Lehner G. Electromagnetic Field Theory for Engineers and Physicists

434 Time Dependent Problems II (Electromagnetic Waves)

[ .

(7.141)

These are Fresnel’s equations for the case of perpendicular polarization. They were

derived only from the boundary conditions of the parallel components of E and H.

The perpendicular components of D vanish, which makes them automatically

continuous. The perpendicular components of B are continuous if

(7.142)

or

.

Using (7.136) gives

or

,

that is, continuity of the perpendicular components of B is ensured by Snell’s law

and by the continuity of the parallel components of E.

Multiplying (7.136) by (7.139) yields

.

(7.143)

In conjunction with (7.72), this gives

.

(7.144)

This equation expresses the conservation of energy principle. The incident field

energy is passed in parts to the reflected wave and the remainder to the transmitted

wave.

Now we approach the case of parallel polarization (Fig. 7.11). Here one has

(7.145)

or

(7.146)

This ensures continuity of the parallel components of H. Those of E have to be

continuous as well

.

(7.147)

The result are Fresnel’s equations for the case of parallel polarization.

E

t0

E

i0

--------





⊥

2

α

1

cos

Z

1

---------------

α

1

cos

Z

1

---------------

α

2

cos

Z

2

---------------+

-------------------------------------

2Z

2

α

1

cos

Z

2

α

1

cos Z

1

α

2

cos+

-------------------------------------------------

==

µ

1

H

i0

H

r0

+()α

1

sin µ

2

H

t0

α

2

sin=

µ

1

α

1

sin

Z

1

--------------------

E

i0

E

r0

+()

µ

2

α

2

sin

Z

2

--------------------

E

t0

=

µ

1

α

1

sin

Z

1

--------------------

µ

2

α

2

sin

Z

2

--------------------=

ε

1

µ

1

α

1

sin ε

2

µ

2

α

2

sin=

E

i0

2

E

r0

2

–()

α

1

cos

Z

1

---------------

E

t0

2

α

2

cos

Z

2

---------------

=

S

i

α

1

cos S

r

α

1

cos– S

t

α

2

cos=

H

i0

H

r0

– H

t0

=

E

i0

E

r0

–()Z

1

⁄ E

t0

Z

2

⁄=

E

i0

E

r0

–()α

1

cos E

t0

α

2

cos=

7.3 Reflection and Refraction 435

,

(7.148)

. (7.149)

The perpendicular components of B vanish and, thus, are continuous. Continuity of

the perpendicular components of D is ensured by Snell’s law and by the continuity

of the parallel components of H:

.

(7.150)

Using (7.146) gives

or

,

in agreement with Snell’s law.

Multiplying (7.146) by (7.147) and in conjunction with (7.143) and (7.144) yields

the conservation of energy, as before.

Fig. 7.11

α

1

p

l

a

n

e

o

f

i

n

c

i

d

e

n

c

e

m

e

d

i

a

b

o

u

n

d

a

r

y

α

1

α

2

E

i0

κ 0=

µ

2

ε

2

,

κ 0=

µ

1

ε

1

,

1

2

E

r0

E

t0

H

t0

H

r0

H

i0

E

r0

E

i0

--------





||

α

1

cos

Z

2

---------------–

α

2

cos

Z

1

---------------+

α

1

cos

Z

2

---------------

α

2

cos

Z

1

---------------+

------------------------------------------

Z

1

α

1

cos Z

2

α

2

cos+–

Z

1

α

1

cos Z

2

α

2

cos+

-----------------------------------------------------

==

E

t0

E

i0

--------





||

2

α

1

cos

Z

1

---------------

α

1

cos

Z

2

---------------

α

2

cos

Z

1

---------------+

-------------------------------------

2Z

2

α

1

cos

Z

1

α

1

cos Z

2

α

2

cos+

-------------------------------------------------

==

ε

1

E

i0

E

r0

–()α

1

sin ε

2

E

t0

α

2

sin=

ε

1

α

1

sin Z

1

ε

2

α

2

sin Z

2

=

ε

1

µ

1

α

1

sin ε

2

µ

2

α

2

sin=

436 Time Dependent Problems II (Electromagnetic Waves)

7.3.3 Nonmagnetic Media

For the special case

,

Fresnel’s equations (7.140), (7.141), (7.148), and (7.149) simplify. Using

(7.151)

now yields

,

(7.152)

, (7.153)

and

,

(7.154)

.

