Lecture Notes 06.5: EM Waves, Wave Propagation in Linear ...

UIUC Physics 436 EM Fields & Sources II

Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

LECTURE NOTES 6.5 Reflection & Transmission of Monochromatic Plane EM Waves at Oblique Incidence at a Boundary Between Two Linear / Homogeneous / Isotropic Media A monochromatic EM plane wave is incident at an oblique angle inc on a boundary between two linear/homogeneous/isotropic media. A portion of this EM wave is reflected at angle  refl , a portion of this EM wave is transmitted, at angle trans . The three angles are defined with respect to the unit normals to the interface nˆ1 , nˆ2 , as shown in the figure below:

nˆ2

nˆ1

The incident EM wave is:

       i k  r t 1   E inc  r , t   E oinc e inc and: Binc  r , t   kînc  E inc  r , t  v1

The reflected EM wave is:

  i k  r t    E refl  r , t   E orefl e refl

  1   and: B refl  r , t   kˆrefl  E refl  r , t  v1

      1     r t  i k The transmitted EM wave is: E trans  r , t   E otrans e trans and: Btrans  r , t   kˆtrans  E trans  r , t  v2

Note that all three EM waves have the same frequency, f   2 This is due to the fact that at the microscopic level, the energy of real photon does not change in a medium, i.e. Evac  Emed  E , and since E  hf for real photons, then hfvac  hfmed  hf . Thus, the frequency of a real photon does not change in a medium, i.e. fvac  fmed  f {n.b. An experimental fact: colors of objects do not change when placed & viewed e.g. underwater}. However, the momentum of a real photon does change in a medium! This is because the momentum of the real photon in a medium depends on index of refraction of that medium nmed via the relation pmed  nmed pvac where nmed  c vmed . Thus the photon momentum depends {inversely} on the speed of propagation in the medium! © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved.

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Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

From the DeBroglie relation between momentum and wavelength of the real photon p  h 

we see that pmed  nmed pvac  nmed  h vac   h  nmed vac   h med and hence med  vac nmed .

Thus, for macroscopic EM waves propagating in the two linear/homogeneous/isotropic media (1) and (2), we have f1  f 2  f , and since   2 f then 1  2   . But since:   kv then: 1  2    k1v1  k2 v2 thus:   kinc v1  krefl v1  ktrans v2    Now: kinc  kinc  2 1 ; krefl  krefl  2 1 ; ktrans  ktrans  2 2

And:

  1  2  2  v1 1   2  v2 2 

Then:   2 f1  2 f1  2 f 2  finc

frefl

f1  f 2  finc  f refl  ftrans

ftrans

Then: 1  o n1

2  o n2

And: v1  c n1

v2  c n2

Thus: k1  n1ko

k2  n2 ko

where: o  vacuum wavelength

c f

where: ko  vacuum wavenumber  2 o   c

From:   kinc v1  krefl v1  ktrans v2 v  v  n  n  We see that: kinc  krefl  k1   2  ktrans   2  k2   1  ktrans   1  k2 Since vi  c ni i  1, 2  v1   v1   n2   n2 

The total (i.e. combined) EM fields in medium 1):             ETot1  r , t   E inc  r , t   E refl  r , t  and: BTot1  r , t   Binc  r , t   B refl  r , t  must be matched (i.e. joined smoothly) to the total EM fields in medium 2):         ETot2  r , t   E trans  r , t  and: BTot2  r , t   Btrans  r , t 

using the boundary conditions BC1) → BC4) at z = 0 (in the x-y plane). At z = 0, these four boundary conditions generically are of the form: (

)e

  i kinc r  t





(

)e

  i krefl r  t





(

)e

  i ktrans r t





These boundary conditions must hold for all (x,y) on the interface at z = 0, and also must hold for arbitrary/any/all times, t. The above relation is already satisfied for arbitrary time, t, since the factor e  it is common to all terms.

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© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved.


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Thus, the following generic relation must hold for any/all (x,y) on the interface at z = 0: 





   i  k r  i  k r  i k r  ( ) e inc  ( ) e refl  ( ) e trans        For z = 0 (i.e. on the interface in the x-y plane) we must have: kinc r  krefl r  ktrans r

More explicitly: kincx x  kincy y  kincz z  kreflx x  krefl y y  kreflz z  ktransx x  ktrans y y  ktransz z    z 0

or:

z 0

z 0

kincx x  kinc y y  kreflx x  krefl y y  ktransx x  ktrans y y @ z = 0 in the x-y plane.

The above relation can only hold for arbitrary (x, y, z = 0) iff ( = if and only if):

kincx x  kreflx x  ktransx x and:

 kincx  kreflx  ktransx

kincy y  krefl y y  ktrans y y  kincy  krefl y  ktrans y

Since this problem has rotational invariance (i.e. rotational symmetry) about the zˆ -axis,  (see above pix on p. 1), without any loss of generality we can e.g. choose kinc to lie entirely within the x-z plane, as shown in the figure below… Then: kincy  krefl y  ktrans y  0 and thus: kincx  kreflx  ktransx .    i.e. the transverse components of kinc , krefl , ktrans are all equal and point in the {same}  xˆ direction.

