Update textbook.fr.md

12.temporary_ins/96.electromagnetism-in-media/20.reflexion-refraction-at-interfaces/10.boundary-conditions/10.main/textbook.fr.md

@@ -231,138 +231,70 @@ conductor. For the perfect conductor, as *σ* , its penetration depth
 transmission can occur, the wave is totally reflected. The incident
 and reflected waves are:
 
-*i*
-
-**[B]{.underline}***i* = *B*0 e*i*(*kz*−*ωt*)**ˆe***y* =
->
-*i* e*i*(*kz*−*ωt*)**ˆe***y*
->
-(3.21)
->
-medium 1 perfect dielectric
->
-$`\overrightarrow{E}`$*~i~*
+![]()   
+![]()   
+$`\overrightarrow{E}`$*~r~* $`\overrightarrow{B}`$*~r~*
+medium 2 perfect conductor
 
-$`\overrightarrow{B}`$*~i~* **k***~i~*
+@@@@@@@@ $`\quad (equ. 3.21)`$
 
-**k***~r~*
+![]()   
+_Figure 3.3: Left: Normal incidence at the boundary between a perfect_
+_dielectric and a perfect conductor. Right the total fields at a few_
+_different times._
 
-\", µ
+and
 
-![](media/image172.png)$`\overrightarrow{E}`$*~r~* $`\overrightarrow{B}`$*~r~*
+@@@@@@@@ $`\quad (equ. 3.22)`$
 
-medium 2 perfect conductor
->
-![](media/image174.png)
->
-$`\overrightarrow{j}`$*~s~* O y **n**ˆ
->
-Figure 3.3: Left: Normal incidence at the boundary between a perfect
-dielec- tric and a perfect conductor. Right the total fields at a few
-different times.
+Using the boundary conditions at the interface (*z* = 0) we have:
 
-and
+@@@@@@@@ $`\quad (equ. 3.23)`$
 
-0 *i*(*kz*+*ωt*) *r*
->
-**[B]{.underline}***r* = *B*0 e−*i*(*kz*+*ωt*)**ˆe***y*
->
-(3.22)
->
-Using the boundary conditions at the interface (*z* = 0) we have:
->
-*E~T~* ~1~ − *E~T~* ~1~ = 0 ⇒ $`\overrightarrow{E}`$*~T~* ~1~ = $`\overrightarrow{E}`$*~i~* + $`\overrightarrow{E}`$*~r~*
-= 0 ⇒ *E*^0^ = −*E*^0^
->
-(3.23)
->
 as the boundary conditions are valid in any point of the surface and
 at any time. We cab therefore write the reflected wave as:
->
-*i* 0
->
-**[B]{.underline}***r* = *B*0e−*i*(*kz*+*ωt*)**ˆe***y* = *i y*
->
-(3.24)
->
+
+@@@@@@@@ $`\quad (equ. 3.24)`$
+
 In medium 1 we have a superposition of the incident and reflected
 waves. The total wave is given by:
->
- $`\overrightarrow{E}`$*~tot~* = $`\overrightarrow{E}`$*~i~* + $`\overrightarrow{E}`$*~r~* = *E*^0^ (*e^ikz^* −
-*e*^−*ikz*^l *e*^−*iωt*^ = 2*iE*^0^ sin(*kz*)*e*^−*iωt*^**ˆe***~x~*
->
-(3.25)
 
-*i* *i*
+@@@@@@@@ $`\quad (equ. 3.26)`$
 
 or in real notation
 
-f $`\overrightarrow{E}`$*~tot~* = 2*E*^0^ sin(*kz*) sin(*ωt*)**ˆe***~x~*
-
-*tot*
+@@@@@@@@@@@
 
-*i kz*
->
-*ωt ~y~*
->
 This wave represents a **stationary wave**1: a wave the oscillates in
 place. The nodes and antinodes of the electric field and magnetic
 field are out of phase
->
-1Be careful not to confuse a stationary wave with an evanescent wave.
->
+
+Be careful not to confuse a stationary wave with an evanescent wave.
+
 by *π/*2. In particular the electric field has a node at the interface
 (imposed by the boundary conditions) and the magnetic field as instead
 an antinode at the interface. A stationary wave does not transport any
 net energy. This can be readily understood as as much energy is
 carried by the incident wave from right to left as the reflected one
 in the opposite direction. Calculating the Poynting vector we have:
->
-$`\overrightarrow{S}`$ = [$`\overrightarrow{E}`$*tot **×***]{.underline} [$`\overrightarrow{B}`$*tot*]{.underline} =
-*E*0*B*0 sin(2
->
-) sin(2
->
-)**ˆe**
->
-(3.27)
->
-*tot µ*0
->
-*[i i]{.underline} kz µ*~0~
->
-*ωt ~z~*
->
+
+@@@@@@@@ $`\quad (equ. 3.27)`$
+
 which at any point *z* changes direction periodically. Its
 time-averaged values is indeed *\<* $`\overrightarrow{S}`$*~tot~ \>~t~*= 0.
->
+
 Due to the discontinuity of the magnetic field, a surface current
 density must
 
