electromagnetic (e-m) theory of waves at a dielectric interface ii rr tt nini ntnt ûnûn x y. z b...
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Electromagnetic (E-M) theory of waves at a dielectric interface
i r
t
ni
nt
ûn
x
y
.z
b
ik
rk
tk
iE
rE
tE
While it is possible to understand reflection and refraction from Fermat’s principle, we need to use E-M theory in order to understand quantitatively the relationship between the incident, reflected, and transmitted radiant flux densities:
We can accomplish this treatment by assuming incident monochromatic light waves which form plane waves with well defined k-vectors as shown in the diagram. The interface is shown with an origin and coordinates (x,y,z).
We will consider E-field polarizations which are (i) in the plane of incidence and (ii) perpendicular to the plane of incidence, as shown below.
),,(),,,(),,,( ttttrrrriiii kEIkEIkEI
Ii
Ir
It
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E-field is perpendicular to the plane-of incidence
E-field is parallel to the plane of incidence
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Maxwell’s Equations for time-dependent fields in matter
00
BSdB
t
DjHSd
t
DjrdH
t
BESd
t
BrdE
DdVSdD
HBEDEgj
HMMBH
EPPED
mo
eo
1
D – Displacement field
H – Magnetic Intensity
P – Polarization
M – Magnetization
- Magnetic permeability
- Permittivity
e - Dielectric Susceptibility
m - Magnetic Susc.
g – Conductivity
j – Current density
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Summary of the boundary conditions for fields at an interface
Boundary
Side 1
Side 2
Maxwell’s equations in integral form allow for the derivation of the boundary conditions for the total fields on both sides of a boundary.
jHHu
uBB
EEu
uDD
n
n
n
n
12
12
12
12
ˆ
0ˆ
0ˆ
ˆ Normal component of D is discontinuous by the free surface charge density
Tangential components of E are continuous
Normal components of B are continuous
Tangential components of H are discontinuous by the free surface current density
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For dielectrics, j = 0. Therefore, the components of E and H that are tangent to the interface must be continuous across it. Since we have Ei, Er, and Et the continuity of E components yield:
i r
t
ni
nt
ûn
x
y
.z
b
ik
rk
tk
iE
rE
tE
i 090
iii EE cossin
tnrin EuEEu
ˆˆ
iiinin EEuEu cossinˆˆ
Note that
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tnrin EuEEu
ˆˆ
tttotn
rrrorniioin
tnrnin
trkEu
trkEutrkEu
EuEuEu
cosˆ
cosˆcosˆ
ˆˆˆ
Consider the expression on the interface (y = b) for all x, z and t. The above relationship must hold at all points and at any instant in time on the interface. Therefore
bytttbyrrrbyii trktrktrk |||
Since then we have tri
byttbyrrbyi rkrkrk |||
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Thus, at the interface plane
cos.|ˆ
ˆ||0ˆ|
rconstrkandkkk
ukandkkurkk
byir
nirnrbyri
knu
r
r co
s
ttiitt
ii
ttii
itntbyti
rirriiirn
nncc
kk
kkurkkAlso
kkkku
sinsin,sin1
sin1
sinsin
0ˆ|,
sinsin0ˆ
which is again Snell’s Law
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Case 1: E Plane of incidence
Continuity of the tangential components of E and H give
otoroi EEE
Cosines cancel
Using H = -1 B, the tangential components are
otoroitott
tioroi
i
i
tttt
iriii
riri
tttrrriii
tt
tr
i
ri
i
i
EEEandEn
EEn
Ev
EEv
vvand
vEBandvEBvEBSince
BBB
coscos)(
cos1
cos)(1
,
/,/,/
coscoscos
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The last two equations give
coscos
cos2
coscos
coscos
t
ti
i
i
ii
i
oi
ot
t
ti
i
i
tt
ti
i
i
oi
or
nn
n
E
Eand
nn
nn
E
E
The symbol means E Plane of incidence. These are called the Fresnel equations; most often i t o.
Let r = amplitude reflection coefficient and t = amplitude transmission coefficient. Then, the Fresnel equations appear as
ttii
ii
oi
ot
ttii
ttii
oi
or
nn
n
E
Etand
nn
nn
E
Er
coscos
cos2
coscos
coscos
Note that t - r = 1
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Case 2: E || Plane of Incidence
Continuity of the tangential components of E:
totrorioi EEE coscoscos
Continuity of the tangential components of -1 B:
it
tt
i
i
ii
i
oi
ot
tt
ti
i
i
ti
ii
t
t
oi
or
riri
ottt
orrr
oiii
nn
n
E
Etand
nn
nn
E
Er
and
Ev
Ev
Ev
coscos
cos2
coscos
coscos
;
111
||
||
||
||
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If both media forming the interface are non-magnetic i t o then the amplitude coefficients become
itti
ii
itti
tiit
nn
ntand
nn
nnr
coscos
cos2
coscos
coscos||||
Using Snell’s law
the Fresnel Equations for dielectric media become
ttii nn sinsin
)cos()sin(
cossin2
)sin(
cossin2
,)tan(
)tan(
)sin(
)sin(
||
||
titi
it
ti
it
ti
ti
ti
ti
tandt
randr
Note that t - r = 1 holds for all i , whereas t|| + r|| = 1 is only true for normal incidence, i.e., i = 0.