(7.155)

If , which means that there is no media boundary and nothing noteworthy

happens, i.e., the incident wave continues and there is no reflection. There is no

reflection because of

,

.

Less self-evident is that in the case of parallel polarization, there is no reflected

wave when

.

(7.156)

That this is true can be derived from (7.154):

µ

1

µ

2

µ

0

==

Z

1

Z

2

-----

µ

0

ε

1

------

µ

0

ε

2

------

----------

ε

2

µ

0

ε

1

µ

0

---------------

α

1

sin

α

2

sin

--------------== =

E

r0

E

i0

--------





⊥

α

2

sin α

1

cos α

1

sin α

2

cos–

α

2

sin α

1

cos α

1

sin α

2

cos+

------------------------------------------------------------------

α

2

α

1

–()sin

α

2

α

1

+()sin

--------------------------------

==

E

t0

E

i0

--------





⊥

2 α

2

sin α

1

cos

α

2

sin α

1

cos α

1

sin α

2

cos+

------------------------------------------------------------------

2 α

2

sin α

1

cos

α

2

α

1

+()sin

---------------------------------

==

E

r0

E

i0

--------





||

α

2

sin α

2

cos α

1

sin α

1

cos–

α

2

sin α

2

cos α

1

sin α

1

cos+

------------------------------------------------------------------

2α

2

sin 2α

1

sin–

2α

2

sin 2α

1

sin+

-----------------------------------------

==

α

2

α

1

–()sin α

2

α

1

+()cos

α

2

α

1

+()sin α

2

α

1

–()cos

-----------------------------------------------------------------

α

2

α

1

–()tan

α

2

α

1

+()tan

--------------------------------==

E

t0

E

i0

--------





||

2 α

2

sin α

1

cos

α

1

sin α

1

cos α

2

sin α

2

cos+

------------------------------------------------------------------

2 α

2

sin α

1

cos

α

1

α

2

+()sin α

1

α

2

–()cos

-----------------------------------------------------------------

==

α

1

α

2

=

E

r0

E

i0

--------





⊥

E

r0

E

i0

--------





||

0==

E

t0

E

i0

--------





⊥

E

t0

E

i0

--------





||

1==

α

1

α

2

+

π

2

---=

7.3 Reflection and Refraction 437

.

(7.157)

Snell’s law states that

.

The thereby defined angle

,

(7.158)

is the so-called Brewster angle, also called the polarization angle. It is a

remarkable angle because, if we shine unpolarized light or any unpolarized

electromagnetic radiation, under this angle onto a boundary, then the polarization

of the reflected radiation is completely perpendicularly polarized because the

parallel polarized radiation passes through the boundary in its entirety without

reflection. Modification of eq. (7.158) are necessary if . Furthermore, in

such case there would also be a polarization angle for the perpendicularly polarized

wave. We will discuss neither case any further.

Let us divide (7.143) by : This gives

.

(7.159)

is the fraction of incident energy which is reflected. Conversely,

is the fraction of incident energy which passes

through. Therefore, one defines the two quantities

(7.160)

and

.

(7.161)

R is called reflectance and T is called transmittance. Their sum is of course

.

(7.162)

Take as an example the perpendicular incidence on a non magnetic material, i.e.,

.

