6. Plane waves reflection from media interface - iitg.ac.in · Lossless medium Good conductor Fig. 6.1 Plane waves reflection from media interface Lossyconducting medium TM Brewster

Dr. Rakhesh Singh Kshetrimayum

6. Plane waves reflection from media

interface

Dr. Rakhesh Singh Kshetrimayum

3/25/20141 Electromagnetic Field Theory by R. S. Kshetrimayum

6.1 Introduction Plane waves reflection from a media interface

Normal incidence

TE

Oblique incidencePerfect

conductor

3/25/2014Electromagnetic Field Theory by R. S. Kshetrimayum2

TE

Lossless medium

Good conductor

Fig. 6.1 Plane waves reflection from media interface

Lossy conducting medium

TMBrewster angle

Total internal reflection

Effect on polarization

6.1 Introduction� Till now, we have studied plane waves in various medium� Let us try to explore how plane waves will behave at a media interface

� In practical scenarios of wireless and mobile communications, � radio wave will reflect from

� walls & � other obstacles on its path

� When a radio wave reflects from a surface, � the strength of the reflected waves is less than that of the incident wave

3/25/20143 Electromagnetic Field Theory by R. S. Kshetrimayum

6.1 Introduction� The ratio of the two (reflected wave/incident wave) is known as the ‘reflection coefficient’ of the surface

� This ratio depends on the � conductivity (σ), � permittivity (ε) and


� permittivity (ε) and � permeability (µ)

� of the material that forms the reflective surface � as well as material properties of the

� air

� from which the radio wave is incident

6.1 Introduction� Some part of the wave will be transmitted through the material

� How much of the incident wave has been transmitted through the material is � also dependent on the material parameters mentioned above


� also dependent on the material parameters mentioned above� It is given by another ratio known as ‘transmission coefficient’

� It is the ratio of the transmitted wave divided by the incident wave

6.1 Introduction� In plane wave reflection from media interface, � what we will be doing is

� basically writing down the � electric and � magnetic field expressions


� magnetic field expressions

� in all the regions of interest and � apply the boundary conditions

� to get the values of the � transmission and � reflection coefficients

6.1 Introduction� One of the possible applications of such an exercise is in wireless communication, � where we have to find the

� transmission and � reflection coefficients of

� multipaths


� multipaths� Electromagnetic waves are often

� reflected or � scattered or � diffracted

� at one or more obstacles before arriving at the receiver

6.2 Plane wave reflection from media

interface at normal incidence

� For smooth surfaces, � EM waves are reflected;

� For rough surfaces, � EM waves are scattered;

� For edges of surfaces,


� For edges of surfaces, � EM waves are diffracted

� Fig. 6.2 depicts multipath for a mobile receiver in the car from � direct, � reflected and � diffracted waves

� of the signal sent from the base station

6.1 Introduction


Fig. 6.2 Illustration of reflected and diffracted wave in mobile communications (this chapter’s study will be very helpful in understanding wireless and mobile communications)



� We will consider the case of normal incidence, � when the incident wave propagation vector is along the normal to the interface between two media

6.2.1 Lossy conducting medium

� We will assume plane waves with electric field vector


� We will assume plane waves with electric field vector oriented along the x-axis and

� propagating along the positive z-axis without loss of generality

� For z<0 (we will refer this region as region I and it is assumed to be a lossy medium)



� Let us do a more generalized analysis by assuming this region I as a lossy medium

� It could be a lossless medium like free space� The incident electric and magnetic fields can be expressed as


1ˆ z

i oE xE e

γ−=r

1

1

1ˆ z

i oH y E e

γ

η−=

r



� where η1 is the medium 1 wave impedance and � Eo is the arbitrary amplitude of the incident electric field� The expression for intrinsic wave impedance and propagation constant


( )1 1 1 1

1 1 1 1 1 1

1 1 1 11 1 1

; 1j j j j

j jjj j

ωµ ωµ ωµ ση γ α β ω µ ε

γ ωε σ ωεωµ ωε σ= = = = + = −

++



iEr

Er

2

tγr

1

iγr

iHr

tEr

tHr

iEr

r


2 2 2, ,ε µ σ

rEr

1 1 1, ,ε µ σ

1

rγr

rHrr

Er

Fig. 6.3 A plane EM wave is incident from region I or medium 1 for z<0



� Convention: � a circle with a dot in the center � an arrow pointing perpendicularly out of the page and

� a circle with a cross � an arrow pointing perpendicularly into the page)


the page)

� Notation for fields: � subscript i� incident, r� reflected and t�transmitted



� Note that the incident wave from region I will be partially reflected and transmitted at the media interface between two regions

