2016 EE438538 Part 1 Optical Waves 5 Waves at Interfaces(2)

7/25/2019 2016 EE438538 Part 1 Optical Waves 5 Waves at Interfaces(2)

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EE438/538

OptoelectronicDevices&Applications

Part1:FundamentalsofOpticalWaves

(5) Waves at interfaces


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WavesatInterfaces

Sofar,wehavestudied

wavespropagatinginvacuum

wavespropagatinginuniformdielectricmaterial

shortenedwavelength

slowerphasevelocity highlossatsomefrequencies

Whathappensattheinterfacebetweentwodielectricmedia?

Usefultoolsfordealingwiththeproblem

Phasefront(=Wavefront)

Ray

Phasematching

Two rays next to each otherTheir phasefronts must be in precise alignment.Otherwise there will be destructive interference andthey will gradually kill each other.


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Reflection:TheFirstExampleofPhaseMatching

Ray View

nt

ni

i

r

Whatwillberelationbetweenthetwoangles?

Wecannotanswerthatinthisrayview


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Since #1 and #2 are originally one, thephase fronts must be in-phase up to AC

In phase

Ray#2

nt

ni

Ray#1

i

A

C


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So this is the right picture!

In phase

Ray#2

nt

ni

Ray#1

i

A

C


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Thereflectedlightcanbedescribedinthesameway

Ray#2

Ray#1

Two rays+wavefront view

i

nt

ni

r

Ray#1

Ray#2

I deliberately set i r


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Ray#2Ray#1

i

r

Ray#1

Ray#2

After reflection, along DB, will they stay in phase or not?

A

C

D

B

In phase Not in phase


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Ray#2Ray#1

i

r

Ray#1

Ray#2

After reflection, along DB, will they stay in phase or not?

A

C

D

B

They must stay in-phase

Otherwise, they will cancel each other to a certain amount

There are an infinite number of this two-ray pairs Even a slight phase mismatch can cause near-total destruction of the reflection


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Ray#2Ray#1

i

r

Ray#1

Ray#2

A

C

D

B

To form a reflection, the two must stay in phase

Requirement for Phase Matching


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Ray#2Ray#1

i

r

Ray#1

Ray#2

A

C

D

B


In this setup, phase matching

requires |AB| = |CD|



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i

r


In this setup, phase matching

requires |AB| = |CD| which, in

turn, requires i = r


Snells Law of

Reflection

i= r


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i

r

Snells Law of

Reflection

So far, ntand ni did not play any role

nt

ni


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ApplyingPhaseMatchingtoRefraction

reflection

refraction

Ray View

nt

ni


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reflection

refraction

ni nt

Two rays + wavefront view

Ray View

i

Ray#2Ray#1

A

C

D

B

t

nt

ni

nt

ni


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A

C

D

B

nt

niIn-phase up to AC

In-phase from BD

Then, can we just use |AB| = |CD|

as the requirement?


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A

C

D

B

nt

ni

Or equivalently, can we just use

sin i = sin t as the requirement?

t

i


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No, because we have to deal with waves

propagating in different media this time

Waves propagate different media withdifferent phase velocities and wavelengths

Different phase change even for the same

distance !

A

C

D

B

nt

ni


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No, because we have to deal with waves propagating in

different media this time

Waves propagate different media with different phase

velocities and wavelengths

Different phase change even for the same distance

n = 1

n = 1.5

A

C

D

B

nt

ni


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No, because we have to deal with waves propagating in

different media this time

Waves propagate different media with different phase

velocities and wavelengths

Different phase change even for the same distance

n = 1

n = 1.5

A

C

D

B

nt

ni

phase change = k distance (2/) distance (2/o) n distance


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Along path |AB|: becomes o/nt

Effectively longerfor phase

calculation by a factor of nt

Similarly, for path |CD|, becomes o/ni

|CD| becomes effectively ni*|CD|

A

C

D

B

nt

ni


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Along path |AB|: becomes o/nt

Effectively longerfor phase calculation

by a factor of nt

Similarly, for path |CD|, becomes o/ni

|CD| becomes effectively ni*|CD|

A

C

D

B

nt

ni

Another important concept Optical Pathlength (OPL)

OPL = refractive index actual length


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A

C

D

B

nt

ni

In this setup,

OPL(AB) = OPL(CD)

only if

ni sin i = nt sin t

i

t

Snells Law for Refraction


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TotalInternalReflection

In any case,as i increases,

t will also increase

nt > ni

i

nt < ni


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i

nt < ni

nt < ni

Lets focus on

case


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i

nt < ni


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Thenwhatwillhappenifi >c Thetransmissionangletbecomescomplex Thelightwillget100%reflected

Total Internal Reflection (TIR)

i

nt < ni

r


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Thenwhatwillhappenifi>c Thetransmissionangletbecomescomplex Thelightwillget100%reflected

Butonthelowindexside,therewillbeasmalltailcalledtheevanescentwave

Total Internal Reflection (TIR)

i

nt < ni

r


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WithSnellslaw,wecanpredictinwhatdirectionsthereflectionandrefractionwilloccur

Whataboutthesplittingratio ofpowerbetweenthetwo?