-![](media/image175.png)![](media/image176.jpeg){width="2.4469674103237096in"
-height="1.3326695100612422in"}E*max* E*min*
-
-Figure 3.4: Left: The incident, reflected and resulting wave at a
-particular time. Right: The superposition of several resulting waves
-at different times. See the video of the simulation on the [moodle
-page.](https://moodle.insa-toulouse.fr/course/view.php?id=621)
->
-be present at the interface. Using the fourth boundary condition we
-get:
->
-**nˆ *× ***($`\overrightarrow{H}`$~1~ − 0) = $`\overrightarrow{j}`$*~s~*
->
-⇒ $`\overrightarrow{j}`$*~s~* = −*n***ˆe***~z~ **×***
->
-[*B~tot~*(*z* = 0)]{.underline}**ˆe** =
->
-*µ*0 *y*
->
-2*E*0 cos( )**ˆe**
->
-*µ c*
+![]()   
+![]()   
+_Figure 3.4: Left: The incident, reflected and resulting wave at a_
+_particular time. Right: The superposition of several resulting waves_
+_at different times. See the video of the simulation on the_
+_[moodle_page.](https://moodle.insa-toulouse.fr/course/view.php?id=621)_
+_be present at the interface. Using the fourth boundary condition we get:_
 
-0
+@@@@@@@@@
 
 ##### Case 2: 2 perfect dielectrics
 
@@ -370,87 +302,48 @@ We are dealing now with two perfect dielectrics. For the discussion we
 will consider that the materials are non-magnetic (*µ* = *µ*~0~) and
 that no charges nor currents exist at the interface (*σ~s~* = 0*,*
 $`\overrightarrow{j}`$*~s~* = 0). The incident ware is given by
->
-f **[E]{.underline}***i* = *E*0 e*i*(*k*~1~*z*−*ωt*)**ˆe***x*
->
-*E*^0^*n*
->
-(3.28)
->
-**[B]{.underline}***i* = *B*0 e*i*(*k*~1~*z*−*ωt*)**ˆe***y* = *i* 1
-e*i*(*k*1*z*−*ωt*)**ˆe***y*
->
-![](media/image177.png) ![](media/image180.png)
->
-Figure 3.5: Left: Normal incidence at the boundary between twp perfect
-dielectrics. Right the total fields in medium 1 at a few different
-times showing a partial standing wave. The minimum amplitude of
-$`\overrightarrow{E}`$*~tot~* is no more 0.
->
-where **k**~1~ and *n*~1~ are the wavevecvtor and refractive index of
-medium 1. The reflected and transmitted waves are:
->
-**[E]{.underline}***r* = *E*0 e−*i*(*k*~1~*z*+*ωt*)**ˆe***x*
-
-0
 
-(3.29)
->
-**[B]{.underline}***r* = *B*0 e−*i*(*k*~1~*z*+*ωt*)**ˆe***y* = − *Er
-n*1
->
-e−*i*(*k*~1~*zωt*)**ˆe***y*
+@@@@@@@@ $`\quad (equ. 3.28)`$
 
-and
+![]() 
+![]()   
+_Figure 3.5: Left: Normal incidence at the boundary between twp perfect_
+_dielectrics. Right the total fields in medium 1 at a few different_
+_times showing a partial standing wave. The minimum amplitude of_
+_$`\overrightarrow{E}`$*~tot~* is no more 0._
 
-*r c*
->
-**[E]{.underline}***t* = *E*0 e*i*(*k*~2~*z*−*ωt*)**ˆe***x*
+where **k**~1~ and *n*~1~ are the wavevecvtor and refractive index of
+medium 1. The reflected and transmitted waves are:
 
-0
+@@@@@@@@ $`\quad (equ. 3.29)`$
 
-(3.30)
+and
 
-**[B]{.underline}** = *B*0 e*i*(*k*~2~*z*−*ωt*)**ˆe***y* = *Et n*2
-e*i*(*k*~2~*z*−*ωt*)**ˆe***y*
+@@@@@@@@ $`\quad (equ. 3.30)`$
 
-*t t c*
->
 We need now two boundary conditions to determine the reflected and
 trans- mitted waves. Using the tangential boundary conditions for
 $`\overrightarrow{E}`$ and $`\overrightarrow{H}`$ we have:
 
-f *E*^0^ + *E*^0^ = *E*^0^
-
-(3.31)
-
-*n*~1~(*Ei* − *Er* ) = *n*~2~*Et*
+@@@@@@@@ $`\quad (equ. 31)`$
 
 Solving for *E*^0^ and *E*^0^ we obtain:
->
-*r t*
-
- *E*0 = *n*1−*n*2 *E*0
 
- *E*0 = 2*n*1 *E*0
+@@@@@@@@@
 
 We can now define the
->
-*t n*~1~+*n*~2~ *i*
 
--   Reflection coefficient *r* = *^E^*0 = [*n* −*n*]{.underline}
+@@@@@@@@@@@@@
+
+*   Reflection coefficient *r* = *^E^*0 = [*n* −*n*]{.underline}
 
 and
->
-*r*0 1 2
 
-*E n*~1~+*n*~2~
+@@@@@@@@@@@@
 
--   the Transmission coefficient *t* = *^E^*0 = [2*n *]{.underline} .
+*   the Transmission coefficient *t* = *^E^*0 = [2*n *]{.underline} .
 
-0 *n* 1
->
-*E* 1+*n*2
+@@@@@@@@@
 
 #### chap2 Reflection and transmission at oblique incidence
 
@@ -458,8 +351,8 @@ We now turn to the more general case of an oblique incidence at an
 arbitrary angle *θ~i~*. Before that we will need a few definitions and
 considerations.
 
-![](media/image182.png)Figure 3.6: Plane of incidence, *s* and *p*
-polarisations.
+![](media/image182.png)    
+_Figure 3.6: Plane of incidence, *s* and *p* polarisations._
 
 ##### chap5 Plane of incidence
 
@@ -469,13 +362,13 @@ the two materials. Let's consider a linearly polarised plane wave
 making an oblique incidence at the interface between medium 1 and 2.
 Two special cases arise:
 
--   if $`\overrightarrow{E}`$*~i~* is contained in the plane of incidence, the wave is
+*  if $`\overrightarrow{E}`$*~i~* is contained in the plane of incidence, the wave is
     said to be *p* (for parallel to the plane of incidence) or again
     Transverse Magnetic (TM) as the magnetic field will be
     perpendicular to the plane of incidence. The symbol "/I" is also
     often used.
 