The First Law of Geometrical Optics (All wavevectors k lie in a common plane):    The above result tells us that the three wave vectors kinc , krefl and ktrans ALL LIE IN A PLANE

known as the plane of incidence (here, the x-z plane) and that: kincx  kreflx  ktransx as shown in the figure below. Note that the plane of incidence also includes the unit normal to the interface {here}, nˆ IF   zˆ -axis. The x-z Plane of Incidence:

trans

,

nˆ IF


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Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

The Second Law of Geometrical Optics (Law of Reflection):

From the above figure, we see that:

kincx  kinc sin inc = kreflx  krefl sin  refl = ktransx  ktrans sin trans But: kinc  krefl  k1  sin inc  sin  refl

 Angle of Incidence = Angle of Reflection inc   refl Law of Reflection! The Third Law of Geometrical Optics (Law of Refraction – Snell’s Law):

For the transmitted angle, trans we see that: kinc sin inc  ktrans sin trans In medium 1): kinc  k1   v1  n1 c  n1ko where ko  vacuum wave number  2 o and o  vacuum wave length In medium 2): ktrans  k2   v2  n2 c  n2 ko Thus:

kinc sin inc  ktrans sin trans  k1 sin inc  k2 sin trans

But since:

kinc  k1  n1ko and ktrans  k2  n2 ko

Then:

k1 sin inc  k2 sin trans

Which can also be written as:

 n1 sin inc  n2 sin trans

Law of Refraction (Snell’s Law)

sin trans n1  n2 sin inc

Since trans refers to medium 2) and inc refers to medium 1) we can also write Snell’s Law as: n1 sin 1  n2 sin  2

sin  2 n1  sin 1 n2

or:

(incident) (transmitted) Because of the above three laws of geometrical optics, we see that:       kinc r z 0  krefl r z 0  ktrans r z 0 everywhere at/on the interface @ z = 0 in the x-y plane. Thus we see that: e

  i kinc  r t





z 0

e

  i krefl  r t





z 0

e

  i ktrans  r  t





z 0

everywhere at/on the interface at

z = 0 in the x-y plane, this relation is also valid/holds for any/all time(s) t, since  is the same in either medium (1 or 2).

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Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Thus, the boundary conditions BC 1) → BC 4) for a monochromatic plane EM wave incident at/ on an interface at an oblique angle inc between two linear/homogeneous/isotropic media become:

 BC 1): Normal (i.e. z-) component of D continuous at z = 0 (no free surface charges):



1 E o  E o incz

refl z



   E 2

otransz



using D   E

 BC 2): Tangential (i.e. x-, y-) components of E continuous at z = 0:

 E

oincx , y

 E orefl

x,y

  E

otransx , y

 BC 3): Normal (i.e. z-) component of B continuous at z = 0:

 B

oincz

 Borefl

z

  B

otransz

 BC 4): Tangential (i.e. x-, y-) components of H continuous at z = 0 (no free surface currents): B  

1

1

oincx , y

 Borefl

x,y

  1 B 2

otransx , y

 1  ˆ  Note that in each of the above, we also have the relation Bo  k  E o v For a EM plane wave incident on a boundary between two linear / homogeneous / isotropic media at an oblique angle of incidence, there are three possible polarization cases to consider:  Case I): Einc  plane of incidence – known as Transverse Electric (TE) Polarization  { Binc  plane of incidence}

 Case II): Einc  plane of incidence – known as Transverse Magnetic (TM) Polarization  { Binc  plane of incidence}  Case III): The most general case: Einc is neither  nor  to the plane of incidence.  {  Binc is neither  nor  to the plane of incidence} i.e. Case III is a linear vector combination of Cases I) and II) above! LP ˆ   sin   ˆ  cos  xˆ  sin  yˆ  cos    LP vector: nînc



CP  xˆ  iyˆ , also the EP (elliptical polarization case. CP vector: nînc

 Simply decompose the polarization components for the general-case incident EM plane wave ˆ  and yˆ  ˆ  vector components – i.e. the E-field components perpendicular to and into its xˆ  parallel to the plane of incidence (TE polarization and TM polarization respectively). Solve these separately, then combine results vectorially… © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2015. All Rights Reserved.

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Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Case I): Electric Field Vectors Perpendicular to the Plane of Incidence: Transverse Electric (TE) Polarization

A monochromatic plane EM wave is incident {from the left} on a boundary located at z = 0 in the x-y plane between two linear / homogeneous / isotropic media at an oblique angle of  incidence. The polarization of the incident EM wave (i.e. the orientation of Einc is transverse    (i.e.  ) to the plane of incidence {= the x-z plane containing the three wavevectors kinc , krefl , ktrans and the unit normal to the boundary/interface, nˆ   zˆ }), as shown in the figure below:

 Note that all three E -field vectors are  yˆ {i.e. point out of the page} and thus all three

 E -field vectors are  to the boundary/interface at z = 0, which lies in the x-y plane.   Since the three B -field vectors are related to their respective E -field vectors by the right   hand rule cross-product relation B  1v kˆ  E then we see that all three B -field vectors lie in the x-z plane {the plane of incidence}, as shown in the figure above.   The four boundary conditions on the {complex} E - and B -fields on the boundary at z = 0 are:  BC 1) Normal (i.e. z-) component of D continuous at z = 0 (no free surface charges)   

0

1  E o

incz

0 0   E orefl    2 E otrans z z  

 0  0  0 {see/refer to above figure}

 BC 2) Tangential (i.e. x-, y-) components of E continuous at z = 0:

 E

oinc y

 E orefl

y

  E

otrans y

 E oinc  E orefl  E otrans

{n.b. All Ex ' s  0 for TE Polarization}

 BC 3) Normal (i.e. z-) component of B continuous at z = 0:

 B

oincz

6

 Borefl

z

  B

otransz



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

 BC 3) {continued}: n.b. Since only the z-components of B ' s on either side of interface are involved here, and all unit wavevectors kînc ,kˆrefl andkˆtrans lie in the plane of incidence (x-y  plane) and all E -field vectors are || to the  yˆ direction for TE polarization, then because of the   1 ˆ  cross-product nature of B  k  E , we only need the x-components of the unit wavevectors, i.e.: v

kînc  kîncx  kîncz

 sin inc xˆ  cos inc zˆ

kˆrefl  kˆreflx  kˆreflz

 sin  refl xˆ  cos  refl zˆ

kˆtrans  kˆtransx  kˆtransz  sin trans xˆ  cos trans zˆ

 B

oincz

 





