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Electromagnetic waves: Interference

Wednesday November 6, 2002

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Haidinger’s Bands: Fringes of equal inclination

dd

nn22

nn11

Beam splitterBeam splitter

ExtendedExtendedsourcesource

PPII PP22

PP

xx

ff

FocalFocalplaneplane

11

11

DielectricDielectricslabslab

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Fizeau Fringes: fringes of equal thickness

Now imagine we arrange to keep cos ’ constant We can do this if we keep ’ small That is, view near normal incidence Focus eye near plane of film Fringes are localized near film since rays diverge

from this region Now this is still two beam interference, but whether

we have a maximum or minimum will depend on the value of t

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Fizeau Fringes: fringes of equal thickness

cos2 2121 IIIII

where, where,

kt

kt

2

'cos2

Then if film varies in thickness we will see fringes as we move our eye.Then if film varies in thickness we will see fringes as we move our eye.

These are termed These are termed Fizeau fringesFizeau fringes..

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Fizeau Fringes

Extended sourceExtended source

Beam splitterBeam splitter

xx nn

nn22

nn

kt

kt

2

'cos2

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Wedge between two plates11 22

glassglassglassglass

airair

DDyy

LL

Path difference = 2yPath difference = 2yPhase difference Phase difference = 2ky - = 2ky - (phase change for 2, but not for 1) (phase change for 2, but not for 1)

Maxima 2y = (m + ½) Maxima 2y = (m + ½) oo/n/n

Minima 2y = mMinima 2y = moo/n/n

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Wedge between two plates

Maxima 2y = (m + ½) Maxima 2y = (m + ½) oo/n/n

Minima 2y = mMinima 2y = moo/n/n

Look at p and p + 1 maximaLook at p and p + 1 maxima

yyp+1p+1 – y – ypp = = oo/2n /2n ΔΔxx

where where ΔΔx = distance between adjacent maximax = distance between adjacent maxima

Now if diameter of object = DNow if diameter of object = D

Then LThen L = D = D

And (D/L) And (D/L) ΔΔx= x= oo/2n or /2n or D = D = ooL/2n L/2n ΔΔxx

airair

DDyy

LL

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Wedge between two plates

Can be used to test the quality of surfacesCan be used to test the quality of surfaces

Fringes follow contour of constant yFringes follow contour of constant y

Thus a flat bottom plate will give straight fringes, otherwise Thus a flat bottom plate will give straight fringes, otherwise ripples in the fringes will be seen. ripples in the fringes will be seen.

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Newton’s ringsUsed to test spherical surfacesUsed to test spherical surfaces

Beam splitterBeam splitter

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Newton’s ringsR- RcosR- Rcos = y = yor,or,RR22=(R-y)=(R-y)22+r+r22

RR22(1-2y/R) + r(1-2y/R) + r22

RR

RR

yy

rr

i.e. ri.e. r22 = 2yR = 2yR

Maxima when,Maxima when,y = (m+1/2) y = (m+1/2) oo/2n/2n

Gives rings,Gives rings,

RRmm22=(m+1/2)=(m+1/2)ooR/nR/n

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Reflection from dielectric layer: Antireflection coatings Important in instruments such as cameras

where reflections can give rise to spurious images

Usually designed for particular wavelength in this region – i.e. where film or eye are most sensitive

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Anti-Reflection coatingsA. A. Determine thickness of filmDetermine thickness of film

filmfilm

glassglass

airairnn11

nn33

nn22nn1 1 < n< n2 2 < n< n33

Thus both rays (1 and 2) are shifted in phase by Thus both rays (1 and 2) are shifted in phase by on reflection. on reflection.

1122

For destructive interference (near normal incidence)For destructive interference (near normal incidence)

2n2n22t=(m+1/2)t=(m+1/2)

Determines the thickness of the filmDetermines the thickness of the film(usually use m=0 for minimum t)(usually use m=0 for minimum t)

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Anti-Reflection coatingsB. Determine refractive index of filmB. Determine refractive index of film

Near normal incidenceNear normal incidence

Amplitude at AAmplitude at A

filmfilm

glassglass

airair

nn11

nn33

nn22

1122

AAA’A’

23

23

23

23

32

2

21

1 22

''

nn

nnE

nn

nn

nn

n

nn

nE

EE

A

A

oA

Since Since ’ ~ 1’ ~ 1

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Anti-reflection coating

12

12' nn

nnEA

Amplitude at A’Amplitude at A’

B. Determine refractive index of filmB. Determine refractive index of film

To get perfect cancellation, we would like ETo get perfect cancellation, we would like EA A = E = E A’A’

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12

23

23

nn

nn

nn

nn

312 nnn should be index of AR filmshould be index of AR film

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Multiple Beam interference

Thus far in looking at reflectivity from a dielectric layer we have assumed that the reflectivity is small

The problem then reduces to two beam interference Now consider a dielectric layer of uniform thickness d

and assume that the reflectivity is large e.g. || > 0.8 This is usually obtained by coating the surface of the

layer with a thin metallic coating – or several dielectric coatings to give high reflectivity

Or, one can put coatings on glass plates , then consider space between plates

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Multiple beam interferenceLet 12 = 21= ’ 12= 21= ’

nn11

nn22

nn11

’’

AA BB CC DD

EEoo

’’’ ’ EEoo

’ ’ EEoo

((’)’)33’E’Eoo

((’)’)22’E’Eoo ((’)’)66’E’Eoo

((’)’)44’E’Eoo

((’)’)55’E’Eoo

((’)’)77’E’Eoo

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Multiple Beam Interference

Assume a (for the time being) a monochromatic source , ’ small ( < 30o) usually Now || = |’| >> , ’ Thus reflected beams decrease rapidly in amplitude (from

first to second) But amplitude of adjacent transmitted beam is about the

same amplitude Amplitude of successfully reflected beams decreases

slowly (from the second) Thus treat in transmission where contrast should be

somewhat higher The latter is the configuration of most applications