� For reflected wave, z<0, wave direction is in the –z axis and the reflected electric field is expressed as


the reflected electric field is expressed as

� where the Γ is the reflection coefficient

1ˆ z

r oE x E e

γ+= Γr



� The reflection coefficient is defined as the ratio of amplitude of the reflected electric field divided by amplitude of the incident electric field as follows:

r

i

E

EΓ =


� The reflected magnetic field can be obtained from the reflected electric field using the Maxwell’s curl equations as

iE



1

1

1 1 1

ˆ ˆ ˆ

0 0

r r

r r

r

z

o

E j H

x y z

E j E jH

j x y z

E eγ

ωµ

ωµ ωµ ωµ+

∇× = −

∇× ∇× ∂ ∂ ∂⇒ = − = =

∂ ∂ ∂

Γ

r r

r rr


( ) ( ) ( )1 1 11

1

1 1 1

0 0

ˆ ˆ ˆ

o

z z z

o o o

E e

j jy E e y E e y E e

z j

γ γ γγγ

ωµ ωµ ωµ+ + +

Γ

∂ = Γ = Γ = − Γ

∂

1

1

ˆ z

r oH y E e

γ

η+Γ

∴ = −r



� Therefore, the Poynting vector for reflected wave of region I is

� shows power is traveling in the -z axis for the reflected wave

2 2* 2

*

1

ˆz

r r r o

zS E H E e

α

η+= × == − Γ

r r r


� shows power is traveling in the -z axis for the reflected wave� For transmitted wave, z>0, wave is propagating in lossymedium 2 (we will refer this region as region II),



where η is the wave impedance of lossy medium 2 and τ is

zot eExE 2ˆ γτ −=

r

zot e

EyH 2

2

ˆ γ

ητ −=

r


� where η2 is the wave impedance of lossy medium 2 and τ is the transmission coefficient

� intrinsic wave impedance and propagation constant are



� transmission coefficient is defined as the ratio of amplitude of the transmitted electric field divided by amplitude of the

( )2 2 2 2

2 2 2 2 2 2

2 2 2 22 2 2

; 1j j j j

j jjj j

ωµ ωµ ωµ ση γ α β ω µ ε

γ ωε σ ωεωµ ωε σ= = = = + = −

++


the transmitted electric field divided by amplitude of the incident electric field as follows:

t

i

E

Eτ =



� Our intention here is to find the transmission and reflection coefficients

� Let us rewrite the fields in the two regions: region I (z<0) and region II (z>0) (see Table 6.1) and

� Apply the boundary conditions to obtain the two unknown


� Apply the boundary conditions to obtain the two unknown coefficients



Region I (lossy medium 1) Region II (lossy medium 2)

Table 6.1 Fields in the two lossy regions (normal incidence)


1ˆ z

i oE xE e

γ−=r

1

1

1ˆ z

i oH y E e

γ

η−=

r

1ˆ z

r oE x E e

γ+= Γr

1

1

ˆ z

r oH y E e

γ

η+Γ

= −r

zot eExE 2ˆ γτ −=

r zot e

EyH 2

2

ˆ γ

ητ −=

r



� This is basically boundary value problem with boundary conditions at z=0

� Note that total electric and magnetic fields (both incident and reflected) are tangential to the interface at z=0

� Similarly, the transmitted electric and magnetic field are also


� Similarly, the transmitted electric and magnetic field are also tangential to the interface at z=0

� Also note that there are no surface current density at the interface

� Hence, the tangential components of electric and magnetic fields must be continuous at the interface z=0



Therefore,

( ) 0 00

1 1i r t

zE E E E Eτ τ

=+ = ⇒ + Γ = ⇒ + Γ =

r r r

0 00

1 2 1 2

1 1i r t

zH H H E E

τ τ

η η η η=

− Γ − Γ+ = ⇒ = ⇒ =

r r r


� Therefore,

� Hence,

( ) ( ) 2 1

2 1

1 2 1 2 2 1

1 1 11 1

η ητη η

η η η η η η

−− Γ − Γ + Γ= ⇒ = ⇒ − Γ = + Γ ⇒ Γ =

+

2 1 2

2 1 1 2

21 1

η η ητ

η η η η

−= + Γ = + =

+ +



6.2.2 Lossless medium

� If the regions are lossless dielectric, � then, σ = 0 and µ and ε are real quantities

� The propagation constant for this case is purely imaginary and can be written as


and can be written as � γ = jβ =jω√(µε)

� The wave impedance of the dielectric is � η= = = = = jωµ

γ

j

j

ωµ

β

ωµ

ω µε

µ

ε r

r

ε

µη0



� For a lossless medium, � η is real, so, � both Γ and τ are real� Electric and magnetic fields are in phase with each otherThe wavelength in the dielectric is


� The wavelength in the dielectric is� = =� And the phase velocity is� = = =� where c is the speed of light in free space

λ2π

β

2π

ω µε

pvω

β µε

1

rr

c

εµ



Table 6.2 Fields in the two lossless regions (normal incidence)

Region I (lossless medium 1) Region II (lossless medium 2)


1ˆ j z

i oE xE e

β−=r

1

1

1ˆ j z

i oH y E e

β

η−=

r

1ˆ j z

r oE x E e

β+= Γr

1

1

ˆ j z

r oH y E e

β

η+Γ

= −r

2ˆ j z

t oE x E e

βτ −=r

2

2

ˆ j zo

t

EH y e

βτη

−=r



� Applying the boundary conditions like before, � we can get the expression for

� reflection and � transmission coefficients,

� we will get the same expression as before



2 1

2 1

;η η

η η

−Γ =

+

2

1 2

2ητ

η η=

+



� Poynting vector in region I

( ) ( ) ( ) ( )1 1 1 1

* ** *

1 1 1 0 0

0 0

1 1

ˆ ˆ ˆ ˆj z j z j z j z

tot tot i r i r

E ES E H E E H H E e x E e x e y e y

β β β β

η η− + −

= × = + × + = + Γ × − Γ

r r r r r r r

2 2 2 2 2 22 2 2j z j zβ β− + Γ Γ Γ Γ


( ) ( )

