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4cd943aa
Commit
4cd943aa
authored
Dec 12, 2022
by
Claude Meny
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Update textbook.fr.md
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4cd943aa
...
@@ -33,12 +33,12 @@ important to find some of the results later in this chapter. The
...
@@ -33,12 +33,12 @@ important to find some of the results later in this chapter. The
boundary conditions can be obtained from the Maxwell equations. The
boundary conditions can be obtained from the Maxwell equations. The
first two equations
first two equations
@@@@@@@@@@@@ $
`\quad (equ. 3.4)`
$
équation 3.4
will give information on the normal components of $
`\overrightarrow{D}`
$ and $
`\overrightarrow{B}`
$,
will give information on the normal components of $
`\overrightarrow{D}`
$ and $
`\overrightarrow{B}`
$,
while the third and fourth
while the third and fourth
@@@@@@@@@@@@ $
`\quad (equ. 3.5)`
$
équation 3.5
will give information on the tangential components of $
`\overrightarrow{E}`
$ and $
`\overrightarrow{H}`
$.
will give information on the tangential components of $
`\overrightarrow{E}`
$ and $
`\overrightarrow{H}`
$.
...
@@ -64,7 +64,7 @@ figure [3.1] which extends from one side to the other on
...
@@ -64,7 +64,7 @@ figure [3.1] which extends from one side to the other on
the separation surface. This box has a base surface A and an
the separation surface. This box has a base surface A and an
infinitesimally small thickness $`
\d
elta
`$. We get:
infinitesimally small thickness $`
\d
elta
`$. We get:
@@@@@@@@@@@@@@@@@@@@@@@@@ $`
\q
uad (equ. 3.6)
`$
équation 3.6
If we now let $`
\d
elta
\l
ongrightarrow 0
`$ symmetrically with respect to the separation
If we now let $`
\d
elta
\l
ongrightarrow 0
`$ symmetrically with respect to the separation
surface such that the cylinder gets "pressed" onto the surface:
surface such that the cylinder gets "pressed" onto the surface:
...
@@ -82,20 +82,20 @@ surface such that the cylinder gets "pressed" onto the surface:
...
@@ -82,20 +82,20 @@ surface such that the cylinder gets "pressed" onto the surface:
We obtain:
We obtain:
@@@@@@@@ $`
\q
uad (equ. 3.7)
`$
équation 3.7
Now, considering that $`
d
\o
verrightarrow{a_2}=-d
\o
verrightarrow{a_1}
`$ we can write:
Now, considering that $`
d
\o
verrightarrow{a_2}=-d
\o
verrightarrow{a_1}
`$ we can write:
@@@@@@@@@@@@@ $`
\q
uad (equ. 3.8)
`$
équation 3.8
Finally, as $`
S_1=S'
`$ and $`
d
\o
verrightarrow{a_2}=
\o
verrightarrow{n}_{2
\r
ightarrow 1}
\,
da_1
`$ we can
Finally, as $`
S_1=S'
`$ and $`
d
\o
verrightarrow{a_2}=
\o
verrightarrow{n}_{2
\r
ightarrow 1}
\,
da_1
`$ we can
write:
write:
@@@@@@@@@ $`
\q
uad (equ. 3.9)
`$
équation 3.9
or
or
@@@@@@@@@@@ $`
\q
uad (equ. 3.10)
`$
équation 3.10
The normal component of the vector $`
\o
verrightarrow{D}
`$ is in general discontinuous.
The normal component of the vector $`
\o
verrightarrow{D}
`$ is in general discontinuous.
It is continuos only if there are no conduction charges at the
It is continuos only if there are no conduction charges at the
...
@@ -107,11 +107,11 @@ The situation is identical for the vector $`\overrightarrow{B}`$, the only diffe
...
@@ -107,11 +107,11 @@ The situation is identical for the vector $`\overrightarrow{B}`$, the only diffe
being that the right hand side of the equation is always 0. We
being that the right hand side of the equation is always 0. We
conclude that:
conclude that:
@@@@@@@@@ $`
\q
uad (equ. 3.11)
`$
équation 3.11
or
or
@@@@@@@@@ $`
\q
uad (equ. 3.12)
`$
équation 3.12
The normal component of $`
\o
verrightarrow{B}
`$ is always conserved.
