−5 2πcλ

2 L2 − L1( )c

= π

That makes that for a wavelength of 5000Å (0.5 µm) yields 2.5 cm (1 in.) L2 − L1( ) =105 λ4

Example: Superpose two waves with the same amplitude, polarization and frequency, but let us allow the phase to randomly jump every T seconds. In the short exposure the integrated intensity yields perfect fringes. For longer exposures (NT seconds) the integrated intensity will yield:

2 2 2 20 0 1 0 1 2 0 1 1max max max max

1 I cos I cos I cos I cos2 2 2 2 2 2 2 2x x x x N

Nk k k kD T T T T

NTθ φ θ φ φ θ φ φ φ θ φ φ φ −⎡ ⎤+ + + + + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

LL

V =Imax − IminImax + Imin



To evaluate coherence we use an interferometer

screen

Light source

l l Δl = c Δt

Δω Δt ≥ ½ → Δt ≈ 1 / Δυ

Δυ Δt

υ

I

t

I

If c Δ t > Δ l light pulses will not overlap and interference will not be visible

Coherence length Δ l = c Δ t ≈ c / Δυ → Δl = λ2 /Δλ

λ Mean value for the wavelength Δλ << λ

Basic idea: * If the source is perfectly monochromatic there will be a single fringe system * If the source is not monochromatic, each λ will generate its own fringe system with different period. Then the superposition will blur the fringes

Some numbers:

Δυ ≈ 108 1/s → Δ l = 3 m

Na spectral lamp

Stabilized laser

Δυ ≈ 104 1/s → Δ l = 30 km


λ/2 Δx

θ

Δx = λ2 sinθ

≅λ2θ

Spatial (transverse) coherence

We use a double slit Young experiment

P1

P2

½ Δθ

R

S

Hence the interference will be visible if the holes are in an area

Coherence area Source area

SRRA

222)( λ

θ ≈Δ≈Δ

We will see interference if λθ ≤ΔΔ S

AΔ : transversal coherence length

The coherence length increases with R. What defines the coherence is actually the solid angle

2RAΔ

SRA 2

2

λ≅

Δ=ΔΩ ´2 ΔΩ= RSreplacing

Solid angle of the source calculated from the holes

´

2

ΔΩ≅ΔλA • Basic idea: two points in the source produces fringes shifted. If we

increase the distance between the holes, the fringe separation is reduced. If we superpose these two fringe patterns the interference fringes get blurred

Example : Thermal source

1 ( )

5002

S mm source size

nmR mλ

Δ =

=

=

223

2292

2

1)10()10500(4 mmm

SRA ===Δ

−

−

λ

Sun (filtering λ = 500 nm) Sun’s angular extension α = 4,65 mrad

sradsrad

5

232

1081.6)1065.4(14.3´

−

−

=

=≅≅ΔΩ απ

235

2252

106.31081.6)105(

´mmcmA −

−

−

≅=ΔΩ

=Δλ

→coherence length ≈ √ΔA = 61 µm

α-Orion (angular extension α = 2.3 10-7 rad )

srad142 1015.4´ −≅≅ΔΩ απ

214

2252

61015.4)105(

´mcmA ≅=

ΔΩ=Δ

−

−λ

→coherence length ≈ √ΔA = 2.45 m

Spatial (transverse) coherence

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