--   if $`\overrightarrow{E}`$*~i~* is perpendicular to the plane of incidence, the wave
+*  if $`\overrightarrow{E}`$*~i~* is perpendicular to the plane of incidence, the wave
     is said to be *s* (for the german word for perpendicular - to the
     plane of incidence) or Transverse Electric (TE). The symbol "⊥" is
     used in this case.
@@ -494,51 +387,45 @@ wavevectors's moduli in the same medium. We will
 >
 demonstrate it here. Let's consider the the general situation as of an
 inci- dent, reflected and refracted wave as depicted in figure
-[3.7.](#_bookmark64) The three plane
+[3.7.]. The three plane
 
-![](media/image183.png)
+![]()   
+_Figure 3.7: General case of reflection and refraction_
 
-Figure 3.7: []{#_bookmark64 .anchor}General case of reflection and
-refraction
->
 monochromatic wave are:
->
-**[E]{.underline}***i* = $`\overrightarrow{E}`$0*ei*(**k***~i~*·**r**−*ωt*)
->
-**[E]{.underline}** = $`\overrightarrow{E}`$0*ei*(**k***r* ·**r**−*ω*^l^*t*)
-**[E]{.underline}** = $`\overrightarrow{E}`$0*ei*(**k***t*·**r**−*ω*^ll^*t*)
->
+
+@@@@@@@@@@@@@@
+
 The boundary conditions are applied to the interface (*z* = 0) and
 must be valid for **all points on the interface** and **for all
 times**.
->
+
 In *z* = 0, for the tangential component of the electric field we
 have:
->
-0 *ei*(**k***~i~*·**r***~s~*−*ωt*) + $`\overrightarrow{E}`$0
->
-*ei*(**k***r* ·**r***s*−*ω*^l^*t*) = $`\overrightarrow{E}`$0
->
-*ei*(**k***t*·**r***s*−*ω*^ll^*t*) (3.35)
->
+
+@@@@@@@@@@@@
+
+@@@@@@@@
+
+@@@@@@@@ $`\quad (equ. 3.35)`$
+
 where **r***~s~* is a vector on the interface.
 
 i.  For this condition to be valid *t* we must have *ω* = *ω*^t^ =
     *ω*^tt^. As a consequence we have
 
-\|**k***~i~*\| = \|**k***~r~*\| = *[ω]{.underline} n*~1~ (3.36)
+@@@@@@@@ $`\quad (equ. 3.36)`$
 
-\|**k***~t~*\| = *[ω]{.underline} n*~2~ (3.37)
+@@@@@@@@ $`\quad (equ. 3.37)`$
 
 ii. For this condition to be valid []{#_bookmark65 .anchor}∀**r***~s~*
     we must have:
 
-**k***~i~* · **r***~s~* = **k***~r~* · **r***~s~* = **k***~t~* ·
-**r***~s~* (3.38)
->
+@@@@@@@@ $`\quad (equ. 3.38)`$
+
 Taking the first two terms we can write
 
-(**k***~i~* − **k***~r~*) · **r***~s~* = 0*.* (3.39)
+@@@@@@@@ $`\quad (equ. 3.39)`$
 
 As **r***~s~* belongs to the interface, (**k***~i~* **k***~r~*) is
 normal to it. This means that the vectors **k***~i~*, **k***~r~* and
@@ -546,21 +433,21 @@ the normal belong to the same plane, i.e. the plane of incidence.
 
 iii. Equation [3.38](#_bookmark65) can be recast as
 
-*x*(*k~i~*)*~x~* + *y*(*k~i~*)*~y~* = *x*(*k~r~*)*~x~* +
-*y*(*k~r~*)*~y~* = *x*(*k~t~*)*~x~* + *y*(*k~t~*)*~y~* ∀*x, y* (3.40)
->
+@@@@@@@@ $`\quad (equ. 3.40)`$
+
 These relations can be satisfied only if all the *x* and *y*
 components are the same. Taking the *x* components we can write
 
-a.  (*k~i~*)*~x~* = \|*k~i~*\| sin *θ~i~*
+a.  @@@@@@@@@
 
-b.  (*k~r~*)*~x~* = \|*k~r~*\| sin *θ~r~*
+b.  @@@@@@@@
 
-c.  (*k~t~*)*~x~* = \|*k~t~*\| sin *θ~t~*
+c.  @@@@@@@
 
 From the first two relations we get *θ~i~* = *θ~r~* as \|*k~i~*\| =
 \|*k~r~*\|.
->
+
+
 From the first and the third we get \|*k~i~*\| sin *θ~i~* = \|*k~t~*\|
 sin *θ~t~* or
 
@@ -568,12 +455,9 @@ sin *θ~t~* or
 
 From the previous section we can write the incident, reflected and
 transmitted waves as:
->
-**[E]{.underline}***i* = $`\overrightarrow{E}`$0*ei*\[**k**~1~(*x* sin *θ*~1~+*z* cos
-*θ*~1~)−*ωt*\] **[E]{.underline}***r* = $`\overrightarrow{E}`$0*ei*\[**k**~1~(*x* sin
-*θ*~1~−*z* cos *θ*~1~)−*ωt*\] **[E]{.underline}***t* =
-$`\overrightarrow{E}`$0*ei*\[**k**~2~(*x* sin *θ*~2~+*z* cos *θ*~2~)−*ωt*\]
->
+
+@@@@@@@@@@
+
 where the indices "1" and "2" indicate the first and second medium
 physical quantities. We will now separate the discussion for a TE and
 a TM wave. Due to the isotropic nature of the media considered here, a
@@ -583,9 +467,7 @@ Likewise for a TM incident wave.
 ##### chap3 TE Wave TM Wave
 