The normal component of $`
\o
verrightarrow{B}
`$ is always conserved.
...
@@ -130,22 +130,22 @@ figure [3.2.]. We chose to integrate the line integral
...
@@ -130,22 +130,22 @@ figure [3.2.]. We chose to integrate the line integral
following the right-hand sense relative to the surface normal
following the right-hand sense relative to the surface normal
$`
\o
verrightarrow{n_a}
`$. By letting $`
\d
elta
\r
ightarrow 0
`$, we get
$`
\o
verrightarrow{n_a}
`$. By letting $`
\d
elta
\r
ightarrow 0
`$, we get
@@@@@@@@@@@ $`
\q
uad (equ. 3.13)
`$
équation 3.13
as the line integral along the sides goes to zero and the flux of the
as the line integral along the sides goes to zero and the flux of the
induction field $`
\o
verrightarrow{B}
`$, which is a finite quantity, approaches 0.
induction field $`
\o
verrightarrow{B}
`$, which is a finite quantity, approaches 0.
Considering that $`
\o
verrightarrow{CD}-
\o
verrightarrow{AB}=d
\v
ec{l}
`$, we get:
Considering that $`
\o
verrightarrow{CD}-
\o
verrightarrow{AB}=d
\v
ec{l}
`$, we get:
@@@@@@@@@@ $`
\q
uad (equ. 3.14)
`$
équation 3.14
or again
or again
@@@@@@@@@@@@ $`
\q
uad (equ. 3.15)
`$
équation 3.15
i.e. the tangential components of the electric field are always
i.e. the tangential components of the electric field are always
conserved at the interface. This condition can also be written:
conserved at the interface. This condition can also be written:
@@@@@@@@@@ $`
\q
uad (equ. 3.16)
`$
équation 3.16
__$`
\o
verrightarrow{H}
`$ vector__
__$`
\o
verrightarrow{H}
`$ vector__
...
@@ -153,7 +153,7 @@ Following the same reasoning as for the $`\overrightarrow{E}`$ vector we write f
...
@@ -153,7 +153,7 @@ Following the same reasoning as for the $`\overrightarrow{E}`$ vector we write f
left hand side when $`
\d
elta
\r
ightarrow 0
`$ :
left hand side when $`
\d
elta
\r
ightarrow 0
`$ :
@@@@@@@@@@@@@ $`
\q
uad (equ. 3.17)
`$
équation 3.17
The right hand side needs more attention. The flux of the vector $`
\o
verrightarrow{D}
`$
The right hand side needs more attention. The flux of the vector $`
\o
verrightarrow{D}
`$
approaches 0. However the flux of the vector $`
\o
verrightarrow{J}
`$ over an infinitesimal surface
approaches 0. However the flux of the vector $`
\o
verrightarrow{J}
`$ over an infinitesimal surface
...
@@ -165,31 +165,31 @@ not $`[A/m*^2]`$ as $`\overrightarrow{J}`$) can. This is typically the case of a
...
@@ -165,31 +165,31 @@ not $`[A/m*^2]`$ as $`\overrightarrow{J}`$) can. This is typically the case of a
perfect conductor where a finite current can flow through a
perfect conductor where a finite current can flow through a
infinitesimally small area. We get:
infinitesimally small area. We get:
@@@@@@@@ $`
\q
uad (equ. 3.18)
`$
équation 3.18
or again
or again
@@@@@@@@ $`
\q
uad (equ. 3.19)
`$
équation 3.19
i.e. the tangential components of the magnetic field $`
\o
verrightarrow{H}
`$ is
i.e. the tangential components of the magnetic field $`
\o
verrightarrow{H}
`$ is
discontinuous unless no surface currents exist. As for the electric
discontinuous unless no surface currents exist. As for the electric
field, this condition can also be written:
field, this condition can also be written:
@@@@@@@@ $`
\q
uad (equ. 3.20)
`$
équation 3.20
!! *Summary*
!! *Summary*
!!
!!
!! *vector form*
!! *vector form*
!! *
@@@@@
!! *
x
!! *
@@@@@
!! *
x
!! *
@@@@@
!! *
x
!! *
@@@@@
!! *
x
!!
!!
