2.717J/MAS.857J Optical Engineering · 2019-09-12 · Reflection & transmission @ dielectric...

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⊥