See/refer to above figure

1 ˆ 1 ˆ kincx  E oinc yˆ  kˆreflx  E orefl yˆ  ktransx  E otrans yˆ zˆ  Borefl zˆ  Botrans zˆ = y y y z z v1 v2







 xˆ  yˆ   zˆ

=

1  1  Eoinc sin inc  xˆ  yˆ  E orefl sin  refl  xˆ  yˆ  Eotrans sin trans  xˆ  yˆ v1 v2

=

1  1 Eoinc sin inc  E orefl sin  refl zˆ  E otrans sin trans zˆ v1 v2









 BC 4) Tangential (i.e. x-, y-) components of H continuous at z = 0 (no free surface currents): n.b. Same reasoning as in BC3 above, but here we only need the z-components of the unit wavevectors, i.e.: B  

1

1

oincx





1  xˆ  Borefl xˆ  Botrans xˆ {n.b. All By ' s  0 for TE Polarization – see above pix} x x

2







=

1 ˆ 1 ˆ kincz  E oinc yˆ  kˆreflz  E orefl yˆ  k  E otrans yˆ y y y 1v1 2v2 transz

=

1  1  Eoinc cos inc  zˆ  yˆ  E orefl cos  refl  zˆ  yˆ  E cos  trans  zˆ  yˆ 1v1 2 v2 otrans

=

1  1  Eoinc   cos inc   E orefl cos  refl xˆ  E   cos trans  xˆ 1v1 2 v2 otrans





Thus, we obtain: E oinc  E orefl  E otrans



 zˆ  yˆ   xˆ







(from BC 2))

 v sin trans Using the Law of Reflection inc   refl on the BC 3) result: E oinc  E orefl   1   v2 sin inc

  Eotrans 

Using Snell’s Law / Law of Refraction: n n 1 1 sin inc  sin trans n1 sin inc  n2 sin trans  1 sin inc  2 sin trans  v1 v2 c c

or:

v2 sin inc  v1 sin trans or:

 v1 sin trans    v2 sin inc

  1 


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 v sin trans  E oinc  E orefl   1   v2 sin inc

Lect. Notes 6.5

Prof. Steven Errede

  Eotrans  E otrans i.e. BC 3) gives the same info as BC 1) ! 

 E

From the BC 4) result:

Fall Semester, 2015

oinc

  v cos  trans  E orefl   1 1    2 v2 cos inc



  Eotrans 

Thus, {again} from BC 1) → BC 4) we have only two independent relations, but three unknowns for the case of transverse electric (TE) polarization: 1) E oinc  E orefl 2)

 E

oinc

 E otrans

  v cos  trans    E orefl   1 1   Eotrans   2 v2 cos inc 

 v  Z Now:    1 1   1   2 v2  Z 2



 cos trans  and we also define:     {n.b. Both α and β > 0 and real}  cos  inc 

Then eqn. 2) above becomes: E oinc  E orefl   E otrans and eqn. 1) is: E oinc  E orefl  E otrans  2 Add equations 1) + 2) to get: 2 E oinc  1    E otrans  E otrans    1  

  Eoinc 

 1   Subtract eqn’s 2)  1) to get: 2 E orefl  1    E otrans  E orefl    2

  Eotrans eqn. (21) 

 1     2 Plug eqn. (2+1) into eqn. (21) to obtain: E orefl     2   1    1   Thus: E orefl    1  

  2   Eoinc and Eotrans     1  

eqn. (1+2)

  1    Eoinc     1  


E orefl  1    E otrans  2     E or: and    oinc  E oinc  1    E oinc  1    

 2  n.b. since both α and β > 0 and purely real quantities then:   > 0 and hence the  1    transmitted wave is always in-phase with the incident wave for TE polarization. The ratios of electric field amplitudes @ z = 0 for transverse electric (TE) polarization are thus:  The Fresnel Equations for E  to Interface @ z = 0  E  to Plane of Incidence = Transverse Electric (TE) Polarization EoTErefl EoTEinc

8

EoTEtrans  2   cos trans   1v1   1      and:      and: TE    with:    Eoinc  1     1     2 v2   cos inc 



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Now because the incident monochromatic plane EM wave strikes the interface (lying in the x-y  plane) at an oblique angle inc , the time-averaged EM wave intensity is I  z  0   S  z  0, t  nˆ

Watts m  where nˆ is the unit normal of the interface, oriented in the direction of energy flow. 2

Because the incident EM wave is propagating in a linear / homogeneous / isotropic medium, we  have the relation: S  z  0, t   v u inc  z  0, t  kˆ . Thus, the time-averaged incident inc

EM

1

inc

intensity (aka irradiance) for an oblique angle of incidence is:  inc inc I inc  z  0   Sinc  z  0, t   zˆ  v1 uEM  z  0, t  kînc  zˆ  v1 uEM  z  0, t  cos inc For TE polarization the incident, reflected and transmitted intensities are:  TE I inc  z  0   Sinc  z  0, t   zˆ



inc  v1 uEM  z  0, t  kînc  zˆ

 12 1v1 EoTEinc

 TE refl I refl  z  0   Srefl  z  0, t    zˆ   v1 uEM  z  0, t  kînc   zˆ   12 1v1 EoTErefl





2

cos inc

 cos 2

 TE trans I trans  z  0   Strans  z  0, t   zˆ  v2 uEM  z  0, t  ktrans  zˆ  12 v2 2 EoTEtrans





2

inc

n.b. used law of reflection: refl = inc

cos trans

Thus the reflection and transmission coefficients for transverse electric (TE) polarization  (with all E -field vectors oriented  to the plane of incidence) @ z = 0 are:

RTE 

I I

TE refl TE inc

1 2



2

inc

 

TE TE I trans z  0  12  2 v2 Eotrans   TTE  TE TE 1 I inc  z  0  2  1v1 Eoinc

But:

 refl

 z  0  1v1  E  cos inc   z  0  12 1v1  EoTE 2 cos inc TE orefl

  v   v  Z   1 1  2 2  1   2 v2   1v1  Z 2

And from above (p. 8):

 EoTErefl Thus: RTE   TE  Eo  inc

 EoTErefl  TE   Eoinc

 

2

cos trans

2

cos inc

   

2

  v   cos trans   2 2   1v1   cos inc

  1         1   

and:

 EoTEtrans and  TE  Eoinc

 EoTEtrans TTE    TE  Eo  inc

  Eotrans   TE   Eoinc TE

   

2

 EoTEtrans  TTE    TE  Eoinc

 cos  trans  and:      cos inc 

2

  1    2     1    

 EoTErefl   TE  Eo  inc

   

2

  2     1    

2

 4    1   2 


9


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Explicit Check: Does RTE  TTE  1 ? (i.e. is EM wave energy conserved?)