!! *component form*
!! *component form*
!! *
@@@@@
!! *
x
!! *
@@@@@
!! *
x
!! *
@@@@@
!! *
x
!! *
@@@@@
!! *
x
! *Remarks*
! *Remarks*
...
@@ -207,7 +207,7 @@ field, this condition can also be written:
...
@@ -207,7 +207,7 @@ field, this condition can also be written:
! of $`
\o
verrightarrow{E}
`$ and $`
\o
verrightarrow{B}
`$ alone using the constitutive relations
! of $`
\o
verrightarrow{E}
`$ and $`
\o
verrightarrow{B}
`$ alone using the constitutive relations
! $`
\o
verrightarrow{D}=
\e
psilon
\,\o
verrightarrow{E}
`$ and $`
\o
verrightarrow{B}=
\m
u
\,\o
verrightarrow{H}
`$.
! $`
\o
verrightarrow{D}=
\e
psilon
\,\o
verrightarrow{E}
`$ and $`
\o
verrightarrow{B}=
\m
u
\,\o
verrightarrow{H}
`$.
&&&&&&&&&&&&&&&&&&&&&&&&&
équation
<br>
<br>
...
@@ -233,7 +233,7 @@ conductor. For the perfect conductor, as *σ* , its penetration depth
...
@@ -233,7 +233,7 @@ conductor. For the perfect conductor, as *σ* , its penetration depth
transmission can occur, the wave is totally reflected. The incident
transmission can occur, the wave is totally reflected. The incident
and reflected waves are:
and reflected waves are:
@@@@@@@@ $`
\q
uad (equ. 3.21)
`$
équation 3.21




...
@@ -243,25 +243,25 @@ _different times._
...
@@ -243,25 +243,25 @@ _different times._
and
and
@@@@@@@@ $`
\q
uad (equ. 3.22)
`$
équation 3.22
Using the boundary conditions at the interface (*z* = 0) we have:
Using the boundary conditions at the interface (*z* = 0) we have:
@@@@@@@@ $`
\q
uad (equ. 3.23)
`$
équation 3.23
as the boundary conditions are valid in any point of the surface and
as the boundary conditions are valid in any point of the surface and
at any time. We cab therefore write the reflected wave as:
at any time. We cab therefore write the reflected wave as:
@@@@@@@@ $`
\q
uad (equ. 3.24)
`$
équation 3.24
In medium 1 we have a superposition of the incident and reflected
In medium 1 we have a superposition of the incident and reflected
waves. The total wave is given by:
waves. The total wave is given by:
@@@@@@@@ $`
\q
uad (equ. 3.26)
`$
équation 3.26
or in real notation
or in real notation
@@@@@@@@@@@
équation
This wave represents a **stationary wave**1: a wave the oscillates in
This wave represents a **stationary wave**1: a wave the oscillates in
place. The nodes and antinodes of the electric field and magnetic
place. The nodes and antinodes of the electric field and magnetic
...
@@ -276,23 +276,23 @@ net energy. This can be readily understood as as much energy is
...
@@ -276,23 +276,23 @@ 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
carried by the incident wave from right to left as the reflected one
in the opposite direction. Calculating the Poynting vector we have:
in the opposite direction. Calculating the Poynting vector we have:
@@@@@@@@ $`
\q
uad (equ. 3.27)
`$
équation 3.27
which at any point *z* changes direction periodically. Its
which at any point *z* changes direction periodically. Its
time-averaged values is indeed *\<* $`
\o
verrightarrow{S}
`$*~tot~ \>~t~*= 0.
time-averaged values is indeed *\<* $`
\o
verrightarrow{S}
`$*~tot~ \>~t~*= 0.
Due to the discontinuity of the magnetic field, a surface current
Due to the discontinuity of the magnetic field, a surface current
density must
density must
be present at the interface. Using the fourth boundary condition we get:




_Figure 3.4: Left: The incident, reflected and resulting wave at a_
_Figure 3.4: Left: The incident, reflected and resulting wave at a_
_particular time. Right: The superposition of several resulting waves_
_particular time. Right: The superposition of several resulting waves_
_at different times. See the video of the simulation on the_
_at different times._
_[moodle_page.](https://moodle.insa-toulouse.fr/course/view.php?id=621)_
_be present at the interface. Using the fourth boundary condition we get:_
@@@@@@@@@

équation
##### Case 2: 2 perfect dielectrics
##### Case 2: 2 perfect dielectrics
...
