Capítulo 6 de libro de Teoría electromagnética

8/16/2019 Capítulo 6 de libro de Teoría electromagnética

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2/39

s 8o/ can see from fig/re +.", the plane perpendic/lar to the vector ' is seen from its

side appearing as a line P(W. The dot prod/ct nB r is the proGection of the radial vector ralong the normal to the plane and will have the constant val/e O( for all points on the

plane. The e=/ation ' ) r constant is the characteristic propert8 of a plane

perpendic/lar to the direction of propagation '.

The e=/iphase e=/ation is

B r C B


3/39

p C P <

power densit8 flow

%efinition:

transverse electromagnetic wave TEM2 > electromagnetic wave having electricfield vectors and magnetic field vectors perpendic/lar to the direction of propagation.

, is perpendic/lar to -, and both - and , are perpendic/lar to the direction of

propagation '. The e


4/39

Find the magnetic field strength.

3olution:

The direction of propagation nB is > aA. The vector amplit/de of the magnetic field is then

given b8

Η Ε

m

x y z

x y

n a a a

a a A m= = − = −

∧β

η η

"6

$

6

5

"

6

"

5**5 $

*note

η µ

ε = "$6J5**K ppendi< D > Table D."2

-/#(P0- 64

The phasor electric field e


5/39

ence, a /nit vector in the direction of propagation nB is given b8

nB C (6.+a


6/39

5. The wavelength 5 is given b8

λ π

β

π = = =

$ $

$ 5 $ *5

. . m

and the fre=/enc8

f = = ∗

=c

GHz λ

5 "6

$ *5 6 ""

7

. .

. The e=/ation of the s/rface of constant phase is

nB r C (6.+< @ 6.78 C constant

The general e


7/39

The fig/re shows an incident wave polariAed with the E field in the plane of incidence

and the power flow in the direction of β i at angle θ i with respect to the normal to the

s/rface of the perfect cond/ctor.

The direction of propagation is given b8 the Po8nting vector and the β i , -8 and , fields

need to be arranged so that β i is in the same direction as Ε Η i i∧ at an8 time. The

magnetic field is o/t of the plane of the paper,Η Η = y ya

for the direction of the electricfield shown. There is no transmitted field within the perfect cond/ctorO however there

will be a reflected field with power flow at the angle θ r with respect to the normal to the

interface. To maintain the power densit8 flowΕ Η r r ∧ will be in the same direction β r as. The e


8/39

( ) , cos cosΕ Ε Ε x mi

i j i r m

r j r r r x z e e= +

− ⋅ − ⋅θ θ β β

( ) , sin sinΕ Ε Ε z mi

i j i r m

r j r r r x z e e= − +

− ⋅ − ⋅θ θ β β

cos cosΕ Ε Ε x at z mi

i j i r m

r j r r r e e=

− ⋅ − ⋅= − =6

6θ θ β β

= − =− ⋅ − ⋅ cos cossin sinΕ Ε m

ii

j xmr j x r

r ie eθ θ

β β θ θ 6 +.72

E=/ation +.7 shows the relationship between the incident and reflected amplit/des for a

perfect cond/ctor the total tangential - field at the s/rface m/st be Aero which satisfies

the bo/ndar8 condition. To be Aero at all val/es of < along the s/rface of the cond/cting plane, the phase terms m/st be e=/al to each other >

θ θ i r = +.#2

E=/ation +.# is known as %nell0s law of reflection.

%efinition:%nell0s !aw is a r/le of Ph8sics that applies to visible light passing from air or

vac//m2 to some medi/m with an inde< of refraction different from air.

