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s 8o/ can see from fig/re +.", the plane perpendic/lar to the vector ' is seen from its
side appearing as a line P(W. The dot prod/ct nB r is the proGection of the radial vector ralong the normal to the plane and will have the constant val/e O( for all points on the
plane. The e=/ation ' ) r constant is the characteristic propert8 of a plane
perpendic/lar to the direction of propagation '.
The e=/iphase e=/ation is
B r C B
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p C P <
power densit8 flow
%efinition:
transverse electromagnetic wave TEM2 > electromagnetic wave having electricfield vectors and magnetic field vectors perpendic/lar to the direction of propagation.
, is perpendic/lar to -, and both - and , are perpendic/lar to the direction of
propagation '. The e
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Find the magnetic field strength.
3olution:
The direction of propagation nB is > aA. The vector amplit/de of the magnetic field is then
given b8
Η Ε
m
x y z
x y
n a a a
a a A m= = − = −
∧β
η η
"6
$
6
5
"
6
"
5**5 $
*note
η µ
ε = "$6J5**K ppendi< D > Table D."2
-/#(P0- 64
The phasor electric field e
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ence, a /nit vector in the direction of propagation nB is given b8
nB C (6.+a
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5. The wavelength 5 is given b8
λ π
β
π = = =
$ $
$ 5 $ *5
. . m
and the fre=/enc8
f = = ∗
=c
GHz λ
5 "6
$ *5 6 ""
7
. .
. The e=/ation of the s/rface of constant phase is
nB r C (6.+< @ 6.78 C constant
The general e
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The fig/re shows an incident wave polariAed with the E field in the plane of incidence
and the power flow in the direction of β i at angle θ i with respect to the normal to the
s/rface of the perfect cond/ctor.
The direction of propagation is given b8 the Po8nting vector and the β i , -8 and , fields
need to be arranged so that β i is in the same direction as Ε Η i i∧ at an8 time. The
magnetic field is o/t of the plane of the paper,Η Η = y ya
for the direction of the electricfield shown. There is no transmitted field within the perfect cond/ctorO however there
will be a reflected field with power flow at the angle θ r with respect to the normal to the
interface. To maintain the power densit8 flowΕ Η r r ∧ will be in the same direction β r as. The e
8/16/2019 Capítulo 6 de libro de Teoría electromagnética
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( ) , cos cosΕ Ε Ε x mi
i j i r m
r j r r r x z e e= +
− ⋅ − ⋅θ θ β β
( ) , sin sinΕ Ε Ε z mi
i j i r m
r j r r r x z e e= − +
− ⋅ − ⋅θ θ β β
cos cosΕ Ε Ε x at z mi
i j i r m
r j r r r e e=
− ⋅ − ⋅= − =6
6θ θ β β
= − =− ⋅ − ⋅ cos cossin sinΕ Ε m
ii
j xmr j x r
r ie eθ θ
β β θ θ 6 +.72
E=/ation +.7 shows the relationship between the incident and reflected amplit/des for a
perfect cond/ctor the total tangential - field at the s/rface m/st be Aero which satisfies
the bo/ndar8 condition. To be Aero at all val/es of < along the s/rface of the cond/cting plane, the phase terms m/st be e=/al to each other >
θ θ i r = +.#2
E=/ation +.# is known as %nell0s law of reflection.
%efinition:%nell0s !aw is a r/le of Ph8sics that applies to visible light passing from air or
vac//m2 to some medi/m with an inde< of refraction different from air.
%/bstit/te e=/ation +.# into e=/ation +.7 >
Ε Ε mi
mr = +."62
Therefore, the total electric field in free space is
( ) ( ) , 2 , ,Ε Ε Ε x z x z a x z a x x z z = +
( )= −− − cos sin cos cosΕ mi i j x i j z i j z i xe e e aθ β θ β θ β θ
( )− +− − sin sin cos cosΕ mi i j x i j z i j z ie e e z aθ β θ β θ β θ
= −
$ j z e ami
i j x i
i x cos sin cos 2
sinΕ θ β θ
β θ +.""2
( )− −$ j z e ami
i i j x i z
sin cos cos sinΕ θ β θ β θ
'otes b8: Debbie Prestridge 7
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[= −$ cos sin cos 2Ε mi i i x j z aθ β θ
( ) ]− −sin cos cos sini i z j x i z a eθ β θ β θ
Take e=/ation +."" and recover the time(domain form of the total electric field
( ) ( )( )Ε Ε r t r e j t , 9e = ϖ
1bserve the variation of the total field with the + variable indicating there is a traveling
wave in the + direction with a phase constant
β β θ x i= sin
nd in the & direction the field forms a standing wave.