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Consider limiting cases for nearly normal incidence: i 0.
In which case, we have:
ti
ti
iirr
sin
sin00|| since 1sintan xxx
ittti
tittiti n
n cossincossincossincossinsin
Also, using the following identity with Snell’s law
Therefore, the amplitude reflection coefficient can be written as:
iitt
iitt
iti
t
iti
t
nn
nn
nnnn
ri
coscos
coscos
coscos
coscos
0
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ti
i
it
it
nn
nttand
nn
nnrr
iiii
20||00||0
In the limiting case for normal incidence i=t = 0, we have:
Note that these equalities occur for near normal incidence as a consequence of the fact that the plane of incidence is no longer specified when i t 0.
Consider a specific example of an air-glass interface:
ni = 1
nt = 1.5
ttii nn sinsin
We will consider a particular angle called the Brewster’s angle: p + t = 90 .tan
i
tp n
n
3.56
1
5.1tan 1
p
At the polarization angle p, only the component of light polarized normal to the incident plane and therefore parallel to the surface will be reflected.
i
t
External reflection nt > ni
Internal reflection ni > nt
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External Reflection (nt > ni) Internal Reflection (nt < ni)
n
i
tcttii n
nnn
tci sinsinsin 90
8.415.1/1sin 1c
7.33
5.1
1tan
'
'
p
i
tp n
n
3.56
1
5.1tan 1
p
1
5.1
i
tti n
nnConsider
5.1
1
i
tti n
nnConsider
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Concept of Phase Shifts () in E-M waves : trkEE o
cos
0)sin(
)sin(
ti
ti
oi
or
E
Er
Since when nt > ni and t < i
as in the Air Glass interface,
we expect a reversal of sign in the electric field for the Ecase when
.1 forei
We need to define phase shift for two cases:
A. When two fields E or B are to the plane of incidence, they are said to be (i) in-phase (=0) if the two E or B fields are parallel and (ii) out-of-phase ( = ) if the fields are anti-parallel.
B. When two fields E or B are parallel to the plane of incidence, the fields are (i) in-phase if the y-components of the field are parallel and (ii) out-of-phase if the y-components of the field are anti-parallel.
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Examples of Phase shifts for two particular cases:
,,
,
0,,
,
phaseofoutBB
EE
phaseinEE
BB
ri
ri
ti
ti
)a( )b(
0,,
,,
phaseinallBBB
EEE
tri
tri
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Glass (n = 1.5) Air (n = 1)
Air (n = 1) Glass (n = 1.5)
Analogy between a wave on a string and an E-M wave traversing the air-glass interface.
= 0 = 0
= = 0
Compare with the case of .0 ior forE
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Examples of phase-shifts using our air-glass interface:
In order to understand these phase shifts, it’s important to understand the definition of .
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Reflected E-field orientations at various angles for the case of External Reflection (ni < nt). It is worth checking and comparing with the various plots for the phase shift on the previous slides.
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Reflected E-field orientations at various angles for the case of Internal Reflection (ni > nt). It is worth checking and comparing with the various plots for the phase shift on the previous slides.
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Reflectance and Transmittance
Remember that the power/area crossing a surface in vacuum (whose normal is parallel to the Poynting vector) is given by .2 BEcS o
The radiant flux density or irradiance (W/m2) is
TToo
TEcEvE
cSI 222
2
/1vPhase velocity
From the geometry and total area A of the beam at the interface, the power (P) for the (i) incident, (ii) reflected and (ii) transmitted beams are:
ttt
rrr
iii
AIPiii
AIPii
AIPi
cos)(
cos)(
cos)(
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Define Reflectance and Transmittance:
2
2
2
2
2
2
cos
cos
cos
cos
cos
cos
2/
2/
cos
cos
tn
n
E
E
n
n
AI
AI
P
PT
RrE
E
Ev
Ev
I
I
AI
AI
P
PR
ii
tt
oi
ot
ii
tt
ii
tt
i
t
oi
or
oiii
orrr
i
r
ii
rr
i
r
Note that
2
2222
22
,
1
2
11
2
1
2
1
2
1
2
1
o
oo
oo
oo
oo
oo
EnITherefore
Ec
nEc
c
vEEvEI
Conservation of Energy at the interface yields:
riritottrorrioii
ttrrii
nnEnEnEn
AIAIAI
,222 ;coscoscos
coscoscos
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Therefore, TRE
E
n
n
E
E
oi
ot
ii
tt
oi
or
22
cos
cos1
We can write this expression in the form of componets and:||
2||||
22||||
2
cos
cos
cos
cos, t
n
nTandt
n
nTrRandrR
ii
tt
ii
tt
11, |||| TRandTRTherefore
1
5.1
i
tti n
nnConsider
We must use the previously calculated values for ttrr ,,, ||||
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1
42
4
,0
2
22
2||
2
||
it
itiitt
it
it
it
iti
nn
nnnnnnTR
nn
nnTTTand
nn
nnRRRWhen
It’s possible to verify for the special case of normal incidence:
Consider the case of Total Internal Reflection (TIR):
ttii nn sinsin
5.1
1
i
tti n
nnConsider
t
i
nt = 1
ni = 1.5
8.41
5.1
1sin,90
c
i
tit n
nWhen
c