EElectromagnetic Fields
ᱥ ᯱ ᰆ ᰆ
8㨰㵜 Plane Wave Oblique Incidence
S
Sodqh#Zdyhv#dw#Lqwhuidfhv#+Reoltxh Lqflghqfh,
• In the phase plane, electric and magnetic fields are the same direction, same magnitude and same phase.
• Interface plane is defined as the boundary between two different media.
• Plane of Incidence contains the propagation vector. It is both perpendicular to the interface plane and phase plane of the wave.
x
y
z
EElectromagnetic Fields
When we have e oblique incidence of plane waves on a dielectric interface, the reflection and transmission characteristics become interface, the reflection and transmission characteris
polarization and angle of incidence dependent
We need to distinguish between the two different polarizations~!!!
-
We need to distinguish between We
- Perpendicular Polarization -
Perpendicular Polariz P
P
-- Parallel Polarization
S
Sodqh#Zdyhv#dw#Lqwhuidfhv#+Reoltxh#Lqflghqfh,
• Perpendicular Polarization
-
Electric field is perpendicularto the plane of incidence.
- Magnetic field is parallel to the plane of incidence.
- The fields are configured as in the Transverse Electric (TE) mode.
•
The fields are configured a Parallel Polarization
-
Electric field is parallelto the plane of incidence.
- Magnetic field is perpendicular to the plane of incidence.
- The fields are configured as in the Transverse Magnetic (TM) mode.
x
y
z
EElectromagnetic Fields
? Polarization Perpendicular Polarization Q: Plane of incidence is ?
x y
x E
H
E is parallel to the plane of incidence
E is perpendicular to the plane of incidence
Sodqh#Zdyhv#dw#Lqwhuidfhv#+Reoltxh#Lqflghqfh,
x y
·
E H
z ·
z ·
XY plane
Parallel Polarization
> @
> sin cos ) @
ˆ exp
) cos sin
( ˆ exp
1 1
1
i i
o
i i
o i
y j x
j E
z
y x
j E
z E
T E
T E
T T
E
& propagating in medium 1
component in –x direction component in –y direction Within exponential term p p p p p p p p p
Outside exponential term p p p p p p p
ˆ cos ˆ sin exp >
1( sin cos ) @
1
r r
o r
r
r
E j x y
y x
H E T T
T K
T *
A&
x and y components of H
r*
A: Perpendicular Reflection Coefficient
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
EElectromagnetic Fields
>
( sin cos )@
ˆ oexp 1 i i
i yE j x z
E
E T
T
?&
Perfect Conductor
z
Hi Ei
ani
Incident
Ti
Perpendicular Incidence İ
0, ȝ
0Reflected? Hr
Er
anr
Tr
Reflected
y
Ti
E1
Ti
zˆE1cosTi
xˆE1sinTi
>
jE
xsinT
i jE
zcosT
i@
exp 1 1
What is propagation direction of incident wave?
z x
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Frqgxfwru
>
( sin cos )@
ˆ oexp 1 i i
i yE j x z
E& E T T Perfect Conductor
Hi
z
Ei
ani
Incident
Ti
>
( sin cos )@
ˆ rexp 1 r r
r yE j x z
E& E T T
Perpendicular Incidence İ
0, ȝ
0Reflected? Hr
Er
anr
Tr
Reflected
i ?
H&
ˆcos ˆsin exp>
1( sin cos )@
1
i i
o i
i E j x z
z
x E T T
T K
T
Hi y
Ei
ani
Ti
Ti
Ti
S 5 . 0 Ti
i
Hi
zˆ sinT
ˆsin i z T
i
Hi
zHˆHHH sinT
i
Hi
xˆ cosT
xHˆHHicosTi
xˆcosTi
xˆE1sinTr
zˆE1cosTr
Hr
Er
anr
Tr
r
Hr
xˆ cosT Tr
Tr
S 5 . 0
z x
r
Hr
zˆ sinT
ˆcos ˆsin exp>
1( sin cos )@
1
r r
r r r
r E j x z
z x
H E T T
T K
T
&
Tr
E1
What is propagation direction of reflected wave?
