Planewave Solutions

8/12/2019 Planewave Solutions

1/19

PLANE WAVE SOLUTIONS OF MAXWELL'S EQUATIONS

I. CHARACTERISTICS OF PLANE WAVE SOLUTIONS:

For the record, let us once again restate the form of the macroscopic Maxwell's equations

in the time domain which is valid in the high frequency or optical regime.

r rE rr ,t( ) = t

rB rr, t( ) = 0 trH rr , t( ) [I-1a]

r

r

Brr, t( ) = 0

r

r

Hrr , t( ) = 0

r

Jrr,t( ) +

t

r

Prr , t( )

+ 0 0

t

r

Err,t( ) [I-1b]

r

r

Err, t( ) =

1

0rr, t( )

r

r

Prr,t( )[ ] [I-1c]

r

r

B

r

r, t( ) = 0r

r

H

r

r , t( ) = 0 [I-1d]

The most general phenomenological tensorial representation of the linear dielectric response

of a given material which incorporates dissipative and anisotropic effects may be

written in the time domain as

Prr, t( ) = 0 d t

rr, t t( )E

rr, t( )

t

[I-2a]

or in the frequency domain as

Prr, ( ) = 0

rr, ( )E

rr , ( )

[I-2b]

where

rr, ( ) = d t

rr, t t( ) exp j t t( )[ ]

t

= d rr, ( ) exp j [ ]

0

[I-3]


2/19

PLANE WAVE SOLUTIONS OF MAXWELL' S EQUATIONS PAGE 2

R. Victor Jones, October 23, 2002

Neglecting anisotropy, the macroscopic, optical regime Maxwell's equations in the

frequency domain for a linear, isotropic media may be written

r

rE

rr , ( ) =

r

1rr,( )

rDrr , ( ) = j

rB

rr, ( ) = j 0

rH

rr, ( ) [I-4a]

r

rBrr, ( ) = 0

r

rH

rr, ( ) = 0

rJrr , ( ) + j 0

rr, ( )

rErr ,( )

= 0rJrr, ( ) + j 0

rDrr, ( )

[I-4b]

r

rr , ( )

rErr, ( ) =

r

rDrr, ( ) =

rr, ( ) [I-4c]

r

rBrr, ( ) = 0

r

rH

rr, ( ) = 0 [I-4d]

Consider the possibility of a plane wave solution within a uniform medium -- i.e. rr , ( ) ( ) .


3/19



of the form

rE

rr, ( ) =

rE ( ) exp j

rr

rk( ) =

rE ( ) exp j x kx +y ky + zkz( )[ ] [I-5]

so that the spatial partial derivatives simplify to

x

r

Err, ( ) =

r

E ( )

xexp j

rr

r

k( ) =r

E ( ) j kx[ ] exp jrr r

k( ) [I-6a]

y

r

Err, ( ) =

r

E ( )y

exp jrr

r

k( ) =r

E ( ) j ky[ ] exp jrr r

k( ) [I-6b]

z

rE

rr, ( ) =

rE ( )

z

exp jrr

rk( ) =

rE ( ) j kz[ ] exp j

rr

rk( ). [I-6c]

Therefore

r

rE

rr, ( ) =

r

rE ( ) exp j

rr

rk( ) = j

rk

rE

rr, ( ) [I-7a]

r

rE

rr , ( ) =

r

rE ( ) exp j

rr

rk( ) = j

rk

rE

rr , ( ) [I-7b]

and Maxwell's equations in a charge/current free regionbecome

jrk rE rr, ( ) = j 0 rH rr, ( ) [I-8a]

j

rk

rH

rr, ( ) =j ( )

rErr, ( ) [I-8b]

j

rk

rErr, ( ) = 0 [I-8c]

j

r

k r

Hrr , ( ) = 0 [I-8d]

Operate through on both sides of Equation [I-8a] with the operator "rk "


4/19



r

k r

k r

Err , ( )[ ] = 0

r

k r

Hrr, ( ) [I-9a]

and using the "bac-cab" rule1and Equation [I-8b] this becomes

rk rk rE rr, ( )[ ] rk rk[ ] rE rr , ( ) =

2 0 ( ) rE rr, ( ) [I-9b]

or finally

r

k r

k[ ]r

Err ,( ) = 2 0 ( )

r

Err, ( ) k2 = 2 0 ( ) [I-9c]

