ECE 3300 Wave Reflection and Transmission – Oblique Incidence

Wave Reflection and Transmission – Oblique Incidence

A plane wave (shown by the solid lines) is propagating so that it hits a boundary at

oblique incidence θ (anything other than normal incidence, when θ would be zero). The

direction of propagation

of the plane wave is

shown by the light arrow.

The Plane of incidence is

defined as the plane that

includes the DOP and the

vector normal to surface.

The plane of incidence is

the paper.

There are two possible polarizations for the electric field, as shown above.

Parallel Polarization: E field is parallel to Plane of incidence (the paper)

Also called Transverse Magnetic (magnetic field is perpendicular

(transverse) to plane of incidence)

Perpendicular Polarization: E field is perpendicular to Plane of incidence (the paper)

Also called Transverse Electric (electric field is perpendicular (transverse)

to plane of incidence)

This lecture is about the perpendicular polarization case (E is y-polarized).

We have three waves of interest (incident, reflected, transmitted)

The angles can be found this way:

θi = θr

k1 sinθr = k2 sinθt (Snell’s law, derived later)

For each wave, we have to define three values:

The polarization vector represents the orientation of the E vector. In the case of

perpendicular polarization, all of the waves have y polarization (out of the paper).

The propagation constants ki = kr and kt are found from…

The direction of propagation for all three waves (this is just geometry):

ttt

rrr

iii

tri

zxx

zxx

zxx

npropagatioofDirectionx

θθ

θθ

θθ

cossin

cossin

cossin

,,

+=

−=

+=

=

Then the waves are specified as:

Ei = |E

i| e

– j ki x

i y

Er = |E

r| e

– j kr x

r y

Et = |E

t| e

– j kt x

t y

Now lets worry about themagnetic fields

The magnetic fields are always perpendicular to both E and the direction of propagation.

Hi = |H

i| e

– j ki x

i yi

Hr = |H

r| e

– j kr x

r yr

Ht = |H

t| e

– j kt x

t yt

Their polarization vectors are given by (geometry, see figure above):

Their magnitudes are given by: |H| = |E| / η (where ηi = ηr and ηt are defined in your text)

ttt

rrr

iii

zxy

zxy

zxy

θθ

θθ

θθ

sinˆcosˆˆ

sinˆcosˆˆ

sinˆcosˆˆ

+−=

+=

+−=

Apply Boundary Conditions:

We are now going to apply the boundary conditions to the TOTAL fields in region 1 and

region 2.

Total field in region 1: sum of incident and reflected fields

Total field in region 2: transmitted field only.

From boundary conditions, we know that

Tangential E fields = across boundary

For the perpendicular polarization case we are doing, all E fields (y-components) are

tangential to the boundary

From boundary conditions, we know that the

Tangential H fields = across boundary (ONLY if there is no current distribution, which is

the case here)

The x-component of H fields is tangential to this boundary for perpendicular polarization.

The only way both the E and H boundary conditions can be met is if

k1 sinθi = k1 sinθr = k2 sinθt

This is called PHASE MATCHING

This gives us a derivation of Snell’s Law:

From the first set of equalities: k1 sinθi = k1 sinθr we get θi =θr

From the second set of equalities: k1 sinθr = k2 sinθt we can calculate θt

tri xjkt

y

xjkr

y

xjki

y

t

y

r

y

i

y

eEeEeE

zatEEE

θθθ sinsinsin 211

0~~~

−

⊥

−

⊥

−

⊥

⊥⊥⊥

=+

==+

tri xjk

t

t

yxjk

r

r

yxjk

i

i

y

t

x

r

x

i

x

eE

eE

eE

zatHHH

θθθ θη

θη

θη

sin

2

sin

1

sin

1

211 coscoscos

0~~~

−⊥−⊥−⊥

⊥⊥⊥

−=+−

==+

We can also calculate the Reflection and Transmission Coefficients. These are used to

find the reflected and transmitted electric field magnitudes (and from there, the reflected

and transmitted magnetic field magnitudes).

Now, how do we use this:

Typical question:

Given a 10 V/m electric field incident at 30 degrees to an interface between air (region 1)

and dionized water (er = 80, sig = 0 S/m), find the electric and magnetic fields in each

region.

Steps

1) Find η and k for each region.

2) Find θ for each region from Snell’s law

3) Find the reflection and transmission coefficients.

4) Find the electric field magnitudes from the reflection and transmission

coefficients.

5) Find the magnetic field magnitudes from the electric field magnitudes and

characteristic impedances.

6) Write the equations for incident, reflected, and transmitted fields.

⊥⊥

⊥

⊥⊥

⊥

⊥⊥

Γ+=

+==

+

−==Γ

1

coscos

cos2

coscos

coscos

12

2

0

0

12

12

0

0

τ

θηθη

θητ

θηθη

θηθη

ti

i

i

t

ti

ti

i

r

E

E

E

E