@@ -301,7 +301,7 @@ will consider that the materials are non-magnetic (*µ* = *µ*~0~) and
...
@@ -301,7 +301,7 @@ will consider that the materials are non-magnetic (*µ* = *µ*~0~) and
that no charges nor currents exist at the interface (*σ~s~* = 0*,*
that no charges nor currents exist at the interface (*σ~s~* = 0*,*
$`
\o
verrightarrow{j}
`$*~s~* = 0). The incident ware is given by
$`
\o
verrightarrow{j}
`$*~s~* = 0). The incident ware is given by
@@@@@@@@ $`
\q
uad (equ. 3.28)
`$
équation 3.28




...
@@ -313,35 +313,35 @@ _$`\overrightarrow{E}`$*~tot~* is no more 0._
...
@@ -313,35 +313,35 @@ _$`\overrightarrow{E}`$*~tot~* is no more 0._
where **k**~1~ and *n*~1~ are the wavevecvtor and refractive index of
where **k**~1~ and *n*~1~ are the wavevecvtor and refractive index of
medium 1. The reflected and transmitted waves are:
medium 1. The reflected and transmitted waves are:
@@@@@@@@ $`
\q
uad (equ. 3.29)
`$
équation 3.29
and
and
@@@@@@@@ $`
\q
uad (equ. 3.30)
`$
équation 3.30
We need now two boundary conditions to determine the reflected and
We need now two boundary conditions to determine the reflected and
trans- mitted waves. Using the tangential boundary conditions for
trans- mitted waves. Using the tangential boundary conditions for
$`
\o
verrightarrow{E}
`$ and $`
\o
verrightarrow{H}
`$ we have:
$`
\o
verrightarrow{E}
`$ and $`
\o
verrightarrow{H}
`$ we have:
@@@@@@@@ $`
\q
uad (equ. 31)
`$
équation 3.31
Solving for *E*^0^ and *E*^0^ we obtain:
Solving for *E*^0^ and *E*^0^ we obtain:
@@@@@@@@@
équation
We can now define the
We can now define the
@@@@@@@@@@@@@
équation
* Reflection coefficient *r* = *^E^*0 = [*n* −*n*]{.underline}
* Reflection coefficient *r* = *^E^*0 = [*n* −*n*]{.underline}
and
and
@@@@@@@@@@@@
équation
* the Transmission coefficient *t* = *^E^*0 = [2*n *]{.underline} .
* the Transmission coefficient *t* = *^E^*0 = [2*n *]{.underline} .
@@@@@@@@@
équation
<br>
<br>
...
@@ -398,7 +398,7 @@ _Figure 3.7: General case of reflection and refraction_
...
@@ -398,7 +398,7 @@ _Figure 3.7: General case of reflection and refraction_
monochromatic wave are:
monochromatic wave are:
@@@@@@@@@@@@@@
équation
The boundary conditions are applied to the interface (*z* = 0) and
The boundary conditions are applied to the interface (*z* = 0) and
must be valid for **all points on the interface** and **for all
must be valid for **all points on the interface** and **for all
...
@@ -407,29 +407,27 @@ times**.
...
@@ -407,29 +407,27 @@ times**.
In *z* = 0, for the tangential component of the electric field we
In *z* = 0, for the tangential component of the electric field we
have:
have:
@@@@@@@@@@@@
équation
@@@@@@@@
équation 3.35
@@@@@@@@ $`
\q
uad (equ. 3.35)
`$
where **r***~s~* is a vector on the interface.
where **r***~s~* is a vector on the interface.
i. For this condition to be valid *t* we must have *ω* = *ω*^t^ =
i. For this condition to be valid *t* we must have *ω* = *ω*^t^ =
*ω*^tt^. As a consequence we have
*ω*^tt^. As a consequence we have
@@@@@@@@ $`
\q
uad (equ. 3.36)
`$
équation 3.36
@@@@@@@@ $`
\q
uad (equ. 3.37)
`$
équation 3.37
ii. For this condition to be valid []{#_bookmark65 .anchor}∀**r***~s~*
ii. For this condition to be valid []{#_bookmark65 .anchor}∀**r***~s~*
we must have:
we must have:
@@@@@@@@ $`
\q
uad (equ. 3.38)
`$
équation 3.38
Taking the first two terms we can write
Taking the first two terms we can write
@@@@@@@@ $`
\q
uad (equ. 3.39)
`$
équation 3.39
As **r***~s~* belongs to the interface, (**k***~i~* **k***~r~*) is
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
normal to it. This means that the vectors **k***~i~*, **k***~r~* and
...
