UPW AT A PLANAR BOUNDARY
Case I: TE Wave
Ei
θi
Hi
kx
x
kz
ki
ki = k o
“Transverse Electric”
E ⊥ Plane of incidence
k o = ω με
θi
θr
y
Er
kz
kr
ε,μ
E
z
εt,μt
k
θt
kt
Et
Trial Solutions:
Incident:
Reflected:
+ j(k ocosθi )x - j(kosinθi )z
Ei = yE
ˆ oe + jk x x− jk z z = yE
ˆ oe
Er = ŷ Γ Eoe − jk x x - jk z z
+jk x - jk z z
Transmitted: Et = yT
ˆ Eo e tx
L10-1
TE WAVE⎯H AT A BOUNDARY
Ei = yE
ˆ oe + jk x x − jk z z
Eo
Hi = −
(x̂ sinθi + zˆ cos θi )e + jk x x − jk z z
ηi
ΓEo
Reflected:
Hr =
( − x̂ sinθi + zˆ cos θi )e − jk x x − jk z z
ηi
TEo
Transmitted: Ht = −
(x̂ sin θt + zˆ cos θt ) e+jk tx x - jk tz z
ηt
Incident:
Ei
θi
Hi
TE Wave
x
ki
ε,μ
εt,μt
θi θi
y
θi
Er
z
Hr
θt
Ht
L10-2
IMPOSE BOUNDARY CONDITIONS
E // and H// are continuous at x = 0:
Ei// + Er // = Et// , and Hi// + Hr// = Ht// for all y, z.
Therefore:
0
0
+ jk x - j(k o sinθi )z
yE
+ yˆ Γ Eoe − jk x x - j(ko sinθr )z
ˆ oe x
0
+jk x - j(k t sinθ t )z
= yT
ˆ Eo e tx
sinθ t k o
(Snell's Law)
⇒ k o sinθi = k o sinθr = k t sinθ t = k z ⇒ θi = θr ,
=
sinθi k t
⇒ Eo (1 + Γ) = Eo T ⇒ 1 + Γ = T
Similarly:
Eo
+ jk x x − jk z z ΓEo
−
+
zˆ cos θi e
zˆ cos θi e − jk x x − jk z z
ηi
ηi
TEo
=−
zˆ cos θt e+jk tx x - jk tz z
ηt
ΓEo
ηi cos θt
Eo
TEo
⇒
cos θi −
cos θi =
cos θt ⇒ 1 − Γ = T
ηi
ηi
ηt
ηt cos θi
L10-3
SOLVE TE BOUNDARY EQUATIONS
We found:
1+ Γ = T
and
Solving yields:
Γ(θi ) =
1− Γ = T
ηi cos θt
ηt cos θi
ηt cos θi − ηi cos θt ηn '− 1
=
where ηn'
ηt cos θi + ηi cos θt ηn '+ 1
ηt cos θi
ηi cos θt
Normal incidence: θ = 0, cosθ = 1, and ηn’ = ηt/ηi
ηn '− 1 where ηn'
Γ(θi ) =
ηn '+ 1
μt / εt
ηt
=
ηi
μi / εi
Glass has ε ≅ 4εo so ηn’ = 0.5 ⇒ Γ = -1/3 when light hits glass
Reflected power fraction = |Γ|2 = 1/9 (from both surfaces)
L10-4
SOLVE TM BOUNDARY EQUATIONS
TM Wave Hi
Ei
x
ki
ε,μ
εt,μt
Er
Hr
θi θi
y
z
θt
Et
Ht
Option A: Repeat method for TE (write field expressions with unknown
Γ and T; impose boundary conditions; solve for Γ and T)
Option B: Use duality to map TE solution to TM case
L10-5
DUALITY OF MAXWELL’S EQUATIONS
By making these
substitutions to
our TE solution:
E → H
H → −E
ε ↔ μ
Become these:
Our original
equations :
∇ × E = −μ ∂H
∂t
∇ × H = ε ∂E
∂t
∇ i εE = 0
∇iμH = 0
∇ × H = ε ∂E
∂t
→ − ∇ × E = μ ∂H
∂t
→
∇iμH = 0
→
→
∇ iεE = 0
Are they valid?
Under what circumstances does the solution to the resulting
set of equations satisfy Maxwell’s Equations?
They ARE Maxwell’s Equations, just reordered, so the
solution to the first set works if ε ↔ μ and ….
L10-6
DUALITY: TM WAVE SOLUTIONS
For TE waves we found:
η '− 1
where ηn'
Γ TE (θi ) = n
ηn '+ 1
ηt cos θ
i
μ
=
η
ηi cos θt and
ε
For TM waves :
Duality swaps μ ↔ ε to yield:
η−t 1 cos θi
− 1
−1
η cos θt
η' − 1
η cos θi
Γ TM (θi ) = i−1
= TMn
where η'TMn = i
η'TMn − 1
ηt cos θt
ηt cos θi
+ 1
ηi−1 cos θt
But this swap requires the boundary conditions to be
dual too. They are dual here because both⎯E// and
⎯H// are continuous across this boundary.
L10-7
TE AND TM WAVE REFLECTIONS
Γ
-1
|Γ|2
1
TE
TM
0
90o
θ 0
|Γ|2
1
TE
TM
90o
θ 0
Critical
angle
TE
TM
90o
θ
Brewster’s angle θB Brewster’s angle θB
1
High power
laser beam
No reflection
at θB
Brewster’s angle
laser window
Horizontally
polarized glasses
cut glare
Ocean wave
L10-8
MIT OpenCourseWare
http://ocw.mit.edu
6.013 Electromagnetics and Applications
Spring 2009
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