Substituting into Equation [I-8a]

r

H

r

r, ( ) = 0( )

1

k

k

r

E

r

r, ( )[ ] = ( ) 0

k

r

E

r

r , ( )[ ] [I-10]

so that the wave impedance is given by

( ) =

r

Err, ( )

r

Hrr, ( ) = 0 ( ) . [I-11]

Thus, the complete expression for an electromagnetic plane wave propagating in a direction

k is

r

Err, t( ) =

r

E ( ) exp jrr r

k t( )[ ] [I-12a]

rH

rr, t( ) = ( )[ ]

1k

rE

rr, ( )[ ] [I-12b]

1 That is

ra

r

b rc( ) =

r

bra

rc( )

rcra

r

b( ) .


5/19



II. ELECTROMAGNETIC INTERFACIAL CONTINUITY CONDITIONS:

The previous section gives a completeplane wave solution within a particularuniform,

linear, isotropic medium. The key remaining problem is to find how that solution may be

extended into a second uniform, linear, isotropic medium. The conditions for extendingthe solution across an interface between two materials are give by consideration of the

appropriate integral forms of Maxwell's equations -- viz.

rE

rr, t( ) d

rl = t

rBrr, t( ) d

rA [II-1a]

rH

rr, t( ) d

rl =

rJrr, t( ) d

rA + t

rD

rr , t( ) d

rA [II-1b]

Applying these equations to the smallthought loopthat spans the interfacial surface, as

illustrated below

Medium 1

Medium 2

Interface1, 1, 1[ ]

2 , 2 , 2[ ]L

l"Thought"loop

r

r

it is seen that Equation [II-1a] yields

rE

rr, t( ) d

rl

rE2

rr, t( )

rE1

rr , t( ){ }

r L 0 [II-2]

unlessrBrr, t( ) is pathologically large over the loop. Similarly, it is seen that Equation

[II-1b] yields


6/19



r

Hrr, t( ) d

r

l r

H2rr , t( )

r

H1rr, t( ){ }

r L 0 [II-3]

unlessr

Jrr, t( ) and/or

r

Drr, t( ) are pathologically large over the loop.

In words and in general, the tangential component of the electric

field strengthrE

rr, t( ) and the magnetic field strength

rH

rr, t( )are

continuous across an interfacial surface between two materials

unless electric current densityrJrr, t( ) , the magnetic flux density

r

Brr, t( ) , or the electric flux density

r

Drr, t( ) are pathologically

large near that interfacial surface.

III. THE FRESNEL EQUATIONS:Consider then a plane wave incident on a planar interfacial surface.

The Spatial Configuration:2

2

Note: In this figure we have taken the plane of reflect ionto be identical to the plane of incidence. Whileassumed here for simplicity, this important identity is establish in the analysis below.


7/19



The Mathematical Representation of F ields :

In abstract vector form, the incident field is given by3

rE inc =

rE

inc 1 kinc

rH||

inc{ } exp jk1 k incrr( )

rH

inc = rH ||inc + 1

1k

inc rEinc{ } exp j k1 k inc rr( ) [III-1a]

the reflected field is given by

rE ref =

rE

ref 1 kref

rH||

ref{ } exp j k1 k refrr( )

r

Href =

r

H ||ref + 1

1k

ref r

Eref{ } exp j k1 kref

rr( )

[III-2]

and the transmitted field is given by

rE tran =

rE

tran 2 ktran

rH||

tran{ } exp j k2 ktranrr( )

rH

tran =rH||

tran + 21

ktran

rE

tran{ } exp j k2 ktranrr( )

[III-3]

In coordinate form these equations become:

rE inc = E

incy 1 cosinc x + sin inc z[ ] H||incy[ ]{ } exp j k1 x cos inc +zsin inc( )[ ]

rHinc = H ||

incy + 11 cos inc x + sininc z[ ] E

incy[ ]{ } exp j k1 x cos inc + zsin inc( )[ ][III-1b]

r

E ref = Erefy 1 cos ref x + sin ref z[ ] H ||

refy[ ]{ } exp j k1 xcos ref +zsin ref( )[ ]rH ref = H||

refy + 11 cos ref x + sin ref z[ ] E

refy[ ]{ } exp j k1 x cosref +zsin ref( )[ ][III-2b]

3 A note on notation: The subscripts and || refer to the polariztion of the electric field taken with respect to the

plane of incidence. The field components are also called transverse electricor TE components and the||

field components are called transverse magnetic or TM components.