( ) ( )

1 1

1 1

2 2 2 2 2 22 2 2

0 0 0 0 0 0 2 22

1 1 1 1 1 1

2 2

0 02

1

1 1

ˆ ˆ1

ˆ1 2 sin 2

j z j z

j z j zE E e E e E E E

z e e z

E Ej z z

β β

β β

η η η η η η

βη η

− +

+ − Γ Γ Γ Γ

= − + − = − Γ + −

Γ = − Γ −



� Poynting vector in region II

( )2 2

* 22

02 * 0

0

2 2

ˆ ˆ ˆj z j z

t t

EES E H x E e e y z

β β τττ

η η− −

= × = × =

r r r


( )

( ) ( ) ( )

22

2 12 2 1 2

2 2 2

12 1 2 1 2 1

4 411 1

η η η η η

ηη η η η η η

− − Γ− Γ = − = ⇒ =

+ + +Q

( )

( ) 222 2 202 22 2

2

2 1 1 2 12 1

12 4 1ˆ

ES z

η η ττ τ

η η η η ηη η

− Γ− Γ= ⇒ = ⇒ = ∴ =

+ +

r



Power conservation

� Compare the time average power flow in the two regions� For z <0, the time average power flow through 1m2 cross section is

( )2

1 1 1− Γr


� And for z >0, the time average power flow through 1m2

cross section is

( )2

21 1

0

1

1 1 1ˆRe

2 2avg

S S z Eη

− Γ= • =

r

( )2

22 2

0

1

1 1 1ˆRe

2 2avg

S S z Eη

− Γ= • =

r



� Hence,

� So real power is conserved6.2.3 Good conductor:

If the region II (z >0) is a good (but not perfect) conductor

1 2

avg avgS S=


� If the region II (z >0) is a good (but not perfect) conductor and

� region I is a lossless medium like free space, � the propagation constant for medium 2 is



� where δ is the skin depth for the medium 2, � which is a good conductorSimilarly, the intrinsic impedance of the conducting medium

( ) ( )2 2

2 2 2 2

11 1

2s

j j jωµ σ

γ α βδ

= + = + = +

2

s


� Similarly, the intrinsic impedance of the conducting medium 2 (see also example 6.2) simplifies to

( )

( )( )( )

( )

( )( )

( )

2

22 2 2

2

2 22

2 2 2 2

2

2 2

1 1(1 )

2

1 1 12 2 21

1 1 2 21

j j j j

j jj

j j j jj

j j j

ωµωµ ωµ ωµη

γ ωµ σ σωµ σ

ωµ ωµ ωµ ωµ

σ σ σ σ

= = = =+ +

+

− + += = = = +

+ − −



� Now, the wave impedance of medium 2 is complex, with a phase angle of 45o so the electric

� and magnetic field will be 45o out of phase and Γ and τ will be complex.

� Now, let us write down the field expressions in region I and


� Now, let us write down the field expressions in region I and II for lossless dielectric (region I) and good conductor (region II) interface



Table 6.4 Fields in the two regions (normal incidence)

Region I (lossless medium) Region II (good conductor)


1ˆ j z

i oE xE e

β−=r

1

1

1ˆ j z

i oH y E e

β

η−=

r

1ˆ j z

r oE x E e

β+= Γr

1

1

ˆ j z

r oH y E e

β

η+Γ

= −r

2ˆ z

t oE x E e

γτ −=r

2

2

ˆ zo

t

EH y e

γτη

−=r



� Applying the boundary conditions like before, � we can get the expression for

� reflection and � transmission coefficients,




2 1

2 1

;η η

η η

−Γ =

+2

1 2

2ητ

η η=

+



� Note that we have to use the previous expressions for intrinsic wave impedances for lossless medium and good conductor

1;

µη

ε= ( ) 2

21

2j

ωµη

σ= +


Power conservation

� let us find out the Poynting vector in the two regions� Noting that the field in region I comprise of incident and reflected wave,

� we can write the Poynting vector for z <0

ε ( )2

2

12

jησ

= +



For z >0, only transmitted fields exist and we can write

( ) ( ) ( ) ( )

( )

1 1 1 1

1 1

* ** *

1 1 1 *0 0

0 0

1 1

2

20 2 2*

1

ˆ ˆ ˆ ˆ

ˆ1

j z j z j z j z

tot tot i r i r

j z j z

E ES E H E E H H E e x E e x e y e y

Ee e z

β β β β

β β

η η

η

− + −

− +

= × = + × + = + Γ × − Γ

= − Γ + Γ − Γ

r r r r r r r


� For z >0, only transmitted fields exist and we can write down the Poynting vector as

( )*

2 2* *

02 * 20

0 * *

2 2

ˆ ˆ ˆz z z

t t

EES E H x E e e y z e

γ γ ατττ

η η− − −

= × = × =

r r r


interface at normal incidence� At z=0,

( )2

201 *

1

ˆ1E

S zη

= − Γ + Γ − Γr

( )( )

2 2 2 2

20 0 2 02 *4

ˆ ˆ ˆ1E E E

S z z zτ η

= = = − Γ + Γ − Γr


� Complex power is conserved� Then the time average power flow through a 1m2 cross section can be calculated as follows

