Electromagnetic (E-M) theory of waves at a dielectric interface ii rr tt nini ntnt ûnûn x y. z b...
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Transcript of Electromagnetic (E-M) theory of waves at a dielectric interface ii rr tt nini ntnt ûnûn x y. z b...
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
E-field is perpendicular to the plane-of incidence
E-field is parallel to the plane of incidence
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
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
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
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 |||
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
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
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
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
||
||
||
||
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.
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
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
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
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.
Examples of Phase shifts for two particular cases:
,,
,
0,,
,
phaseofoutBB
EE
phaseinEE
BB
ri
ri
ti
ti
)a( )b(
0,,
,,
phaseinallBBB
EEE
tri
tri
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
Examples of phase-shifts using our air-glass interface:
In order to understand these phase shifts, it’s important to understand the definition of .
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.
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.
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)(
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
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 ,,, ||||
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