%/bstit/te e=/ation +.# into e=/ation +.7 >

Ε Ε mi

mr = +."62

Therefore, the total electric field in free space is

( ) ( ) , 2 , ,Ε Ε Ε x z x z a x z a x x z z = +

( )= −− − cos sin cos cosΕ mi i j x i j z i j z i xe e e aθ β θ β θ β θ

( )− +− − sin sin cos cosΕ mi i j x i j z i j z ie e e z aθ β θ β θ β θ

= −

$ j z e ami

i j x i

i x cos sin cos 2

sinΕ θ β θ

β θ +.""2

( )− −$ j z e ami

i i j x i z

sin cos cos sinΕ θ β θ β θ

'otes b8: Debbie Prestridge 7


9/39

[= −$ cos sin cos 2Ε mi i i x j z aθ β θ

( ) ]− −sin cos cos sini i z j x i z a eθ β θ β θ

Take e=/ation +."" and recover the time(domain form of the total electric field

( ) ( )( )Ε Ε r t r e j t , 9e = ϖ

1bserve the variation of the total field with the + variable indicating there is a traveling

wave in the + direction with a phase constant

β β θ x i= sin

nd in the & direction the field forms a standing wave.

The total magnetic field is

( ) ( ) ( ) ( ) , , , ,Η Η Η Η x z x z a x z a x z a y y yi

y yr

y= = +

se the relation Η Ε

=∧nβ

η for each of the incident and reflected fields to emplo8 the

e


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( )= − +

sin cosΕ mi

j i x i z ye aη

β θ θ

The reflected magnetic fields is given b8

( ) sin cosΗ Ε r mi

j i x i z ye a= − −η β θ θ

The total magnetic field Η ∧


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Normal Incident:

i x aveθ = =6 6, ,Ρ Power flow in the + direction is Aero2

verage power flow perpendic/lar to the cond/cting s/rface is Aero, beca/se the average

Po8nting Lector is Aero in that direction

( ) z ave x y P , 9e = =∗"

$ 6Ε Η

Wh8Q &eca/se Ε x is m/ltiplied b8 , therefore Ε Η x yand are o/t of phase b8 ;


12/39

The plane of the Aero Ε x field occ/r at m/ltiples of λ $ along the direction of propagation, and the8 are located at integer m/ltiples of λ z $ along the A(a


13/39

where ( )β β θ θ i i ir x z ⋅ = −sin cos . ss/me that the reflected field is also in the $ directionso the magnetic field m/st be perpendic/lar to both - and the Po8nting Lector P - > ,,

Ε Ε r mr j r

ye ar = − ⋅β

( )

cos sinΗ

Ε Ε r r

mr

r x r z j

r r

n

a a er

=

∧

= +

∧

− ⋅

β β

η η θ θ

Where ( )β β θ θ r r r r x z ⋅ = −sin cos . Determine the angle of reflection θ r and the

amplit/de of the reflected electric field Ε mr b8 /sing the bo/ndar8 conditions at A C 6.

This also incl/des Aero val/es of the tangential electrical field - and the normal

component of the magnetic field ,.

( ) , Ε Ε Ε y yi

yr

x z = + = 6 at A C 6

Therefore,

( ) , sin sinΕ Ε Ε y mi j x i m

r j x r x e e6 6= + =− −β θ β θ

nd

'otes b8: Debbie Prestridge

β i

Ε yr β r

θ i θ r

1

a&

a+

Perfect

;ond/ctor

Ε yi

Η r

7i"ure 6?

"5

iΗ


14/39

( ) , sin sinsin sinΗ Ε Ε z mi

i j x i m

r i

j x r x e e6 " "

6= + =− −η

θ η

θ β θ β θ

'ote: These two conditions will provide the same res/lts for the /nknowns r mr

and θ Ε

, and be tr/e for ever8 val/e of + along &


15/39

There is a standing(wave in the & direction beca/se the reflected and incident waves

travel in the opposite direction along the A(a


16/39

The sheet c/rrent J in ampere per meter2 is determined b8 the total tangential

magnetic field at the s/rface. From the bo/ndar8 condition,

J n= ∧ Η

where the normal n to the s/rface for the geometr8 of Fig/re +.) is n C (aA. The magneticfield in this case has two components:

( )

cos cos cos sinΗ Ε

xmi

i i j x i z e=

− −$η

θ β θ β θ

( )

sin sin cos sinΗ Ε

z mi

i i j x i

j z e=

− −$η

θ β θ β θ

The s/rface is then c/rrent is then

cos sin J a a eat z

z y m

i

i j x i

= =

−∧ =

∧

−

6

$Η

Ε

η θ β θ

nd the peak val/e of the s/rface c/rrent at A C 6 is given b8

cos 2 cos

. J x A m peakvaluemi

i= = = −$ $ "6 M)

5**5 *) "6 $

Ε η

θ

-/#(P0-:

The electric field associated with a plane wave propagating in an arbitrar8 direction isgiven b8

. . 2 . . 2Ε = + − − +* 75 M M ) * 6 ) 6 7* x y z j x z a a a e

If this incident on a perfectl8 cond/cting plane oriented perpendic/lar to the A a


17/39

$. Total electric field in region in front of the perfect cond/ctor.

