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Update textbook.fr.md

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    ......@@ -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**
    @@@@@@@@
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