@@ -438,18 +436,18 @@ the normal belong to the same plane, i.e. the plane of incidence.
...
@@ -438,18 +436,18 @@ the normal belong to the same plane, i.e. the plane of incidence.
!! *First law of reflexion*
!! *First law of reflexion*
!! ....
!! ....
iii. Equation [3.38]
(#_bookmark65)
can be recast as
iii. Equation [3.38] can be recast as
@@@@@@@@ $`
\q
uad (equ. 3.40)
`$
équation 3.40
These relations can be satisfied only if all the *x* and *y*
These relations can be satisfied only if all the *x* and *y*
components are the same. Taking the *x* components we can write
components are the same. Taking the *x* components we can write
a.
@@@@@@@@@
a.
équation
b.
@@@@@@@@
b.
équation
c.
@@@@@@@
c.
equation
From the first two relations we get *θ~i~* = *θ~r~* as \|*k~i~*\| =
From the first two relations we get *θ~i~* = *θ~r~* as \|*k~i~*\| =
\|*k~r~*\|.
\|*k~r~*\|.
...
@@ -474,7 +472,7 @@ sin *θ~t~* or
...
@@ -474,7 +472,7 @@ sin *θ~t~* or
From the previous section we can write the incident, reflected and
From the previous section we can write the incident, reflected and
transmitted waves as:
transmitted waves as:
@@@@@@@@@@
equation
where the indices "1" and "2" indicate the first and second medium
where the indices "1" and "2" indicate the first and second medium
physical quantities. We will now separate the discussion for a TE and
physical quantities. We will now separate the discussion for a TE and
...
@@ -496,16 +494,16 @@ From figure [3.8](#_bookmark67) we can write using the two tangential
...
@@ -496,16 +494,16 @@ From figure [3.8](#_bookmark67) we can write using the two tangential
boundary conditions and the relations $`
\o
verrightarrow{B}
`$ = *µH*, \|$`
\o
verrightarrow{B}
`$\| =
boundary conditions and the relations $`
\o
verrightarrow{B}
`$ = *µH*, \|$`
\o
verrightarrow{B}
`$\| =
^[\|$`
\o
verrightarrow{E}
`$\|]{.underline}^ *n*:
^[\|$`
\o
verrightarrow{E}
`$\|]{.underline}^ *n*:
@@@@@@@@
equation
As before, we define the reflection and transmission coefficient for a
As before, we define the reflection and transmission coefficient for a
TE ( or
TE ( or
@@@@@@@@ $`
\q
uad (equ. 3.42)
`$
equation 3.42
which by solving the previous equations can be evaluated to:
which by solving the previous equations can be evaluated to:
@@@@@@@@ $`
\q
uad (equ. 3.43)
`$
equation 3.43
! *Remarks*
! *Remarks*
! ....
! ....
...
@@ -517,12 +515,12 @@ transverse. From figure [3.8](#_bookmark67) we can write using the two
...
@@ -517,12 +515,12 @@ transverse. From figure [3.8](#_bookmark67) we can write using the two
tangential boundary conditions and the relations $`
\o
verrightarrow{B}
`$ = *µH*,
tangential boundary conditions and the relations $`
\o
verrightarrow{B}
`$ = *µH*,
\|$`
\o
verrightarrow{B}
`$\| = ^[\|$`
\o
verrightarrow{E}
`$\|]{.underline}^ *n*:
\|$`
\o
verrightarrow{B}
`$\| = ^[\|$`
\o
verrightarrow{E}
`$\|]{.underline}^ *n*:
@@@@@@@@
equation
Solving the system we obtain the reflection and transmission
Solving the system we obtain the reflection and transmission
coefficients for a TM (/I or *p*) wave as:
coefficients for a TM (/I or *p*) wave as:
@@@@@@@@ $`
\q
uad (equ. 3.44)
`$
equation 3.44
##### Brewster's angle
##### Brewster's angle
...