5. Total magnetic field.

3olution

&eca/se a vector in the direction of propagation and a /nit vector normal to the reflectings/rface are contained in the


18/39

The magnit/de of the incident electric field Ε mi is therefore C *.75N6.7* C # or .)N6.) C

#. ence, the electric field associated with the parallel polariAation case can be e


19/39

= − +

− +M M * 6 ) 6 7*η

θ η

θ cos sin . . 2i x i z j x z a a e

&eca/se, for the case,

Ε Ε⊥ ⊥=

r i

cos sin 2 . . 2Η⊥ − += − +r x z

j x za a eM

56 56 * 6 ) 6 7*η

The total reflected magnetic field is then

cos sin 2 . . 2Η r x y z j x z a a a e= − − + − − +"

M 56 # M 56 * 6 ) 6 7*

η

6@ Reflection and Refraction at Plane Interface betAeen TAo (edia:

Oblique Incidence

Fig/re +.* shows two media with electrical properties ε "and µ " in medi/m ", and ε $

and µ $ in medi/m $. ere a plane wave incident angleθ i on a bo/ndar8 between the

two media will be partiall8 transmitted into and partiall8 reflected at the dielectrics/rface. The transmitted wave is reflected into the second medi/m, so its direction of

propagation is different from the incidence wave. The fig/re also shows two ra8s for

each the incident, reflected, and transmitted waves. ra8 is a line drawn normal to thee=/iphase s/rfaces, and the line is along the direction of propagation.

'otes b8: Debbie Prestridge "#


20/39

7i"ure 6B

The incident ra8 $ travels the distance ;&, while on the contrar8 the reflected ra8 "

travels the distance E. For both ; and &E to be the incident and reflected wave frontsor planes of e=/iphase, the incident wave sho/ld take the same time to cover the distance

E. The reason being that the incident and reflected wave ra8s are located in the same

medi/m, therefore their velocities will be e=/al,

CBV

AE V " $

=

19

AB ABi r sin sinθ θ =

With this being the case then it follows that

θ θ i r =

What is the relationship between the angles of incidence θ i and refractionθ r Q

It takes the incident ra8 the e=/al amo/nt of time to cover distance ;& as it takes the

refracted ra8 to cover distance D >

CB

V

A

V " $=


!

4

9eflected

ra8s

!

4

ε µ $ $,

ε µ " ",

Incident

ra8s

θ i

#

θ t

θ r

C -

$6

9eflectedra8s


21/39

nd the magnit/de of the velocit8 L" in medi/m " is:

"" "

"V =

∗ µ ε

nd in medi/m $:

$

$ $

"V =

∗ µ ε

lso,

CB AB

A AB

i

i

==

sin

sin

θ

θ

Therefore,

CB

A

V

V

i

t

= = = ∗

∗

sin

sin

θ

θ

µ ε

ε µ "

$

$ $

" "

For most dielectrics µ µ µ $ "= =

Therefore,sin

sin

i

t

θ

θ

ε

ε µ µ µ

== =

$

"" $

+."$2

E=/ation +."$ is known as %nell0s !aw of 9efraction.

6@! Parallel Polari&ation Case - is in Plane of Incidence

'otes b8: Debbie Prestridge $"


22/39

The /nknown amplit/des of the reflected and transmitted electric fields RR RRr t and Ε Ε can be determined b8 simpl8 appl8ing the bo/ndar8 conditions at the dielectric interface.

The electric fields RR RRr t and Ε Ε will now be /sed in the anal8sis to emphasiAe the case of

parallel polariAation, instead of /sing the electric fields m mt r and Ε Ε .