 ![](media/image184.png) ![](media/image187.png)    
->
-Figure 3.8: []{#_bookmark67 .anchor}Configuration for a TE and TM
-incidence.
+_Figure 3.8: Configuration for a TE and TM incidence._
 
 __**chap5 TE Wave**__
 
@@ -593,37 +475,17 @@ $`\overrightarrow{E}`$ is tangential and $`\overrightarrow{B}`$ is contained in
 From figure [3.8](#_bookmark67) we can write using the two tangential
 boundary conditions and the relations $`\overrightarrow{B}`$ = *µH*, \|$`\overrightarrow{B}`$\| =
 ^[\|$`\overrightarrow{E}`$\|]{.underline}^ *n*:
->
- *n*1 (*E*^0^ − *E*^0^) cos *θ*~1~ = *n*2 *E*^0^ cos *θ*~2~
->
-As before, we define the reflection and transmission coefficient for a
-TE ( or
-
-s)  wave as:
 
-*E*0 *E*0
+@@@@@@@@
 
-*r*~⊥~ = *^[r]{.underline}^ , t*~⊥~ = *^[t]{.underline}^ ,* (3.42)
->
-0 0
+As before, we define the reflection and transmission coefficient for a
+TE ( or
 
-*i* *i*
+@@@@@@@@ $`\quad (equ. 3.42)`$
 
 which by solving the previous equations can be evaluated to:
->
- *n*1 cos *θ*~1~ − *n*2 cos *θ*~2~
->
-cos *θ*~1~ + cos *θ*~2~ *µ*~1~ *µ*~2~
->
-[2*n*1]{.underline} cos *θ*~1~
->
-*µ*~1~
 
-*.* (3.43)
-
-*t*⊥ = *n*~1~ *n*~2~
->
-cos *θ*~1~ + cos *θ*~2~ *µ*~1~ *µ*~2~
+@@@@@@@@ $`\quad (equ. 3.43)`$
 
 __**chap5 TM Wave**__
 
@@ -632,29 +494,15 @@ transverse. From figure [3.8](#_bookmark67) we can write using the two
 tangential boundary conditions and the relations $`\overrightarrow{B}`$ = *µH*,
 \|$`\overrightarrow{B}`$\| = ^[\|$`\overrightarrow{E}`$\|]{.underline}^ *n*:
 
- *n*1 (*E*^0^ − *E*^0^) = *n*2 *E*^0^ *.*
+@@@@@@@@
 
 Solving the system we obtain the reflection and transmission
 coefficients for a TM (/I or *p*) wave as:
 
- *n*1 cos *θ*~2~ − *n*2 cos *θ*~1~
-
-cos *θ*~1~ + cos *θ*~2~ *µ*~2~ *µ*~1~
->
-[2*n*1]{.underline} cos *θ*~1~
->
-*µ*~1~
-
-*.* (3.44)
-
-*t*I/ = *n*~2~ *n*~1~
-
-
+@@@@@@@@ $`\quad (equ. 3.44)`$
 
 chap4 Brewster's angle
 
-cos *θ*~1~ + cos *θ*~2~ *µ*~2~ *µ*~1~
->
 Let's make a few considerations on the consequences of the Fresnel
 relations by plotting the reflection coefficients for *s* and *p*
 polarisations. We, first of all, separate the two cases according to
@@ -662,9 +510,9 @@ the relative value of the two media refrac- tive indices *n*~1~ and
 *n*~2~. By using Snell-Descartes law *n*~1~ sin(*θ*~1~) = *n*~2~
 sin(*θ*~2~) we have that
 
-a.  If *n*~1~ *\< n*~2~ −→ *θ*~1~ *\θ*~2~
+a.  If @@@@@@@
 
-b.  If *n*~1~ *\n*~2~ −→ *θ*~1~ *\< θ*~2~
+b.  If @@@@@@@@
 
 chap5 Case a)
 
@@ -672,218 +520,124 @@ In this case, we consider the range 0 *θ*~1~ *π/*2 for the incident
 angle *θ*~1~. Correspondingly, the refraction angle *θ*~2~ will vary
 in the range 0 *θ*~2~ *θ*2*max*. Using Snell we have:
 
-i.  For *θ*~1~ = 0 → *θ*~2~ = 0, =⇒ *r*~⊥~ = *r*I/ =
-    [*n*1−*n*2]{.underline} *\<* 0.
-
-n~1~ \< n~2~ n~1~ \n~2~
-
-![](media/image190.png)
-
-Figure 3.9: The snell law for *n*~1~ *\< n*~2~ and *n*~1~ *\n*~2~.
->
-1.0
->
-n~1~ = 1
->
-1.0
->
-0.8
->
-0.5 n~2~ = 1.5 ✓*B*
->
-0.0
->
-0.6
->
-0.4
->
--0.5
->
--1.0
->
-r*~?~*
->
-r*~k~*
-
-0 20 40
+@@@@@@@@@@@
 
-60 80
->
-0.2
->
-0.0
-
-0 20 40
+![]()   
+_Figure 3.9: The snell law for *n*~1~ *\< n*~2~ and *n*~1~ *\n*~2~._
 
-60 80
->
 Angle of incidence (deg) Angle of incidence (deg)
 
-1.0
-
-0.5
-
-0.0
-
--0.5
-
--1.0
-
-0 20
-
-40 60 80
->
-1.0
->
-0.8
->
-0.6
->
-0.4
->
-0.2
->
-0.0
-
-0 20
 
-40 60 80
->
-Angle of incidence (deg) Angle of incidence (deg)
->
-Figure 3.10: []{#_bookmark68 .anchor}Plot of the reflection
-coefficients (*r*~⊥~ and *r*I/) and the correspond- ing reflectivities
-(*R*~⊥~ and *R*I/) for case a (top) and case b (bottom).
+_Figure 3.10: Plot of the reflection_
+_coefficients (*r*~⊥~ and *r*I/) and the correspond- ing reflectivities_
+_(*R*~⊥~ and *R*I/) for case a (top) and case b (bottom)._
 
-ii. For *θ*~1~ = *π/*2 → *θ*~2~ = *θ*2*max* =⇒ *r*~⊥~ = −1 and *r*I/ =
-    +1
+@@@@@@@@@
 
 This signifies that for the particular angle *θ~B~* (the Brewster's
 angle) *r*I/(*θ~B~*) = 0, as shown in figure [3.10.](#_bookmark68) By
-plotting the reflectance (*R* = \|*r*\|2), we see that
->
-no reflection occurs for *p*-polarised wave for *θ*~1~ = *θ~B~* and
+plotting the reflectance (*R* = \|*r*\|2), we see that no reflection 
+occurs for *p*-polarised wave for *θ*~1~ = *θ~B~* and
 the reflection for *p* waves is, in any case, much smaller than the
 reflections for *s* waves for angles close to *θ~B~*. This phenomenon
 suggests different applications:
 
--   The possibility of polarising unpolarised waves by reflection for an
+*  The possibility of polarising unpolarised waves by reflection for an
     inci- dence angle *θ* = *θ~B~*: the reflected wave will be 100%
     linearly polarised while the transmitted one only partially
     polarised (see figure [3.11).](#_bookmark69)
 
--   The possibility of removing unwanted reflections by using polarisers
+*  The possibility of removing unwanted reflections by using polarisers
     (Po- laroid sun-glasses for instance, figure
     [3.12);](#_bookmark70)
 
--   The possibility of transmitting 100% of the intensity of an incident
+*  The possibility of transmitting 100% of the intensity of an incident
     wave thought an interface using a proper polarisation and
     incidence angle.
 
-![](media/image191.png){width="2.991770559930009in"
-height="2.5861450131233594in"}
-
-Figure 3.11: []{#_bookmark69 .anchor}An illustration of the
-polarisation of light that is incident on an interface at Brewster's
-angle.
-[https://en.wikipedia.org/wiki/Brewster'](https://en.wikipedia.org/wiki/Brewster%27s_angle)
-[s_angle](https://en.wikipedia.org/wiki/Brewster%27s_angle)
+![](media/image191.png)    
+_Figure 3.11: An illustration of the_
+_polarisation of light that is incident on an interface at Brewster's_
+_angle._
+_[https://en.wikipedia.org/wiki/Brewster'](https://en.wikipedia.org/wiki/Brewster%27s_angle)_
+_[s_angle](https://en.wikipedia.org/wiki/Brewster%27s_angle)_
 
 chap5 Case b)
 
 In this case, total reflections will occur for angles *θ*~1~
 *θ*1*lim*. The considera- tions on the Brewster's angle stated above
 will still hold.
->
+
 The condition *r~p~* = 0 inserted in the Fresnel equation gives
->
-*n*~1~ cos *θ*~2~ = *n*~2~ cos *θ~B~* (3.45) while from Snell's law we
-have
->
-*n*~2~ sin *θ*~2~ = *n*~1~ sin *θ~B~.* (3.46)
->
-![](media/image4.jpeg){width="3.0679155730533685in"
-height="0.9943744531933508in"}
->
-Figure 3.12: []{#_bookmark70 .anchor}Photographs of a window taken
-with a camera polariser filter rotated to two different angles. In the
-picture at left, the polariser is aligned with the polarisation angle
-of the window reflection. In the picture at right, the polariser has
-been rotated 90° eliminating the heavily polarised reflected sunlight.
-[https://en.wikipedia.org/wiki/Brewster's_angle](https://en.wikipedia.org/wiki/Brewster%27s_angle)
->
+
+@@@@@@@@ $`\quad (equ. 3.45)`$
+
+while from Snell's law we have
+ 
+@@@@@@@@ $`\quad (equ. 3.46)`$
+
+![]()    
+_Figure 3.12: Photographs of a window taken_
+_with a camera polariser filter rotated to two different angles. In the_
+_picture at left, the polariser is aligned with the polarisation angle_
+_of the window reflection. In the picture at right, the polariser has_
+_been rotated 90° eliminating the heavily polarised reflected sunlight._
+_[https://en.wikipedia.org/wiki/Brewster's_angle](https://en.wikipedia.org/wiki/Brewster%27s_angle)_
+
 By multiplying the left and the right terms of the previous equations
 together we get
->
-sin 2*θ~B~* = sin 2*θ*~2~ (3.47)
->
+
+@@@@@@@@ $`\quad (equ. 3.47)`$
+
 which gives as only possible solution *θ~B~* + *θ*~2~ = *π/*2 ( i.e.
 the transmitted and reflected beams are perpendicular, see figure
-[3.11)](#_bookmark69) and *θ~B~* = arctan [*n*2]{.underline} . It is
->
-easy to show that *θ~B~ \< θ*1*lim*, i.e. the Brewster's angle is
+[3.11)](#_bookmark69) and *θ~B~* = arctan [*n*2]{.underline} . It is easy 
+to show that *θ~B~ \< θ*1*lim*, i.e. the Brewster's angle is
 always smaller than the limit angle for total reflection.
 