� Since

( )( )2*

2 11 2

ˆ ˆ ˆ1S z z zη ηη η

= = = − Γ + Γ − Γ+



� is purely an imaginary number, therefore,

1 12 2* j z j ze e

β β− +−Γ + Γ

( )2

21 1

0

11 1ˆRe

2 2avg

S S z Eη

− Γ= • =

r


� which shows power balance at z=0 only

( ) 0

1

ˆRe2 2

avgS S z E

η= • =

( ) z

avg eEzSSα

η2

2

2

2

0

221

2

1ˆRe

2

1 −Γ−=•=

r



� Note that the Poynting vector for z>0, � this power density in the lossy conductor decays exponentially according to the attenuation factor

6.2.4 Perfect conductor:

� Now assume that the region II (z >0) contains a perfect


� Now assume that the region II (z >0) contains a perfect conductor and region I is a lossless medium

� Let us rewrite the expressions for propagation constant and wave impedance of a good conductor and deduce the expressions for perfect conductor



Perfect conductor implies σ →∝, and

( ) ( )2 2

2 2 2 2

11 1

2s

j j jωµ σ

γ α βδ

= + = + = +

( ) 2

2

2

12

jωµ

ησ

= +


� Perfect conductor implies σ2→∝, and � correspondingly, α2→∝, δs2→0, η2→0� Note that

� Therefore, τ→0 and Γ→-1

2 2 1

1 2 2 1

2,

η η ητ

η η η η

−= Γ =

+ +



Table 6.5 Fields in the two regions (normal incidence)

Region I (lossless medium) Region II (perfect conductor)


1ˆ j z

i oE xE e

β−=r

1

1

1ˆ j z

i oH y E e

β

η−=

r

1ˆ j z

r oE x E e

β+= Γr

1

1

ˆ j z

r oH y E e

β

η+Γ

= −r

0t

E ≅r

0t

H ≅r



� The field for z >0 thus decay infinitely fast and � are identically zero in the perfect conductor as we have seen in example 6.2 as well

� The perfect conductor can be thought of as ‘shorting out’ the incident electric field


incident electric field� Now let us write down the field expressions in region I and II for lossless dielectric and perfect conductor interface (see Table 6.5)



Power conservation

� In order to see the power conservation, � let us find out the Poynting vector in the two regions� Noting that the field in region I comprise of

� incident and


� incident and � reflected wave,

� we can write the Poynting vector for z <0 (Γ=-1)

( )1 11

0 0 1ˆ ˆ2 sin

j z j z

tot i rE E E xE e e x jE z

β β β−= + = − = −r r r



� Therefore, the Poynting vector for z<0 is

( )1 11 0 0

1

1 1

ˆ ˆ2 cosj z j z

tot i r

E EH H H y e e y z

β β βη η

−= + = + =r r

( )2

4 Er r


� which has a zero real part and � thus indicates that no real power is delivered to the perfect conductor� Fields in region II is anyway zero and hence power is also zero

( )2

*01 1 1

1 1

1

4ˆ sin cos

tot tot

ES E H zj z zβ β

η= × = −

r r


interface at oblique incidence

� We will consider the problem of a plane wave � obliquely incident on a plane interface � between two lossy conducting regions

� We will first consider two particular cases of this problem as follows:


follows:� the electric field is in the xz plane (parallel polarization)� the electric field is in normal to the xz plane (perpendicular polarization)



� Any arbitrary incident plane wave can be expressed � as a linear combination of these two principal polarizations

� The plane of incidence is that plane containing � the normal vector to the interface and � the direction of propagation vector of the incident wave


� the direction of propagation vector of the incident wave



� For Fig. 6.4, this is the xz plane� For perpendicular polarization (TE),

� electric field is perpendicular to the plane of incidence � For parallel polarization (TM),

� electric field is parallel to the plane of incidence


� electric field is parallel to the plane of incidence



tt HErr

,

z

x

rr HErr

,

rθ tθ


Fig. 6.5 Oblique incidence of plane EM wave at a media interface

1 1 1, ,ε µ σ

2 2 2, ,ε µ σ

z

Region II

ii HErr

,

Region I

iθrθ tθ

n



6.3.1 Perpendicular polarization (TE):

� In this case, electric field vector is perpendicular to the xzplane,

� Hence, it will have component along the y-axis� Since the electric field is transversal to the plane of incidence


� Since the electric field is transversal to the plane of incidence� They are also known transverse electric (TE) waves

iθ rθn n z

xx

z

iθγ cos1

iθγ sin1

rθγ cos1

rθγ sin11

iγr

1

rγr

x


tθn z

2cos

tγ θ

2sin tγ θ 2

tγr

Fig. 6.6 Wave propagation vector for (a) incident (b) reflected and (c) transmitted EM waves at oblique incidence



� Let us assume that the incident wave propagates in the first quadrant of xz plane without loss of generality and

� (incident propagation vector) makes an angle θi with the normal (see Fig. 6.6 (a))

1

iγr


( ) ( ) ( )'

1 1 1 1 1 1ˆ ˆˆ ˆcos sin cos sin cos sin

i

i i i i i iz z x zz xx z x z xγ γ θ γ θ γ θ γ θ γ θ θ• = + • + = + = +

r r

( )1 cos sin

0î iz x

iE E e y

γ θ θ− +=

r

1

1

i

i i i

EE j H H

jωµ

ωµ

∇×∇× = − ⇒ =

−

rr r r

Q



( )

( ) ( )

( ) ( )