The tangential component of , sho/ld be contin/o/s across the bo/ndar8. Therefore,

RRΗ i j i

r ye a

− ⋅ +β RRΗ r j i r ye a

− ⋅ =β RRΗ t j i r ye a

− ⋅β

There is no need to carr8 the a$ vector, beca/se the magnetic fields onl8 have onecomponent in the $ direction. 9ecall that this relation is valid at A C 6,

RR

sin 2Η i j i i xe− +β θ RR sin 2Η r j

ii r x

e− =β θ RRsin 2Η t j i t xe− β θ +."52

β " β $" are the magnit/des of β in regions " $, respectivel8. In order for this to be

valid at an8 val/e of + at an8 point on the interface, and knowingθ θ i r = :

β θ β θ " $sin sini t =

1r


β i

β r

θ i θ r

19egion $

R Rr Η

1/t of

paper2

θ t

β t

RRt

Ε

RRi

Ε

9egion "

Η R Ri

Η RRt

$$

2

/

RRr

Ε

7i"ure 6;


23/39

sin

sini

t

V V

V

V

θ

θ

β

β

ω

ω = = =

$

"

$

"

"

$

S This is the same relation that was determined earlier from %nell0s !aw. %/bstit/tesin

sin

i

t

V

V

θ

θ = "

$ into e=/ation +."5 to obtain

RRΗ i

+ RRΗ r

= RRΗ t t A C 6 +."2

E and are related b8 η , so e=/ation +." can be rewritten as

RRΕ i + RRΕ

r =

"

$

η

η

RRΕ t

+.")2

Tangential components of E m/st be contin/o/s across the bo/ndar8, therefore

cosRRΕ i

iθ − cosRRΕ r

r θ = cosRRΕ t

t θ t A C 6 +."+2

S9emember the e

RRΕ r

=

cos cos

cos cos

RR

T

Ε i i t

i t

" $

" $

η θ η θ

η θ η θ

−

+nd

RRΕ t =

+

cos

cos cosRR TΕ i i

i t

$

" $

$η θ

η θ η θ +."*2

SMaking /se of the fact thatθ θ i r = . Define the reflection coefficient RRΓ and the

transmission RRΤ :

RR

RR

RR

Γ Ε

Ε =

r

i= −

+ =

−

+

= =

$ "

$ "

$

"

$

"" $

η θ η θ η θ η θ

θ ε

ε

θ

θ ε

ε θ

µ µ µ

cos coscos cos

cos cos

cos cos

t i

t i

t i

t i

nd

'otes b8: Debbie Prestridge $5


24/39

RR

RR

RR

ΤΕ

Ε = =

t

i

$ $ $"

$ " $

"" $

η θ η θ

η θ η θ

θ

θ ε

ε θ

µ µ µ

cos cos

cos cos

cos

cos cos

t t

t t

i

i i

−

+=

+

= =

The total electric field in region " is

RRΕ t!t

=

RRΕ i

+ RRΕ r

= cos sin 2Ε mi

i x i z j i r a a eθ θ β − − ⋅ @ cos sin 2Ε m

r r x r z

j r r a a e− − − ⋅θ θ β

= ∧

−cos sini mi j x ieθ β θ Ε

cos− j z ie β θ +

∧

RR 2cosΓ j z ie a x

β θ

+ −

−

sin sini mi j x ie

Travelin" #ave part

θ β θ Ε

( )− +−e e a j z j z z S din" plu$travelin"#ave$

i iβ θ β θ cos cos

tan

RRΓ

+."72

%/bstit/ted β β i r r r ⋅ ⋅, from e


25/39

The transmitted fields in medi/m $ are

( ) cos sinRRΕ Ε i

mt

t x t z j r

a a e t = − − ⋅θ θ β

C ( ) cos sinRRΤ Ε mi

t x t z j r

a a e t θ θ β − − ⋅

nd

RR

RRΗ Η Τ Ε t

mt

y j r m

i j r

ya e e at t t = =− ⋅∧

− ⋅⋅β β η $

Where ( )β β θ θ t t t r x z ⋅ = +$ sin cos and RRΕ Ε Τmt

mi = .