1     4 2 2 1    1    2



1  2   2  2  4

1   

2



1  2   2  2

1   

2

1    2 1   

2



 1 Yes !!!

Note that at normal incidence: inc  0   refl  0 and trans  0 {See/refer to above figure}  cos trans Then:     cos inc

 cos 0 1   1   cos 0

Thus, at normal incidence: RTE

Note that these results for RTE

inc  0

inc  0

 1      1  

and TTE

2

and: TTE

inc  0

inc  0



4

1   

2

are the same/identical to those we obtained

previously for a monochromatic plane EM wave at normal incidence on interface!!! In the special/limiting-case situation of normal incidence, where inc   refl  trans  0 , the plane of incidence collapses into a line (the zˆ axis), the problem then has rotational invariance about the zˆ axis, and thus for normal incidence the polarization direction associated with the  spatial orientation of Einc no longer has any physical consequence(s).

10



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Case II): Electric Field Vectors Parallel to the Plane of Incidence: Transverse Magnetic (TM) Polarization

A monochromatic plane EM wave is incident {from the left} on a boundary located at z = 0 in the x-y plane between two linear / homogeneous / isotropic media at an oblique angle of  incidence. The polarization of the incident EM wave (i.e. the orientation of Einc is now parallel    (i.e.  ) to the plane of incidence {= the x-z plane containing the three wavevectors kinc , krefl , ktrans and the unit normal to the boundary/interface, nˆ   zˆ }), as shown in the figure below:

 For TM polarization, all three E -field vectors lie in the plane of incidence.   Since the three B -field vectors are related to their respective E -field vectors by the right   hand rule cross-product relation B  1v kˆ  E then we see that all three B -field vectors are  yˆ {i.e. either point out of or into the page} and thus are  to the plane of incidence {hence the  origin of the name transverse magnetic polarization}; hence note that all three B -field vectors are  to the boundary/interface at z = 0, which lies in the x-y plane as shown in the figure above.   The four boundary conditions on the {complex} E - and B -fields on the boundary at z = 0 are:  BC 1) Normal (i.e. z-) component of D continuous at z = 0 (no free surface charges)

    E

1 E o  E o incz

1

oinc

refl z

   E 2

otransz





sin inc  E orefl sin  refl   2  E otrans sin trans



{n.b. see/refer to above figure}

 BC 2) Tangential (i.e. x-, y-) components of E continuous at z = 0:

 E  E

oincx oinc

 E orefl

x

  E

otransx



cos inc  E orefl cos  refl  E otrans cos trans {n.b. see/refer to above figure}


11


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

 BC 3) Normal (i.e. z-) component of B continuous at z = 0: 0  0  Boincz  Boreflz 

0    Botransz 

 00  0

{n.b. see/refer to above figure}

 BC 4) Tangential (i.e. x-, y-) components of H continuous at z = 0 (no free surface currents):  Borefl

B  

1  yˆ  Borefl yˆ  Botrans yˆ

1



  1  B 

B   1

oinc y

1

1

oinc y

y

2

y



{n.b. All Bx ' s  0 for TM Polarization}

otrans y

 

y

2

  n.b. Can use full cross-product(s) B  1v kˆ  E here!



   1 ˆ 1 ˆ ˆ    k  E  k  E  k  E = 1v1 inc oinc refl orefl 2v2 trans otrans









1  1  Eoinc yˆ  E orefl yˆ  E yˆ 1v1 2 v2 otrans

=





B   1

1

oinc y

 Borefl

y

  1  B  2

otrans y







Use right-hand rule for all cross-products {n.b. see/refer to above figure}





1  1  Eoinc  E orefl  E 1v1 2 v2 otrans



1 E o sin inc  E o sin  refl   2  E o

From BC 1) at z = 0: Redundant info – both BC’s give same relation





inc

refl

sin trans

trans

c c , n2  v1 v2 n v  1  Snell's Law   2 n2 v1

But:

inc   refl (Law of Reflection) and: n1 

And:

n1 sin 1  n2 sin  2 



sin  2 sin trans  sin 1 sin inc



 n   v  E oinc  E orefl   2 1  E otrans   2 2  E otrans   E otrans  1 n2   1v1 

From BC 4) at z = 0:

 v  E oinc  E orefl   1 1  E otrans   E otrans  2 v2 

From BC 2) at z = 0:

 E

cos trans cos inc  E orefl cos  refl  E otrans cos trans but:   cos inc

 E

 cos trans  E orefl    cos inc

oinc



oinc

  v   v  where:    1 1    2 2   2 v2   1v1 





  Eotrans   E otrans 

Thus for the case of transverse magnetic (TM) polarization:   v   v  cos trans E oinc  E orefl   E otrans and E oinc  E orefl   E otrans with    1 1    2 2  and   cos inc  2 v2   1v1 

12



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Solving these two above equations simultaneously, we obtain:  2  2 E oinc      E otrans  E otrans    Eoinc          and: 2 E orefl      E otrans  E orefl    Eo  2  trans 

  E orefl     


The real / physical electric field amplitudes for transverse magnetic (TM) polarization are thus:  The Fresnel Equations for B  to Interface  B  to Plane of Incidence = Transverse Magnetic (TM) Polarization  EoTM refl  TM  Eo  inc