EElectromagnetic Fields
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Frqgxfwru
Perfect Conductor
z
Hi Ei
ani
Ti
> @
ˆcos ˆsin exp>
( sin cos )@
) cos sin
( ˆ exp
1 1
1
r r
r r r
r
r r
r r
z x
E j z
x H
z x
j E
y E
T T
K E T T
T T
E
&
&
1
*
i r
E 0 E
r
o E
E
Apply the boundary condition @ z=0
; Snell’s law of reflection
İ
0, ȝ
0Hr
Er
anr
Tr
y
>
( sin cos )@
ˆ oexp 1 i i
i yE j x z
E& E T T
>
( sin cos )@
ˆ exp>
( sin cos )@
ˆ exp 1 1
1
r r
r i
i o
r i
z x
j E
y z
x j E
y
E E E
T T
E T
T
E
&
&
&
>
( sin )@
ˆ exp>
( sin )@
ˆ exp 1 1
1 0 o i r r
z
x j E
y x
j E
y
E& E T E T
tGX tGY l k i o
s
s
t
w
j
H4w@#H5w Î H4w @#3
G4q @#G5q.㼪v Î G4q @#㼪v
E4q@#E5q Î E4q @#3
K4w@#K5w.Mv Î K4w@#Mv
r
i
T
T It must be satisfied for all x.
>
( sin )@
exp>
( sin )@
0exp 1 1
?Ey Eo jE x Ti Er jE x Ti
x
z=0 și
Hi
))&
Ei
)&
•
Hr
))&
Er
)&
x
z Tr
Perfect Conductor
Incident wave Reflected wave
Parallel Incidence
KZ=#Reoltxh#Lqflghqfh#dw#d#Sodqh#Frqgxfwru
y
EElectromagnetic Fields
Perpendicular Incidence
Mathematical expression for fields
Incident wave Reflected wave
Transmitted wave
x
y
z
xˆE1sinTi
Ti yˆE1cosTi
H
iTi
i
Hi
xˆ cosT
i
Hi
yˆ sinT
E1
> @
ˆ cos ˆ sin exp > ( sin cos ) @
cos sin
ˆ exp
1 1
1 1
i i
o i i
i
i i
o i
y x
E j y
x H
y j x
j E
z E
T T
K E T T
T E
T E
&
&
> @
ˆ cos ˆ sin exp > ( sin cos ) @
) cos sin
( ˆ exp
1 1
1
i i
o i i
i
i i
o i
y x
E j y
x H
y x
j E
z E
T T
K E T T
T T
E
&
&
> @
ˆ cos ˆ sin exp > ( sin cos ) @
) cos sin
( ˆ exp
1 1
1
r r
r r r
r
r r
r r
y x
E j y
x H
y x
j E
z E
T T
K E T T
T T
E
&
&
> @
ˆ cos ˆ sin exp > ( sin cos ) @
) cos sin
( ˆ exp
2 2
2
t t
t t t
t
t t
t t
y x
E j y
x H
y x
j E
z E
T T
K E T T
T T
E
&
&
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
EElectromagnetic Fields
> @
ˆ cos ˆ sin exp > ( sin cos ) @
) cos sin
( ˆ exp
1 1
1
i i
o i i
i
i i
o i
y x
E j y
x H
y x
j E
z E
T T
K E T T
T T
E
&
&
> @
ˆ cos ˆ sin exp > ( sin cos ) @
) cos sin
( ˆ exp
1 1
1
r r
o r
r r
r r
o r
y x
E j y
x H
y x
j E
z E
T T
K E T
T
T T
E
*
*
A A
&
&
> @
ˆ cos ˆ sin exp > ( sin cos ) @
) cos sin
( ˆ exp
2 2
2
t t
o t
t t
t t
o t
y x
E j y
x H
y x
j E
z E
T T
K E T W
T
T T
E W
A A
&
&
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
* *
AE
oW
AE
ooR
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
tGX tGY l k i o
s
s
t
s
s
t
H4w @#H5w G4q @#G5q E4q @#E5q K4w @#K5w
E
E tangential components ( s (z-component) are continuous at the boundary.
At y=0,
0 0 0
0
0
21
E y E y
y E
y E y
E
t r
i
o
>
i i@
o ii
y z E j x z E j x
E & 0 ˆ exp E
1( sin T 0 cos T ) ˆ exp E
1sin T
r
>
r r@
r rr
y z E j x z E j x
E & 0 ˆ exp E
1( sin T 0 cos T ) ˆ exp E
1sin T
t
>
t t@
t tt
y z E j x z E j x
E & 0 ˆ exp E
2( sin T 0 cos T ) ˆ exp E
2sin T
EElectromagnetic Fields
i r r t to
j x E j x E j x
E exp E
1sin T exp E
1sin T exp E
2sin T
?
t r
i
E T E T
T
E
1sin
1sin
2sin Î Phase matching:
Arguments of the exponentials must be equal~!!!
r
i
T
T
sin sin
2 1
i
t T
E T E
Snell el ˅ l˅s Law
t
r
E
E E
0 E Z PHH T P Z
H P
Z sin
2 2
1
1 i
i r i t io
j x E j x E j x
E exp E
1sin T exp E
1sin T exp E
1sin T
?
