Inter Diffr

8/8/2019 Inter Diffr

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Interference and diffraction

Interference and diffraction phenomena are both results of thesuperposition of EM waves from sources that are coherent to

each other

Interference: combined effects from discrete sources

Diffraction: combined effects from a continuousdistribution of sources

Interference from two coherent sources -

Youngs experiment :

At point Q, the distance traveled for EM wave from sourceA is

different from that from sourceB, : path difference.

phase difference = 2period

traveloftimeextra

=T

1

c2

=

2

Phase difference of0, 2, 4, are equivalent as far asinterference is concerned


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Assuming sourcesA andB gives out EM waves of the same

intensity, so that

ET= EA + EB = E0 cos t + E0 cos(t+)

Using complex notation,

TE~

= E0 eit + E0 e

i(+) = E0 eit(1 + ei)

= E0 eitei/2(e-i/2+ ei/2)

ET = 2 E0 cos

2

cos (t + 2

)

Since IT < ET2> and I0 < E0

2> ,

IT = 4I0 cos2

2

= 0 when = , 3, 5

= 4I0 when = 0, 2, 4

If the screen is located far from the sources,

d sin d tan =L

dx

IT = 4I0 cos2

Ldx

= 0 when x =

d2L ,

d2L3 ,

d2L5

= 4I0 when x = 0,d

L,

d

L2

The well-known Youngs interference bright and dark fringeson the screen


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What happen if the two sources are not completely coherentwith respect to each other?

A wave is perfectly coherent if the phase has a definite value byany delay on the wave itself

A perfectly coherent wave must be monochromatic, i.e. aperfect sinusoidal wave with no beginning and ending

An EM wave with a finite duration must consist of a band offrequency components

e.g. A wave with finite duration of2t:

E(t) =ti 0e

-t < t < t

= 0 otherwise

This can be written as an integral over the frequency:

( )

=

deEtE ti)(~

2

1

The spectral distributionE() is given by the Fourier transform

ofE(t) :

( )

==t

t

titidtedtetEE

)( 0)(~

= t2t

tsin

0

0

)(

])[(


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The power spectrumI() is proportional to the square of ( )E~

:

( ) ( )2~

EI

The power spectrum of our e.g. look like this:

Since ])[( tsin 0 = 0 when = 0t

The bandwidth of this spectral distribution, , is given by

t =

This uncertainty principle tell us that a finite wave durationrequires a non-zero spread of frequency

Atoms emit light by jumping down from an excited state. Finiteexcited state lifetime finite pulse duration frequencyspread

A collection of uncorrelated atoms gives an incoherent source


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Coherent length lc : average length of the light wave within

which the phase has a meaning

The coherent length is related to the coherent time by

lc ~ ctc

So that it is related to the spectral bandwidth by

lc ~ c

In a Youngs experiment, for a path difference l,

Ifl > lc : no interference fringes

Ifl lc : reduced contrast of the fringes


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Interference from multiple beams

e.g. multiple reflection from two interfaces

In a medium : E ~ eikz =zink0e , k0 = /c

Define the optical path difference in a medium of refractive index

n by OPD = nl l : physical path difference

So that eik(z+l) = eikzlink0e = eikz

OPDik0e

= eikz

)/( OPD2i

e , = 2/k0

Constructive interference for OPD = mDestructive interference for OPD = (m+1/2)

Consider two adjacent transmitted beams:


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Optical path difference between the two beams:

OPD =

2

cos

nd2

h sin0

=2cos

nd2

( )2

2

2 insncos

insd2

= 2nd cos2

phase difference =

OPD2 = 2cosnd4

To obtain the reflectivity and transmittivity of light at a non-absorbing boundary, we use the following argument:

For light incident from medium 1:

incident beam amplitude =A ,

reflected beam amplitude = rA ,

transmitted beam amplitude = tA


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For light incident from medium 2:

reflected beam amplitude = rA ,

transmitted beam amplitude = tA

Consider the following:

If we allow time to run backward:


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By principle of reversibility, we require

Art + Atr = 0

i.e. r = r (12.1)

and Ar2 + Att = A

i.e. tt = 1 r2 (12.2)

Eqs. (12.1) and (12.2) are called Stokes relations

Stokes relations are consistent with Fresnels equations

Stokes relations are still correct if there is a thin, non-absorbing conducting(i.e. highly reflective) layer at the interface.