@@ -534,9 +532,9 @@ the relative value of the two media refrac- tive indices *n*~1~ and
...
@@ -534,9 +532,9 @@ 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~
*n*~2~. By using Snell-Descartes law *n*~1~ sin(*θ*~1~) = *n*~2~
sin(*θ*~2~) we have that
sin(*θ*~2~) we have that
a. If
@@@@@@@
a. If
equation
b. If
@@@@@@@@
b. If
equation
##### Case a)
##### Case a)
...
@@ -545,7 +543,7 @@ In this case, we consider the range 0 *θ*~1~ *π/*2 for the incident
...
@@ -545,7 +543,7 @@ 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
angle *θ*~1~. Correspondingly, the refraction angle *θ*~2~ will vary
in the range 0 *θ*~2~ *θ*2*max*. Using Snell we have:
in the range 0 *θ*~2~ *θ*2*max*. Using Snell we have:
@@@@@@@@@@@
equation




...
@@ -561,7 +559,7 @@ _Figure 3.10: Plot of the reflection_
...
@@ -561,7 +559,7 @@ _Figure 3.10: Plot of the reflection_
_coefficients (*r*~⊥~ and *r*I/) and the correspond- ing reflectivities_
_coefficients (*r*~⊥~ and *r*I/) and the correspond- ing reflectivities_
_(*R*~⊥~ and *R*I/) for case a (top) and case b (bottom)._
_(*R*~⊥~ and *R*I/) for case a (top) and case b (bottom)._
@@@@@@@@@
equation
This signifies that for the particular angle *θ~B~* (the Brewster's
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
angle) *r*I/(*θ~B~*) = 0, as shown in figure [3.10.](#_bookmark68) By
...
@@ -600,11 +598,11 @@ will still hold.
...
@@ -600,11 +598,11 @@ will still hold.
The condition *r~p~* = 0 inserted in the Fresnel equation gives
The condition *r~p~* = 0 inserted in the Fresnel equation gives
@@@@@@@@ $`
\q
uad (equ. 3.45)
`$
equation 3.45
while from Snell's law we have
while from Snell's law we have
@@@@@@@@ $`
\q
uad (equ. 3.46)
`$
equation 3.46
![]()
![]()
_Figure 3.12: Photographs of a window taken_
_Figure 3.12: Photographs of a window taken_
...
@@ -617,7 +615,7 @@ _[https://en.wikipedia.org/wiki/Brewster's_angle](https://en.wikipedia.org/wiki/
...
@@ -617,7 +615,7 @@ _[https://en.wikipedia.org/wiki/Brewster's_angle](https://en.wikipedia.org/wiki/
By multiplying the left and the right terms of the previous equations
By multiplying the left and the right terms of the previous equations
together we get
together we get
@@@@@@@@ $`
\q
uad (equ. 3.47)
`$
equation 3.47
which gives as only possible solution *θ~B~* + *θ*~2~ = *π/*2 ( i.e.
which gives as only possible solution *θ~B~* + *θ*~2~ = *π/*2 ( i.e.
the transmitted and reflected beams are perpendicular, see figure
the transmitted and reflected beams are perpendicular, see figure
...
@@ -630,44 +628,44 @@ always smaller than the limit angle for total reflection.
...
@@ -630,44 +628,44 @@ always smaller than the limit angle for total reflection.