     EoTM trans  and  TM        Eoinc 

  2  cos trans   with   cos inc     

  v   v  Z and    1 1    2 2   1  2 v2   1v1  Z 2

The Fresnel relations for TM polarization are not identical to the Fresnel relations for TE polarization:  EoTErefl  TE  Eo  inc

  1     EoTEtrans  and  TE   1     Eoinc 

  2     1    

We define the incident, reflected & transmitted intensities at oblique incidence on the interface @ z = 0 for the TM case exactly as we did for the TE case:

 TM I inc  z  0   SincTM  z  0, t   zˆ



 12 1v1 EoTM inc



2

cos inc

 TM 2 TM I refl  z  0   Srefl  z  0, t    zˆ   12 1v1 EoTMrefl cos inc  ref  inc   TM 2 TM I trans  z  0   Strans  z  0t   zˆ  12  2v2 EoTMtrans cos trans









The reflection and transmission coefficients for transverse magnetic (TM) polarization  (with all B -field vectors oriented  to the plane of incidence @ z = 0) are:

RTM 

TTM

TM I refl  z  0 TM I inc

 EoTM refl   TM   z E 0    oinc

2

     2         

TM I trans  z  0     2v2   cos trans  TM   I inc  z  0   1v1   cos inc

  Eotrans   TM   Eoinc TM

2

  EoTM trans     TM   Eo   inc

2

 4        2 


13


i.e.

RTM

 EoTM refl   TM  Eo  inc

Fall Semester, 2015

2

     2         

and:

TTM

Lect. Notes 6.5

 EoTM trans    TM  Eo  inc

Prof. Steven Errede

2

 4        2 

Again, note that the reflection and transmission coefficients for transverse magnetic (TM) polarization are not identical/the same as those for the transverse electric case:  EoTErefl RTE   TE   Eoinc

2

  1    2       1   

 EoTEtrans TTE    TE  Eo  inc

and:

2

 4    1   2 

Explicit Check: Does RTM + TTM = 1? (i.e. is EM wave energy conserved?)    4  2  2   2  4  2  2   2           1 Yes !!!  2 2 2 2                      2

RTM  TTM

2

Note again at normal incidence: inc  0   refl  0 and trans  0 {See/refer to above figure}  cos trans  cos 0 1   1 Then:      cos inc  cos 0

Thus at normal incidence: RTM

inc  0

 1      1  

2

and TTM

inc  0



4

1   

2

These are identical to those for the TE case at normal incidence, as expected  due to rotational invariance / symmetry about the zˆ axis: At normal incidence: RTE

14

inc  0

 1      1  

2

and TTE

inc  0



4

1   

2



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

The Fresnel Relations TE Polarization

TM Polarization

 EoTErefl  TE   Eoinc

  1         1   

 EoTM refl  TM  Eo  inc

           

 2   1     cos trans  cos inc

 EoTM trans  TM  Eo  inc

 2       

 EoTEtrans  TE  Eo  inc



v1  c

1v1  2 v2 1n2  2 n1 Z1     2v2 1v1 2 n1 1n2 Z 2

v2  c

n1 n2

 1  1

11  2 2

Reflection and Transmission Coefficients R & T R+T=1 TE Polarization

RTE 

TE I refl TE I inc

 EoTErefl   TE  Eo  inc

TM Polarization 2

  1    2     1    

TE  EoTEtrans  I trans  TTE   TE     TE  Eo  I inc   inc

 

TM I refl

RTM 

2

 4    1   2 

cos trans cos inc

TTM

TM I inc

2

     2         

TM  EoTM  I trans  trans   TM     TM  I E  inc   oinc

v1  c

1v1  2 v2 1n2  2 n1 Z1     2v2 1v1 2 n1 1n2 Z 2

 EoTM refl   TM  Eo  inc

n1

v2  c

n2

 1  1

2

 4   2      

11  2 2

Note that since Eo21,2  n1,2  t  E 1,2 , the reflection coefficient/reflectance R can thus be seen as the statistical/ensemble average probability that at the microscopic scale, individual



photons will be reflected at the interface: R  Eorefl Eoinc



2

 n refl  t 

n inc  t   Prefl ,

and since R  T  1 then T  1  R  1  Prefl  Ptrans , since we must have Prefl  Ptrans  1 !!! Now we want to explore / investigate the physics associated with the Fresnel relations and the reflection and transmission coefficients – comparing results for TE vs. TM polarization for the cases of external reflection (n1 < n2) and internal reflection n1 > n2) Just as  can be written several different but equivalent ways (see above), so can the Fresnel relations, as well as the expressions for R & T using various relations including Snell’s Law.


15


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Starting with the Fresnel relations as given above, explicitly writing these out alternate versions: Fresnel Relations TE Polarization

TM Polarization

E  TE  Eo  inc

 n1  n  cos inc   2  cos trans     1   2     n1     cos inc   n2  cos trans  1   2 

E  TM  Eo  inc

 n2  n  cos inc   1  cos trans     2   1     n2     cos inc   n1  cos trans  2   1 

E   E 

n  2  1  cos inc   1    n   n2  1    cos inc    cos trans  1   2 

E  E 

n  2  1  cos inc   1    n   n1  2    cos inc    cos trans  2   1 

TE orefl

TE otrans TE oinc

TM orefl

TM otrans TM oinc

If we now neglect / ignore the magnetic properties of the two media – e.g. if paramagnetic / diamagnetic such that  m  1 then 1  2  o the Fresnel relations then become: TE Polarization

TM Polarization

 EoTErefl  TE  Eo  inc

 n cos   n cos  2 inc trans  1  n1 cos inc  n2 cos trans 


 n cos   n cos  1 inc trans  2  n2 cos inc  n1 cos trans 


 2n1 cos inc   n1 cos inc  n2 cos trans 

 EoTM trans  TM  Eo  inc

 2n1 cos inc   n2 cos inc  n1 cos trans 

Using Snell’s Law n1 sin 1  n2 sin  2  ninc sin inc  ntrans sin trans and various trigonometric identities, the above Fresnel relations can also equivalently be written as: TE Polarization

TM Polarization


 sin inc  trans    sin inc  trans  


 tan inc  trans    tan inc  trans  


 2 cos inc sin trans   sin inc  trans  

 EoTM trans  TM  Eo  trans

 2 cos inc sin trans   sin inc  trans  cos inc  trans  

 n.b. the signs correlate to the TE & TM E -field vectors as shown in the above figures!

