Today’s summary - MITweb.mit.edu/2.710/Fall06/2.710-wk7-b-sl.pdf · • Reflection and refraction...
Transcript of Today’s summary - MITweb.mit.edu/2.710/Fall06/2.710-wk7-b-sl.pdf · • Reflection and refraction...
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
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 Optics10/19/05 wk7-b-24
tiit
ti
i
t
tiit
tiit
i
r
nnn
EEt
nnnn
EEr
θθθ
θθθθ
coscoscos2
coscoscoscos
||0
0||
||0
0||
+=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+−
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
Following a similar procedure ...
Reflection & transmission @ dielectric interface
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.
MIT 2.71/2.710 Optics10/19/05 wk7-b-31
Why does Brewster happen?
elementaldipoleradiatorexcited bythe incidentfield
MIT 2.71/2.710 Optics10/19/05 wk7-b-32
Why does Brewster happen?
MIT 2.71/2.710 Optics10/19/05 wk7-b-33
Why does Brewster happen?
MIT 2.71/2.710 Optics10/19/05 wk7-b-34
Why does Brewster happen?
MIT 2.71/2.710 Optics10/19/05 wk7-b-35
Turning the tables
rt
1
r’ t’
1
Is there a relationship between r, t and r’, t’ ?
MIT 2.71/2.710 Optics10/19/05 wk7-b-36
Relation between r, r’ and t, t’
a
arat
air glass
a’
a’r’
a’t’
12 =′+
−=′
ttrrr Proof: algebraic from the Fresnel coefficients
or using the property of preservation of thepreservation of thefield properties upon time reversalfield properties upon time reversal
glass air
Stokes relationships
MIT 2.71/2.710 Optics10/19/05 wk7-b-37
Proof using time reversal
a
arat
air glassatt’
arat
air glass
atr’
artar2
( )( ) 1
0 22 =′+⇒=′+
′−=⇒=′+
ttrattrarrtrra
⇔
MIT 2.71/2.710 Optics10/19/05 wk7-b-38
Total Internal Reflection
no energy transmitted
Happens when1sin >iin θ
Substitute into Snell’s law
1sinsin >= it
it n
n θθ
ok if θt complex
y
MIT 2.71/2.710 Optics10/19/05 wk7-b-39
Total Internal Reflection
no energy transmitted
Propagating component[ ]ttt yikE θcosexp∝
wherey
1sincos 2
22
−±=t
iit n
ni θθ
so
⎥⎥⎦
⎤
⎢⎢⎣
⎡−−∝ 1sinexp 2
22
t
iitt n
nykE θ
MIT 2.71/2.710 Optics10/19/05 wk7-b-40
Total Internal Reflection
no energy transmitted
y⎥⎥⎦
⎤
⎢⎢⎣
⎡−−∝ 1sinexp 2
22
t
iitt n
nykE θ
Pure exponential decay≡≡ evanescentevanescent wave
It can be shown that:
0≈tS
MIT 2.71/2.710 Optics10/19/05 wk7-b-41
Phase delay upon reflection
Phase delay is 0
Phase delay is π
ni=1 (air)nt=1.5 (glass)
MIT 2.71/2.710 Optics10/19/05 wk7-b-42
Phase delay upon TIR
⊥== ⊥δδ ii rr e and e ||
||
y
E0i
E0r= rE0i
where
it
tiii
nnnn
θθδ
cossin
2tan 2
222|| −
−=
ii
tii
nnn
θθδ
cossin
2tan
222 −−=⊥
MIT 2.71/2.710 Optics10/19/05 wk7-b-43
Phase delay upon TIR
y
E0i
E0r= rE0i
δ||
δ⊥
MIT 2.71/2.710 Optics10/19/05 wk7-b-44
Interference
MIT 2.71/2.710 Optics10/19/05 wk7-b-45
Optical path delay
z
t = 0
z
t = T/4z = 2.875λ
MIT 2.71/2.710 Optics10/19/05 wk7-b-46
Optical path delay
z
t = 0
z
t = T/4
0E 022 E
022 E−
0
z = 2.875λ
MIT 2.71/2.710 Optics10/19/05 wk7-b-47
Optical path delay
z
t = T/2
z
t = 3T/4
0E− 022 E−
022 E
0
z = 2.875λ
MIT 2.71/2.710 Optics10/19/05 wk7-b-48
Optical path delay
z
t( )tE ω−cos0 ⎟
⎠⎞
⎜⎝⎛ +−
4cos0
πωtE
z = 2.875λ
t
MIT 2.71/2.710 Optics10/19/05 wk7-b-49
Optical path delay
z
t( )tE ω−cos0 ⎟
⎠⎞
⎜⎝⎛ +−
47cos0πωtE
z = 2.875λ
t
( ) ⎟⎠⎞
⎜⎝⎛⇒−
λπω ziEtkzE 2expcos 00In general,
phasor dueto propagation(path delay)
MIT 2.71/2.710 Optics10/19/05 wk7-b-50
Plane wave propagation
z=0
plane of observation
x path delay increaseslinearly with xx
θz
λ
MIT 2.71/2.710 Optics10/19/05 wk7-b-51
Plane wave propagation
zz=0
plane of observation
x path delay increaseslinearly with xx
θ⎟⎠⎞
+⎜⎝⎛
θλ
π
θλ
π
cos2
sin2exp0
zi
xiEλ
MIT 2.71/2.710 Optics10/19/05 wk7-b-52
Plane wave propagation
x
path delayφ
at fixed z
π2π3π
– π–2π– 3π
λ 2λ 3λ– 3λ – 2λ –λ
( ) θλ
πθλ
πϕϕ cos2sin2 whereexp0zxiE +=