Let's consider the oblique incidence shown in figure
Let's consider the oblique incidence shown in figure
[3.8.](#_bookmark67) The transmitted field is given by
[3.8.](#_bookmark67) The transmitted field is given by
@@@@@@@@ $`
\q
uad (equ. 3.48)
`$
equation 3.48
Using the boundary conditions we have previously found that
Using the boundary conditions we have previously found that
@@@@@@@@ $`
\q
uad (equ. 3.49)
`$
equation 3.49
Using this relation we can write
Using this relation we can write
@@@@@@@@ $`
\q
uad (equ. 3.50)
`$
equation 3.50
When the incidence occurs at the limit angle *θ*~1~ = *θ*1*lim*, Snell
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
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~
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
sin *θ*~1~ or *n*^2^ − *n*^2^ sin2 *θ*~1~ *\<* 0. Plugging this into
equation [3.49,]
(#_bookmark71)
we
equation [3.49,] we
obtain:
obtain:
@@@@@@@@ $`
\q
uad (equ. 3.51)
`$
equation 3.51
i.e. (*k*~2~)*~z~* is purely imaginary and the transmitted wave
i.e. (*k*~2~)*~z~* is purely imaginary and the transmitted wave
becomes:
becomes:
@@@@@@@@ $`
\q
uad (equ. 3.52)
`$
equation 3.52
The wave has a propagating character in the *x* direction and an
The wave has a propagating character in the *x* direction and an
evanescent character in the *z* direction. Let's consider the case of
evanescent character in the *z* direction. Let's consider the case of
a TE wave. The Fresnel relation gives
a TE wave. The Fresnel relation gives
@@@@@@@@ $`
\q
uad (equ. 3.53)
`$
equation 3.53
As (*k*~2~)*~z~* is purely imaginary we have, as expected,
As (*k*~2~)*~z~* is purely imaginary we have, as expected,
@@@@@@@@
equation
but also
but also
@@@@@@@@ $`
\q
uad (equ. 3.55)
`$
equation 3.55
i.e. the reflected wave is totally reflected with a phase shift.
i.e. the reflected wave is totally reflected with a phase shift.
...
@@ -697,14 +695,14 @@ incident and reflected fields.
...
@@ -697,14 +695,14 @@ incident and reflected fields.
Considering the situation depicted in figure [3.13,] the
Considering the situation depicted in figure [3.13,] the
incident wavevector is given by:
incident wavevector is given by:
@@@@@@@@ $`
\q
uad (equ. 3.56)
`$
equation 3.56
Inserting this into the incident fields, and calculating the components of the
Inserting this into the incident fields, and calculating the components of the
$`
\o
verrightarrow{E}
`$ and $`
\o
verrightarrow{B}
`$ fields form the picture (or using
$`
\o
verrightarrow{E}
`$ and $`
\o
verrightarrow{B}
`$ fields form the picture (or using
the plane wave rule $`
\o
verrightarrow{B}_i=
\d
frac{
\o
verrightarrow{k}_i
\t
imes
\o
verrightarrow{E}_i}{
\o
mega}
`$)
the plane wave rule $`
\o
verrightarrow{B}_i=
\d
frac{
\o
verrightarrow{k}_i
\t
imes
\o
verrightarrow{E}_i}{
\o
mega}
`$)
, we obtain :
, we obtain :
@@@@@@@@@@@@
equation
2there is "-" sign in front of the fields as the chosen electric field
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.
drawn in figure [3.13](#_bookmark74) is antiparallel to the *x* axis.
...
@@ -716,42 +714,42 @@ _Figure 3.13 : Configuration des champs e.m pour le mode TE._
...
@@ -716,42 +714,42 @@ _Figure 3.13 : Configuration des champs e.m pour le mode TE._
and
and
@@@@@@@@ $`
\q
uad (equ. 3.58)
`$
equation 3.58
##### Reflected fields
##### Reflected fields
@@@@@@@@@@@@
equation
The reflected electric field **[E]{.underline}***~r~* = *E*~0~ e*i*
The reflected electric field **[E]{.underline}***~r~* = *E*~0~ e*i*
(*ky* cos *θ* + *kz* sin *θ* − *ωt*)**ˆe***~x~* is obtained by
(*ky* cos *θ* + *kz* sin *θ* − *ωt*)**ˆe***~x~* is obtained by
repeating the same procedure. Considering that the incident and
repeating the same procedure. Considering that the incident and
reflected angles are the same, we have:
reflected angles are the same, we have:
@@@@@@@
equation
In addition, applying the boundary condition to the tangential
In addition, applying the boundary condition to the tangential
component of the electric field (which is here the total electric
component of the electric field (which is here the total electric
field) we have *E~r~x* + *E~i~x* = 0. We finally obtain:
field) we have *E~r~x* + *E~i~x* = 0. We finally obtain:
@@@@@@@@ $`
\q
uad (equ. 3.59)
`$
equation 3.59
and
and
@@@@@@@@ $`
\q
uad (equ. 3.60)
`$
equation 3.60
##### Total fields
##### Total fields
@@@@@@@@@@
equation 3.61
We calculate now the total fields **[E]{.underline}**
We calculate now the total fields **[E]{.underline}**
@@@@@@@@@@
equation 3.62
existing in medium 1 for a TE wave:
existing in medium 1 for a TE wave:
@@@@@@@@ $`
\q
uad (equ. 3.61)
`$
equation 3.61
@@@@@@@@ $`
\q
uad (equ. 3.62)
`$
equation 3.62
field is parallel to the interface.