16



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

We now use Snell’s Law ninc sin inc  ntrans sin trans to eliminate trans : TE Polarization

TM Polarization 2


   

E   E 

   

TE otrans TE oinc

2

n  cos inc   2   sin 2 inc  n1 


2

n  cos inc   2   sin 2 inc  n1  2 cos inc 2

n  cos inc   2   sin 2 inc  n1 

 Eo The functional dependence of  refl  Eo  inc

E  E 

  Eotrans  ,   Eo   inc

TM otrans TM oinc

2

n  n    2  cos inc   2   sin 2 inc  n1   n1 

  2 2   n2    n2  2   cos inc     sin inc n n  1  1 n  2  2  cos inc  n1 

  2 2    n2   n2  2   cos inc     sin inc  n1   n1 

 Eorefl   , the reflection coefficient R    Eo   inc  2

   

2

2

 Eotrans   n2 n1   sin 2 inc  Eotrans  and the transmission coefficient T         Eo   Eo   cos inc  inc   inc  as a function of the angle of incidence inc for external reflection (n1 < n2) and internal reflection (n1 > n2) for TE & TM polarization are shown in the figures below: 2


17


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

External Reflection (n1 = 1.0 < n2 = 1.5):

Internal Reflection (n1 = 1.5 > n2 = 1.0):

18



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Comment 1): When  Erefl Einc   0 , Eorefl is 180o out-of-phase with Eoinc since the numerators of the original Fresnel relations for TE & TM polarization are 1    and     respectively.

Comment 2): For TM Polarization (only), there exists an angle of incidence where  Erefl Einc   0 , i.e. no reflected wave occurs at this incident angle for TM polarization! This incidence angle is known as Brewster’s angle  B (also known as the polarizing angle  P - because e.g. an incident wave that is a linear combination of TE and TM polarizations will have a reflected wave which is 100% pure-TE polarized for an incidence angle inc   B   P !!). * n.b. Brewster’s angle  B exists for both external (n1 < n2) & internal reflection (n1 > n2) for TM polarization (only). * Brewster’s Angle  B / the Polarizing Angle   for Transverse Magnetic (TM) Polarization

   From the numerator of EoTM EoTM   of the {originally-derived} expression for TM refl inc     polarization, this numerator = 0 at Brewster’s angle  B (aka the polarizing angle   ), which





occurs when      0 , i.e. when    . But:



cos trans cos inc

 v  n n and    1 1   1 2  2  2 v2  2 n1 n1

for 1  2  o

n  Now: cos trans  1  sin 2 trans and Snell’s Law: n1 sin inc  n2 sin trans  sin trans   1  sin inc  n2   at Brewster’s angle inc   B = polarizing angle   , where    , this relation becomes: 2



or:

cos trans n n   1 2  2 2 n1 n1 cos inc 1

1



2

for 1  2  o

n  1   1  sin 2 inc n   n2   2    cos inc  n1 

sin 2 inc   2 cos 2 inc   2 1  sin 2 inc  ← Solve for sin 2 inc

1      1  1  2  1   2   2   2  sin 2 inc  sin 2 inc  1 2 2 1   4    2

But:

1        1   1    1   2





1   4  1   2 1   2  sin inc 2

2

2

2

2

2

2

 sin inc 

 1  2


19


Geometrically: sin inc 

 1 

cos inc 

2

1 1 

2

Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

=

opp. side hypotenuse

1  2

=

adjacent hypotenuse

θinc = θB

β

n  opp. side 1  2  adjacent  n1  Thus, at an angle of incidence inc   Binc   Pinc = Brewster’s angle / the polarizing angle for a TM polarized incident wave, where no reflected wave exists, we have:

tan inc  

=

n  tan  Binc  tan  Pinc   2  for 1  2  o  n1 

From Snell’s Law: n1 sin inc  n2 sin trans we also see that: tan  Binc 

sin  Binc n2  cos  Binc n1

or: n1 sin  Binc  n2 cos  Binc for 1  2  o . Thus, from Snell’s Law we see that: cos  Binc  sin trans when inc   Binc   Pinc . So what’s so interesting about this??? 0   Well: cos  Binc  sin  2   Binc   sin  2  cos  Binc  cos  2  sin  Binc  sin trans i.e. sin    Binc   sin trans 2 

 When inc   Binc   Pinc for an incident TM-polarized EM wave, we see that trans   2   Binc Thus:  Binc  trans   2 , i.e.  Binc   inc and trans are complimentary angles !!! TM Polarized EM Wave Incident at Brewster’s Angle  Binc :

20



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

Thus, e.g. if an unpolarized EM wave (i.e. one which contains all polarizations/random polarizations) or an EM wave which is a linear combination of TE and TM polarization is incident on the interface between two linear/homogeneous/isotropic media at Brewster’s angle  Binc   inc , the reflected beam will be 100% pure TE polarization!! Hence, this is why Brewster’s angle  B is also known as the polarizing angle  P . Comment 3): inc For internal reflection (n1 > n2) there exists a critical angle of incidence  critical past which no transmitted beam exists for either TE or TM polarization. The critical angle does not depend on polarization – it is actually dictated / defined by Snell’s Law:

n  n    inc max inc inc  n2 sin trans  n2 sin    n2 or: sin  critical n1 sin  critical   2  or:  critical  sin 1  2  2  n1   n1 

inc , no transmitted beam exists → incident beam is totally internally reflected. For inc   critical inc For inc   critical , the transmitted wave is actually exponentially damped – it becomes a so-called