The total magnetic field is
( ) ( ) ( ) ( ) , , , ,Η Η Η Η x z x z a x z a x z a y y yi
y yr
y= = +
se the relation Η Ε
=∧nβ
η for each of the incident and reflected fields to emplo8 the
e
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( )= − +
sin cosΕ mi
j i x i z ye aη
β θ θ
The reflected magnetic fields is given b8
( ) sin cosΗ Ε r mi
j i x i z ye a= − −η β θ θ
The total magnetic field Η ∧
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Normal Incident:
i x aveθ = =6 6, ,Ρ Power flow in the + direction is Aero2
verage power flow perpendic/lar to the cond/cting s/rface is Aero, beca/se the average
Po8nting Lector is Aero in that direction
( ) z ave x y P , 9e = =∗"
$ 6Ε Η
Wh8Q &eca/se Ε x is m/ltiplied b8 , therefore Ε Η x yand are o/t of phase b8 ;
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The plane of the Aero Ε x field occ/r at m/ltiples of λ $ along the direction of propagation, and the8 are located at integer m/ltiples of λ z $ along the A(a
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where ( )β β θ θ i i ir x z ⋅ = −sin cos . ss/me that the reflected field is also in the $ directionso the magnetic field m/st be perpendic/lar to both - and the Po8nting Lector P - > ,,
Ε Ε r mr j r
ye ar = − ⋅β
( )
cos sinΗ
Ε Ε r r
mr
r x r z j
r r
n
a a er
=
∧
= +
∧
− ⋅
β β
η η θ θ
Where ( )β β θ θ r r r r x z ⋅ = −sin cos . Determine the angle of reflection θ r and the
amplit/de of the reflected electric field Ε mr b8 /sing the bo/ndar8 conditions at A C 6.
This also incl/des Aero val/es of the tangential electrical field - and the normal
component of the magnetic field ,.
( ) , Ε Ε Ε y yi
yr
x z = + = 6 at A C 6
Therefore,
( ) , sin sinΕ Ε Ε y mi j x i m
r j x r x e e6 6= + =− −β θ β θ
nd
'otes b8: Debbie Prestridge
β i
Ε yr β r
θ i θ r
1
a&
a+
Perfect
;ond/ctor
Ε yi
Η r
7i"ure 6?
"5
iΗ
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( ) , sin sinsin sinΗ Ε Ε z mi
i j x i m
r i
j x r x e e6 " "
6= + =− −η
θ η
θ β θ β θ
'ote: These two conditions will provide the same res/lts for the /nknowns r mr
and θ Ε
, and be tr/e for ever8 val/e of + along &
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There is a standing(wave in the & direction beca/se the reflected and incident waves
travel in the opposite direction along the A(a
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The sheet c/rrent J in ampere per meter2 is determined b8 the total tangential
magnetic field at the s/rface. From the bo/ndar8 condition,
J n= ∧ Η
where the normal n to the s/rface for the geometr8 of Fig/re +.) is n C (aA. The magneticfield in this case has two components:
( )
cos cos cos sinΗ Ε
xmi
i i j x i z e=
− −$η
θ β θ β θ
( )
sin sin cos sinΗ Ε
z mi
i i j x i
j z e=
− −$η
θ β θ β θ
The s/rface is then c/rrent is then
cos sin J a a eat z
z y m
i
i j x i
= =
−∧ =
∧
−
6
$Η
Ε
η θ β θ
nd the peak val/e of the s/rface c/rrent at A C 6 is given b8
cos 2 cos
. J x A m peakvaluemi
i= = = −$ $ "6 M)
5**5 *) "6 $
Ε η
θ
-/#(P0-:
The electric field associated with a plane wave propagating in an arbitrar8 direction isgiven b8
. . 2 . . 2Ε = + − − +* 75 M M ) * 6 ) 6 7* x y z j x z a a a e
If this incident on a perfectl8 cond/cting plane oriented perpendic/lar to the A a
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$. Total electric field in region in front of the perfect cond/ctor.