1 1

1

1 1

cos sin cos sin

0

1 1

cos sin

0

cos sin cos sin

0

ˆ ˆ ˆ

1ˆ ˆ

0 0

ˆ ˆ

i i i i

i i

i i i i

z x z x

z x

z x z x

x y z

E e ex z

j x y z j z x

E e

E e ex z

γ θ θ γ θ θ

γ θ θ

γ θ θ γ θ θ

ωµ ωµ

− + − +

− +

− + − +

∂ ∂ ∂ ∂ ∂ = = − + − ∂ ∂ ∂ − ∂ ∂

∂ ∂ = −

( ) { }1 cos sin0 1 ˆ ˆcos sini iz xEe x z

γ θ θγθ θ− +

= − +


0

1

ˆ Ê e e

x zj z xωµ

∂ ∂ = − ∂ ∂

( ) { }

( ) ( )

1

1

cos sin0 1

1

cos sin0

1

ˆ ˆcos sin

ˆ ˆcos sin

i i

i i

z x

i i

z x

i i

Ee x z

j

Ee x z

γ θ θ

γ θ θ

γθ θ

ωµ

θ θη

− +

− +

= − +

= − +



� Let us assume that the reflected wave propagates in the second quadrant of xz plane and

� (reflected propagation vector) makes an angle θr with the normal (see Fig. 6.6 (b))

1

rγr


( ) ( ) ( )1 1 1 1 1 1ˆ ˆˆ ˆ' cos sin cos sin cos sin

r

r r r r r rz z x zz xx z x z xγ γ θ γ θ γ θ γ θ γ θ θ• = − + • + = − + = − +

r r

( )1 cos sin

0ˆr rz x

r TEE E e y

γ θ θ− − += Γ

r



� Note that and will have the same magnitude � since both the waves are still in the same region I, � only their direction changes

� Since the Poynting vector must be negative like the previous case of normal incidence,

1

rγr

1

iγr


case of normal incidence,

� You could also use the Maxwell’s curl equation below to find this

( ) ( )1 cos sin0

1

ˆ ˆcos sinr rz x

r TE r r

EH e x z

γ θ θ θ θη

− − += Γ +

r

1ωµj

EH r

r−

×∇=

rr



� The transmitted fields will have similar expression with the incident fields except� that now the θi should be replaced by θt (angle that transmitted propagation vector makes with the normal),

� γ1 should be replaced by γ2 (wave is in region II now) and


� γ1 should be replaced by γ2 (wave is in region II now) and� multiplication by (transmission coefficient)

� The transmitted fields are ( )2 cos sin

0ˆ t tz x

t TEE yE e

γ θ θτ − +=

r

( ) ( )2 cos sin0

2 2

ˆ ˆcos sint tz xt TE

t t t

E EH e x z

j

γ θ θτθ θ

ωµ η

− +∇×= = − +

−

rr



Table 6.5 Fields in two regions (oblique incidence: perpendicular polarization)



( )1 cos sin

0î iz x

iE E e y

γ θ θ− +=

r

( ) ( )1 cos sin0

1

ˆ ˆcos sini iz x

i i i

EH e x z

γ θ θ θ θη

− += − +

r

( )1 cos sin

0ˆr rz x

r TEE E e y

γ θ θ− − += Γ

r

( ) ( )1 cos sin0

1

ˆ ˆcos sini iz xTE

r r r

EH e x z

γ θ θ θ θη

− +Γ= +

r

( )2 cos sin

0ˆ t tz x

t TEE yE e

γ θ θτ − +=

r

( ) ( )2 cos sin0

2

ˆ ˆcos sint tz xTE

t t t

EH e x z

γ θ θτθ θ

η

− += − +

r



� Equating the tangential components of electric field � (electric field has only Ey component and it is tangential at the interface z=0) and

� magnetic field � (magnetic field has two components: Hx and Hz and only Hx is


� (magnetic field has two components: Hx and Hz and only Hx is tangential at the interface z=0)

� at z=0 gives1 21sin sinsini trx xx

TE TEe e e

γ θ γ θγ θ τ− −−+ Γ =

1 21sin sinsin

1 1 2

1cos cos cosi trx xxTE TE

i r te e e

γ θ γ θγ θ τθ θ θ

η η η− −−Γ−

+ = −



� If Ex and Hy are to be continuous at the interface z = 0 for all x,

� then, this x variation must be the same on both sides of the equations (also known as phase matching condition)


tri θγθγθγ sinsinsin 211 ==

1 2; sin sin

i r i tθ θ γ θ γ θ⇒ = =



� The first is Snell’s law of reflection � which states that the angle of incidence equals the angle of reflection

� The second result is the Snell’s law of refraction � (refraction is the change in direction of a wave due to change in


� (refraction is the change in direction of a wave due to change in velocity from one medium to another medium)

� Also note that refractive index of a medium is defined as

0 0

0 0

r r

r r

p

cn

v

µ ε µ εµ ε

µ ε= = =



� hence, for a lossless dielectric media,

Now we can simplify above two equations by applying Snell’s

2 2 22 2 1 2

1 1 2 11 1 1

sin

sin

i

t

v n

v n

µ ε εθ γ β

θ γ β µ ε ε= = = = = =


� Now we can simplify above two equations by applying Snell’s two laws as follows

1TE TE

τ+ Γ =

1 1 2

cos coscosi r TE

TE t

θ θ τθ

η η η− + Γ = −



� The above two equations has two unknowns τTE and ГTE and it can be solved easily as follows