Letsgetbacktotheplanewaveview

FresnelFormula:QuantifyingReflection&Transmission


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Ourgoalistoderiveexpressionsfor

Whatkindofinformationdowealreadyhave?

AdditionalinformationcomesfromBoundaryConditions

Thecontinuityofthetangentialcomponentofelectricfieldacrosstheboundary

Thecontinuityofthetangentialcomponentofmagneticfieldacrosstheboundary

Toapplytheseconditions,wemustseparatethefollowingtwowaveconfigurations

Snells Law

Reflection coefficient: =

Transmission coefficient: =

io

ro

E

E

io

to

E

E

sin sin


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StateofPolarizationofEMWaves

Note:thispolarizationisdifferentfrom

butrelated

Thestateofpolarizationrepresentstheoscillationdirectionoftheelectricfieldportion

ofEMwaves(sometimesthedirectionvariesovertime).

E

test charge

E

induceddipole

And this is important because thedipole radiation pattern is notsymmetric


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TE:TransverseElectric


field Skims the interface

The direction ofoscillation of fieldis orthogonal to theplane of incidence

a.k.a. S polarization

nt

niH

E


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TM:TransverseMagnetic


The direction ofoscillation of fieldis parallel to the

plane of incidence

nt

niE

H


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field Pokes the interface

The direction ofoscillation of fieldis parallel to the

plane of incidence

a.k.a. P polarization

nt

niE

H

F l F l Q if i R fl i & T i i


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TE:TransverseElectric


Anyarbitrarystateofpolarizationcanbesynthesizedbycombiningthesetwo

Ofcourse,fornormalincidence,thedistinctionbetweenTEandTMvanishes

So,weneedtoderive4quantities:rTE,rTM,tTE,tTMor r,r//,t,t//


field Skims the interface

field Pokes the interface

The direction ofoscillation of fieldis orthogonal to theplane of incidence

a.k.a. S polarization

The direction ofoscillation of fieldis parallel to theplane of incidence

a.k.a. P polarization

nt

niH

E

nt

niE

H


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F l F l Q tif i R fl ti & T i i


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Quitecomplicated.Butthesecanbemuchsimplifiedbydefininganewparameter


Pnn iit 22 sin

PPr

i

i

coscos

Pt

i

i

cos

cos2

iit

iit

nnPnnPr

cos)/(cos)/(

2

2

//

iit

iit

nnP

nnt

cos)/(

cos)/(22//

Fresnel Formula: Quantifying Reflection & Transmission


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Quitecomplicated.Butthesecanbemuchsimplifiedbydefininganewparameter

ItisimportanttoseethatPcanbeimaginary!

Therecanbephaseshiftsassociatedwithreflectionandrefraction!

Pbecomesimaginarywhentheincidenceanglereaches:

i.e.,thecriticalangle,theonsetofTIR


Pnn iit 22 sin

P

Pr

i

i

cos

cos

Pt

i

i

cos

cos2

iit

iit

nnPnnPr

cos)/(cos)/( 2

2

//

iit

iit

nnP

nnt

cos)/(

cos)/(22//

sin

Practical Example: Glass/Air Interface


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Letshavealookatthereflectioncoefficientsofglass/airinterface

PracticalExample:Glass/AirInterface

TE

1cos

costan2

2

21

i

cic

amplitude response

1.0

i

-

c

|r| phase response

iglass ( = 1.44) air ( = 1)

High-to-LowInternal reflection

Practical Example: Glass/Air Interface


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PracticalExample:Glass/AirInterface Letshavealookatthereflectioncoefficientsofglass/airinterface

TM

1

cos

cos

sin

1tan2

1

2

2

2

1

//c

c

iglass ( = 1.44) air ( = 1)

amplitude response |r//| phase response //

ic

1.0

starting point:same as TE

p

cpi

High-to-LowInternal reflection

HowaboutAir/GlassInterface?


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Whatwillhappenifwereversethedirectionofpropagation?