Evanescent Wave:

  n    i  k2 x sin inc  1  t    z  n2     Etrans  r , t   Eotrans e e

Exponential damping in z

2

n    k2  1  sin 2 inc  1  n2  Oscillatory along interface in x-direction


21


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

inc For Total Internal Reflection inc   critical , n1 > n2:

inc , n1 > n2), the reflected wave is actually displaced laterally, For total internal reflection ( inc   critical along the interface (in the direction of the evanescent wave), relative to the prediction from geometrical optics. The lateral displacement is known as the Goos-Hänchen effect [F. Goos and H. Hänchen, Ann. Phys. (Leipzig) (6) 1, 333-346 (1947)] and is different for TE vs. TM polarization:

DTE 

sin inc 1  sin 2   n n inc

2

1

DTM

sin inc   1  sin 2   n n inc

2

1



2

where: 1  o n1 and: o  c f = vacuum wavelength

    n2 n1



2

sin 2  inc 

n2 n1

2 2

cos 2 inc  

 DTE 

    n2 n1

sin 2  inc 

n2 n1

2 2

cos 2 inc  

inc Experiment to demonstrate that the transmitted EM wave for inc   critical is exponentially damped:

Use two 45o prisms, separated by a small distance d as shown in the figure below – (e.g. use glass prisms {for light}; can use paraffin prisms {for microwaves} !!) UIUC Physics 401 Experiment # 34 !!!

See e.g. D.D. Coon, Am. J. Phys. 34, 240 (1966)

 Microscopically, this experiment is an example of quantum mechanical barrier penetration / quantum mechanical tunneling phenomenon (using real photons) !!! 22



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

The above lateral displacement(s) for TE vs. TM polarization are also correlated with phase inc shifts that occur in the reflected wave when inc   critical for total internal reflection (n1 > n2): Using the (last) version of Fresnel Equations (p. 17 of these lecture notes): TE Polarization  EoTErefl  TE  Eo  inc

 cos inc     cos inc 

TM Polarization

    n2 n1

n2 n1

2

2

 sin 2 inc


 sin 2 inc

   

2

n   n12 cos inc   2  n2  cos inc  n1

   sin     sin  2

n2 n1

n2 n1

2

2

inc

2

inc

inc inc When inc   critical , Snell’s Law is: sin  critical   n2 n1  {since sin trans  sin 90o  1 }

The above E-field amplitude ratios become complex for internal reflection, because for when: sin 2 inc 

   1, n2 n1

   1 , then:  

n2 2 n1

inc  sin 2 inc becomes imaginary. Thus, for inc   critical  sin 1  for n1 > n2 (internal reflection), we can re-write the above E -field ratios as: n2 n1

TE Polarization  EoTErefl  TE  Eo  inc

TM Polarization

 cos inc  i sin 2 inc     cos inc  i sin 2  inc 

    n2 n1

n2 n1

2

2


   

   

2

n n   n12 cos inc  i sin 2 inc  n12  2 2  n2  cos inc  i sin 2 inc  nn12 n1

2

It is easy to verify that these ratios lie on the unit circle in the complex plane – simply multiply them by their complex conjugates to show AA  1 , as they must for total internal reflection. These formulae imply a phase change of the reflected wave (relative to the incident wave) inc that depends on the angle of incidence inc   critical  sin 1  n2 n1  for total internal reflection.  Eo We set:   refl  Eo  inc

  i ae  i   e   i    2 and tan  2   tan    ae 

Where δ = phase change (in radians) of the reflected wave relative to the incident wave. Thus, we see that (from the numerators of the above formulae) that:  tan  TE  2

sin 2 inc    cos inc 

  n2 n1

2

and:

 

sin 2 inc  nn12   TM  tan   2 n2  2  cos inc n

2

  1


23

  n2 n1


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

The relative phase difference between total internally-reflected TM vs. TE polarized waves    TM   TE can also be calculated:     TE  tan    tan  TM 2 2 

2  cos inc sin inc   n2 n1    sin 2 inc 

2

Phase shifts of the reflected wave relative to the incident wave for external, internal reflection and for TE, TM polarization are shown in the following graphs: Phase Shifts Upon Reflection: External Reflection (n1 = 1.0 < n2 = 1.5): Internal Reflection (n1 = 1.5 > n2 = 1.0):

Note that a phase shift of 180o is equivalent to a phase shift of +180o. 24



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

An Example of the {Clever} Use of Internal Reflection Phase Shifts - The Fresnel Rhomb:

From last graph of the internal reflection phase shifts (above), we see that the relative difference in TM vs. TE phase shifts for total internal reflection at a glass-air interface (n1 = 1.5 {glass}, n2 = 1.0 {air}) is    TM   TE   4  45o when inc  54.6o Fresnel used this TM vs. TE relative phase-shift fact associated with total internal reflection and developed / designed a glass rhomb-shaped prism that converted linearly-polarized light to circularly-polarized light, as shown in the figure below. He used light incident on the glass rhomb-shaped prism with polarization angle at 45o with respect to face-edge of the glass rhomb (thus the incident light was a 50-50 mix of TE and TM polarization). Note that the transmitted wave actually undergoes two total internal reflections before emerging from rhomb at the exit face, with a 45o relative phase TM-TE phase shift occurring at each total internal reflection. Thus, the first total internal reflection converts a linearly polarized wave into an elliptically polarized wave, the second total internal reflection converts the elliptically polarized wave into a circularly polarized wave!!! The total phase shift (for 2 internal reflections):  tot  2  2  TM   TE    2  90o (for rhomb apex angle A = 54.6o, nair = 1.0 and nglass = 1.5)

NOTE: By time-reversal invariance of the EM interaction, we can also can see from the above that Fresnel’s rhomb can also be used to convert circularly-polarized incident light into linearly polarized light !!!