 chap4 Total internal reflection
 
 Let's consider the oblique incidence shown in figure
 [3.8.](#_bookmark67) The transmitted field is given by
->
-$`\overrightarrow{E}`$*t* = $`\overrightarrow{E}`$0*ei*\[(*k*~2~)*~x~x*+(*k*~2~)*~z~z*−*ωt*\] (3.48)
->
+
+@@@@@@@@ $`\quad (equ. 3.48)`$
+
 Using the boundary conditions we have previously found that
->
-[]{#_bookmark71 .anchor}( 2 *ω*2 2
 
-*k*~2~)*~x~* = (*k*~1~)*~x~* = *k*~1~ sin *θ*~1~ and *k*2 = *n*2*.*
-(3.49)
+@@@@@@@@ $`\quad (equ. 3.49)`$
 
-*c*2
->
 Using this relation we can write
 
-(*k*~2~)2 = *k*^2^ − (*k*~2~)2 = *[ω]{.underline}* (*n*^2^ − *n*^2^ sin2
-*θ*~1~) *.* (3.50)
+@@@@@@@@ $`\quad (equ. 3.50)`$
 
 When the incidence occurs at the limit angle *θ*~1~ = *θ*1*lim*, Snell
 law gives *n*~2~ = *n*~1~ sin *θ*1*lim*. If now we increase the
 incidence angle, we have *θ*~1~ *\θ*1*lim* and thus *n*~2~ *\< n*~1~
 sin *θ*~1~ or *n*^2^ − *n*^2^ sin2 *θ*~1~ *\<* 0. Plugging this into
 equation [3.49,](#_bookmark71) we
->
+
 obtain:
->
-2 1
 
-(*k*~2~)
+@@@@@@@@ $`\quad (equ. 3.51)`$
 
-= *i[ω]{.underline}* j*n*^2^ sin2 *θ*~1~ − *n*^2^
->
-(3.51)
->
 i.e. (*k*~2~)*~z~* is purely imaginary and the transmitted wave
 becomes:
->
-$`\overrightarrow{E}`$*t* = $`\overrightarrow{E}`$0*ei*\[(*k*~2~)*~x~x*−*ωt*\]*e*−(*k*~2~)*~z~z* (3.52)
->
+
+@@@@@@@@ $`\quad (equ. 3.52)`$
+
 The wave has a propagating character in the *x* direction and an
 evanescent character in the *z* direction. Let's consider the case of
 a TE wave. The Fresnel relation gives
->
-*r* = [*n*~1~ cos *θ*~1~ − *n*~2~ cos *θ*~2~]{.underline} =
-[(*k*~1~)*~z~* − (*k*~2~)*~z~*]{.underline}
->
-(3.53)
  
-⊥ *n*~1~ cos *θ*~1~ − *n*~2~ cos *θ*~2~ (*k*~1~)*~z~* + (*k*~2~)*~z~*
+@@@@@@@@ $`\quad (equ. 3.53)`$
 
 As (*k*~2~)*~z~* is purely imaginary we have, as expected,
->
-\|*r*~⊥~\| = 1 (3.54)
->
+
+@@@@@@@@
+
 but also
 
-*r*~⊥~ = \|*r*~⊥~\|*e^iφ^* = *e^iφ^* (3.55)
+@@@@@@@@ $`\quad (equ. 3.55)`$
 
 i.e. the reflected wave is totally reflected with a phase shift.
 
@@ -905,244 +659,101 @@ chap5 TE Wave Incident fields
 Considering the situation depicted in figure [3.13,](#_bookmark74) the
 incident wavevector is given by:
 
-0
-
-1
+@@@@@@@@ $`\quad (equ. 3.56)`$
 
-**k***~i~* = −*k* cos *θ*
-
-*,* (3.56)
->
-1 *k* sin *θ*
->
-$`\overrightarrow{E}`$ and $`\overrightarrow{B}`$ fields form the picture (or using the plane wave rule
-**[B]{.underline}***~i~*
->
-we obtain[2](#_bookmark73):
->
-= [**k***~i~ **×***]{.underline} $`\overrightarrow{E}`$*~i~* ),
+we obtain :
 
-*ω*
+@@@@@@@@@@@@ 
  
-**[E]{.underline}***i* = []{#_bookmark73 .anchor}−*E*0 e*i* (−*ky* cos
-*θ* + *kz* sin *θ* − *ωt*)**ˆe***x* (3.57)
->
 2there is "-" sign in front of the fields as the chosen electric field
 drawn in figure [3.13](#_bookmark74) is antiparallel to the *x* axis.
 
 chap3 TE Wave TM Wave
 
-![](media/image192.png)![](media/image194.png)
+![]()
+![]()   
+_Figure 3.13 : Configuration des champs e.m pour le mode TE._
 
-Figure 3.13: []{#_bookmark74 .anchor}Configuration des champs ´e.m
-pour le mode TE.
->
 and
->
-$`\overrightarrow{B}`$ = 1
->
-*i* 1
->
-0
->
-− *nE*0 sin *θ* e*i* (−*ky* cos *θ* + *kz* sin *θ* − *ωt*)
->
-*,* (3.58)
+
+@@@@@@@@ $`\quad (equ. 3.58)`$
 
 chap5 Reflected fields
 
-1 − *nE*0 cos *θ* e*i* (−*ky* cos *θ* + *kz* sin *θ* − *ωt*)
->
+@@@@@@@@@@@@
+
 The reflected electric field **[E]{.underline}***~r~* = *E*~0~ e*i*
 (*ky* cos *θ* + *kz* sin *θ* − *ωt*)**ˆe***~x~* is obtained by
 repeating the same procedure. Considering that the incident and
 reflected angles are the same, we have:
 
