MIT 2.71/2.710 Optics 10/19/05 wk7-b-1 Today’s summary • 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 • Two-beam interference
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MIT 2.71/2.710 Optics10/19/05 wk7-b-1
Today’s summary
• 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
• Two-beam interference
MIT 2.71/2.710 Optics10/19/05 wk7-b-2
Energy
MIT 2.71/2.710 Optics10/19/05 wk7-b-3
The Poynting vector
E
B
S
BEBES ×=×= 02
0
1 εµ
c
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 Optics10/19/05 wk7-b-4
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 Optics10/19/05 wk7-b-5
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 Optics10/19/05 wk7-b-6
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 Optics10/19/05 wk7-b-7
Relationship between E and B
EB
k
( )
EkB
k
xEBE rk
×=⇒
−≡∂∂
×≡∇×⇒
=∂∂
−=×∇ −⋅
ω
ω
ω
1
and
eˆ where 0
it
i
Et
ti
Vectors k, E, B form aright-handed triad.
Note: free space or isotropic media only
MIT 2.71/2.710 Optics10/19/05 wk7-b-8
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 Optics10/19/05 wk7-b-9
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
( )φεε φ cos22
...e2 2010
220
210
02
20100 EEEEcEEc i ++==+ −
and the quantity
MIT 2.71/2.710 Optics10/19/05 wk7-b-10
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
( )φεε φ cos22
...e2 2010
220
210
02
20100 EEEEcEEc i ++==+ −
and the quantity
= !!
MIT 2.71/2.710 Optics10/19/05 wk7-b-11
Poynting vector and irradiance
SummarySummary
20
0
;1 ESBES εµ
c=×=
∫+
=Tt
t
tT
d1 SS
Poynting vector
Irradiance (or intensity)
220 phasor or phasor2
∝= SS εc
(free space or isotropic media)
MIT 2.71/2.710 Optics10/19/05 wk7-b-12
Reflection / RefractionFresnel coefficients
MIT 2.71/2.710 Optics10/19/05 wk7-b-13
Reflection & transmission @ dielectric interface
MIT 2.71/2.710 Optics10/19/05 wk7-b-14
Ei ki kiEi
Reflection & transmission @ dielectric interface
MIT 2.71/2.710 Optics10/19/05 wk7-b-15
Reflection & transmission @ dielectric interface
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 ==
MIT 2.71/2.710 Optics10/19/05 wk7-b-16
Reflection & transmission @ dielectric interface
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
MIT 2.71/2.710 Optics10/19/05 wk7-b-17
Reflection & transmission @ dielectric interface
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
=⇔=
=
MIT 2.71/2.710 Optics10/19/05 wk7-b-18
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 Optics10/19/05 wk7-b-19
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 Optics10/19/05 wk7-b-20
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 Optics10/19/05 wk7-b-21
Reflection & transmission @ dielectric interface
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
=
×=
MIT 2.71/2.710 Optics10/19/05 wk7-b-22
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 Optics10/19/05 wk7-b-23
Reflection & transmission @ dielectric interface
I. Polarization normal to plane of incidenceI. Polarization normal to plane of incidence
II. Polarization parallel to plane of incidenceII. Polarization parallel to plane of incidence
MIT 2.71/2.710 Optics10/19/05 wk7-b-25
Reflection & transmission @ dielectric interface
n=1.5
“Brewsterangle”
MIT 2.71/2.710 Optics10/19/05 wk7-b-26
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 Optics10/19/05 wk7-b-27
Energy conservation
1coscos i.e. ,1 2t2 =+=+ t
nnrTR
ii
t
θθ
MIT 2.71/2.710 Optics10/19/05 wk7-b-28
Reflection & transmission of energyenergy@ dielectric interface
MIT 2.71/2.710 Optics10/19/05 wk7-b-29
Normal incidence
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
MIT 2.71/2.710 Optics10/19/05 wk7-b-30
Brewster angle
.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
Recall Snell’s Law ttii nn θθ sinsin =
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.