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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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Reflection:TheFirstExampleofPhaseMatching
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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Reflection:TheFirstExampleofPhaseMatching
So this is the right picture!
In phase
Ray#2
nt
ni
Ray#1
i
A
C
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Reflection:TheFirstExampleofPhaseMatching
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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Reflection:TheFirstExampleofPhaseMatching
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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Reflection:TheFirstExampleofPhaseMatching
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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Reflection:TheFirstExampleofPhaseMatching
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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Reflection:TheFirstExampleofPhaseMatching
Ray#2Ray#1
i
r
Ray#1
Ray#2
A
C
D
B
To form a reflection, the two must stay in phase
In this setup, phase matching
requires |AB| = |CD|
Requirement for Phase Matching
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Reflection:TheFirstExampleofPhaseMatching
i
r
To form a reflection, the two must stay in phase
In this setup, phase matching
requires |AB| = |CD| which, in
turn, requires i = r
Requirement for Phase Matching
Snells Law of
Reflection
i= r
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Reflection:TheFirstExampleofPhaseMatching
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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ApplyingPhaseMatchingtoRefraction
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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ApplyingPhaseMatchingtoRefraction
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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ApplyingPhaseMatchingtoRefraction
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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ApplyingPhaseMatchingtoRefraction
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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ApplyingPhaseMatchingtoRefraction
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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ApplyingPhaseMatchingtoRefraction
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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ApplyingPhaseMatchingtoRefraction
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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ApplyingPhaseMatchingtoRefraction
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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ApplyingPhaseMatchingtoRefraction
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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TotalInternalReflection
i
nt < ni
nt < ni
Lets focus on
case
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TotalInternalReflection
i
nt < ni
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TotalInternalReflection
Thenwhatwillhappenifi >c Thetransmissionangletbecomescomplex Thelightwillget100%reflected
Total Internal Reflection (TIR)
i
nt < ni
r
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TotalInternalReflection
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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FresnelFormula:QuantifyingReflection&Transmission
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
FresnelFormula:QuantifyingReflection&Transmission
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
FresnelFormula:QuantifyingReflection&Transmission
The direction ofoscillation of fieldis parallel to the
plane of incidence
nt
niE
H
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TM:TransverseMagnetic
FresnelFormula:QuantifyingReflection&Transmission
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
TM:TransverseMagnetic
Anyarbitrarystateofpolarizationcanbesynthesizedbycombiningthesetwo
Ofcourse,fornormalincidence,thedistinctionbetweenTEandTMvanishes
So,weneedtoderive4quantities:rTE,rTM,tTE,tTMor r,r//,t,t//
FresnelFormula:QuantifyingReflection&Transmission
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
FresnelFormula:QuantifyingReflection&Transmission
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
FresnelFormula:QuantifyingReflection&Transmission
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
HowaboutAir/GlassInterface?
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/
Whatwillhappenifwereversethedirectionofpropagation?
Willitbethesame?NO.Letshavealookatthereflectioncoefficients
iglass ( = 1.44) air ( = 1)
Start from aNEGATIVE valueA built-in phase shift
rTE
i
1
0
-1
rTM
/2
p
Low-to-HighExternal reflection
HowaboutAir/GlassInterface?
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/
Whatwillhappenifwereversethedirectionofpropagation?
Willitbethesame?NO.Letshavealookatthereflectioncoefficients
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
Low-to-HighExternal reflection
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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
HowaboutPowerSplittingRatio?
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Forreflection,
Reflectance
R
|I
r|/|I
i|
=
|r|
2
HowaboutPowerSplittingRatio?
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Fortransmission,weneedtoconsiderthefactthatthe
twowavesareindifferentmaterials!
TransmittanceT |It|/|I
i|=(n
i/n
t)|t|2
HowaboutPowerSplittingRatio?
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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
//,
//,//
PeculiaritiesinReflection:(1)BrewsterAngle
i f l (i f iki di )
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Sometimesuseful(imagefromwikipedia)
TM,TE
TM,TE
TEp
PeculiaritiesinReflection:(1)BrewsterAngle
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Also
useful
for
making
Sunglasses
Makingsunglasseswithdarkenedglassissimple.BUT
Nobodywantstowearapairofsunglasseswhichisalwaysdark.