8/19



rE tran = E

trany 2 cos tran x + sin tran z[ ] H ||trany[ ]{ } exp j k2 x costran +zsin tran( )[ ]

rH tran = H||

trany + 21 cos tran x + sin tran z[ ] E

tran y[ ]{ } exp j k2 x cos tran +z sintran( )[ ][III-3b]

Or expanding out the cross-products:

rE

inc = Einc

y + 1 H ||inc( ) cos inc z + sininc x[ ]{ } exp j k1 xcos inc +zsin inc( )[ ]

rH inc = H ||

incy 11 E

inc( ) cos inc z + sininc x[ ]{ } exp j k1 xcosinc +zsin inc( )[ ][III-1c]

rE

ref = Eref

y + 1 H||ref( ) cos ref z + sin ref x[ ]{ } exp j k1 xcos ref +zsin ref( )[ ]

rH ref = H||

refy 11 E

ref( ) cosref z+ sinref x[ ]{ } exp j k1 x cosref +zsin ref( )[ ][III-2c]

r

E tran = Etrany + 2 H||

tran( ) costran z + sin tran x[ ]{ } exp j k2 x cos tran +zsin tran( )[ ]rHtran = H||

trany 21 E

tran( ) cos tran z +sin tran x[ ]{ } exp j k2 xcos tran +zsin tran( )[ ][III-3c]

Applying any kind of continuity conditions at the interface requires that

ref = inc Law of Sinus [III-4a]

k2 sin tran = k1 sininc Law of Snell [III-4b]

Applying, in particular, the continuity conditions discussed in the previous section -- viz.

rE

1[ ]tang

=rE

2[ ]tang

andrH

1[ ]tang

=rH

2[ ]tang

[III-5]

at the interface, requires that

Einc + E

ref = Etran

1

1cos

inc

E

inc E

ref

[ ]=

2

1cos

tran

E

tran

[ ]

[III-6]


9/19



and that

H||inc + H ||

ref = H||tran

1 cos inc H||inc H||

ref[ ] = 2 costran H ||

tran[ ]

[III-7]

These two sets of equations yield the Fresnel Reflection Equations

E ref

Einc =

11cos inc 2

1 cos tran1

1cosinc +2

1cos tran

[III-8a]

and

H||ref

H||inc =

1 cosinc 2 costran1 cosinc +2 costran

[III-9a]

Since 11

sin inc = 21

sin tran

E ref

Einc =

cos inc sintran cos tran sin inccos inc sin tran + cos tran sin inc

=sin tran inc( )sin tran + inc( )

[III-8b]

and

H||ref

H||inc =

cos inc sininc cos tran sin trancos inc sin inc + cos tran sintran

=tan inc tran( )tan inc + tran( )

[III-9b]

These equations taken together with first equations from Equations [III-6] and [III-7] yield

the Fresnel Transmission Equations -- i.e.

Etran

Einc =

2 cosinc sintrancosinc sintran + cos tran sininc

[III-10]

and

H||tran

H||inc =

2 cos inc sininccos incsin inc + costran sin tran

[III-11]


10/19



FAMOUS FRE SNEL REFLECTION CURVES ( n2 n1 = 1 n2 = 2 1 =1.5 )

The minimum (zero) in H ||ref

H||inc

occurs at the Brewster angle where

tan incBrewster + tran

Brewster( ) [III-12a]

or tranBrewster = 2 inc

Brewster[III-12b]

or (from Snell's equation)

tan incBrewster = 1 2 = n2 n1 = 2 1 . [III-12c]


11/19



Total Internal Reflection

Reconsider Equation [III-3c] and use Snell's law to write the exponential factors in the

form

rE

tran = Etran

y + 2 H||tran( ) costran z + sin tran x[ ]{ } exp j x k2

2 k12

sin2 inc[ ] exp j z k1sin inc[ ] [III-13]

When sininc > k2 k1 = n2 n1 sin inccrit

,rE

tran, the solution in medium 2, is

attenuated!

incident

beam

attenuated

f ie ld

Reconsideration of Equation [III-8a] and [III-9a] shows that the magnitude of the

reflection coefficients are onewhen sininc > sininccrit

-- viz.