%efinition:

&rewster ngle > from &rewster0s !aw2, the polariAing angle of which whenlight is incident2 the reflected and refracted inde< is e=/al to the tangent of the polariAing

angle. In other words, the angle of incidence of which there is no reflection.

From the reflection coefficient e


26/39

This condition is important, beca/se it is /s/all8 satisfied b8 the materials often /sed in

optical applications.

E=/ation +."# will take the form >

i t cos cosθ ε ε θ = "

$+.$62

%=/are both sides of e=/ation +.$6 and /se %nell0s !aw for the special case of µ µ µ " $= = for the following res/lt:

$ "

$cos iθ

ε

ε ( )= = −$ " $

$

"cos sint t θ ε

ε θ

( )= −"$

" $ε ε

θ sin i

The last s/bstit/tion was based on %nell0s !aw of refraction. Therefore,

( )" $− =sin iθ "

$

ε

ε − "

$

$$

$ε

ε θ sin i

" "

$− =

ε

ε $sin iθ "

"$

$$

−

ε

ε

nd

$ $

$ "sin iθ

ε

ε ε =

++.$"2

The &rewster angle of incidence is

sin iθ ε

ε ε =

+

$

$ "+.$$2

specific val/e of Hi can be obtained from e=/ation +.$" (

'otes b8: Debbie Prestridge $+


27/39

" $ $

$ "− =

+cos iθ

ε

ε ε

1r

$ $

$ ""cos iθ ε

ε ε = −

+ =

"

$ "

ε

ε ε + =

cos iθ ε

ε ε =

+

"

$ "+.$52

From e=/ations +.$$ +.$5 >

tan iθ ε

ε =

$

"

This specific angle of incidence θ i is called the &rewster angleθ β .

β θ ε

ε = −"

$

"tan

6@4 Perpendicular Polari&ation case - Normal to Plane of Incidence

s shown in fig/re +."6 is a perpendic/lar polariAed wave incident at angleθ i a dielectric

medi/m $. %nell0s !aw states that a reflected wave will be at the same angleθ θ r i= , andthe transmitted wave in medi/m $ at angle θ t can be calc/lated /sing this law. The

amplit/de of the reflected and transmitted waves can be determined b8 appl8ing the

contin/it8 of the tangential components of E at the bo/ndar8.

This is given b8 >

cosΗ⊥i

iθ − ⊥ cosΗ

r iθ C cosΗ⊥

t t θ

'otes b8: Debbie Prestridge $*


28/39

%ince E are related b8 η ,

cosΕ ⊥

i

i"

η θ − =

⊥

cosΕ r

i"

η θ

cosΕ ⊥

t

t $

η θ +.$2

Ε ⊥

i+

⊥Ε

r =

⊥Ε

t t A C 6 +.$)2

S'ote: The e


29/39

Τ⊥ =

Ε

Ε

⊥

⊥

t

i C

$

$ "

$η θ

η θ η θ

cos

cos cos

i

i t +

For nonmagnetic material,

cos

cos cos

Τ⊥ =$

$

"

i

i t

θ

θ ε

ε θ

6D Comparison betAeen Reflection Coefficients RRΓ and Γ ⊥ for Parallel

and Perpendicular Polari&ations

The significant differences between the two will be ill/strated in the following e


30/39

β θ = −" 7"tan C 75.*?

b2 Water into air:ε ε r r and " $7" "= =

ence,

β θ = −" "

7"tan C +.5?

To relate the &rewster angles in both cases, let /s calc/late the angle of

refraction.

sin

sin

i

t

θ

θ

ε

ε = $

"

Therefore, in case a,

sin

sin

Βθ

θ t = 7"

Therefore,

t sinsin .

.θ = =75 *

# 6 ""

1r θ t = + 5M. , which is the same as the &rewster angle for case b. lso, the angle of

refraction in case b is given b8 %nell0s !aw as:

sin

sin

Βθ

θ

ε

ε t =

7"=

"

7"

Therefore,

t sinsin .

.θ = =+ 5M

"

7"

6 ##

1r t θ = 75 *. , which is the &rewster angle for case a.