λθπ sin2slope
MIT 2.71/2.710 Optics10/19/05 wk7-b-53
Spherical wave propagation
z=0
plane of observation
x path delay increaseswith x x as
z
λ ( )
zx
zxz
2
222
λπλπ
≈
−+
for x << x << zz
MIT 2.71/2.710 Optics10/19/05 wk7-b-54
Spherical wave propagation
z=0
plane of observation
x path delay increaseswith x x as
z
λ ( )
zx
zxz
2
222
λπλπ
≈
−+
quadraticquadraticnear the z axis
MIT 2.71/2.710 Optics10/19/05 wk7-b-55
Spherical wave propagation
z=0
plane of observation
x
z
λ
⎟⎠⎞
+⎜⎜⎝
⎛
λπ
λπ
zi
zxiE
2
exp2
0
MIT 2.71/2.710 Optics10/19/05 wk7-b-56
Spherical wave propagation
x
path delayφ
at fixed z
π2π3π
– π–2π– 3π
λ 2λ 3λ– 3λ – 2λ –λ
( )zyxziE
λπ
λπϕϕ
22
0 2 whereexp ++=
MIT 2.71/2.710 Optics10/19/05 wk7-b-57
Optical path delays matter
L
L
nn
path delay inmaterial ofindex n :
λπ nL2
compare withfree space
propagation :
λπ L2
MIT 2.71/2.710 Optics10/19/05 wk7-b-58
Optical path delays matter
L
L
nn path differencedifference:
( )λ
π Ln 12 −
MIT 2.71/2.710 Optics10/19/05 wk7-b-59
Phase delays matter
• Direct measurement does not work: light waves oscillate too fast for any instrument to follow
• We need an indirect method• Solution: interferometersinterferometers “map” phase onto light
intensity which can be measured directly
Can we measure them and how?
MIT 2.71/2.710 Optics10/19/05 wk7-b-60
Wave interference
λ1λ2
observation screen
MIT 2.71/2.710 Optics10/19/05 wk7-b-61
Interference: extreme cases
+ =Waves in-phase
Constructiveinterference
+ =Waves out-of-phase
Destructiveinterference
MIT 2.71/2.710 Optics10/19/05 wk7-b-62
Interference vs phase delay & contrast
φ
Intensity
averageintensity
fringevisibility
0 π 2π 3π
22
21
212aa
aam+
=
( ) ( )φφ cos10 mII +=
22
210 aaI +=
( )( )φωω
+==
tata cos 2 Field
cos 1 Field
2
1
0I
m akacontrast
relative phase delay
I
MIT 2.71/2.710 Optics10/19/05 wk7-b-63
Interference vs phase delay & contrast
Intensity
Intensity
φ
φ
a1=a2perfect contrast
a1<<a2no interference
Intensity
φ
a1≠a2imperfect contrast
m=1 0<m<1
m≈0Highest contrast / fringe visibility is obtained
by interfering beams ofequal amplitudes
MIT 2.71/2.710 Optics10/19/05 wk7-b-64
Polarization and interference
+φ
-polarized wavesinterfere
-polarized wavesdo not interfere
+φ
I
I
MIT 2.71/2.710 Optics10/19/05 wk7-b-65
Michelson interferometer
incominglaserbeam
l1
l2
( )2122 ll −=λπφ
path difference:
MIT 2.71/2.710 Optics10/19/05 wk7-b-66
Michelson with variable phase-delay
incominglaserbeam
l1
l1
Gas cell
( ) cell 122 tn −∆=λπφ
∆n
MIT 2.71/2.710 Optics10/19/05 wk7-b-67
Mach-Zehnder interferometer
incominglaserbeam
incominglaserbeam
interferencepattern
spatialperiod ~ θ
λsin
θ
brightfringe
(matchedpaths)
MIT 2.71/2.710 Optics10/19/05 wk7-b-68
Young interferometer
incominglaserbeam opaque
screen
x
d1
d2
l
a
MIT 2.71/2.710 Optics10/19/05 wk7-b-69
Young interferometer
incomingplane wave
opaquescreen
x´
d1
d2
l
a
intensity
alλ
=Λx=–a/2
x=a/2
MIT 2.71/2.710 Optics10/19/05 wk7-b-70
Two point sources interfering: math…
x´
intensity
alλ
=Λ
( ) ( )
( ) ( ).2cos12cos4
cos42exp2),(),(:Intensity
.cos42exp2
22exp
22exp42exp1
2/exp2/exp2exp1),(
:Amplitude
22
2
2
22
2
2
22
2
22
2
2222
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎭⎬⎫
⎩⎨⎧ ′⎟
⎠⎞
⎜⎝⎛+=⎟
⎠⎞
⎜⎝⎛ ′
=
=⎟⎠⎞
⎜⎝⎛ ′
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧ ′++′+=′′=′′
⎟⎠⎞
⎜⎝⎛ ′
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧′++′
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎭⎬⎫
⎩⎨⎧ ′
+⎭⎬⎫
⎩⎨⎧ ′−
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧′++′
+=
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎭⎬⎫
⎩⎨⎧ ′++′
+⎭⎬⎫
⎩⎨⎧ ′+−′
⎭⎬⎫
⎩⎨⎧−=′′
xl
all
xal
lxa
l
yaxili
liyxeyxI
lxa
l
yaxili
li
lxai
lxai
l
yaxili
li
lyaxi
lyaxili
liyxe
λπ
λλπ
λ
λπ
λπ
λπ
λ
λπ
λπ
λπ
λ
λπ
λπ
λπ
λπ
λ
λπ
λπ
λπ
λ
Paraxial analysis: