1 Traveling Standing Wave

1 Traveling Standing Wave

E(r, t) =~ E_osin(βy)e^−iβze^iωtx.ˆ Relationship betweenβ and λ: ω =kc, wherek=q

k²_x+k_y²+k_z²,λ= ^2π_k . kx = 0,ky =β,kz =β, therefore k=p

β²+β²+ 0 =√ 2β λ= ^2π_k = √^2π

2β = √ 2π

β

From Faraday,∇ ×~ E~ =−^∂B_∂t,B~ =µ_oH~ so, _µ¹

o

∇ ×~ E~ = ⁻_∂t^{∂ ~H} =iω ~H. Therefore, H~ = i

µ_oω∇ ×~ E.~ As

∇ ×~ E~ =



 xˆ yˆ zˆ

∂_x ∂_y ∂_z E_x E_y E_z





∇ ×~ E~ = ˆx₍₍₍₍(∂yEz−⁽⁽⁽∂zEy⁽) + ˆy(∂zEx−∂xEz) + ˆz(∂xEy −∂yEx)

= ˆy∂_zE_x+ ˆz∂_yE_x=iβyEˆ _x−zβˆ cos(βy)e^iβz. Thus,

H~ = Eoiβ

µ_oω [iˆysin (βy)−zˆcos (βy)]e^i(βz−ωt). Since ^ω_c =k=√

2β, thenβ= √^ω 2c

β

µ_oω = 1 µ_oω

ω c√ 2

=

√_oµ_o

√2µ_o = 1

√2 r_o

µ_o. Z_o=q

µo

o =376 Ω, and this is called Vacuum impedance.

H~ = i

√2 E_o

Zo[isin(βyy)ˆ −cos(βy)ˆz]e^i(βz−ωt), H~^∗ = −i

√2 E_o

Z_o[isin(βy)ˆy−cos(βy)ˆz]e^i(βz−ωt). The power flow is directed along the Poynting vectorS, where~

S~ = 1 2Ren

E~ ×H~^∗o

(time average power flow). Since,

E~ ×H~ =



 xˆ yˆ zˆ E_x E_y E_z H_x H_y H_z





= ˆx(EyHz −EzHy) + ˆy(EzHx −ExHz) + ˆz(ExHy−EyHx) E~ ×H~^∗ =−yEˆ xH_z^∗+ ˆzExH_y^∗

=−E_osinβy iE_o

√2Z_o cos(βy)ˆy+E_osin(βy) −E_o

√2Z_osin(βy)ˆz E~ ×H~^∗ = E_o²

√2Z_osin(βy) (−icosβyyˆ−sin(βy)ˆz) Ren

E~ ×H~^∗o

=− E_o²

√2Zo

sin²(βy)ˆz S~ =− 1

2√ 2

E_o²

Z sin²(βy)ˆz.

Traveling standing wave as the superposition of two plane waves:

sin(βy) = e^iβy−e^−iβy 2i , E~ = E_oxˆ

2i h

e^iβy+iβz−e^iβz−iβyi . For the polarization direction:

E~ ⇥H~ = 0

@ ˆ

x yˆ zˆ Ex Ey Ez

H_x H_y H_z 1 A

= ˆx(⇠⇠⇠E_yH_z ⇠⇠⇠E_zH_y) + ˆy(⇠⇠⇠E_zH_x E_xH_z) + ˆz(E_xH_y ⇠⇠⇠E_yH_x) E~ ⇥H~^⇤ = yEˆ _xH_z^⇤+ ˆzE_xH_y^⇤

= E_osin y iEo

p2Z_o cos( y)ˆy+E_osin( y) Eo

p2Z_osin( y)ˆz E~ ⇥H~^⇤ = E_o²

p2Z_osin( y) ( icos yyˆ sin( y)ˆz) Ren

E~ ⇥H~^⇤o

= E_o²

p2Z_o sin²( y)ˆz S~ = 1

2p 2

E_o²

Z sin²( y)ˆz.

sin( y) = exp^{i y} exp ^{i y}

2i ,

E~ = E_oxˆ 2i

hexp^{i y+i z} exp^{i z i y}i .

P~ = ✏0 E~ a~r ·E ✏~ 0 E~ r ·~ E~ r ·~ E~ = 0 P~ =✏₀ E~ Figure 1: Polarization direction

2 Dielectric Media

• P~ = 0χ ~E −a~∇ ·E ~ 0χ ~E is linear with the EF vector. ∇ ·~ E~ must be zero, because the medium is isotropic, i.e. ∇ ·~ E~ = 0 Thus, the equation, P~ =₀χ ~E, represents linear medium.

• P~+a ~P² =₀E ~~ P stands for a linear medium, whereasP~²represents a non-linear medium.

Hence, this equation describes a non-linear medium.