E ref

Einc =

cos inc j sin2 inc k2 k1( )

2

cos inc + j sin2 inc k2 k1( )

2

= exp j 2 tan1sin2 inc k2 k1( )

2

cos inc

[III-14a]

and

H||ref

H||inc =

cos inc j k1 k2( )2

sin2 inc 1

cos inc + j k1 k2( )2

sin2 inc 1

= exp j 2 tan1k1 k2( )

2sin2 inc 1

cos inc

. [III-14b]


12/19



IV. A DIFFERENT "TAKE" ON THE FRESNEL EQUATIONS:

Equations [III-1c] through [III-3c] may be rewritten in the following form:

rE inc = E

inc y + H||inc 1 cosinc( ) z + H||

inc 1 sin inc( ) x} exp jk1 x cosinc +z sininc( )[ ]rH inc = H||

incy Einc 1

1 cosinc( ) z Einc 1

1 sin inc( ) x{ } exp j k1 x cosinc +zsin inc( )[ ] [IV-1a]

r

E ref = Erefy H||

ref 1 cos ref( ) z + H||ref 1 sin ref( ) x{ } exp j k1 x cos ref +z sinref( )[ ]

rH ref = H ||

refy + Eref 1

1 cosref( ) z Eref 1

1 sin ref( ) x{ } exp j k1 x cos ref +zsin ref( )[ ][IV-1b]

r

E tran = Etrany + H ||

tran 2 cos tran( ) z + H ||tran 2 sin tran( ) x{ } exp j k2 x costran +zsin tran( )[ ]

rHtran = H||

trany Etran 2

1 cos tran( ) z Etran 2

1 sin tran( ) x{ } exp j k2 xcos tran +zsin tran( )[ ][IV-1c]

Notice: If = cos is interpreted as the

characteristic impedance of a wave polarized perpendicular to

the plane of incidence (TE wave) which is propagating at an

angle with respect to the normal of an interfacial plane and

if || = cos is interpreted as the characteristic

impedance of a wave polarized parallel to the plane ofincidence (TM wave) which is propagating at an angle with

respect to the normal of an interfacial plane, then the whole

analysis of specular reflection fits neatly into transmission line

theory .

In particular, a re-interpretation the famous transmission line equation for the

voltage reflection coefficient


13/19



V z,( ) =Z z,( ) Zc ( )Z z,( ) + Zc ( )

[IV-2]

yields an equation for the electric field strength reflection coefficient at the interfacial plane

for TE waves

=

tran inc

tran +

inc =2 costran 1 cosinc2 cos tran + 1 cosinc

[IV-3]

which is in precise agreement with Equation [III-8a] -- i.e.

E ref

Einc =

11cos inc 2

1 cos tran1

1cosinc +2

1cos tran

.

Similarly, a re-interpretation the equally famous transmission line equation for the current

reflection coefficient

I z ,( ) =Y z ,( ) Yc ( )Y z,( ) + Yc ( )

[IV-4]

yields an equation for the magnetic field strength reflection coeficient at the interfacial plane

for TM waves

|| =1 ||

tran 1 ||inc

1 ||tran +1 ||

inc =1 2 costran 1 1 cosinc1 2 cos tran +1 1 cosinc

[IV-5]

which is in precise agreement with Equation [III-9a] -- i.e.

H||ref

H||inc =

1 cosinc 2 costran1 cosinc +2 costran

.


14/19



V. PARALLEL PLATE WAVEGUIDE:

Consider the propagation of a plane wave between two parallel perfectly conducting planes.

Perspective view Side view

Combining Equations [III-1c] and [III-2c], the electric field strength of the TE wave in the

region between the plates may be written

rE = y E

incexp j x k1 cosinc( ) + E

refexp j x k1 cosinc( )[ ] exp j z k1 sin inc( ) [V-1]

