2.717j/mas.857j optical engineering · 2019-09-12 · reflection & transmission @ dielectric...
TRANSCRIPT
Today’s summary• Polarization• Energy / Poynting’s vector• Reflection and refraction at a dielectric interface:
– wave approach to derive Snell’s law– reflection and transmission coefficients– total internal reflection (TIR) revisited
MIT 2.71/2.710 Optics10/13/04 wk6-b-1
Polarization
MIT 2.71/2.710 Optics10/13/04 wk6-b-2
Propagation and polarization
x
y
kwave-vector
z
Eelectric field vector
const.efrontplanar wav
=⋅rk
) toparallelnot have could
one crystals, e.g.media, canisotropi
in :(reminder0
generally, More
i.e.0
etc.) glass, amorphousspace, free (e.g.
media isotropicIn
DE
Dk
EkEk
=⋅
⊥=⋅
MIT 2.71/2.710 Optics10/13/04 wk6-b-3
Linear polarization (frozen time)
zy
x t=0
E(z=0,t=0)
E(z=λ,t=0)
phase=constant
on the plane
E(z=λ/2,t=0)
Φ=0
Φ=2π
MIT 2.71/2.710 Optics10/13/04 wk6-b-4
Linear polarization (fixed space)
ty
x z=0
E(z=0,t=0)
E(z=0,t=2π/ω)
phase=constant
on the plane
E(z=0,t=π/ω)
Φ=0
Φ=2π
MIT 2.71/2.710 Optics10/13/04 wk6-b-5
Circular polarization (frozen time)
zy
x t=0
E(z=0,t=0)E(z=λ/2,t=0)
phase=constant
on the plane
E(z=λ,t=0)
Φ=0
Φ=2π
MIT 2.71/2.710 Optics10/13/04 wk6-b-6
Circular polarization: linear components
+
Ex
Ey
z
z
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MIT 2.71/2.710 Optics10/13/04 wk6-b-8
Circular polarization (fixed space)
z y
x
E(z=0,t=0)
E(z=0,t=π/ω)
E(z=0,t=π/2ω) E(z=0,t=3π/2ω)
rotationdirection
λ/4 plate
z
LinearLinearpolarizationpolarization
CircularCircularpolarizationpolarizationbirefringent
λ/4 plate
MIT 2.71/2.710 Optics10/13/04 wk6-b-9
λ/2 plate
z
LinearLinearpolarizationpolarization
Linear (90Linear (90oo--rotated)rotated)polarizationpolarizationbirefringent
λ/2 plate
MIT 2.71/2.710 Optics10/13/04 wk6-b-10
Think about that
birefringentλ/4 plate
LinearLinearpolarizationpolarization
z
mirror
IncomingIncoming
????????
OutgoingOutgoing
MIT 2.71/2.710 Optics10/13/04 wk6-b-11
Relationship between E and B
( )
EkB
k
xEBE rk
×=⇒
−≡∂∂
×≡∇×⇒
=∂∂
−=×∇ −⋅
ω
ω
ω
1
and
eˆ where 0
it
i
Et
ti
EB
k
Vectors k, E, B form aright-handed triad.
Note: free space or isotropic media only
MIT 2.71/2.710 Optics10/13/04 wk6-b-12
Energy
MIT 2.71/2.710 Optics10/13/04 wk6-b-13
The Poynting vector
BEBES ×=×= 02
0
1 εµ
cE
B
S so in free space
kS ||
S has units of W/m2
so it representsenergy flux (energy perunit time & unit area)
MIT 2.71/2.710 Optics10/13/04 wk6-b-14
Poynting vector and phasors (I)
20
02
1 ESEEBEkB
BESε
ωω
εc
ck
c=⇒
==⇒×=
×=
For example, sinusoidal field propagating along z
( ) ( )tkzEctkzE ωεω −=⇒−= 22000 coscosˆ SxE
Recall: for visible light, ω~1014-1015Hz
MIT 2.71/2.710 Optics10/13/04 wk6-b-15
Poynting vector and phasors (II)
Recall: for visible light, ω~1014-1015Hz
So any instrument will record the averageaverage incident energy flux
∫+
=Tt
t
tT
d1 SS where T is the period (T=λ/c)
S is called the irradianceirradiance, aka intensityintensityof the optical field (units: W/m2)
MIT 2.71/2.710 Optics10/13/04 wk6-b-16
Poynting vector and phasors (III)
21d)(cos)(cos 22 =−=− ∫
+
ttkztkzTt
t
ωω
2002
1 Ecε=S
For example: sinusoidal electric field,
Then, at constant z:
( ) ( )tkzEctkzE ωεω −=⇒−= 22000 coscosˆ SxE
MIT 2.71/2.710 Optics10/13/04 wk6-b-17
Poynting vector and phasors (IV)
Recall phasor representation:
( ) ( )( ) ( ) ( )
e : phasor""or amplitudecomplex sincos,ˆ
cos,
φ
φωφω
φω
iAtkziAtkzAtzf
tkzAtzf
−
−−+−−=
−−=
Can we use phasors to compute intensity?