The intensity reflection and transmission coefficientsR and Tare

given by

R = r2 = r2

T = 1 R = 1 r2 = tt

Consider multiple reflections between the two boundaries:


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E1 = t t E0 0i

e

0 : phase shift in the first trip

E2 = t tr2 E0 0

ie

ie , = 2cosnd

4

E3

= t t r4 E0 0

ie

i2e , etc.

Eout = E1 + E2 + E3 +

= t t E0 0i

e

( 1 + r2 ie + r4 i2e +)

The infinite geometric series is easy to sum, and gives

Eout =

i2

i0

er1

eEtt 0

'

'

=

i

i0

Re1

eTE 0

Intensity |E|2

in

out

I

I=

2in

2out

E

E

||

||=

2i

2

Re1

T

||

2iRe1 ||

= |1 Rcos iRsin|2 = (1 Rcos)2 + (Rsin)2

= 1 + R2 2Rcos = 1 + R2 2R

2sin21

2

= (1 R)2 + 4R

2

sin2

in

out

I

I =

2sinR4R1

R1

22

2

+

)(

)(

=

2sin1

1

2 +, =

2R1

R4

)(


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Maxima occur when /2 = m i.e. 2cosnd4

2

1

= m

2cosnd2 = m

Maximum ofIout = Iin

Minimum of Iout =+1

1

=2

R1

R4

)(

whenR1

Contrast between maxima and minima becomes very sharp asRgets close to 1:

This can be compared with that for two-beam interference(e.g.

Youngs interference fringes) : Iout = 4I0 cos2

2


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Two beams interference:

Multiple beams interference:


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The reflectivityR can be enhanced by coating a layer of highlyreflective metal on glass

This configuration is called a Fabry-Perot interferometer. It canbe applied to detect very small differences in wavelength

Suppose that for two separate wavelengths 1 and 2, their halfintensity points are overlapped :

The two wavelengths can be considered as just resolvable by theinterferometer

Half-intensity occurs when

Iout =2

I0 =)/( 2sin1

I2

0

+

i.e. when )/( 2sin2 = 1 i.e. 22)/( 1/

=

2


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When deviated from 2m by

2, the intensity drop to one-

half.

Recalled that = 2cosnd

4

For fixed 2, the variation ofdue to the variation ofis given by

|| =22

cosnd4

||

Suppose at a particular angle 2and wavelength 1,

= 21

cosnd4

= 2m

If wavelength 2is barely resolvable from 1, so that at the

same angle 2, for 2is deviated from 2mby22 =

4

4 =

22cosnd4

|| i.e.

|| =

2cosnd

ForR = 90% , = 2901

904

).(

.

= 360.

For = 550 nm (green light), 2 0, n = 1, d= 1cm, we have

|| =

360010143

105509

..= 0.6 10-6

The resolution is about one part in a million, i.e. wavelengths of550 nm and 550.0005 nm can be resolved by this simple device.


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Interference from multiple coherent sources

An array of coherent sources:

Etotal = E0 + E0i

e + E0i2

e + + E 0)( 1Ni

e

= E0 (1 +i

e + i2e + + )( 1Nie ), where =

insd2 = kdsin

The geometric series can be easily summed:

Etotal = E0

1e

1ei

iN

= E0

2i

2iN

2i2i

2iN2iN

e

e

ee

ee/

/

//

//

= E021Nie

2sin

2Nsin

/)(


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Other local maxima occur near

= 5/N, 7/N etc

However, for = 2N/N= 2, Itotal = I0

2

2

sin

Nsin= N2 I0,

which is again an absolute maximum.

Plot ofxsinN

Nxsin22

2

:


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We have, forNcoherent emitters

Itotal = I0

2

2

kdsinsin

2

Nkdsinsin

,

where I0 = intensity resulting from one emitter only, and

d= separation between adjacent emitters

First principal maximum occur when = 0, second principalmaximum occur when

kdsin= 2, i.e. dsin=

which is not possible ifd<

For d


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where =2

kDsin

when = 0 ,= 0 ,2

2sin

1 finite

We may therefore write

I() = I(0)2

2sin

Interference from a continuous source

diffraction


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Diffraction

Huygens principle: every point on a propagating wavefrontserves as the source of spherical secondary wavelets, such that the

wavefront at a later time is the envelop of these wavelets

To deal with diffraction problems:

Huygen-Fresnel principle: The Huygen wavelets serves as

spherical light sources, with the amplitude of the optical field at

any point beyond equals to superposition of the E-field from the

wavelets


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dEp = dE0r

e

ikr

, dE0EL da

EL: E-field amplitude of the light on the aperture

The E-field amplitude at the aperture is due to a spherical wave

generated from the source :

EL ~ Es'

'

r

eikr

Es: amplitude of the source field

Therefore,

Ep ~ darr

errik

'

)'(

aperture

+

(12.3)

The diffracted field is calculated from a surface integral over anarea defined by the aperture

One major problem with this expression is that it predicts adiffracted field even for an observation point on the same side

of the source (i.e. by placing P before the aperture)


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Fresnel proposed to get around this problem by introducing anobliquity factor O(), with O() = )cos( +1

2

1.