(7.163)

Then we use (7.140) and (7.148) together with (7.151) to obtain

E

r0

E

i0

--------





||

0= for α

1

α

2

+

π

2

---=

α

1

sin

α

2

sin

--------------

ε

2

ε

1

-----

n

2

n

1

-----

α

1

sin

π

2

--- α

2

–





cos

------------------------------

α

1

sin

α

1

cos

--------------- α

1

tan=== = =

α

1

α

1

tan

n

2

n

1

-----

ε

2

ε

1

-----==

µ

1

µ

2

≠

E

i0

2

α

1

cos Z

1

⁄

1

E

r0

E

i0

--------





2

–

E

t0

E

i0

--------





2

Z

1

α

2

cos

Z

2

α

1

cos

---------------------

=

E

r0

E

i0

⁄()

2

E

t0

E

i0

⁄()

2

Z

1

α

2

cos Z

2

α

1

cos⁄()

R

E

r0

E

i0

--------





2

=

T

E

t0

E

i0

--------





2

Z

1

α

2

cos

Z

2

α

1

cos

---------------------

=

TR+1=

α

1

α

2

0==

α

1

cos α

2

cos 1==

438 Time Dependent Problems II (Electromagnetic Waves)

(7.164)

and using (7.141) and (7.149) together with (7.151) to obtain

.

(7.165)

Consequently for the reflectance we get

(7.166)

and for the transmittance

.

(7.167)

Of course, in case of it must be and . The graph of R and T

is illustrated in Fig. 7.12 for perpendicular incidence.

E

r0

E

i0

--------





⊥

E

r0

E

i0

--------





||

Z

2

Z

1

–

Z

2

Z

1

+

------------------

1

Z

1

Z

2

-----–

1

Z

1

Z

2

-----+

---------------

1

ε

2

ε

1

-----–

1

ε

2

ε

1

-----+

-------------------====

E

t0

E

i0

--------





⊥

E

t0

E

i0

--------





||

2Z

2

Z

1

Z

2

+

------------------

2

1

Z

1

Z

2

-----+

---------------

2

1

ε

2

ε

1

-----+

-------------------====

R

⊥

R

||

R

ε

2

ε

1

-----1–

ε

2

ε

1

-----1+

-------------------







2

===

T

⊥

T

||

T

4

ε

2

ε

1

-----⋅

ε

2

ε

1

-----1+







2

---------------------------===

ε

1

ε

2

= R 0= T 1=

1

0

Rx()

Ax()

64

0.016 x

R

T

H

2

H

1

-----

1

2

-- -

1

4

-- -

1

8

-- -

1

16

----- -

1

32

----- -

1

64

----- -

1 2 4 8 16 32 64

0

1

R

T

1

0

Fig. 7.12 Reflectivity and transmissivity

7.3 Reflection and Refraction 439

7.3.4 Total Reflection

Snell’s law states

.

(7.168)

The medium with the relatively lower speed of light is a higher-index medium, the

one with the greater speed of light is a lower-index medium.

The lower-index medium belongs to the larger angle, while the higher-index

medium corresponds to the smaller angle. When light travels into a higher-index

medium, refraction is towards the normal direction, while when travelling into a

lower-index medium, refraction is away from the normal (Fig. 7.13).For the

refraction away from the normal we have

.

(7.169)

Now, there are certain angles which require . This is not possible for

real angles. The result is, that in this case the entire incident energy is reflected.

There is no refracted plane wave. The limit between ordinary reflection and this so-

called total reflection, i.e., the maximum possible angle for which regular

reflection and refraction is possible is given by

i.e.,

.

(7.170)

This angle is called critical angle of total reflection.

It must be emphasized that this does not at all mean that there are no waves in

medium 2. However this wave in medium two is not a plane (homogeneous) wave,

α

2

sin

α

1

sin

--------------

ε

1

µ

1

ε

2

µ

2

---------------

n

1

n

2

-----

c

2

c

1

-----===

Fig. 7.13 Refraction at boundary

α

2

α

1

boundary

ε

1

µ,

1

c

1

,

c

2

c

1

>

µ

2

ε

2

µ

1

ε

1

<

α

2

α

1

boundary

c

2

c

1

<

µ

2

ε

2

µ

1

ε

1

>

refraction towards the normal refraction away from the normal

n

1

n

1

n

2

n

1

> n

2

n

1

<

ε

1

µ,

1

c

1

,

α

2

sin

c

2

c

1

-----

α

1

sin α

1

sin>=

α

1

α

2

sin 1>

α

1

α

2

sin 1

c

2

c

1

-----

α

c

sin==

α

c

sin

c

1

c

2

-----=

α

c

440 Time Dependent Problems II (Electromagnetic Waves)

but an inhomogeneous wave. This wave is necessary to satisfy the boundary

conditions. It travels parallel to the boundary surface and decreases exponentially