-0
+@@@@@@@
 
-1
-
-**k***~r~* = *k* cos *θ ,*
-
-1
-
-1 *k* sin *θ*
->
 In addition, applying the boundary condition to the tangential
 component of the electric field (which is here the total electric
 field) we have *E~r~x* + *E~i~x* = 0. We finally obtain:
 
-**[E]{.underline}***r* = *E*0e*i* (*ky* cos *θ* + *kz* sin *θ* −
-*ωt*)**ˆe***x* (3.59)
+@@@@@@@@ $`\quad (equ. 3.59)`$
 
 and
 
-$`\overrightarrow{B}`$ = 1
->
-*r* 1
->
-0
->
-*nE*0 sin *θ* e*i* (*ky* cos *θ* + *kz* sin *θ ωt*) *c*
->
-*.* (3.60)
+@@@@@@@@ $`\quad (equ. 3.60)`$
 
 chap5 Total fields
 
-1 − *nE*0 cos *θ* e*i* (*ky* cos *θ* + *kz* sin *θ* − *ωt*)
->
-We calculate now the total fields **[E]{.underline}**
->
-= **[E]{.underline}** + **[E]{.underline}** , **[B]{.underline}** =
-**[B]{.underline}** + **[B]{.underline}**
->
-existing in
->
-medium 1 for a TE wave:
->
-⊥ *i r* ⊥ *i r*
-
-**[E]{.underline}** = *E*0 (e*i* (*ky* cos *θ* + *kz* sin *θ* − *ωt*) −
-e*i* (−*ky* cos *θ* + *kz* sin *θ* − *ωt*)l **ˆe***x* =
+@@@@@@@@@@
 
-= 2*iE*~0~ sin (*ky* cos *θ*) e*i* (*kz* sin *θ* − *ωt*)
->
-**ˆe***~x~*
->
-*.* (3.61)
-
-\'- amp..li,.tude
+We calculate now the total fields **[E]{.underline}**
 
-., \'-
->
-propa..g,.ation
->
-., pola\'-r..is,.a.,tion
->
-The total magnetic field is given by:
->
-0
->
-1
+@@@@@@@@@@
 
-$`\overrightarrow{B}`$ = 1
+existing in medium 1 for a TE wave:
 
-2*inE*0 sin *θ* sin (*ky* cos *θ*) e*i* (*kz* sin *θ* − *ωt*)
->
-*.* (3.62)
+@@@@@@@@ $`\quad (equ. 3.61)`$
 
-**-- [B]{.underline}**~⊥~
+@@@@@@@@ $`\quad (equ. 3.62)`$
 
-= [2*nE*~0~]{.underline} *\_u*
->
-*c*
->
-cos *θ* e*i* (*kz* sin *θ* − *ωt*), i.e. the magnetic
->
 field is parallel to the interface.
 
 -   The fields have a standing wave character along the *y* axis (see
-    figure [3.14).](#_bookmark75) The positions of the nodes and
+    figure [3.14] The positions of the nodes and
     antinodes plane for the electric field are given by:
 
-    -   $`\overrightarrow{E}`$ = **0** for *y~n~*
+@@@@@@@@@@@@@
 
-*nπ k* cos *θ*
+![]()
 
-, *n* integer
+![]()
 
--   **[E]{.underline}** \| is maximum for *y~p~*
 
-= [(2*p* + 1)*π*]{.underline} , *p* integer.
->
-*k θ*
->
-![](media/image196.jpeg)same *z* components
->
-![](media/image198.png) propagating wave // *z*
->
-antiparallel *y* components
->
- standing wave // *y*
->
-Figure 3.14: []{#_bookmark75 .anchor}The total electric field
-amplitude of a TE wave upon oblique incident onto a perfect conductor.
-The field has a propagation character along the *z* axis and a
-standing wave character along the *y* axis. A node and an antinode
-plane are displayed. The blue and red colours represent the minimum
-and maximum of the field amplitude. You can easily reproduce this
-pattern with water waves.
+
+_Figure 3.14: The total electric field_
+_amplitude of a TE wave upon oblique incident onto a perfect conductor._
+_The field has a propagation character along the *z* axis and a_
+_standing wave character along the *y* axis. A node and an antinode_
+_plane are displayed. The blue and red colours represent the minimum_
+_and maximum of the field amplitude. You can easily reproduce this_
+_pattern with water waves._
 
 chap5 Poynting vector
 
 To this aim it is convenient to revert to the real notation for the
 fields:
 
-$`\overrightarrow{E}`$~⊥~ = −2*E*~0~ sin (*ky* cos *θ*) sin (*kz* sin *θ* − *ωt*)
-**ˆe***~x~,* (3.63)
+@@@@@@@@ $`\quad (equ. 3.63)`$
 
 and
 
-$`\overrightarrow{B}`$ = 1
->
-⊥ 1
->
-0
->
-−2 *c* sin *θ* sin (*ky* cos *θ*) sin (*kz* sin *θ* − *ωt*)
->
-*.* (3.64)
->
-1 −2 *c* cos *θ* cos (*ky* cos *θ*) cos (*kz* sin *θ* − *ωt*)
->
-Calculating $`\overrightarrow{S}`$ = [$`\overrightarrow{E}`$⊥ ***×***]{.underline} [$`\overrightarrow{B}`$⊥]{.underline}
-
-*µ*~0~
-
-1
+@@@@@@@@ $`\quad (equ. 3.64)`$
 
 we obtain:
 