Willitbethesame?NO.Letshavealookatthereflectioncoefficients

rTE

iglass ( = 1.44) air ( = 1)

i

1

0

-1

rTM

/2

p

Low-to-HighExternal reflection



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/



iglass ( = 1.44) air ( = 1)

Start from aNEGATIVE valueA built-in phase shift

rTE

i

1

0

-1

rTM

/2

p




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/



iglass ( = 1.44) air ( = 1)

No TIR

No incidence angle-dependent,

smooth change in phase

One abrupt phase change for TM No phase change for TE at all!

rTE

i

1

0

-1

rTM

/2

p



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Onethingincommon?


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Onethingthatscommontobothinternalandexternalreflections?

Incidenceangle,reflectedlight

Hmm, the floor isnot shiny.

Hey, it is shiny!

ic

1.0

ic

1.0

p

rTE

i

1

0

-1

rTM

/2

p


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HowaboutTransmission?


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Changesintransmissioncoefficientsasfunctionsofincidenceanglewillbesimplerthan

thoseofreflectioncoefficients

Unliker,wecanseethattwillalwaysbereal!

Ofcourse,Pcanstillbeimaginary

Butbythetime,therewillbenotransmission,duetoTIR

So,twillalwaysbearealnumber

Itwillalwaysbeapositivenumberaswell Nophasechangeatall

Justamplitudechangesinaccordancewithreflectioncoefficientchanges

P

P

ri

i

cos

cos

Pt

i

i

cos

cos2

iit

iit

nnP

nnPr

cos)/( cos)/( 2

2

//

iit

iit

nnP

nnt

cos)/(

cos)/(22//

HowaboutPowerSplittingRatio?


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Sofar,wehavedealtwiththeamplitudesoftheelectricfieldcomponent

Whenenergyandpowerareconcerned,theintensityisamorerelevant

quantity

Intensityisproportionalto|E|2/ andhasaunitof[W/m2] =(/)

ItisarealnumberwithNOphaseinformationincluded

A complex number possesses phase information



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Forreflection,

Reflectance

R

|I

r|/|I

i|

=

|r|

2



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Fortransmission,weneedtoconsiderthefactthatthe

twowavesareindifferentmaterials!

TransmittanceT |It|/|I

i|=(n

i/n

t)|t|2



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Remeber: r+t 1

|r|

2

+

(ni/nt)|t|2

=

1

Reflectance (R) Transmittance (T)

PeculiaritiesinReflection:(1) P


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Whatisit?RecaptheTMreflection

Glass/Air Air/Glass

There is one point in the incidence angle at which the TM reflection becomes zero!

Brewster Angle or Polarizing Angle

ic

1.0

p

i

1

0

-1

rTM

/2

p

TM,TE

TM,TE

TEp

PeculiaritiesinReflection:(1)BrewsterAngle


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Whatisit?RecaptheTMreflection

Glass/Air Air/Glass

There is one point in the incidence angle at which the TM reflection becomes zero!

The point can be easily found from the Fresnel formula

ic

1.0

p

i

1

0

-1

rTM

/2

p

i

tPn

n1tan iitiit

iitiit

io

ro

nnnn

nnnn

E

Er

cossin

cossin

222

222

//,

//,//


i f l (i f iki di )


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Sometimesuseful(imagefromwikipedia)

TM,TE

TM,TE

TEp



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Also

useful

for

making

Sunglasses

Makingsunglasseswithdarkenedglassissimple.BUT

Nobodywantstowearapairofsunglasseswhichisalwaysdark.

When

do

people

need

sunglasses

most?


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Also

useful

for

making

Sunglasses

Makingsunglasseswithdarkenedglassissimple.BUT

Nobodywantstowearapairofsunglasseswhichisalwaysdark.

Whendopeopleneedsunglassesmost?

Nowweknowthatduringsunset:



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Also

useful

for

making

Sunglasses

Applycoating tothelensessothatonlytheTEportioncanbeblocked

YoucanstillseewiththetinyTMportion

Duringdaytime,whensunishighaboveyourhead,TE:TM~50:50

Youhavenoproblemwatchingaround

Price for non-polarized sunglasses (i.e., darkened glass) < Price for polarizing ones.

But it can cause safety problems. So reserve non-polarizing ones for ski trips only



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Whydoesthispolarizingreflectionhappen?WhyonlytoTM?

Toanswerthat,letsgetoutofthistypicalview

TME



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Letsmagnifytheareaneartheimpingingpoint

http://pixabay.com/en/magnifying-glass-magnifier-glass-189254/

TME



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YouwillseealotofdipolesinducedbytheelectricfieldcomponentEwhich

pokestheinterface

TME

h d f h d l h fl d !



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Thereradiationfromthedipolesconstitutethereflectionandtransmission!

Th di ti f th di l tit t th fl ti d t i i !



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Thereradiationfromthedipolesconstitutethereflectionandtransmission!

Dipoles along the interface are aligned with their dipole axis along Et, not Ei


R Di l di ti tt NOT h i ll t i !