5. Total magnetic field.
3olution
&eca/se a vector in the direction of propagation and a /nit vector normal to the reflectings/rface are contained in the
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The magnit/de of the incident electric field Ε mi is therefore C *.75N6.7* C # or .)N6.) C
#. ence, the electric field associated with the parallel polariAation case can be e
8/16/2019 Capítulo 6 de libro de Teoría electromagnética
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= − +
− +M M * 6 ) 6 7*η
θ η
θ cos sin . . 2i x i z j x z a a e
&eca/se, for the case,
Ε Ε⊥ ⊥=
r i
cos sin 2 . . 2Η⊥ − += − +r x z
j x za a eM
56 56 * 6 ) 6 7*η
The total reflected magnetic field is then
cos sin 2 . . 2Η r x y z j x z a a a e= − − + − − +"
M 56 # M 56 * 6 ) 6 7*
η
6@ Reflection and Refraction at Plane Interface betAeen TAo (edia:
Oblique Incidence
Fig/re +.* shows two media with electrical properties ε "and µ " in medi/m ", and ε $
and µ $ in medi/m $. ere a plane wave incident angleθ i on a bo/ndar8 between the
two media will be partiall8 transmitted into and partiall8 reflected at the dielectrics/rface. The transmitted wave is reflected into the second medi/m, so its direction of
propagation is different from the incidence wave. The fig/re also shows two ra8s for
each the incident, reflected, and transmitted waves. ra8 is a line drawn normal to thee=/iphase s/rfaces, and the line is along the direction of propagation.
'otes b8: Debbie Prestridge "#
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7i"ure 6B
The incident ra8 $ travels the distance ;&, while on the contrar8 the reflected ra8 "
travels the distance E. For both ; and &E to be the incident and reflected wave frontsor planes of e=/iphase, the incident wave sho/ld take the same time to cover the distance
E. The reason being that the incident and reflected wave ra8s are located in the same
medi/m, therefore their velocities will be e=/al,
CBV
AE V " $
=
19
AB ABi r sin sinθ θ =
With this being the case then it follows that
θ θ i r =
What is the relationship between the angles of incidence θ i and refractionθ r Q
It takes the incident ra8 the e=/al amo/nt of time to cover distance ;& as it takes the
refracted ra8 to cover distance D >
CB
V
A
V " $=
'otes b8: Debbie Prestridge
!
4
9eflected
ra8s
!
4
ε µ $ $,
ε µ " ",
Incident
ra8s
θ i
#
θ t
θ r
C -
$6
9eflectedra8s
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nd the magnit/de of the velocit8 L" in medi/m " is:
"" "
"V =
∗ µ ε
nd in medi/m $:
$
$ $
"V =
∗ µ ε
lso,
CB AB
A AB
i
i
==
sin
sin
θ
θ
Therefore,
CB
A
V
V
i
t
= = = ∗
∗
sin
sin
θ
θ
µ ε
ε µ "
$
$ $
" "
For most dielectrics µ µ µ $ "= =
Therefore,sin
sin
i
t
θ
θ
ε
ε µ µ µ
== =
$
"" $
+."$2
E=/ation +."$ is known as %nell0s !aw of 9efraction.
6@! Parallel Polari&ation Case - is in Plane of Incidence
'otes b8: Debbie Prestridge $"
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The /nknown amplit/des of the reflected and transmitted electric fields RR RRr t and Ε Ε can be determined b8 simpl8 appl8ing the bo/ndar8 conditions at the dielectric interface.
The electric fields RR RRr t and Ε Ε will now be /sed in the anal8sis to emphasiAe the case of
parallel polariAation, instead of /sing the electric fields m mt r and Ε Ε .
The tangential component of , sho/ld be contin/o/s across the bo/ndar8. Therefore,
RRΗ i j i
r ye a
− ⋅ +β RRΗ r j i r ye a
− ⋅ =β RRΗ t j i r ye a
− ⋅β
There is no need to carr8 the a$ vector, beca/se the magnetic fields onl8 have onecomponent in the $ direction. 9ecall that this relation is valid at A C 6,
RR
sin 2Η i j i i xe− +β θ RR sin 2Η r j
ii r x
e− =β θ RRsin 2Η t j i t xe− β θ +."52
β " β $" are the magnit/des of β in regions " $, respectivel8. In order for this to be
valid at an8 val/e of + at an8 point on the interface, and knowingθ θ i r = :
β θ β θ " $sin sini t =
1r
'otes b8: Debbie Prestridge
β i
β r
θ i θ r
19egion $
R Rr Η
1/t of
paper2
θ t
β t
RRt
Ε
RRi
Ε
9egion "
Η R Ri
Η RRt
$$
2
/
RRr
Ε
7i"ure 6;
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sin
sini
t
V V
V
V
θ
θ
β
β
ω
ω = = =
$
"
$
"
"
$
S This is the same relation that was determined earlier from %nell0s !aw. %/bstit/tesin
sin
i
t
V
V
θ
θ = "
$ into e=/ation +."5 to obtain
RRΗ i
+ RRΗ r
= RRΗ t t A C 6 +."2
E and are related b8 η , so e=/ation +." can be rewritten as
RRΕ i + RRΕ
r =
"
$
η
η
RRΕ t
+.")2
Tangential components of E m/st be contin/o/s across the bo/ndar8, therefore
cosRRΕ i
iθ − cosRRΕ r
r θ = cosRRΕ t
t θ t A C 6 +."+2
S9emember the e
RRΕ r
=
cos cos
cos cos
RR
T
Ε i i t
i t
" $
" $
η θ η θ
η θ η θ
−
+nd
RRΕ t =
+
cos
cos cosRR TΕ i i
i t
$
" $
$η θ
η θ η θ +."*2
SMaking /se of the fact thatθ θ i r = . Define the reflection coefficient RRΓ and the
transmission RRΤ :
RR
RR
RR
Γ Ε
Ε =
r
i= −
+ =
−
+
= =
$ "
$ "
$
"
$
"" $
η θ η θ η θ η θ
θ ε
ε
θ
θ ε
ε θ
µ µ µ
cos coscos cos
cos cos
cos cos
t i
t i
t i
t i
nd
'otes b8: Debbie Prestridge $5
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RR
RR
RR
ΤΕ
Ε = =
t
i
$ $ $"
$ " $
"" $
η θ η θ
η θ η θ
θ
θ ε
ε θ
µ µ µ
cos cos
cos cos
cos
cos cos
t t
t t
i
i i
−
+=
+
= =
The total electric field in region " is
RRΕ t!t
=
RRΕ i
+ RRΕ r
= cos sin 2Ε mi
i x i z j i r a a eθ θ β − − ⋅ @ cos sin 2Ε m
r r x r z
j r r a a e− − − ⋅θ θ β
= ∧
−cos sini mi j x ieθ β θ Ε
cos− j z ie β θ +
∧
RR 2cosΓ j z ie a x
β θ
+ −
−
sin sini mi j x ie
Travelin" #ave part
θ β θ Ε
( )− +−e e a j z j z z S din" plu$travelin"#ave$
i iβ θ β θ cos cos
tan
RRΓ
+."72
%/bstit/ted β β i r r r ⋅ ⋅, from e
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The transmitted fields in medi/m $ are
( ) cos sinRRΕ Ε i
mt
t x t z j r
a a e t = − − ⋅θ θ β
C ( ) cos sinRRΤ Ε mi
t x t z j r
a a e t θ θ β − − ⋅
nd
RR
RRΗ Η Τ Ε t
mt
y j r m
i j r
ya e e at t t = =− ⋅∧
− ⋅⋅β β η $
Where ( )β β θ θ t t t r x z ⋅ = +$ sin cos and RRΕ Ε Τmt
mi = .
%efinition:
&rewster ngle > from &rewster0s !aw2, the polariAing angle of which whenlight is incident2 the reflected and refracted inde< is e=/al to the tangent of the polariAing
angle. In other words, the angle of incidence of which there is no reflection.
From the reflection coefficient e
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This condition is important, beca/se it is /s/all8 satisfied b8 the materials often /sed in
optical applications.
E=/ation +."# will take the form >
i t cos cosθ ε ε θ = "
$+.$62
%=/are both sides of e=/ation +.$6 and /se %nell0s !aw for the special case of µ µ µ " $= = for the following res/lt:
$ "
$cos iθ
ε
ε ( )= = −$ " $
$
"cos sint t θ ε
ε θ
( )= −"$
" $ε ε
θ sin i
The last s/bstit/tion was based on %nell0s !aw of refraction. Therefore,
( )" $− =sin iθ "
$
ε
ε − "
$
$$
$ε
ε θ sin i
" "
$− =
ε
ε $sin iθ "
"$
$$
−
ε
ε
nd
$ $
$ "sin iθ
ε
ε ε =
++.$"2
The &rewster angle of incidence is
sin iθ ε
ε ε =
+
$
$ "+.$$2
specific val/e of Hi can be obtained from e=/ation +.$" (
'otes b8: Debbie Prestridge $+
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" $ $
$ "− =
+cos iθ
ε
ε ε
1r
$ $
$ ""cos iθ ε
ε ε = −
+ =
"
$ "
ε
ε ε + =
cos iθ ε
ε ε =
+
"
$ "+.$52
From e=/ations +.$$ +.$5 >
tan iθ ε
ε =
$
"
This specific angle of incidence θ i is called the &rewster angleθ β .
β θ ε
ε = −"
$
"tan
6@4 Perpendicular Polari&ation case - Normal to Plane of Incidence
s shown in fig/re +."6 is a perpendic/lar polariAed wave incident at angleθ i a dielectric
medi/m $. %nell0s !aw states that a reflected wave will be at the same angleθ θ r i= , andthe transmitted wave in medi/m $ at angle θ t can be calc/lated /sing this law. The
amplit/de of the reflected and transmitted waves can be determined b8 appl8ing the
contin/it8 of the tangential components of E at the bo/ndar8.
This is given b8 >
cosΗ⊥i
iθ − ⊥ cosΗ
r iθ C cosΗ⊥
t t θ
'otes b8: Debbie Prestridge $*
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%ince E are related b8 η ,
cosΕ ⊥
i
i"
η θ − =
⊥
cosΕ r
i"
η θ
cosΕ ⊥
t
t $
η θ +.$2
Ε ⊥
i+
⊥Ε
r =
⊥Ε
t t A C 6 +.$)2
S'ote: The e
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Τ⊥ =
Ε
Ε
⊥
⊥
t
i C
$
$ "
$η θ
η θ η θ
cos
cos cos
i
i t +
For nonmagnetic material,
cos
cos cos
Τ⊥ =$
$
"
i
i t
θ
θ ε
ε θ
6D Comparison betAeen Reflection Coefficients RRΓ and Γ ⊥ for Parallel
and Perpendicular Polari&ations
The significant differences between the two will be ill/strated in the following e
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β θ = −" 7"tan C 75.*?
b2 Water into air:ε ε r r and " $7" "= =
ence,
β θ = −" "
7"tan C +.5?
To relate the &rewster angles in both cases, let /s calc/late the angle of
refraction.
sin
sin
i
t
θ
θ
ε
ε = $
"
Therefore, in case a,
sin
sin
Βθ
θ t = 7"
Therefore,
t sinsin .
.θ = =75 *
# 6 ""
1r θ t = + 5M. , which is the same as the &rewster angle for case b. lso, the angle of
refraction in case b is given b8 %nell0s !aw as:
sin
sin
Βθ
θ
ε
ε t =
7"=
"
7"
Therefore,
t sinsin .
.θ = =+ 5M
"
7"
6 ##
1r t θ = 75 *. , which is the &rewster angle for case a.
6? Total Reflection at Critical #n"le of Incidence
'otes b8: Debbie Prestridge 56
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7i"ure 6!@ ill/strates the fact thatθ θ ε ε t i if > >, " $ . The critical angle θ c is defined as
the val/e of θ i at which θ t C N$.
Envision a beam of light impinging on an interface between two transparent media where
n ni t < . t normal incidence θ i C 62 most of the incoming light is transmitted into theless dense medi/m. s θ i increases, more and more light is reflected back into the dense
medi/m, while θ t increases. When θ t C #6?, θ i is defined to be θ c and the
transmittance becomes Aero. For θ i V θ c all of the light is totall8 internall8 reflected,
remaining in the incident medi/m.
-/#(P0-3:
• se %nell0s !aw to derive an e
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• Determine the critical angle for a water n# C".552 >glass n " C".)62 interface.
We have
sin θ c C nti
1r θ c C sin −" "55")6..
C sin −"6.77* C +$.)?
66 -lectroma"netic 3pectrum
'otes b8: Debbie Prestridge 55
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5 3A 5 < "6"$ A 5
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npolariAed light > light in which the wave orientation is random aro/nd the a entirel8 transmitted
thro/gho/t the interfaces
• More glass elements and the transmitted light co/ld be essentiall8 completel8
polariAed, - parallel to the plane of incidence
6B4 reAster WindoAs or reAster Cuts in 0#3-R In a normal sit/ation there are more electrons in the gro/nd state level "2 than in the
e
8/16/2019 Capítulo 6 de libro de Teoría electromagnética
36/39θ β $
photons co/ld be the res/lt if this sit/ation co/ld be inverted. %/ch a condition is called
pop/lation inversion. This in fact is the f/ndamental principle involved in the operation
of a laser. Fig/re +.7 ill/strates this principle.
%efinition:
!aser !ight mplification b8 %tim/lated Emission2 > device that prod/cescoherent radiation in the visible(light range, between *)66 and 5#66 angstroms
%/mmariAed steps leading to !%E9 action in three(level r/b8 laser material:
". The laser material is in the shape of a long rod that is s/bGected to radiation from
an e
8/16/2019 Capítulo 6 de libro de Teoría electromagnética
37/39
n
7i"ure 6!B 0i"ht polari&ations b$ multiple reflections.
Fig/re +."7 is a schematic diagram ill/strating the se=/ence of events.The role of the &rewster angle:
FnoAn 7actors
• The o/tp/t of man8 lasers is linearl8 polariAed
• The ratio of the light polariAed in one direction e
8/16/2019 Capítulo 6 de libro de Teoría electromagnética
38/39
7i"ure 6!G 3equence of events occurrin" in laser action
Fig/re +."# is a schematic ill/strating the /se of &rewster windows in a gas dischargelaser. The &rewster angle makes s/re that light in one polariAation direction is
transmitted o/t of the medi/m of the laser to the reflecting mirrors and back into the
medi/m of the laser with no loss. Where the light is polariAed perpendic/lar to the planeof incidence a large loss at the &rewster s/rface will take place d/e to the reflection o/t
of the medi/m of the laser. The preferred polariAation case linear polariAation2 will lase
emit coherent light2 that will acco/nt for the high degree of polariAation taking place atthe o/tp/t.
The device in Fig/re +."# e
8/16/2019 Capítulo 6 de libro de Teoría electromagnética
39/39
Hi
Ht
H$
%mallest critical angle
9eflected point
n Z
7i"ure 64< %chematic ill/strating the principle of light propagation in optical fibers.
%nell0s !aw of refraction is the relationship between Hi and Ht as the wave enter the fiber is
sin
sin
i
t n
θ
θ
ε
ε = =
$
"ε ε " = +.$*2
If θ $ is s/ppose to be larger thanθ c , then
sin θ $ C cos θ t Z sin θ c +.$72
9efraction from fiber to air sin θ c C "Nn, therefore, from e=/ation +.$* +.$7 >
sin cos sin sin$ $
$$" "
" "θ θ θ θ = = − = −
≥t t in n
+.$#2
%olve for n,
n i$ $"≥ + sin θ +.562
For e=/ation +.56 to be tr/e then C N$, all incident light will be passed b8 the fiber
re=/iring
n$ Z $ or n Z $
Most t8pes of glass have n [ ".)O therefore, we have a valid e=/ation.
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