� Therefore,

2

1 1

cos cos

cos

i r

TE TE

t

θ θ ητ

η η θ

= − Γ

1TE TE

τ+ Γ =


� Therefore,

22 2 2

1 1 1 1

1 2 2 1 2 1

1 1 1 2

cos coscos cos1 1 1

cos cos cos cos

cos cos cos cos cos cos

cos cos cos cos

i ir r

TE TE TE

t t t t

t r i t i t

TE TE

t t t r

θ η θη η θ η θ

η θ η θ η θ η θ

η θ η θ η θ η θ η θ η θ

η θ η θ η θ η θ

+ Γ = − Γ ⇒ Γ + = −

+ − −⇒Γ = ⇒Γ =

+



� Hence, the reflection and transmission coefficients for oblique incidence (Fresnel coefficients) for perpendicular

2 1 1 2 2 1 2

1 2 1 2 1 2

cos cos cos cos cos cos 2 cos1 1

cos cos cos cos cos cos

i t t r i t i

TE TE

t r t r t r

η θ η θ η θ η θ η θ η θ η θτ

η θ η θ η θ η θ η θ η θ

− + + −∴ = + Γ = + = =

+ + +


oblique incidence (Fresnel coefficients) for perpendicular polarization are given as follows:

� For normal incidence, it is a particular case and put

2 1

2 1

cos cos

cos cos

i t

TE

i t

η θ η θ

η θ η θ

−Γ =

+2

2 1

2 cos

cos cos

i

TE

i t

η θτ

η θ η θ=

+

0=== tri θθθ



6.3.2 Parallel Polarization (TM):

� In this case, electric field vector lies in the xz plane� Since the magnetic field is transversal to the plane of incidence such waves are also called as transverse magnetic (TM) waves


(TM) waves� So let us start with H vector which will have only y component,

( )ii xzi e

EyH

θθγ

ηsincos

1

0 1ˆ +−=r



1

1

i

i

HH j E E

jωε

ωε

∇×∇× = ⇒ =

rr r r

Q

( ) ( )1 1cos sin cos sin

ˆ ˆ ˆ

1 i i i iz x z x

x y z

E e eγ θ θ γ θ θ− + − + ∂ ∂ ∂ ∂ ∂


( )

( ) ( )

( ) { } ( )

1 1

1

1 1

cos sin cos sin

0

1 1 1

cos sin0

1

cos sin cos sin0 1

0

1 1

1ˆ ˆ

0 0

ˆ ˆˆ ˆcos sin cos si

i i i i

i i

i i i i

z x z x

z x

z x z x

i i i

E e ex z

j x y z j z x

Ee

Ee x z E e x z

j

γ θ θ γ θ θ

γ θ θ

γ θ θ γ θ θ

ωε ωε η

η

γθ θ θ

ωε η

− + − +

− +

− + − +

∂ ∂ ∂ ∂ ∂ = = − + ∂ ∂ ∂ ∂ ∂

= − = −( )ni

θ



� Similar to the previous case of parallel polarization, we can write the reflected magnetic field vector as

� The transmitted fields will have similar expression with the

( )1 cos sin0

1

ˆ ˆ r rz xTM

r

EH y ye

γ θ θ

η

− − +Γ= −

r


� The transmitted fields will have similar expression with the incident fields except that � now the θi should be replaced by θt (angle that transmitted propagation vector makes with the normal),

� γ1 should be replaced by γ2 (wave is in region II) and � multiplied by a factor (transmission coefficient for TM case)



Table 6.6 Fields in two regions (oblique incidence: parallel

( ) ( )2cos sin

0ˆ ˆcos sint tz x

t TM t tE E e x z

γ θ θτ θ θ

− += −

r

( )2 cos sin0

2

ˆ t tz xTM

t

EH y e

γ θ θτ

η

− +=

r


Table 6.6 Fields in two regions (oblique incidence: parallel polarization) {next page}




( ) ( )1 cos sin

0ˆ ˆcos sini iz x

i i iE E e x z

γ θ θ θ θ− += −

r

( )Er

( ) ( )2cos sin

0ˆ ˆcos sint tz x

t TM t tE E e x z

γ θ θτ θ θ

− += −

r


( )ii xzi e

EyH

θθγ

ηsincos

1

0 1ˆ +−=r

( ) { }1 cos sin

0ˆ ˆcos sinr rz x

r TM r iE E e x z

γ θ θ θ θ− − += Γ +

r

( )1 cos sin0

1

ˆ ˆ r rz xTM

r

EH y ye

γ θ θ

η

− − +Γ= −

r

( )2 cos sin0

2

ˆ t tz xTM

t

EH y e

γ θ θτ

η− +

=r



� Equating the tangential components of magnetic field � (magnetic field has only Hy component and it is tangential at the interface z=0) and

� electric field � (electric field has two components: Ex and Ez and only Ex is


� (electric field has two components: Ex and Ez and only Ex is tangential at the interface z=0)

