Electromagnetic (E-M) theory of waves at a dielectric interface
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:
 
 
 
Ii (Ei , ki ,i ), I r (Er , kr ,r ), It (Et , kt ,t )
y
We can accomplish this treatment by
assuming incident monochromatic light
.
waves which form plane waves with well z
defined k-vectors as shown in the diagram.
The interface is shown with an origin and
coordinates (x,y,z).

Er

Ei
Ii

ki
i
Ir

kr
r
ûn
ni
nt
b
x
t

Et

kt
It
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.
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
 
 
D  
 D  dS   dV


 


B
B
 E  dr   t  dS   E   t


 
   D
  D  
 H  dr    j  t   dS   H  j  t
 
 
B  0
 B  dS  0

 
D  oE  P

1  
H
BM
o


j  gE


D  E


P  e E


M  m H


B  H
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
Side 1
Boundary
Side 2




D2  D1  uˆ n  




uˆ n  E2  E1  0




B2  B1  uˆ n  0





uˆ n  H 2  H 1  j
Maxwell’s equations in integral form
allow for the derivation of the
boundary conditions for the total
fields on both sides of a boundary.
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:

Ei sin   Ei cosi

  900  i
Ei



Er

 

uˆn  Ei  Er  uˆn  Et

ki
Note that

uˆn  Ei  uˆn E i sin   Ei cosi
i

kr
r
ûn
y
.
z
ni
nt
b
x
t

Et

kt


 

uˆn  Ei  Er  uˆn  Et



 uˆn  Ei  uˆn  Er  uˆn  Et
 
 


 uˆn  Eoi cos ki  r  i t  uˆn  Eor cos kr  r  r t   r
 

 uˆn  Eot cos kt  r  t t   t






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






 
 
 
ki  r  it | y b  kr  r  r t   r | y b  kt  r  t t   t | y b
Since
i  r  t
 

then we have



 
 
 
ki  r | y  b  k r  r   r | y  b  kt  r   t | y  b
Thus, at the interface plane




  
 

k i  k r  r | y b   r
 uˆn  kr  ki  0 and k || uˆn
  

ˆ
k  kr  ki
and k  r | y b  const.  r cos
uˆn


r cos 

k


r
 
uˆn  k r  ki  0  ki sin  i  k r sin  r
 i   r
  
 
Also,
ki  k t  r | y b   t
uˆn  kt  ki  0


 ki sin  i  kt sin  t

c 1
 i
sin  i 

c 1
 t
sin  t , ni sin  i  nt sin  t
which is again Snell’s Law
E  Plane of incidence
Case 1:
Continuity of the tangential components of E and H give



Eoi  Eor  Eot
Cosines cancel
Using H = -1 B, the tangential components
are

Bi
i
cos i 
Since
Br
i
Bi  Ei / vi ,
vi  vr ,
and
cos r  
Bt
t
cos t
Br  E r / vr ,
and
Bt  Et / vt
i   r
1
1

( Ei  Er ) cos i 
Et cos t
i vi
t vt

ni
i
( Eoi  Eor ) cos i 
nt
t
Eot cos t
and Eoi  Eor  Eot
The last two equations give
 Eor

 Eoi
ni
cos i 
nt
cos t


t
  i
  ni cos i  nt cos
i
t
E
and  ot
 Eoi
2
ni
cos i

i
 
  ni cos i  nt cos
i
t
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
 Eor
r  
 Eoi

n cos  i  nt cos  t
  i
  ni cos  i  nt cos  t
Note that t - r = 1
and
 Eot
t   
 Eoi

2ni cos  i
 
  ni cos  i  nt cos  t
Case 2: E || Plane of Incidence
Continuity of the tangential components of E:
Eoi cosi  Eor cosr  Eot cost
Continuity of the tangential components of -1 B:
1
1
1
Eoi 
Eor 
Eot
i vi
 r vr
t vt
and
i   r ;  i   r
nt
cos i 
ni
cos t
E 

i
r||   or   t
 Eoi || ni cos i  nt cos t
i
t
2
ni
cos i
E 
i
and t||   ot  
 Eoi || ni cos t  nt cos i
i
t
If both media forming the interface are non-magnetic i  t  o then
the amplitude coefficients become
nt cosi  ni cost
r|| 
ni cost  nt cosi
Using Snell’s law
2ni cosi
and t|| 
ni cost  nt cosi
ni sin i  nt sin t
the Fresnel Equations for dielectric media become
r  
sin( i   t )
sin( i   t )
2 sin  t cos i
t 
sin( i   t )
and r|| 
tan( i   t )
,
tan(i   t )
2 sin t cos i
and t|| 
sin(i   t ) cos( i   t )
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:
r 
||  0
i
  r i 0
sin  i  t 