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Recap:DipoleradiationpatternwasNOTsphericallysymmetric!

Now it will make a real problem (or fun stuff)


These reflection and transmission directions are determined by the phase matching


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These reflection and transmission directions are determined by the phase-matching

condition There is nothing the dipoles can do about it

So this type of awkward situation is possible!



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So,thistypeofawkwardsituationispossible!

The dipole axis coincides with the direction of reflection? Of course, there will be no reflection!


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When = the angle between the reflection and transmission becomes 90



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Wheni=p,theanglebetweenthereflectionandtransmissionbecomes90

p r= p

t

90 p

90 t

(90 p) + (90 t) = 90 p = 90 t

When i = the angle between the reflection and transmission becomes 90



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Wheni p,theanglebetweenthereflectionandtransmissionbecomes90

(90 p) + (90 t) = 90 p = 90 tBrewster

condition

Plugging the relation into Snells law

ni sin i = nt sin t

ni sin p = nt sin (90 p)

p = tan-1(nt /ni)

Exactly the same Brewster angle that

we obtained from Fresnel formula!



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Maxwells Eq. Wave Eq.

EM Potential Dipole Radiation Pattern

Wave solution Boundary Conditions Fresnel Formula

Two totally separate routes Same results!


WhyonlytoTM?WhynoBrewstereffecttoTE?


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y y y

TEE

Dipole axis orthogonal to the plane of incidence Radiation pattern is perfectly symmetricAll angles available for re-radiation!

Lets have a more quantitative look

PeculiaritiesinReflection:(2)EvanescentWave


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Let shaveamorequantitativelook

Transmitted wave



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This vector determinesthe spatial properties ofthe wave including:

- propagation direction- transversal extent

Transmitted wave



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Totally determined by the material

Just the position vector

2



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one -dependent term one -dependent term



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Note: When sin 1 !

> 1 for internalreflection

between 0 and 1

sin sin

sin

sin

Whensin 1,whatwillhappentocos ?



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, pp

Itcontrolshowthebehavesinthedirection

cos s in

Whensin 1,whatwillhappentocos ?



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Itwillbeimaginarybecausecos 1 sin

cos s in

Inotherwords



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When

real

imaginary


Beyondthecriticalangle,thefieldacrosstheinterfacecanbewrittenas


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Beyondthecriticalangle,thefieldacrosstheinterfacecanbewrittenas


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c

The amplitude and phase front of the transmitted wave will look like these:



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c

whilepropagatingin z-direction

decays in y-direction

Basically, on the transmission side, there will be a wave which



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c

Along the transversal direction:

The waves amplitude will decay exponentially: e-y

1/ gives the 1/e point of the amplitude penetration depth 1sin

2 22

t

o

n



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Along the direction of propagation:

Since sin t > 1, the waves phase velocity will be lower than co/nt

The node-to-node distance will get shorter than o/nt

c



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Of course, this propagation wont last long!

c



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As you can see, the phase is not matched across the interface

The propagation wont be sustained

The energy cannot flow either in or direction It has to return to the only phase-matched path:

Someofyoumayhavewonderedwhenyousawthese

PeculiaritiesinReflection:(3)GoosHanschenShift


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//

cpi

1

coscos

sin1tan2

1

2

2

2

1

//c

c

1coscostan2

2

2

1

i

c

i

-

c

Someofyoumayhavewonderedwhenyousawthese


//


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What

causes

these

phase

shift

during

TIR?

//

cpi

1

coscos

sin1tan2

1

2

2

2

1

//c

c

1coscostan2

2

2

1

i

c

i

-

c

Thequantitativesolutionalreadyshowedthat



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There will be penetration of

the light into the other sideeven under TIR!

Thequantitativesolutionalreadyshowedthat



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Sothecorrectphysicalpictureis:



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Thisroundtriptakestime Thatsthecauseofthephasedelay

ThiseffectiscalledGoosHanschenShift

Ingeneral:Astheincidenceangle,theGHshiftaswell.

PeculiaritiesinReflection:(4)FrustratedTIR

Thepenetrationisveryshallowingeneral

e.g. ni = 1.46, nt = 1.43, o = 850 nm


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e.g. ni 1.46,nt 1.43,o 850nm

y~500nm

Butnotanimpossibledistanceformechanicalcontrol

Whatifwebroughtanotherpieceofhighindexmaterialwithinthepenetrationdepth?

WewillbeabletogetsomeleakagefromTIR

FrustratedTIR

Application: Tap coupler

2016 EE438538 Part 1 Optical Waves 5 Waves at Interfaces(2)

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Transcript of 2016 EE438538 Part 1 Optical Waves 5 Waves at Interfaces(2)