MIT 2.71/2.710 Optics10/13/04 wk6-b-18
Poynting vector and phasors (V)
Consider the superposition of two two fields of the samesame frequency:
( ) ( )( ) ( )φω
ω−−=
−=tkzEtzEtkzEtzE
cos,cos,
202
101
( ) ( )φεε cos22
...d 2010220
210
0221
0 EEEEctEET
c Tt
t
++==+= ∫+
S
Now consider the two corresponding phasorsphasors:
φiE
E−e20
10
and the quantity
( )φεε φ cos22
...e2 2010
220
210
02
20100 EEEEcEEc i ++==+ −
MIT 2.71/2.710 Optics10/13/04 wk6-b-19
Poynting vector and phasors (V)
Consider the superposition of two two fields of the samesame frequency:
( ) ( )( ) ( )φω
ω−−=
−=tkzEtzEtkzEtzE
cos,cos,
202
101
( ) ( )φεε cos22
...d 2010220
210
0221
0 EEEEctEET
c Tt
t
++==+= ∫+
S
φiE
E−e20
10
Now consider the two corresponding phasorsphasors:
MIT 2.71/2.710 Optics10/13/04 wk6-b-20
( )φεε φ cos22
...e2 2010
220
210
02
20100 EEEEcEEc i ++==+ −
= !!and the quantity
Poynting vector and irradiance
MIT 2.71/2.710 Optics10/13/04 wk6-b-21
SummarySummary (free space or isotropic media)
20
0
;1 ESBES εµ
c=×=
∫+
=Tt
t
tT
d1 SS
Poynting vector
Irradiance (or intensity)
220 phasor or phasor2
∝= SS εc
Reflection / RefractionFresnel coefficients
MIT 2.71/2.710 Optics10/13/04 wk6-b-22
Reflection & transmission @ dielectric interface
MIT 2.71/2.710 Optics10/13/04 wk6-b-23
MIT 2.71/2.710 Optics10/13/04 wk6-b-24
Ei ki kiEi
Reflection & transmission @ dielectric interface
Reflection & transmission @ dielectric interface
MIT 2.71/2.710 Optics10/13/04 wk6-b-25
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence
( )[ ]( )[ ]txykiE
tiE
iiii
iii
ωθθω
−+−==−⋅=
)sincos(expˆ expˆ
0
0
zrkzE
Incident electric field:
( )[ ]( )[ ]txykiE
tiE
rrir
rrr
ωθθω
−++==−⋅=
)sincos(expˆ expˆ
0
0
zrkzE
Reflected electric field:
( )[ ]( )[ ]txykiE
tiE
tttt
ttt
ωθθω
−+−==−⋅=
)sincos(expˆ expˆ
0
0
zrkzE
Transmitted electric field:
where:vacuumvacuum
2 ,2λπ
λπ t
ti
inknk ==
Reflection & transmission @ dielectric interface
MIT 2.71/2.710 Optics10/13/04 wk6-b-26
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence
( )[ ]( )[ ]( )[ ]txykiE
txykiEtxykiE
EEE
tttt
rrir
iiii
t
ri
ωθθωθθωθθ
−+−=−+++−+−⇒
==+
)sincos(exp )sincos(exp
)sincos(expl)(tangentia
l)(tangential)(tangentia
0
0
0
Continuity of tangential electric fieldat the interface:
But at the interface y=0 so( )[ ]( )[ ]
( )[ ]txkiEtxkiEtxkiE
ttt
rir
iii
ωθωθωθ
−=−+−
sinexp sinexp
sinexp
0
0
0
Reflection & transmission @ dielectric interface
MIT 2.71/2.710 Optics10/13/04 wk6-b-27
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidenceContinuity of tangential electric fieldat the interface:
( )[ ]( )[ ]
( )[ ]txkiEtxkiEtxkiE
ttt
rir
iii
ωθωθωθ
−=−+−
sinexp sinexp
sinexp
0
0
0
Since the exponents must be equalfor all x, we obtain
tt
ii
ttii
ri
nnkk θλπθ
λπθθ
θθ
sin2sin2sinsin
and
vacuumvacuum
=⇔=
=
Reflection & transmission @ dielectric interface
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidenceContinuity of tangential electric fieldat the interface:
ri θθ =
ttii nn θθ sinsin =
law of reflection
Snell’s law of refraction
so wave description is equivalentto Fermat’s principle!! ☺