sin  i   t 
since
tan x  sin x
x  1
Also, using the following identity with Snell’s law
nt
sin i  t   sin i cost  sin t cosi  sin t cost  sin t cosi
ni
Therefore, the amplitude reflection coefficient can be written as:
 r  0
i
nt
cos t  cos i
ni
nt cos t  ni cos i


nt
cos t  cos i nt cos t  ni cos i
ni
In the limiting case for normal incidence i=t = 0, we have :
 r  0  r||  0
i
i
nt  ni

nt  ni
and
t  0  t||  0
i
i
2ni

ni  nt
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:
i
t
ni = 1
nt = 1.5
ni sin i  nt sin t
We will consider a particular angle called the
Brewster’s angle: p + t = 90  tan p 
 1.5 
 p  tan    56.3
 1 
nt
.
ni
1
External reflection nt > ni
Internal reflection ni > nt
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.
External Reflection (nt > ni)
Consider nti 
Internal Reflection (nt < ni)
ni sin i    nt sin t  90  sin c  nt
nt 1.5

ni
1
i
c
t
n
1
Consider nti  t 
ni 1.5
t an p' 
nt
1

ni 1.5
  p'  33.7
n
 1.5 
  56.3
 1 
 p  tan1 
c  sin 1 1/ 1.5  41.8
ni
Concept of Phase Shifts () in E-M waves:
Since

 
 
E  Eo cos k  r  t  

E 
sin( i   t )
r   or  
 0 when nt > ni and t < i
 Eoi   sin( i   t )
as in the Air  Glass interface,
we expect a reversal of sign in the electric field for the Ecase when
ei  1
for    .
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-ofphase ( = ) 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-ofphase if the y-components of the field are anti-parallel.
Examples of Phase shifts for two particular cases:
 
(a)  Bi , Bt 
(b)




in

phase
,



0


  
E
,
E
 Ei , Er , Et 
 i t
 
     all in  phase
Ei , Er 
 Bi , Br , Bt 
   out  of  phase,    
 Bi , Br 
  0
Analogy between a wave on a string and an E-M
wave traversing the air-glass interface.
Glass (n = 1.5)
Air (n = 1)
 = 0
Air (n = 1)
 = 0
Glass (n = 1.5)
 = 
 = 0
  for
Compare with the case of Eor
i  0 .
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
 in2 vacuum
  (whose normal
is parallel to the Poynting vector) is given by S  c  o E  B.
The radiant flux density or irradiance (W/m2) is
   2
c o 2
2
 E
I S T 
Eo  v E
 c

v  1 / 
T
2
 
Phase 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:
(i)
Pi  I i A cosi
(ii)
Pr  I r A cos r
(iii)
Pt  I t A cost
T
Define Reflectance and Transmittance:
2
Pr I r A cos r I r vr  r E / 2  Eor 
  r 2  R
R

 
 
Pi I i A cos i I i vi i E / 2  Eoi 
2
or
2
oi
Pt I t A cos t nt cos t
T 

Pi I i A cos i ni cos i
Note that I 
 Eot

 Eoi
2
  nt cos t
  
  ni cos i
2
t

1
1  2 1  o 2 1 1 c 2 1 1 2
2
vEo 
Eo 
Eo 
Eo  n
Eo
2
2
2 o
2 o
2 v o c
2 o c
Therefore, I  n Eo2
Conservation of Energy at the interface yields:
I i A cosi  I r A cos r  I t A cos t
 ni Eoi2 cosi  nr Eor2 cos r  nt Eot2 cost ; ni  nr ,  i   r
Therefore,
 Eor
1  
 Eoi
2
  nt cos t
  
  ni cos i
 Eot

 Eoi
2

  R  T

We can write this expression in the form of componets  and ||:
R  r
2

 nt cost
and R||  r|| , T  
 ni cosi
Therefore,
2
2
t

 nt cost
and T||  
 ni cosi
R  T  1 and R||  T||  1
nt 1.5
Consider nti  
ni
1
We must use the previously calculated
values for r|| , r , t|| , t
2
t||

It’s possible to verify for the special case of normal incidence:
 nt  ni 

When  i  0, R  R||  R  
 nt  ni 
4nt ni
and T  T||  T 
nt  ni 2
2
nt2  2nt ni  ni2  4nt ni
 R T 
1
2
nt  ni 
ni sin i  nt sin t
Consider the case of Total Internal Reflection (TIR):
nt
1
Consider nti  
ni 1.5
When t  90,
  c  41.8
sin i 
t
nt = 1
c
nt
1
 
ni 1.5
ni = 1.5
i