� at z=0 gives

1 21sin sinsincos cos cosi trx xx

i TM r TM te e e

γ θ γ θγ θθ θ τ θ− −−+ Γ =

21 1 sinsin sin

1 1 2

ti r xx x

TM TMe ee

γ θγ θ γ θ τ

η η η

−− −Γ− =



� Therefore,

which is Snell’s law of reflection and refraction and this

tri θγθγθγ sinsinsin 211 ==

1 2; sin sin

i r i tθ θ γ θ γ θ⇒ = =


� which is Snell’s law of reflection and refraction and this implies that

cos cos cosi TM r TM t

θ θ τ θ+ Γ =

( )1 2

11 TM

TM

τ

η η− Γ =



� The above two equations has two unknowns τTM and ГTM and it can be easily solved as follows

� Therefore,

cos cos

cos

i TM r

TM

t

θ θτ

θ

+ Γ= ( )2

1

1TM TM

ητ

η= − Γ


� Therefore,

( ) ( )2 2

1 1

2 1 1 22 2

1 1 1 1

2 1

2

cos cos cos cos1 1

cos cos cos

cos cos cos cos coscos

cos cos cos cos

cos cos

cos

i TM r i TM r

TM TM

t t t

i t r i tr

TM TM

t t t t

t i

TM

θ θ θη η θ

θ η η θ θ

θ η θ η θ η θ η θη θ η

η θ θ η η θ η θ

η θ η θ

η θ

+ Γ Γ= − Γ ⇒ − Γ = +

− − −⇒ Γ − − = − ⇒ Γ =

−⇒ Γ = 2 1

1 2 1

cos cos

cos cos cos

t i

t r t i

η θ η θ

η θ η θ η θ

−=

+ +



� Hence, the reflection and transmission coefficients for oblique

( ) 2 1 2 1 2 12 2 2

1 1 2 1 1 2 1

1 22

1 2 1 2 1

cos cos cos cos cos cos1 1

cos cos cos cos

2 cos 2 cos

cos cos cos cos

t i t i t i

TM TM

t i t i

i i

t i t i

η θ η θ η θ η θ η θ η θη η ητ

η η η θ η θ η η θ η θ

η θ η θη

η η θ η θ η θ η θ

− + − +∴ = − Γ = − =

+ +

= =

+ +


Hence, the reflection and transmission coefficients for oblique incidence (Fresnel coefficients) for parallel polarization are given as follows:

� Note that cosine terms multiplication in the above equations is different from the previous expression for TE waves

2 1

2 1

cos cos

cos cos

t i

TM

t i

η θ η θ

η θ η θ

−∴Γ =

+

2

2 1

2 cos

cos cos

i

TM

t i

η θτ

η θ η θ=

+



6.3.3 Brewster angle:

� For Γ=0, the angle of incidence is known as Brewster angle (θB)

� This means at this angle of incidence of the EM wave from region I,


region I, � there will be no reflection from region II, � all the EM waves will be absorbed



� For parallel polarization,

2 1

2 1

cos cos0

cos cos

t i

TM

t i

η θ η θ

η θ η θ

−Γ = =

+

TM

i Bθ θ=


� Also we have from 2nd Snell’s law

2 1cos cos

TM

t Bη θ η θ⇒ =

t

TM

B θγθγ sinsin 21 =



� Using the above two equations, we can get (see textbook)

−

=22

2

2

2

1

2

1

sinγη

η

η

θ TM

B


� For lossless media, we have,1

1 1 1 1

1 1

2

2 2 22 22 2 2 2 1 21 1 1 2

2 2 2 2 2

2 12 1 2 1 1 2 1 2 2 2 2

1 2 2 2

1

; sinTM

B

µ ε

µ γ µ µ ε µ ε µ εη γ ε µθ

µ ε µ εη γ µ µ ε µ ε µ γ ε µ

ε µ ε µ

−

= = = = ∴ =

−

Q

−

2

2

2

1

2

2

2

1

γ

γ

η

ηB



� For magnetic media,

� is of the form 1/0 and Brewster angle does not exist

21εε =

TM

Bθ2sin


exist� But, for dielectric media, it exist and can be calculated as follows

2

1 2 2 2

2 1 2 1 1 1

1 2

1

sin tanTM TM

B B

n

n

ε

ε ε εθ θ

ε ε ε ε εε ε

−

⇒ = = ⇒ = =+−



� For perpendicular polarization,

� Similarly, Brewster’s angle doesn’t exist for perpendicular polarization at oblique incidence for a non-magnetic or dielectric medium (see textbook for derivation)

� For magnetic medium


� For magnetic medium

( )

( ) ( )

2

2 1 22 2 2 2 2

2 2

1 2 1 2 1 11 2

sin sin ; tanTE TE TE

B B B

n

n

µ µ µ µ µ µθ θ θ

µ µ µ µ µµ µ

−∴ = = ⇒ = = =

+ +−


interface at oblique incidence� Brewster angle exist for oblique incidence (perpendicular polarization) for interface between two magnetic media

� An incident wave consisting of both polarizations will have reflected wave for only one polarization for Brewster angle

� Brewster angle is also known as Polarizing angle (used in optics and lasers)


Polarizing angle optics and lasers)

� In other words, an incident wave which is composed of both TE and TM waves will have reflected TE (TM) waves only for dielectric (magnetic) media interface

� Thus, a circularly polarized incident wave at polarizing angle will become a linearly polarized wave



6.3.4 Total internal reflection:

� Used in optical fibers� Light strikes an angle of incidence greater than critical angle,

� all of the light energy will be reflected back, and � none of them will be absorbed


� none of them will be absorbed

� This critical angle is related to the refractive index of the two media and is given by