MIT 2.71/2.710 Optics10/13/04 wk6-b-28
Reflection & transmission @ dielectric interface
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence
( )[ ]txykiE iiiii ωθθ −+−= )sincos(expˆ 0zE
Incident electric field:
( )[ ]txykiE rrirr ωθθ −++= )sincos(expˆ 0zEReflected electric field:
( )[ ]txykiE ttttt ωθθ −+−= )sincos(exp0zETransmitted electric field:
Need to calculate the reflected and transmitted amplitudes E0r, E0t
i.e. need twotwo equations
MIT 2.71/2.710 Optics10/13/04 wk6-b-29
Reflection & transmission @ dielectric interface
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidenceContinuity of tangential electric fieldat the interface gives us one equation:
( )[ ]( )[ ]
( )[ ]txkiEtxkiEtxkiE
ttt
rir
iii
ωθωθωθ
−=−+−
sinexp sinexp
sinexp
0
0
0
which after satisfying Snell’s law becomes
tri EEE 000 =+
MIT 2.71/2.710 Optics10/13/04 wk6-b-30
Reflection & transmission @ dielectric interface
MIT 2.71/2.710 Optics10/13/04 wk6-b-31
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence
l)(tangentia l)(tangential)(tangentia
t
ri
BBB=
=+
The second equation comes from continuity of tangential magnetic fieldat the interface:
Recall
0000cossinˆˆˆ
1
1
Ekk θθ
ω
ωzyx
EkB
=
×=
Reflection & transmission @ dielectric interface
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidenceSo continuity of tangential magnetic field Bx at the interface y=0 becomes:
tttrriiii
tttrriiii
EnEnEnEkEkEk
θθθθθθ
coscoscoscoscoscos
000
000
=−⇔=−
MIT 2.71/2.710 Optics10/13/04 wk6-b-32
Reflection & transmission @ dielectric interface
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence
ttii
ii
i
t
ttii
ttii
i
r
nnn
EEt
nnnn
EEr
θθθ
θθθθ
coscoscos2
coscoscoscos
0
0
0
0
+=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+−
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
⊥
⊥
⊥
⊥
Solving the 2×2 system of equations:
tttrriiii EnEnEn θθθ coscoscos 000 =−tri EEE 000 =+
we finally obtain
MIT 2.71/2.710 Optics10/13/04 wk6-b-33
Following a similar procedure ...
tiit
ti
i
t
tiit
tiit
i
r
nnn
EEt
nnnn
EEr
θθθ
θθθθ
coscoscos2
coscoscoscos
||0
0||
||0
0||
+=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+−
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
Reflection & transmission @ dielectric interface
II. Polarization parallel to plane of incidenceII. Polarization parallel to plane of incidence
MIT 2.71/2.710 Optics10/13/04 wk6-b-34
Reflection & transmission @ dielectric interface
n=1.5
“Brewsterangle”
MIT 2.71/2.710 Optics10/13/04 wk6-b-35
Reflection & transmission of energyenergy@ dielectric interface
Recall Poynting vector definition:
20 ES εc=
different on the two sides of the interface
incvacuum
tncvacuum
22
0
0
22
0
0
coscos
coscos t
nn
EE
nnT
rEER
ii
tt
i
t
ii
tt
i
r
θθ
θθ
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
MIT 2.71/2.710 Optics10/13/04 wk6-b-36
Energy conservation
MIT 2.71/2.710 Optics10/13/04 wk6-b-37
1coscos i.e. ,1 2t2 =+=+ t
nnrTR
ii
t
θθ
Reflection & transmission of energyenergy@ dielectric interface
MIT 2.71/2.710 Optics10/13/04 wk6-b-38