sin
2 2
1
1 Ti
H P
H P
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
tGX tGY l k i o
s
s
t
s
s
t
H4w @#H5w G4q @#G5q E4q @#E5q K4w @#K5w
ˆcos ˆsin exp>
1( sin cos )@
1
i i
o i i
i E j x y
y x
H E T T
T K
T
&
ˆcos ˆsin exp>
1( sin cos )@
1r r r
r r
r E j x y
y x
H E T T
T K
T
&
ˆcos ˆsin exp>
2( sin cos )@
2t t t
t t
t E j x y
y x
H E T T
T K
T
&
H fields have x and y components.
H
H fields have x
H tangential components ( mpon
s (x-component) are continuous at the boundary.
At y=0,
0 0
20 0 0
1
H y H y
y H
y H y
H
tx rx
ix
x x
0 cos exp( 1 sin )1o i
i
ix E j x
y
H E T
T K
0 cos exp( 1 sin )
1r r
r
rx E j x
y
H E T
T K
0 cos exp( 2 sin )2t t
t
tx E j x
y
H E T
T K
EElectromagnetic Fields
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
t t
t r
r r
i i
o E j x
x E j
x E j
T E
K T T
E K T
T E
K
1cos T exp
1sin
1cos exp
1sin
2cos exp
2sin
t t r
r i
o E E
E
T
T K T K
K
1cos
1cos
2cos
(E1sinTi E1sinTr E2sinTt)We need to solve two simultaneous equations.
1
t r
o E E
E
2 cos cos
cos
2 1
1
t t r
r i
o E E
E T
T K T K
K
0 0 0
0
0
21
H y H y
y H
y H y
H
tx rx
ix
x x
From,
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
1t r
o E E
E
t t r r i
o E E
E T
T K T K
K1 cos 1 cos 2 cos
¸¸¹
¨¨ ·
©
§
¸¸¹
¨¨ ·
©
§
¸¸¹
¨¨ ·
©
§
t t
t
t r
r
t i
o E E
E
T T K K T
T K K T T K
K cos cos
cos cos
cos cos 2
2 2
1 2
1
3
cos cos cos
cos
1 2 1
2
t r t i o
t
i E E E
T
T K K T
T K K
(1)+(3)Î
1 coscos 1 coscos 01 2 1
2 ¸¸¹
·
¨¨©
§
¸¸
¹
·
¨¨©
§ r
t i o
t
i E E
T T K K T
T K K
i t
i t
t i t i
o r
E E
T K T K
T K T K T T K
K T
T K K
cos cos
cos cos
cos 1 cos
cos 1 cos
2 1
2 1
1 2 1 2
t i
t i
r i
r
E E E E
T K
T K
T K
T K
cos cos
cos cos
1 2
1 2
0
*
?
A ; Fresnel Reflection Coefficientfor Perpendicular Incidence
EElectromagnetic Fields A
*
AW
1
t r
o E E
E
Next, let’s find a transmission coefficient.
From ,
: Fresnel Transmission Coefficient
for Perpendicular Incidence
t i
i t
i t
E E E E
T K T K
T W K
cos cos
cos 2
1 2
2
0
? A
cos cos
cos cos
0 1
2
1 2
t t
i
t i
o E E
E
T K T K
T K T K
0 1
2
1 2
cos cos
cos
cos E
E
t i
t i
r
K T K T
T K T K
t t
i
t i
t
i E E E
T K T K
T K T K T
K T K
cos cos
cos cos
cos cos
1 2
0 1
2 0 1
2
t t
i
t i
t
i E E
0 1
2
1 2
1 2
cos cos
cos cos
cos cos
T K T K
T K T K T K T K
t t
i
i
i E E
0 1
2
2 2
cos cos
cos cos
T K T K
T K T K
t t
i
i E E
1 0
2 2
cos cos
cos 2
T K T K
T
K
¸¸¹¨¨ ·
©
* §
*
i t
i t
i i t
i t
cf
T W T
T K T K
T W K
T K T K
T K T K
cos 1 cos
cos cos
cos 2
cos cos
cos cos
.
||
||
1 2
|| 2
1 2
1
|| 2
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
2 1
sin sin
E E T T
i t
Index of Refraction (n)
: Ratio of the speed of light in free space to that in the medium
v
pn c
t
i
n
n
1sin T
2sin T
?