� where are the refractive index of medium 2 and 1 respectively

1 2

1

sinc

n

nθ −

=



� In fact, this can be observed for both the polarizations we have discussed above

� From Snell’s Second law, we have,

2

1 2

sinsin sin sin t

i t i

γ θγ θ γ θ θ

γ= ⇒ =


� If we consider non-magnetic materials (dielectrics) of , in that case,

1 2

1

sin sin sini t i

γ θ γ θ θγ

= ⇒ =

2

1

sinsin

t

i

ε θθ

ε⇒ =



� Hence, for

� (it means the transmitted wave travels along the interface, for our case, it is x-axis) which implies that

,2

i c t

πθ θ θ= =


our case, it is x-axis) which implies that

2 1 2

11

sin sinc c

ε εθ θ

εε

−

⇒ = ⇒ =



� We can write the above relation as

Note that to have a real value of critical angle

1 1 1 10 2 0 22 2 2 2

1 0 1 0 1 1 1 1

sin sin sin sinr r r r

c

r r r r

n

n

µ µ ε εε µ εθ

ε µ µ ε ε µ ε− − − −

= = = =

ε ε>


� Note that to have a real value of critical angle� Besides, we have assumed that for a dielectric,

1 2ε ε>

1 21

r rµ µ= =


interface at oblique incidence� What will happen when the angle of incidence is greater than the critical angle?

� For all angles greater than the critical angle, � using Snell’s second law, and � noting that waves are moving from a denser to a rarer dielectric medium,


medium,

� The value of should be imaginary

1 1 2

2 2

sinsin 1 cos 1 sin

i

i c t t t

µ ε θθ θ θ θ θ

µ ε> ∴ = > ⇒ = ± −Q

cost

θ

1 21

r rµ µ= =

1 2ε ε>



� It can be shown that (see textbook for derivation)

� For z>0, since is an imaginary number, is also purely imaginary

2 2

2 2

0 02 cos 2 cos* *

2 2

ˆ ˆsin cost tj z j z

t t

E ES E H x e z e

β θ β θτ τθ θ

η η− −∴ = × = +

r r r

cost

θ *cos

tθ


also purely imaginary� Hence, the second term is imaginary, therefore, no real power is flowing along z-axis in region II

� The first term which has direction along x-axis is real and is exponentially decaying with z



� Hence,� When a wave is incident at an angle of incidence greater than or equal to the critical angle from region I,

� the wave will be total internally reflected in region I� Surface waves exist in the region II,


� Surface waves exist in the region II, � which is propagating along x-axis (direction of power flow), and

� attenuating very fast along z-axis � People have used this for communications using submarines



6.3.5 Effects on polarization:

� Since any arbitrary wave incident obliquely from region I can be decomposed into linear combinations of TE and TM waves and

� we also know that the expressions for reflection and


� we also know that the expressions for reflection and transmission coefficients for parallel and perpendicular polarizations are different

2 1

2 1

cos cos;

cos cos

i t

TE

i t

η θ η θ

η θ η θ

−Γ =

+2 1

2 1

cos cos

cos cos

t i

TM

t i

η θ η θ

η θ η θ

−Γ =

+



� So, any circularly polarized incident wave will � become elliptically polarized after reflection or transmission

� Similarly, a linearly polarized wave (note that linear polarization could be expressed as linear combination of RHCP and LHCP) may be


RHCP and LHCP) may be � rotated after reflection



� For instance, � If the region II is a perfect conductor, η2=0, Г=-1 for both TE and TM cases, � this means a RHCP wave after reflection will become LHCP


� this means a RHCP wave after reflection will become LHCP wave and vice versa

� This is why circular polarization is not used for indoor communication systems since we have many reflections inside a room

6.4 SummaryPlane waves reflection from a media interface

Normal incidenceTE

Lossless

Oblique incidence

Lossy conducting medium

Perfect conductor

TM Brewster angle

Effect on polarization

2 1

2 1

;η η

η η

−Γ =

+2

1 2

2ητ

η η=

+

τ→0 and Γ→-1

2 1

2 1

cos cos

cos cos

i t

TE

i t

η θ η θ

η θ η θ

−Γ =

+

2

2 1

2 cos

cos cos

i

TE

i t

η θτ

η θ η θ=

+

cos cosη θ η θ−CP�EP


Lossless medium

Good conductor

Fig. 6.7 Plane waves reflection from media interface in a nutshell

angle Total internal reflection

( )2

21 1

0

1

1 1 1ˆRe

2 2avg

S S z Eη

− Γ= • =

r

1 2

avg avgS S=

2 1

2 1

cos cos

cos cos

t i

TM

t i

η θ η θ

η θ η θ

−∴Γ =

+

2

2 1

2 cos

cos cos

i

TM

t i

η θτ

η θ η θ=

+

CP�EP

= −

1

21sin

ε

εθc

1

2

1

2tann

nTM

B ==ε

εθ

1

2

1

2tann

nTE

B ==µ

µθ

1 1 2

2 2

sinsin 1 cos 1 sin

i

i c t t t

µ ε θθ θ θ θ θ

µ ε> ∴ = > ⇒ = ± −Q( ) z

avg eEzSSα

η2

2

2

2

0

21

2

1ˆRe

2

1 −+ Γ−=•=

r

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6. Plane waves reflection from media interface - iitg.ac.in · Lossless medium Good conductor Fig. 6.1 Plane waves reflection from media interface Lossyconducting medium TM Brewster