• a1∂²P~

∂t² +a2∂ ~P

∂t +P~ =0χ ~EUpon Fourier transformation in the frequency domain, we can write the above equation as−a₁ω²P~ +iωa₂P~ +P~ =₀χ(ω)E~ and obtain P~ =₀χ(ω)E.~ Therefore, the equation represents a linear dispersive medium with a frequency-dependent susceptibilityχ(ω) = 1

√1 +iωa2−ω²a1

.

• P =₀[a₁+a₂ e^−(x2+y2)]E~ This equation describes a linear spatial dispersive medium. P~ is linearly dependent onE, but the term~ 0[a1+a2e^−(x2+y2)]E~ depends also on the spatial coordinatesxandy. This is the case, for example, of a spatial dependent refractive index.

3 EM Waves Polarizations

=_oR =





2.25 0 0 0 2.13 0 0 0 2.02





From the lecture_R= 1 +χ. In this case

R=1+χ, where

1=





1 0 0 0 1 0 0 0 1





is the identity matrix

χ=





1.25 0 0 0 1.13 0 0 0 1.02



 susceptibility tensor (Diagonal)

P~ =_oχ ~E.

We consider a linearly polarized plane wave withλ=600 nm.

(1) The wave is polarized alongxˆ and propagates alongz.ˆ E~ =E_oe^i(kz−ωt)xˆ P~ =_oχ ~E=_o



χxx

χ_yx χ_zx



E_oe^i(kz−ωt) P~ =_oχ_xxE_oe^i(kz−ωt)xˆ

For this arrangement, _R = 1 +χ_xx, _R=2.25. Therefore, the refractive index is n = √_R = 1.5. Then, the wavelength is: λ= ^λ_n^o = ⁶⁰⁰_1.5 w 400 nm.

(2) The wave is polarized alongyˆand propagates along z.ˆ E~ =E_oe^i(kz−ωt)yˆ P~ =_oχ_yyE_oe^i(kz−ωt)yˆ _R= 1 +χyy,n=√_R =1.46,λ= _1.46⁶⁰⁰ w411 nm.

(3) The wave is polarized alongyˆand propagates along x.ˆ E~ =E_oe^i(ky−ωt)xˆ P~ =_oχ_xxE_oe^i(ky−ωt)xˆ _R= 1 +χxx, like the first case.

(4) The wave is polarized alongzˆand propagates alongy.ˆ E~ =E_oe^i(ky−ωt)zˆ P~ =_oχ_zzE_oe^i(ky−ωt)z,ˆ _R= 1 +χ_zz,n=√_R = 1.42,λ= _1.42⁶⁰⁰ w423 nm.

What if the wave is polarized 45^◦ with respect to x?ˆ We assume propagation alongzˆ

✏_R = 1 + _xx ✏_R n =p✏_R =

= _n^o = ⁶⁰⁰_1.5 w ˆ

y zˆ

E~ =Eoe^{i(kz !t)}yˆ P~ =✏_{o yy}E_oe^{i(kz !t)}yˆ

✏_R= 1 + _yy n=p✏_R = _1.46⁶⁰⁰ w ˆ

y xˆ

E~ =Eoe^{i(ky !t)}xˆ P~ =✏_{o xx}E_oe^{i(ky !t)}xˆ

✏_R= 1 + _xx

ˆ

z yˆ

E~ =E_oe^{i(ky !t)}zˆ P~ =✏_{o zz}E_oe^{i(ky !t)}z,ˆ

✏_R= 1 + _zz n=p✏_R = _1.42⁶⁰⁰ w

ˆ x ˆ

z

Figure 2: Polarized wave at 45^◦ alongz-axis

E~ = (xˆ+ ˆy

√2 )e^i(kz−ωt)E_o cos 45^◦= sin 45^◦= 1

√2 P~ =_oχ ~E =_oE_o 1

√2



χxx

χ_yy 0



e^i(kz−ωt)

_R=1+



χxx 0 0 0 χ_yy 0 0 0 χ_zz





Decompose the wave into two components:

ˆ

x: (_R)_xx = 1 +χ_xx= 2.25 ˆ

y : (_R)_yy = 1 +χ_yy = 2.13.

Accordingly, we have two refractive indices: n_x w1.5 andn_y w1.46.

E~ = (xˆ+ ˆy

p2 ) exp^{i(kz !t)}E_o cos 45 = sin 45 = 1

p2 P~ =✏_o E~ =✏_oE_o 1

p2 0

@ ^xx_yy 0

1

Aexp^{i(kz !t)}

✏_R= + 0

@ ^xx

0 0

0 _yy 0

0 0 zz

1 A

ˆ

x: (✏_R)_xx= 1 + _xx= 2.25 ˆ

y: (✏R)yy = 1 + yy = 2.13.

n_xw n_y w

Figure 3: Polarization rotation

Exercises selected fromFundamentals of Photonics, chapter 5, by B.E.A Saleh and M.C. Teich.

andPhotonic Devices, chapter 1, by Jia-ming Liu.