r
i
T
T
Snell’s Law of Reflection:
Snell’s Law of Refraction
2 1
n n
2 1
p p
v v
Z Z
v
pE Z
2 1
p p
v c v
c
v
pn c
Î Snell’s Laws are independent of wave polarization
EElectromagnetic Fields
1 2
K K
2 2
1 1 2
1
sin sin
H P Z
H P Z E E T T
i t
1 2
K K
Reoltxh#Lqfllghqfh#dw#d#S Sodqh#Glhohfwulf#Phglxp
2 2 2
1 1 1
H K P
H K P
For or nonmagnetic materials ( s (ȝ (ȝ= ȝ=ȝ ȝ
o),
1 0 2 0
1 2
H P H P
K K
2 0 1 0
H P H P
2 1 0
1 2 0
H H P
H H P
Let’s see Snell’s law of refraction.
2 1
sin sin
H H T T
i
t
: Snell’s law of refraction for nonmagnetic materials
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp Let’s see the reflection coefficient.
t i
t i
T K T K
T K T K
cos cos
cos cos
1 2
1 2
*A
i i
i i
H T T H
H T T H
2 1 2
2 1 2
sin cos
sin cos
t i
t i
K T T K
K T T K
cos cos
cos cos
2 1 2 1
t i
t i
H T T H
H T T H
cos cos
cos cos
1 2 1 2
K
K
2 1
K K
2 1
sin cos
sin 1 cos
sin 1 sin
1 cos
sin sin
sin
2 1 2 1
2
2 2 1 1 2 1
2
2 2 2 1
2 1 2
2 1 1
i t
i t
i t
t
i i
t
H T T H H H
H T H H T H H H
H T T H
T
H T T H H P
H T P
?
sin
cos 2
1 2 1
2
i
t T
H T H H
H
Tt
H cos H
1 2
Tt
H cos H
1 2
Ti
H
H 2
1 2 sin
Ti
H
H 2
1 2 sin
1 2 2
1
H H K
K
EElectromagnetic Fields
If İ
2>İ
1,
i i
i i
H T T H
H T T H
2 1 2
2 1 2
sin cos
sin cos
*A
real.
A
is
*
. 1
2
1
!
H If İ
1>İ
2(the wave in medium 1 is incident on a less dense medium 2), H
then the square root is positive.
¸ ¸
¹
·
¨ ¨
©
§
2 1
sin sin
H H T
T
i
From Snell’s law of refraction ,
tsin 1 sin !
i t
T T
i
t
T
T !
; Transmission angle is higher than incidence angle.
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Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
2
T
tS
,sin From sin
2 1
H H T T
i t
1
sin 2
H Ti H
Total Reflection
2 1
sin sin 2
H H T
S¸
¹
¨ ·
©
§
i
¸¸¹
·
¨¨©
§
1 1 2 1
1 2 sin
sin n
n
ic H
T H
Critical Angle
¸¸¸¸¹
¸¸
·¸¸
¨¨¨¨©
¨¨
§¨¨
1 1 2 1
1 2 sin
sin n
n
ic H
T H
Critical Angle
0 1 cos
0 cos sin
cos
sin cos
, sin
At
2 1 2
2 1 2
1 2
*A
i i
i i
i i
i T
T H T
T H H T T H
H T H 2? become )
angle(
ion transmiss
Can the
T
tS
When?EElectromagnetic Fields
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
. sin sin
,
When
T
i !T
icT
i !T
ic. sin sin
, sin sin
From
2 1 2 1
ic t
i t
H T T H
H T T H
!
When the incident angle is s larger than the critical angle,
. 1 sin , sin
Because
1
1 2 !
t
ic T
H T H
1 sin 1
sin
cos 2
2
2 r 1
r t i
t j j T
H T H
T
t
t T
T 1 sin2
cos
; imaginary
i
t T
H T H sin sin
2
t 1 Ti
H T H sin sin
2
1
sin2Tt sin2
2 1
Ti
H H
R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
>
( sin cos )@
ˆ oexp 2 t t
t z E j x y
E& WA E T T
When the incident angle is s larger than the critical angle, what is a transmitted electric field?