in the direction perpendicular to the boundary. The wave is of the kind discussed in

Sect. 7.2.3, eq. (7.115). One can calculate amplitude, phase constant, and damping

constant for the wave in medium 2, when starting from appropriate statements,

including the boundary conditions. This is not difficult but tedious and therefore,

will not be presented here. A formal solution would start from Snell’s law. The

angle becomes imaginary when and the refracted wave becomes

inhomogeneous.

Of fundamental interest is a slightly modified case. This is illustrated in

Fig. 7.14, which shows a wave in medium 1, incident on a thin layer (medium 2).

Because of ( ), this should be a case of total reflection. However, a certain

fraction of energy is still passing through. What is happening here is that the

incident wave causes in the medium 2, a wave which decreases exponentially in the

direction perpendicular to the boundary. When the decaying wave has reached the

other end of the thin layer, it has not decayed enough to be actually zero. Under

certain conditions, this remainder of the wave can be the source of another

propagating wave into medium 3. Depending on the thickness of the layer, the

amplitude of this wave may be very small, however. This wave is necessary to

satisfy the boundary conditions on the boundary between medium 2 and 3.

Formally, this is analogous to Quantum Mechanics famous tunnel effect, which is

highly important also for electrical engineering because of its significance for the

properties of semiconductors. Fig. 7.14 sketches the behavior of the refracted wave

for

,

α

2

α

2

sin 1>

Fig. 7.14 Tunneling wave

α

2

α

1

E

1

E

2

E

3

123

α

1

α

1c

>

ε

3

µ

3

ε

1

µ

1

---------------

α

1

sin

ε

2

µ

2

ε

1

µ

1

---------------

>>

7.3 Reflection and Refraction 441

that is, is greater than the critical angle for medium 2, but is smaller than the

critical angle for medium 3. Then, Snell’s law applies for media 1 and 3, just as if

there were no intermediate layer

.

7.3.5 Reflection at a Conducting Medium

The case of another inhomogeneous wave, will not be calculated, but only touched

upon briefly. Consider a wave emerging from an insulator that is incident on a

conductor. Here too, besides the reflected wave back into the insulator, we also

need an inhomogeneous wave in the conductor to satisfy the boundary conditions

between an insulator and the conductor. Fig. 7.15 illustrates this. The

inhomogeneous wave is such that the propagation direction is given by Snell’s law,

but at the same time decays exponentially in the direction perpendicular to the

media boundary. The locus of constant phase is perpendicular to the propagation

direction.

With a coordinate system as defined by Fig. 7.15, the following Ansatz may

serve to solve this problem:

(7.171)

(7.172)

. (7.173)

Besides the material constants and , the quantities and

have to be considered as given. The task is to find the appropriate , which

α

1

α

3

sin

α

1

sin

--------------

ε

1

µ

1

ε

3

µ

3

-----------=

Fig. 7.15 Reflection at a Conducting Medium

m

e

d

i

a

b

o

u

n

d

a

r

y

α

2

κ 0=

µ

1

ε

1

,

1 – insulator

z

const. phase

const. amplitude

plane of incidence

y

µ

2

ε

2

,

κ 0≠

α

1

α

1

2 – conductor

x

E

i

E

i0

i ωtk

1

y α

1

sin– k

1

z α

1

cos–()[]exp=

E

r

E

r0

i ωtk

1

y α

1

sin– k

1

z α

1

cos+()[]exp=

E

t

E

t0

i ωt β

2

y α

2

sin– β

2

z α

2

cos–()[]exp γ

2

z–[]exp=

µ

1

ε

1

µ

2

ε

2

,,, κ

2

ω k

1

,α

1

α

2

β

2

γ

2

,,

442 Time Dependent Problems II (Electromagnetic Waves)

is possible by means of the dispersion relation of the wave in eq. (7.173) and by

means of the following phase relation which was obtained from the phases above

.