At x = 0 the field parallel to the surface of a perfect conductor must be zero so thatE

ref = Einc

and, therefore,

r

E = y E inc exp j x k1 cos inc( ) exp j x k1 cos inc( )[ ] exp j z k1 sininc( )= y 2j E

inc sin x k1 cos inc( ) exp j z ( )[V-2]

where = k1 sininc . At the upper surface -- i.e. x = d-- the field parallel to the surface of

a perfect conductor must also be zero so that

d k1 cos inc = n where n = 1,2,3,K [V-3]

and, therefore,


15/19



TE n = k1 sin inc = k1

2 k12

cos2 inc = k1

2 n d( )2

where n =1, 2,3, K [V-4]

which is the dispersion relationship for TE waves in a parallel plate

waveguide with "cutoff" frequencies at

n

cutoff = n d( ) 0 1( )1

where n = 1,2,3,K [V-5]

Again combining Equations [III-1c] and [III-2c], the electric field strength of the TM wave

in the region between the plates may be written

rE|| = z 1 cos inc( ) H||

incexp j x k1 cos inc( ) H||

refexp j x k1 cos inc( ){ } exp j z k1 sininc( )

+ x 1 sin inc( ) H ||inc exp j x k1 cosinc( ) + H||

ref exp j x k1 cos inc( ){ } exp j z k1 sin inc( )[V-6]

At x = 0 the field parallel to the surface of a perfect conductor must be zero so thatH||

ref = H ||inc

and, therefore,


16/19



rE|| = z 2j H ||

inc 1 cosinc( ) sin x k1 cosinc( ) exp j z ( )

+x 2 H||inc 1 sin inc( ) cos x k1 cos inc( ) exp j z ( )

[V-5]

where = k1

sininc

. At the upper surface -- i.e. x = d-- again the field parallel to the

surface of a perfect conductor must also be zero so that

d k1 cos inc = n where n = 0,1, 2,3,K [V-7]

and, therefore,

TM n = k1 sin inc = k1

2 k12

cos2 inc = k1

2 n d( )2

where n = 0,1,2,3,K [V-8]

which is the dispersion relationship for TM waves in a parallel platewaveguide.

Note that the TM0 mode is a bona fide mode of propagation which does not

have a "cutoff" frequency!


17/19



VI. DIELECTRIC SLAB WAVEGUIDE:

Consider the propagation of waves "trap in" or "guided by" a dielectric slab of thickness d.

In its full generality this is moderately complicated problem, but a rather simple ray optics

model of the propagation is sufficient to yield dispersion relationships for the various

possible modes of propagation. To obtain such relationships, consider the total internal

reflection of a sequence of plane waves as illustrated below.

x

z

k inc

d


18/19



In order for the multiple reflected wave to be self-consistence the following, relatively

obvious, phase condition must hold:4

x =d + x =0 + 2 k1 dcos inc = m 2 where m = 0,1,2, 3,K [VI-1]

where x =d and x =0 are, respectively, the phase shifts associated with the reflections at

the upper and lower dielectric boundaries.

For TE-modes of propagationEquation [III-14a] gives the phase shift at the boundary

(called in the trade the TE Goos-Hnchen shift) and Equation [VI-1] becomes

2 tan1sin2 inc sin

2 inccrit

cos inc

+ k1 dcosinc = m [VI-2]

where sininccrit k2 k1 = n2 n1 = 2 1 . Therefore, the self-consistence relationship

for TE is given by

sin2 inc sin2 inc

crit

cos inc= tan

n1 k0 dcos inc2

m2

[VI-3]

where k0 = c = 2 0 . This is a transcendental equation in the single variable cosinc .

Its solutions yield the allowed bounce angles, inc( )m , of possible modes and, hence, theallowed propagation constants since = k1 sininc The left and right sides of this equationmay be plot as a function of cosinc with n1 k0 d= n1 2 d 0( ) and sininc

crit = n2 n1 as a

parameters. The intersections of such curves yield the allowed bounce angles as illustrated

below

4 This equation is a direct generalization of Equations [V-3] and [V-7] which figure in our analysis of parallel

plane waveguides.


19/19



LHS and RHS of Equation [VI-3] LHS and RHS of Equation [VI-3]

for n1 d 0( ) = 0.5 and cosinccrit = 0.5 for n1 d 0( ) = 1.0 and cosinccrit = 0.5

LHS and RHS of Equation [VI-3] LHS and RHS of Equation [VI-3]

for n1 d 0( ) = 1.5 and cosinccrit = 0.5 for n1 d 0( ) = 2.0 and cosinc

crit = 0.5

Date post:	03-Jun-2018
Category:	Documents
Upload:	ramana-varala
View:	220 times
Download:	0 times

Planewave Solutions

Documents