Normal incidence
MIT 2.71/2.710 Optics10/13/04 wk6-b-39
tiit
ti
i
t
tiit
tiit
i
r
nnn
EEt
nnnn
EEr
θθθ
θθθθ
coscoscos2
coscoscoscos
||0
0||
||0
0||
+=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+−
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
ttii
ii
i
t
ttii
ttii
i
r
nnn
EEt
nnnn
EEr
θθθ
θθθθ
coscoscos2
coscoscoscos
0
0
0
0
+=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+−
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
⊥
⊥
⊥
⊥
0 and 0 == ti θθ
it
i
it
it
nnntt
nnnnrr
+==
+−
==
⊥
⊥
2||
||
( )2||
2
||
4
it
it
it
it
nnnnTT
nnnnRR
+==
⎟⎟⎠
⎞⎜⎜⎝
⎛+−
==
⊥
⊥
Note: Note: independent of polarization
Brewster anglettii nn θθ sinsin =Recall Snell’s Law
.0 ),tan( 2
When . allfor 0 , If
)tan()tan(
coscoscoscos
)sin()sin(
coscoscoscos
||ti
||
===−≠≠
+−
+=+−
=
+−
−=+−
=
⊥
⊥
rnnrnn
nnnnr
nnnnr
i
tiiti
ti
ti
tiit
tiit
it
ti
ttii
ttii
θπθθθ
θθθθ
θθθθ
θθθθ
θθθθ
This angle is known as Brewster’s angle. Under such circumstances, for an incoming unpolarized wave, only the component polarized normal to the incident plane will be reflected.
MIT 2.71/2.710 Optics10/13/04 wk6-b-40
MIT 2.71/2.710 Optics10/13/04 wk6-b-41
Why does Brewster happen?
elementaldipoleradiatorexcited bythe incidentfield
Why does Brewster happen?
MIT 2.71/2.710 Optics10/13/04 wk6-b-42
Why does Brewster happen?
MIT 2.71/2.710 Optics10/13/04 wk6-b-43
Why does Brewster happen?
MIT 2.71/2.710 Optics10/13/04 wk6-b-44
Turning the tables
rt
1
r’ t’
1
Is there a relationship between r, t and r’, t’ ?
MIT 2.71/2.710 Optics10/13/04 wk6-b-45
Relation between r, r’ and t, t’
a
arat
air
a’
a’t’
a’r’
glass airglass
Proof: algebraic from the Fresnel coefficientsor using the property of preservation of thepreservation of thefield properties upon time reversal12 =′+
−=′
ttrrr
field properties upon time reversal
Stokes relationships
MIT 2.71/2.710 Optics10/13/04 wk6-b-46
Proof using time reversal
a
arat
air glassatt’
arat
air
MIT 2.71/2.710 Optics10/13/04 wk6-b-47
glass
atr’art
ar2
⇔
( )( ) 1
0 22 =′+⇒=′+
′−=⇒=′+
ttrattrarrtrra
Total Internal Reflection
Happens when1sin >iin θ
ySubstitute into Snell’s law
1sinsin >= it
it n
n θθ
ok if θt complexno energy transmitted
MIT 2.71/2.710 Optics10/13/04 wk6-b-48
Total Internal Reflection
Propagating component[ ]ttt yikE θcosexp∝
ywhere
1sincos 2
22
−±=t
iit n
ni θθ
so
⎥⎥⎦
⎤
⎢⎢⎣
⎡−−∝ 1sinexp 2
22
t
iitt n
nykE θno energy transmitted
MIT 2.71/2.710 Optics10/13/04 wk6-b-49
Total Internal Reflection
⎥⎥⎦
⎤
⎢⎢⎣
⎡−−∝ 1sinexp 2
22
t
iitt n
nykE θ
y
Pure exponential decay≡≡ evanescentevanescent wave
It can be shown that:no energy transmitted
0≈tS
MIT 2.71/2.710 Optics10/13/04 wk6-b-50
Phase delay upon reflection
Phase delay is 0
Phase delay is π
ni=1 (air)nt=1.5 (glass)
MIT 2.71/2.710 Optics10/13/04 wk6-b-51
Phase delay upon TIR
⊥== ⊥δδ ii rr e and e ||
||
MIT 2.71/2.710 Optics10/13/04 wk6-b-52
E0i
E0r= rE0i
y where
it
tiii
nnnn
θθδ
cossin
2tan 2
222|| −
−=
ii
tii
nnn
θθδ
cossin
2tan
222 −−=⊥
Phase delay upon TIR
MIT 2.71/2.710 Optics10/13/04 wk6-b-53
y
E0i
E0r= rE0i
δ||
δ⊥