(7.174)

This represents Snell’s law in the here applicable from. Calculating the amplitudes

and requires one to invoke the boundary conditions, applicable for the

various field components, whereby the proceedings are analogous to Sect. 7.3.2.

We shall pass on the tedious details of this calculation.

7.4 Potentials and their Wave Equations

7.4.1 The Inhomogeneous Wave Equation for A and ϕ

Starting point are again Maxwell’s equations

(7.175)

(7.176)

(7.177)

. (7.178)

Because of (7.177), one may define as

,

(7.179)

where is a vector depending on space and time

,

(7.180)

But A is not unique. For instance, another vector potential may provide the

same B-field.

,

therefore

.

Consequently, by means of an arbitrary scalar function we may write

.

(7.181)

Inserting (7.179) in (7.175) gives

or

.

(7.182)

k

1

α

1

sin β

2

α

2

sin=

E

r0

E

t0

E∇×

t∂

∂B

–=

H∇× g

t∂

∂D

+=

B∇• 0=

D∇• ρ=

BA∇×=

A

AArt,()=

A

e

BA∇× A

e

∇×==

AA

e

–()∇× 0=

ψ

AA

e

– ψ∇–=

E∇×

t∂

∂

A∇×+ E

t∂

∂A

+





∇× 0==

E

t∂

∂A

+ ϕ r t,()∇–=

7.4 Potentials and their Wave Equations 443

Consequently, one may calculate B and E from A and ϕ, and considering (7.179)

and (7.182) write:

(7.183)

. (7.184)

ϕ is not unique either. Instead of calculating B and E from A and ϕ, we might as

well choose and as starting point,

inasmuch as

.

(7.185)

Therefore, the relation between A, ϕ and , is given by (7.181) and (7.185).

is a function that may be chosen arbitrarily. Thus, there is substantial freedom

when choosing the potentials A and ϕ. We use this freedom to require that the

following condition be met:

.

(7.186)

This condition is called the Lorentz gauge. For time independent problems, this

reverts into the previously introduced Coulomb gauge (5.9). Should the potentials

in a given problem not correspond to the Lorentz gauge, then by means of an

appropriate function , it is always possible to find potentials that do so.

The fields given by (7.183), (7.184) automatically satisfy two of Maxwell’s

equations, namely (7.175) and (7.177). To satisfy the other two requires some

work. We start with (7.176):

With (7.186) we get

.

(7.187)

On the other hand, using (7.178) yields

BA ∇×=

E ϕ r t,()∇–

t∂

∂A

–=

A

e

ϕ

e

E ϕ

e

∇–

t∂

∂A

e

– ϕ∇–

t∂

∂A

–==

ϕ∇–

t∂

∂A

e

–

t∂

∂

ψ∇+=

ϕϕ

e

–

t∂

∂ψ

=

A

e

ϕ

e

ψ

A∇• µε

t∂

∂ϕ

+0 =

ψ

H∇×

1

µ

---

A∇×()∇× g ε∇

t∂

∂

ϕ– ε

∂

2

A

∂t

2

----------

–==

A∇×()∇× εµ∇

t∂

∂

ϕεµ

∂

2

A

∂t

2

----------

++ µg=

A∇•()∇∇

2

A– εµ∇

t∂

∂

ϕεµ

∂

2

A

∂t

2

----------

++ µg .=

∇

2

A εµ

∂

2

A

∂t

2

----------

– µg –=

Lehner G. Electromagnetic Field Theory for Engineers and Physicists

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