25


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

The Fresnel relations for TE / TM polarization for internal / external reflection are in fact useful for any type of polarized EM wave – linear polarization with nˆ  cos  xˆ  sin  yˆ , elliptic or circular polarization – these are all vectorial linear combinations of the two orthogonal polarization cases - TE and TM polarization. For each case, the results associated with the TE and TM components of that EM wave, but then have to be combined vectorially. Finally, for the limiting case of normal incidence (where the plane of incidence collapses) the reflection coefficient R (valid for both TE and TM polarization) at inc  0 is:

     

 1  R  1  

n2 n1

2

2

n2 n1

2

 n n    1 2   4% for  n1  n2 



n1 1.0 (air) n2 1.5 (glass)



Can There be a Brewster’s Angle  Binc for Transverse Electric (TE) Polarization Reflection / Refraction at an Interface? The Fresnel Equations: TE Polarization  EoTErefl r   TE  Eo  inc

TM Polarization

  1       1    

 TE 0 r   TErefl  0  inc

           

 cos trans  1v1  2 v2 1n2  2 n1 Z1     with:     and:   v v n n Z2     cos  inc  2 2 1 1 2 1 1 2 

v1  c

n1

 1

11

v2  c

n2

 1

 2 2

n1 

11

 0 0

n2 

 2 2

 0 0

 We saw P436 lecture notes above (p. 19-21) that for TM polarized EM Waves (where B   plane of incidence {i.e. B  to plane of the interface}, with unit normal to the plane of incidence defined as nˆ  kˆ  kˆ ) that when    inc = Brewster’s angle (aka  inc = polarizing angle), inc

TM orefl

that E

inc

refl

inc

B

P

 0 because the numerator of r , ,      0 i.e.    when inc   Binc   Pinc

Thus, for an incident TM polarized monochromatic plane EM wave, when:

26



 

inc  Binc

Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

TM  cos trans  v  v n  n  1 1  2 2  1 2  2 1    TM   cos inc TM 2 v2 1v1 2 n1 1n2

or:   For non-magnetic media:

1n2 1  2 2    11  2 n1   2 1  2  2 n1 2 11 12 1  2 2 1n2

m  1

TM  cos trans   n  2  2 then:  TM  1 n1  cos inc 

i.e. 1  2  0

We also derived the Brewster angle relation for TM polarization: tan  Binc  tan  Pinc 

n2 n1

For the case of TE polarization, we see that: EoTErefl  0 when the numerator of r , 1     0 i.e. when:   1 or:   1  . What does this mean physically??

For TE Polarization:   1  

 cos trans   cos inc   2 1 1    1 2  cos inc   cos trans 

 cos trans For non-magnetic media where:  m  1 i.e. 1  2  o then:   cos inc

  2 n2   n1  1 

2

 n2  cos 2 inc cos 2 inc 1  sin 2 inc   Thus:    2 2 2  n1  cos trans 1  sin trans 1  sin trans From Snell’s Law:

n1 sin 1  n2 sin  2 sin trans

or: n1 sin inc  n2 sin trans 2

n    1  sin inc  n2 

1  sin 2 inc    n2   Then:   2   n1    n1  1    sin 2 inc    n2     2

or: sin trans 2

n    1  sin 2 inc  n2 

2

or:

 n2  2 2    sin inc  1  sin inc  n1 

2



 n2  inc inc    1 or: n1  n2  can get  B   P for TE Polarization only when n1  n2  n1 

However, when n1  n2 , this corresponds to no interface boundary, at least for non-magnetic material(s), where  m  1 and 1  2  0 . Is there a possibility of a Brewster’s angle for incident TE polarization for magnetic materials???


27


Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

For incident TE polarization, we still need to satisfy the condition   1  . B B B B   2 1  cos 2 inc 1  sin 2 inc 1  sin 2 trans  2 1 cos inc    or:   2 2 12 cos trans  12  cos trans 1  sin trans 1   n1  sin 2  B   inc  n2 

i.e.

2

B  n1   11    2 1  1  sin 2 inc   but: thus:        12  1   11  sin 2  B  n2    2 2    inc   2 2        2 1   1 1  2 B B   sin inc  1  sin 2 inc thus:  2 1     12   1 2    2 2 

2

  2 1   1   2  2 B 2 B      sin inc  1  sin inc  multiply both sides of this eqn. by    12    2   1    2   1  2 B  2   2  2 B      sin inc       sin inc  1   2   1   1    2   2    1   2   2 B              sin inc  1   1    2   1      2   2         2 B  sin inc    1   1     1   2           2   1  

Define:

B sin inc

  2   2         1   1   A i.e.  1   2      2   1 

B  sin 1 Brewster’s angle for TE polarization: inc TE

B inc  0o

B inc  90o

B 1 Note: 0  sin 2 inc

   2   2         A  1   1    1   2           2   1  

  2   2      1   1   sin 1 A  1   2      2   1 

Let us assume that 1 and  2 are fixed {i.e. electric properties of medium 1) and 2) are fixed} but that we can engineer/design/manipulate the magnetic properties of medium 1) and 2) in such a way as to obtain a ratio  1 2   1 to give 0 ≤ A ≤ 1!!!

28



Fall Semester, 2015

Lect. Notes 6.5

Prof. Steven Errede

B Then if inc  sin 1 A can be achieved, it might also be possible to engineer the magnetic TE

B properties  1 2  such that A < 0 – i.e. inc becomes imaginary!!!

Note also that in the above formula that  1 2   1 does not mean sin inc   because the original formula for  1 2   1 was:  n2     n1 

2

  n 2  1   1  sin 2 inc   1  sin 2 inc    n2    

which is perfectly mathematically fine/OK for  n2 n1   1 .


29

Lecture Notes 06.5: EM Waves, Wave Propagation in Linear ...

Lecture Notes 06.5: EM Waves, Wave Propagation in Linear ...

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