6? Total Reflection at Critical #n"le of Incidence



31/39


32/39

7i"ure 6!@ ill/strates the fact thatθ θ ε ε t i if > >, " $ . The critical angle θ c is defined as

the val/e of θ i at which θ t C N$.

Envision a beam of light impinging on an interface between two transparent media where

n ni t < . t normal incidence θ i C 62 most of the incoming light is transmitted into theless dense medi/m. s θ i increases, more and more light is reflected back into the dense

medi/m, while θ t increases. When θ t C #6?, θ i is defined to be θ c and the

transmittance becomes Aero. For θ i V θ c all of the light is totall8 internall8 reflected,

remaining in the incident medi/m.

-/#(P0-3:

• se %nell0s !aw to derive an e


33/39

• Determine the critical angle for a water n# C".552 >glass n " C".)62 interface.

We have

sin θ c C nti

1r θ c C sin −" "55")6..

C sin −"6.77* C +$.)?

66 -lectroma"netic 3pectrum



34/39

5 3A 5 < "6"$ A 5


35/39

npolariAed light > light in which the wave orientation is random aro/nd the a entirel8 transmitted

thro/gho/t the interfaces

• More glass elements and the transmitted light co/ld be essentiall8 completel8

polariAed, - parallel to the plane of incidence

6B4 reAster WindoAs or reAster Cuts in 0#3-R In a normal sit/ation there are more electrons in the gro/nd state level "2 than in the

e


36/39θ β $

photons co/ld be the res/lt if this sit/ation co/ld be inverted. %/ch a condition is called

pop/lation inversion. This in fact is the f/ndamental principle involved in the operation

of a laser. Fig/re +.7 ill/strates this principle.

%efinition:

!aser !ight mplification b8 %tim/lated Emission2 > device that prod/cescoherent radiation in the visible(light range, between *)66 and 5#66 angstroms

%/mmariAed steps leading to !%E9 action in three(level r/b8 laser material:

". The laser material is in the shape of a long rod that is s/bGected to radiation from

an e


37/39

n

7i"ure 6!B 0i"ht polari&ations b$ multiple reflections.

Fig/re +."7 is a schematic diagram ill/strating the se=/ence of events.The role of the &rewster angle:

FnoAn 7actors

• The o/tp/t of man8 lasers is linearl8 polariAed

• The ratio of the light polariAed in one direction e


38/39

7i"ure 6!G 3equence of events occurrin" in laser action

Fig/re +."# is a schematic ill/strating the /se of &rewster windows in a gas dischargelaser. The &rewster angle makes s/re that light in one polariAation direction is

transmitted o/t of the medi/m of the laser to the reflecting mirrors and back into the

medi/m of the laser with no loss. Where the light is polariAed perpendic/lar to the planeof incidence a large loss at the &rewster s/rface will take place d/e to the reflection o/t

of the medi/m of the laser. The preferred polariAation case linear polariAation2 will lase

emit coherent light2 that will acco/nt for the high degree of polariAation taking place atthe o/tp/t.

The device in Fig/re +."# e


39/39

Hi

Ht

H$

%mallest critical angle

9eflected point

n Z

7i"ure 64< %chematic ill/strating the principle of light propagation in optical fibers.

%nell0s !aw of refraction is the relationship between Hi and Ht as the wave enter the fiber is

sin

sin

i

t n

θ

θ

ε

ε = =

$

"ε ε " = +.$*2

If θ $ is s/ppose to be larger thanθ c , then

sin θ $ C cos θ t Z sin θ c +.$72

9efraction from fiber to air sin θ c C "Nn, therefore, from e=/ation +.$* +.$7 >

sin cos sin sin$ $

$$" "

" "θ θ θ θ = = − = −

≥t t in n

+.$#2

%olve for n,

n i$ $"≥ + sin θ +.562

For e=/ation +.56 to be tr/e then C N$, all incident light will be passed b8 the fiber

re=/iring

n$ Z $ or n Z $

Most t8pes of glass have n [ ".)O therefore, we have a valid e=/ation.

Documents

Capítulo 6 de libro de Teoría electromagnética