So that in the forward direction O(0) = 1, and in the backward

direction O() = 0.

Eq. (12.3) becomes

Ep ~ daOrr

errik

)('

)'(

aperture

+

(12.4)

Eq.(12.4) is called the Fresnel-Kirchoff diffraction integral. It isa scalar diffraction theory and can be derived from the wave

equation.

A restricted condition of diffraction, called Fraunhoferdiffraction, assumes

1) The light source is very far away, essentially plane waveincident on the aperture

2) The light is detected at distances very far away from the

aperture. the emerging light from the secondary sources canbe regarded parallel

e.g. A single slit


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Amplitude at P due to an element on the aperture

dEp =r

dE0 ])([ trki 0e

+, dE0 = EL ds

r0 : optical path length for light from the centre of the aperture

EL: amplitude per unit width of the aperture

Ep =

dseer

E iktkri

0

L 0 )( ( rr0 as ris very large)

Now, = sins , k = ks sin

dseik =

2b

2b

siniksdse

/

/

= sinikee 2sinikb2sinikb /)( //

= 2

ksin

2

kbsin

sin

We are only concerned with the intensity, and the variation ofthe intensity with angle

Ip() ~ |Ep|2 ~ 22

sin

, where =

2

kbsin

i.e.Ip() = I(0) 2

2sin

I(0)sinc2

First minimum of sinc2 occurs when=

i.e. when2

kbsin = bsin =


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Second minimum of sinc2 occur when= 2

i.e. when bsin = 2.

In general, minima occur when bsin = m

The maximum between the first and second minima has an

approximate value of

2

2

23

2

3sin

)/(

=

2

3

2

= 0.045


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Essentially all the light energy concentrated within the centralpeak

Angular spread of the light beam due to diffraction is usually

defined by

first minimum = sin-1

b

b

radian

By the same reason, you cannot focus a parallel beam to a spot

smaller than b

f

in radius, wheref is the focal length of the lens

For a rectangular aperture with sidesa andb:

R2

= X2

+ Y2

+ Z2

r2 = X2 + (Y y)2 + (Z z)2


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r R R

ZzYy +(keep terms linear iny,z)

ikre = RZzYyikikRee /)( +

Ep ~

+2a

2a

2b

2b

RZzYyikdydze

/

/

/

/

/)( (12.5)

~ sincsinc

where = R

Z

2

ka

, = R

Y

2

kb

I = I(0)sinc2sinc2


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For a circular aperture :

We have, similar to Eq. (12.5),

Ep ~

ddeRyxik

circle

/)( +

(12.6)

Using polar coordinates for and at the aperture :

=cos , =sin

Also, use polar coordinates for the observation point (x, y) :

x = ' cos ' , y = ' sin '

For an observation point far away from the aperture,

RR

x '= cos ' = sin cos ' cos '

RRy '= sin ' sin '


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(x+ y) / R (cos cos '+ sin sin ' )

= cos( ')

The integral of Eq. (12.6) have a cylindrical symmetry. The resultshould depend on'only and independent of'. We can therefore

choose P to be on the x-axis and set '= 0.

Ep ~

a

0

2

0

ikded

cos (12.7)

Eq. (12.7) cannot be evaluated by elementary functions. Letsmake a power series expansion and evaluate the integral term

by term :

2kik1e2ik /)cos(coscos

=

The

2

0

d integral ofcos is zero. The same is true for cos3,

cos5 etc. Therefore, the integral of Eq. (12.7) gives a real

value.

The first few terms of the results of Eq. (12.7) are :

=a

0

2

2

0

add

a

0

2

0

2

d2

kd

)cos(

=8

ak422

16

ua2

22 , where u ka


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The diffraction pattern for a circular aperture consists of acentral bright disk (called the Airy disk) surrounded by rings of

much reduced intensity. It should be noted that the Airy disk is

not the projection of the circular aperture.

The first zero ofJ1(u) / u occurs at u 3.83. That is, whenka = 3.83,

=D

2

2833

. , D = 2a = diameter

= D221

. .

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