field is parallel to the interface.
...
@@ -780,19 +778,19 @@ _pattern with water waves._
...
@@ -780,19 +778,19 @@ _pattern with water waves._
To this aim it is convenient to revert to the real notation for the
To this aim it is convenient to revert to the real notation for the
fields:
fields:
@@@@@@@@ $`
\q
uad (equ. 3.63)
`$
equation 3.63
and
and
@@@@@@@@ $`
\q
uad (equ. 3.64)
`$
equation 3.64
we obtain:
we obtain:
@@@@@@@@ $`
\q
uad (equ. 3.65)
`$
equation 3.65
while its time average is:
while its time average is:
@@@@@@@@@@ $`
\q
uad (equ. 3.66)
`$
equation 3.66
Power is therfore carried along the positive *z* direction and is null
Power is therfore carried along the positive *z* direction and is null
on the nodal planes.
on the nodal planes.
...
@@ -803,36 +801,36 @@ The discussion is similar to the TE case with the role of $`\overrightarrow{E}`$
...
@@ -803,36 +801,36 @@ The discussion is similar to the TE case with the role of $`\overrightarrow{E}`$
$`
\o
verrightarrow{B}
`$ exchanged. Referring to figure [3.13](#_bookmark74) we have
$`
\o
verrightarrow{B}
`$ exchanged. Referring to figure [3.13](#_bookmark74) we have
again for the wavevectors:
again for the wavevectors:
@@@@@@@ $`
\q
uad (equ. 3.67)
`$
equation 3.67
and
and
@@@@@@@@ $`
\q
uad (equ. 3.68)
`$
equation 3.68
Using the boundary condition for the tangential component of the
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
electric field *E~r~z* + *E~i~z* = 0 and the fact the field experience
total reflection, $`
\o
verrightarrow{E}
`$*~i~* = $`
\o
verrightarrow{E}
`$*~r~* we obtain:
total reflection, $`
\o
verrightarrow{E}
`$*~i~* = $`
\o
verrightarrow{E}
`$*~r~* we obtain:
@@@@@@@@@@@ $`
\q
uad (equ. 3.69)
`$
equation 3.69
and
and
@@@@@@@@@@ $`
\q
uad (equ. 3.70)
`$
equation 3.70
Finally, the total fields existing in medium 1 for a TM ( or *p*) wave
Finally, the total fields existing in medium 1 for a TM ( or *p*) wave
are given by:
are given by:
@@@@@@@@@@@@
equation
and
and
@@@@@@@@@@@@
equation
The same remarks made for the total TE wave can repeated here for the
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 $`
\o
verrightarrow{E}
`$ and
total TM wave we the appropriate changes for the role of the $`
\o
verrightarrow{E}
`$ and
$`
\o
verrightarrow{B}
`
$ fields.
$`
\o
verrightarrow{B}
`
$ fields.
@@@@@@@@@@@@@
equation
, i.e. the magnetic field is
, i.e. the magnetic field is
...
@@ -842,9 +840,9 @@ parallel to the interface.
...
@@ -842,9 +840,9 @@ parallel to the interface.
axis (see figure
[
3.14).
](
#_bookmark75
)
The positions of the nodes
axis (see figure
[
3.14).
](
#_bookmark75
)
The positions of the nodes
and antinodes planes for the magentic fields are given by:
and antinodes planes for the magentic fields are given by:
@@@@@@@@@@
equation
Finally, the time-averaged Poytining vector is given by:
Finally, the time-averaged Poytining vector is given by:
@@@@@@@@
equation
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