Chapter 4   Electromagnetic Waves in a Dielectric

• Feynman, Chapter 32/33
• Dobbs, Chapter 11
• Duffin, Chapter 14

4.1   Electromagnetic Waves in HIL Dielectrics

• |E|/|H|=µ/e=Z
  where Z is the impedence of the medium
• <N>=c_m<U>k

Figure 4.1 - Maxwell's Equations at Boundaries

4.2   Boundary Conditions at the Interface of Two Dielectrics

1. Gauss' Law ( Ñ .D=0 ), we use D because e changes across the boundary and so at the boundary the equations are not being applied to a HIL material.
   
   Figure 4.2 - Gauss' Law at a Boundary
   consider a pill box lying in the plane of the boundary, length d and of area A .

   D.dS=0

   if we now let d® 0 then we obtain
   (D_n1-D_n2)A=0
   so
   D_n1=D_n2     (4.5)
   from this we find that the normal component of D is continuous
2. Faraday's Law ( _cE.dl=-¶/¶ tòB.dS )
   
   Figure 4.3 - Faraday's Law at a Boundary
   if we let d® 0 then the right-hand-side also tends to zero and so we obtain
   (E_t1-E_t2)l=0
   and so in general
   E_t1=E_t2     (4.6)
3. No Monopoles ( Ñ×B=0 )
   Þ B_n1=B_n2     (4.7)
4. Ampere-Maxwell Equation
   H_t1=H_t2     (4.8)

4.3   The Laws of Optics (Reflection and Refraction)

Figure 4.4 - Reflected and Refracted Waves

1. Incident Wave

   EI=E

   I0e

   (kI.r-wIt)

   we set
   k_I=(k_Ix,0,k_Iz)
   this means it lies in the x-z plane
2. Reflected Wave

   ER=E

   R0e

   (kR.r-wRt)

3. Transmitted Wave

   ET=E

   T0e

   (kT.r-wTt)

1. apply equation 4.6 with boundary conditions at x=0
   (E_I+E_R)_t=(E_T)_t
   or

   E

   I0e

   [(kI-kT).r-(wI-wT)t]

   +E

   R0e

   [(kR-kT).r-(wR-wT)t]

   =(ET0)t=(constant)

   so
   w_I=w_R=w_I=w     (4.9)
   this shows that there is no frequency change. We also obtain that
   k_I.r=k_R.r=k_T.r  (at x=0)     (4.10)
2. k_I.r=k_R.r at x=0 so that k_Iz=k_Rz . We have that w²=k_R²c²=k_I²c² so we obatin k_R²=k_I² and k_Rx²=k_Ix² . So we conclude k_Rx=± k_Ix . When a minus sign is used is provides the reflection
   k_I=k_Ixx+k_Izz
   k_R=-k_Ixx+k_Izz
   we now define an angle of incidence and reflection (as shown in figure 4.4)

   sinqI=

   kIz kI

   sinqR=

   kRz kR

   hence
   q_I=q_R     (4.11)
   [(angle of incidence)=(angle of relection)]
3. we have that k_I.r=k_T.r and also at the boundary x=0
   k_Iz=k_Tz     (4.12)
   so we write this as
   k_Isinq_I=k_Tsinq_T

   kI2=

   w2 c12

   =

   w2 c22

   c22 c12

   =kT2

   n12 n22

   where c₁ is the speed of light on side one and c₂ is the speed of light on side two. So

   kT=±

   n2 n1

   kI

   we take the positive side and obtain
   n₂sinq_T=n₁sinq_I     (4.13)
   this is known as Snell's Law.

4.4   Fresnel Equations - Incident Electric Field Is Normal to The Plane of Incidence

1. Incident Wave

   EI=E

   I0e

   (kI.r-w t)

2. Reflected Wave

   ER=E

   R0e

   (kR.r-w t)

3. Transmitted Wave

   ET=E

   T0e

   (kT.r-w t)

1. if µ =µ₀ then Z₁/Z₂=n₁/n₂ and so we obtain

   ER0 EI0

   =

   n1cosqI-n2cosqT n1cosqI+n2cosqT

   (4.17)

   ET0 EI0

   =

   2n1cosqI n1cosqI+n2cosqT

   (4.18)

2. if n₁cosq_I<n₂cosq_T then the reflected wave has a p phase shift
3. the phase of the transmitted wave does not change

4.5   Reflection and Transmission

4.6   Total Internal Reflection and the Evanescent Wave

Figure 4.5 - Total Internal Reflection

• on side two, there is a wave
• it penetrates by about |k_Tx|^-1 , for visible light this is about 10³ atoms
• there is no Poynting vector in the x direction (see problem sheet 5)
• the Poynting vector only propagates in the z direction
  
  Figure 4.6 - The Evanescant Wave
  NOTE: The Critical angle is represented by q_C
  
  Figure 4.7 - Distibution of Tranmitted/Reflected Light, Electric Ey (for n₂>n₁ )
  
  Figure 4.8 - Distibution of Tranmitted/Reflected Light, Electric Ey (for n₂<n₁ )

4.7  

4.8  

Figure 4.9 - Distibution of Tranmitted/Reflected Light, Magnetic Hy (for n₂>n₁ )

Figure 4.10 - Distibution of Tranmitted/Reflected Light, Magnetic Hy (for n₂<n₁ )