-0
-
-$`\overrightarrow{S}`$ = − *µ c θ*
->
-*ky θ*
->
-*kz θ* − *ωt*
->
-*.* (3.65)
-
-*nE*^2^
-
-4 sin *θ µ*~0~*c*
->
-sin2
->
-(*ky*
->
-cos
-
-*θ*) sin2
-
-(*kz*
->
-sin
-
-*θ* − *ωt*)
+@@@@@@@@ $`\quad (equ. 3.65)`$
 
 while its time average is:
 
-($`\overrightarrow{S}`$)*~t~* =
+@@@@@@@@@@ $`\quad (equ. 3.66)`$
 
-2
->
-[0]{.underline} sin *θ* sin2
->
-*µ*~0~*c*
->
-(*ky* cos *θ*) **ˆe***~z~* (3.66)
->
 Power is therfore carried along the positive *z* direction and is null
 on the nodal planes.
 
@@ -1151,90 +762,37 @@ chap5 TM wave
 The discussion is similar to the TE case with the role of $`\overrightarrow{E}`$ and
 $`\overrightarrow{B}`$ exchanged. Referring to figure [3.13](#_bookmark74) we have
 again for the wavevectors:
->
-0 0
->
-1 1
->
-**k***~i~* = −*k* cos *θ*
-
-1 *k* sin *θ*
-
-*,* **k***~r~* = *k* cos *θ ,*
->
-1 *k* sin *θ*
 
-$`\overrightarrow{E}`$*~i~* = 1
-
-1
-
-0
->
-*E*0 sin *θ* e*i* (−*ky* cos *θ* + *kz* sin *θ* − *ωt*)
->
-*.* (3.67)
+@@@@@@@ $`\quad (equ. 3.67)`$
 
 and
 
-1 *E*0 cos *θ* e*i* (−*ky* cos *θ* + *kz* sin *θ* − *ωt*)
->
-**[B]{.underline}** = − [*nE*0]{.underline} e*i* (−*ky* cos *θ* + *kz*
-sin *θ* − *ωt*)**ˆe**
-
-*,* (3.68)
+@@@@@@@@ $`\quad (equ. 3.68)`$
 
 Using the boundary condition for the tangential component of the
 electric field *E~r~z* + *E~i~z* = 0 and the fact the field experience
 total reflection, $`\overrightarrow{E}`$*~i~* = $`\overrightarrow{E}`$*~r~* we obtain:
 
-$`\overrightarrow{E}`$*~r~* = 1
-
-1
+@@@@@@@@@@@ $`\quad (equ. 3.69)`$
 
-0
->
-*E*0 sin *θ* e*i* (*ky* cos *θ* + *kz* sin *θ* − *ωt*)
->
-*.* (3.69)
->
 and
->
-1 −*E*0 cos *θ* e*i* (*ky* cos *θ* + *kz* sin *θ* − *ωt*)
->
-**[B]{.underline}** = − [*nE*0]{.underline} e*i* (*ky* cos *θ* + *kz*
-sin *θ* − *ωt*)**ˆe**
 
-*.* (3.70)
+@@@@@@@@@@ $`\quad (equ. 3.70)`$ 
 
 Finally, the total fields existing in medium 1 for a TM ( or *p*) wave
 are given by:
->
-$`\overrightarrow{E}`$ = 1
->
-I/ 1
->
-0
->
-2*E*~0~ sin *θ* cos (*ky* cos *θ*) e*i* (*kz* sin *θ* − *ωt*)
->
-*.* (3.71)
->
+
+@@@@@@@@@@@@
+
 and
->
-1 −2*iE*~0~ cos *θ* sin (*ky* cos *θ*) e*i* (*kz* sin *θ* − *ωt*)
 
-**[B]{.underline}**~I/~
+@@@@@@@@@@@@
 
-= −2 [*nE*0]{.underline} cos (*ky* cos *θ*) e*i* (*kz* sin *θ* −
-*ωt*)**ˆe** *.* (3.72)
->
 The same remarks made for the total TE wave can repeated here for the
 total TM wave we the appropriate changes for the role of the $`\overrightarrow{E}`$ and
 $`\overrightarrow{B}`$ fields.
 
-**-- [B]{.underline}**~I/~
-
-= − [2*nE*0]{.underline} e*i* (*kz* sin *θ* − *ωt*)**ˆe**
+@@@@@@@@@@@@@
 
 , i.e. the magnetic field is
 
@@ -1244,33 +802,9 @@ parallel to the interface.
     axis (see figure [3.14).](#_bookmark75) The positions of the nodes
     and antinodes planes for the magentic fields are given by:
 
-**-- [B]{.underline}**~I/~
-
-= **0** for *y~n~*
->
-= [(2*n* + 1)*π*]{.underline} , *n* integer.
->
-*k θ*
->
-**--** \|**[B]{.underline}**~I/~\| is maximum for *y~p~*
->
-= *[pπ]{.underline}* , *p* integer.
->
-*k θ*
+@@@@@@@@@@
 
 Finally, the time-averaged Poytining vector is given by:
 
-2*E*^2^*n*
-
-[0]{.underline} 2
-
-($`\overrightarrow{S}`$)*~t~* =
-
-sin *θ* cos
->
-*µ*~0~*c*
->
-(*ky* cos *θ*) **ˆe***~z~* (3.73)
->
-[]{#_bookmark76 .anchor}**Chapter 4**
+@@@@@@@@