»»
¼ º
««
¬
ª r
A exp sin sin 1
ˆ 2
2 1 2 2
1
2 i i
o j x y
E
z T
H E H H T
E H W
»»
¼ º
««
¬ ª
¸¸
¹
·
¨¨
©
§ r
A exp sin sin 1
ˆ 2
2 1 2
1
2 i i
o j x jy
E
z T
H T H
H E H
W cos sin 1
sin sin
2 2 1 2 1
r i
t
i t
j T
H T H
H T T H sinTt cosTt)
sin
2 1
Ti
H H
¼ º
¸¸
¸¸¹
¸¸
·¸¸
1 sin2
2 1
Ti
H H
»»
¼ º
««
¬ ª
»»
¼ º
««
¬
ªr
AEo y i j x i
z T
H E H H T
E H
W exp sin 1 exp sin
ˆ
2 1 2 2
2 1 2
> @
»»
¼ º
««
¬
ª
AEo y j x i
z T
H E H D
W exp exp sin
ˆ
2 1
2 sin2 1
2 1
2 Ti
H E H
D sin2 1
2 1
2 Ti
H E H D
¼ º
««
««¬««
ª««r y Ti
H
E H sin2 1
2 1 2
> @
yDyy; Total Internal Reflection (TIR)
EElectromagnetic Fields cos 0
cos
cos cos
1 2
1 2
*A
t i
t i
T K T K
T K T
K2cosTi K1cosTt K
: Incident angle of no Reflection (ș
Bŏ)
T T
Brewster Angle
t
B K T
T
K2cos A 1cos
t
B K T
T
K2cos A 1 1sin2
t
B T
K
T K 2
2
1 1 sin
cos A
tB T
K
T K 2
2
2
2 1 1 sin
cos ¸¸¹
¨¨ ·
©
§
A
¸¸
¹
·
¨¨
©
§
¸¸¹
¨¨ ·
©
§
¸¸¹
¨¨ ·
©
BA § BA
n
n T
K
T K 2
2
2 1 2
2
2 1 1 sin
sin 1
i t
i t
n n n n
T T
E E T T
2 2
2 2 1
2 1 2 1
sin sin
sin sin
¸¸¹
¨¨ ·
©
? §
Tt
sin2
¹
·
A
TB
sin2
¨¨¨¨©
¨¨§
¨¨
¸¸¹
¨¨ ·
©
§
BA TBA
H P
H P H
P H
T P 2
2 2
1 1 1
2 2
2 1 1 sin
sin 1
1 2
2 1
2 2 1 1
2 1
H P
H P H P H P K K
H P
H P
1 2
2 1
¸¸¸¸¹
¸¸·
¨¨¨¨ ¸¸
©¨¨
§¨¨
©K K 2
2 1
2 2
1 1 2 1 2 1
H P
H P E E n n
H P
H P
2 2
1 1
¸¸¸¸¹
¸¸·
¨¨¨¨ ¸¸
©¨¨
§¨¨
n n 2
2 1
A
A ¸¸¹
¨¨ ·
©
§
B TB
P P H P
H
T P 2
2
2 1 1 2
2
2 1 sin
sin 1
: Incident angle of no Reflection (ș
Bŏ) R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp Brewster Angle
A
A ¸¸¹
¨¨ ·
©
§
B TB
P P H P
H
T P 2
2
2 1 1 2
2
2 1 sin
sin 1
1 2
2 2 1
2
2
1 sin 1
1 PH
H T P
P
P
°¿
°¾
½
°¯
°®
¸¸¹
¨¨ ·
©
§ BA
2
2 1 1 2
2 1 2
1 1 sin
¸¸¹
¨¨ ·
©
§
? A
P P H P
H P TB
For nonmagnetic materials (ȝ=ȝo),
f
A 1 1
1
sin 1
2
2 H
H TB
Î Brewster angle does s not exist for perpendicular incidence.
EElectromagnetic Fields
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
Parallel Incidence
xz=0 și O
Hi
))&
Ei
)&
•
Hr
))&
Er
)&
șt
x
z
Mathematical expression for the incident fields
> @
>
( sin cos )@
ˆ exp
) cos sin
( exp ˆsin
ˆcos
1 1
1
i i
o i
i i
o i i
i
z x
E j y H
z x
j E
z x
E
T T
K E
T T
E T
T
&
&
Tr
Ti
xˆE1sinTi
E
iTi zˆE1cosTi E i
xˆ 0cosT
și
z x
E i
zˆ 0sinT
H&t
E&t
•
z x
y E
1R
Reoltxh#Lqflghqfh#dw#d#Sodqh#Glhohfwulf#Phglxp
Parallel Incidence
xz=0 și O
Hi
))&
Ei
)&
•
Hr
))&
Er
)&
șt
x
z Tr
> @
>
( sin cos )@
ˆ exp
) cos sin
( exp ˆsin
ˆcos
1 1
1
r r
r r
r r
r r r
r
z x
E j y H
z x
j E
z x
E
T T
K E
T T
E T
T
&
&
Mathematical expression for the reflected fields
Tr
xˆE1sinTr
E
rzˆE1cosTr
r
Er
xˆ cosT
r
Er
zˆ sinT
Tr
Tr
Tr
H&t
E&t