When
do
people
need
sunglasses
most?
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PeculiaritiesinReflection:(1)BrewsterAngle
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Also
useful
for
making
Sunglasses
Makingsunglasseswithdarkenedglassissimple.BUT
Nobodywantstowearapairofsunglasseswhichisalwaysdark.
Whendopeopleneedsunglassesmost?
Nowweknowthatduringsunset:
PeculiaritiesinReflection:(1)BrewsterAngle
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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
PeculiaritiesinReflection:(1)BrewsterAngle
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Whydoesthispolarizingreflectionhappen?WhyonlytoTM?
Toanswerthat,letsgetoutofthistypicalview
TME
PeculiaritiesinReflection:(1)BrewsterAngle
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Letsmagnifytheareaneartheimpingingpoint
http://pixabay.com/en/magnifying-glass-magnifier-glass-189254/
TME
PeculiaritiesinReflection:(1)BrewsterAngle
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YouwillseealotofdipolesinducedbytheelectricfieldcomponentEwhich
pokestheinterface
TME
h d f h d l h fl d !
PeculiaritiesinReflection:(1)BrewsterAngle
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Thereradiationfromthedipolesconstitutethereflectionandtransmission!
Th di ti f th di l tit t th fl ti d t i i !
PeculiaritiesinReflection:(1)BrewsterAngle
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Thereradiationfromthedipolesconstitutethereflectionandtransmission!
Dipoles along the interface are aligned with their dipole axis along Et, not Ei
PeculiaritiesinReflection:(1)BrewsterAngle
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)
PeculiaritiesinReflection:(1)BrewsterAngle
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!
PeculiaritiesinReflection:(1)BrewsterAngle
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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
PeculiaritiesinReflection:(1)BrewsterAngle
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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
PeculiaritiesinReflection:(1)BrewsterAngle
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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!
PeculiaritiesinReflection:(1)BrewsterAngle
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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!
PeculiaritiesinReflection:(1)BrewsterAngle
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
PeculiaritiesinReflection:(2)EvanescentWave
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This vector determinesthe spatial properties ofthe wave including:
- propagation direction- transversal extent
Transmitted wave
PeculiaritiesinReflection:(2)EvanescentWave
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Totally determined by the material
Just the position vector
2
PeculiaritiesinReflection:(2)EvanescentWave
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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 ?
PeculiaritiesinReflection:(2)EvanescentWave
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, pp
Itcontrolshowthebehavesinthedirection
cos s in
Whensin 1,whatwillhappentocos ?
PeculiaritiesinReflection:(2)EvanescentWave
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Itwillbeimaginarybecausecos 1 sin
cos s in
Inotherwords
PeculiaritiesinReflection:(2)EvanescentWave
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When
real
imaginary
PeculiaritiesinReflection:(2)EvanescentWave
Beyondthecriticalangle,thefieldacrosstheinterfacecanbewrittenas
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PeculiaritiesinReflection:(2)EvanescentWave
Beyondthecriticalangle,thefieldacrosstheinterfacecanbewrittenas
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c
The amplitude and phase front of the transmitted wave will look like these:
PeculiaritiesinReflection:(2)EvanescentWave
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c
whilepropagatingin z-direction
decays in y-direction
Basically, on the transmission side, there will be a wave which
PeculiaritiesinReflection:(2)EvanescentWave
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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
PeculiaritiesinReflection:(2)EvanescentWave
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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
PeculiaritiesinReflection:(2)EvanescentWave
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Of course, this propagation wont last long!
c
PeculiaritiesinReflection:(2)EvanescentWave
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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
PeculiaritiesinReflection:(3)GoosHanschenShift
//
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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
PeculiaritiesinReflection:(3)GoosHanschenShift
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There will be penetration of
the light into the other sideeven under TIR!
Thequantitativesolutionalreadyshowedthat
PeculiaritiesinReflection:(3)GoosHanschenShift
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Sothecorrectphysicalpictureis:
PeculiaritiesinReflection:(3)GoosHanschenShift
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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