Demostración Ecuación 3.19
editar
ε
i
=
ε
0
ε
r
{\displaystyle \varepsilon _{i}=\varepsilon _{0}\varepsilon _{r}}
μ
=
μ
0
μ
r
{\displaystyle \mu =\mu _{0}\mu _{r}}
z
<
0
=
ε
1
,
G
1
,
μ
1
{\displaystyle z<0=\varepsilon _{1},G_{1},\mu _{1}}
z
>
0
=
ε
2
,
G
2
,
μ
2
{\displaystyle z>0=\varepsilon _{2},G_{2},\mu _{2}}
ε
i
=
H
i
{\displaystyle \varepsilon _{i}=H_{i}}
H
i
=
B
i
sin
(
90
−
θ
i
)
{\displaystyle H_{i}=B_{i}\ \sin {(90\ -\ \theta _{i})}}
B
i
=
ω
μ
i
ε
i
{\displaystyle B_{i}=\omega {\sqrt {\mu _{i}\varepsilon _{i}}}}
ε
r
=
H
r
{\displaystyle \varepsilon _{r}=H_{r}}
H
r
=
B
r
sin
(
90
−
θ
r
)
{\displaystyle H_{r}=B_{r}\ \sin {(90\ -\ \theta _{r})}}
B
r
=
ω
μ
r
ε
r
{\displaystyle B_{r}=\omega {\sqrt {\mu _{r}\varepsilon _{r}}}}
ε
t
=
H
t
{\displaystyle \varepsilon _{t}=H_{t}}
H
t
=
B
t
sin
(
90
−
θ
t
)
{\displaystyle H_{t}=B_{t}\ \sin {(90\ -\ \theta _{t})}}
B
t
=
ω
μ
t
ε
t
{\displaystyle B_{t}=\omega {\sqrt {\mu _{t}\varepsilon _{t}}}}
ε
i
+
ε
r
=
ε
t
{\displaystyle \varepsilon _{i}+\varepsilon _{r}=\varepsilon _{t}}
ε
i
=
ε
r
+
ε
t
{\displaystyle \varepsilon _{i}=\varepsilon _{r}+\varepsilon _{t}}
ε
r
=
ε
i
−
ε
t
{\displaystyle \varepsilon _{r}=\varepsilon _{i}\ -\ \varepsilon _{t}}
ε
t
=
ε
i
−
ε
r
{\displaystyle \varepsilon _{t}=\varepsilon _{i}\ -\ \varepsilon _{r}}
ε
i
=
ε
i
−
ε
t
+
ε
i
−
ε
r
{\displaystyle \varepsilon _{i}=\varepsilon _{i}\ -\ \varepsilon _{t}+\varepsilon _{i}\ -\ \varepsilon _{r}}
ε
r
+
ε
t
=
ε
i
−
ε
t
+
ε
i
−
ε
r
{\displaystyle \varepsilon _{r}\ +\ \varepsilon _{t}=\varepsilon _{i}\ -\ \varepsilon _{t}+\varepsilon _{i}\ -\ \varepsilon _{r}}
B
r
sin
(
θ
r
)
+
B
t
sin
(
θ
t
)
=
B
i
sin
(
θ
i
)
−
B
t
sin
(
θ
t
)
+
B
i
sin
(
θ
i
)
−
B
r
sin
(
θ
r
)
{\displaystyle B_{r}\sin {(\theta _{r})}+B_{t}\sin {(\theta _{t})}=B_{i}\sin {(\theta _{i})}-B_{t}\sin {(\theta _{t})}+B_{i}\sin {(\theta _{i})}-B_{r}\sin {(\theta _{r})}}
θ
=
−
B
r
sin
(
θ
r
)
+
B
t
sin
(
θ
t
)
+
B
i
sin
(
θ
i
)
−
B
t
sin
(
θ
t
)
+
B
i
sin
(
θ
i
)
−
B
r
sin
(
θ
r
)
{\displaystyle \theta =-B_{r}\sin {(\theta _{r})}+B_{t}\sin {(\theta _{t})}+B_{i}\sin {(\theta _{i})}-B_{t}\sin {(\theta _{t})}+B_{i}\sin {(\theta _{i})}-B_{r}\sin {(\theta _{r})}}
θ
r
=
B
i
θ
r
=
θ
i
{\displaystyle \theta _{r}=\ B_{i}\theta _{r}\ =\ \theta _{i}}
B
t
sin
(
θ
t
)
−
B
i
sin
(
θ
i
)
=
−
ω
μ
r
ε
r
sin
(
θ
r
)
−
ω
μ
t
ε
t
sin
(
θ
t
)
+
ω
μ
i
ε
i
sin
(
θ
i
)
−
ω
μ
t
ε
t
sin
(
θ
t
)
{\displaystyle B_{t}\sin {(\theta _{t})}-B_{i}\sin {(\theta _{i})}=-\omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin {(\theta _{r})}-\omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})}-\omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}}
ω
μ
t
ε
t
sin
(
θ
t
)
−
ω
μ
i
ε
i
sin
(
θ
i
)
=
−
ω
μ
r
ε
r
sin
(
θ
r
)
−
ω
μ
t
ε
t
sin
(
θ
t
)
+
ω
μ
i
ε
i
sin
(
θ
i
)
−
ω
μ
r
ε
r
sin
(
θ
r
)
{\displaystyle \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}-\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})}=-\omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin {(\theta _{r})}-\omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})}-\omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin {(\theta _{r})}}
ω
(
μ
t
ε
t
sin
(
θ
t
)
−
ω
μ
i
ε
i
sin
(
θ
i
)
)
=
−
2
ω
μ
r
ε
r
sin
(
θ
r
)
−
(
ω
μ
t
ε
t
sin
(
θ
t
)
+
ω
μ
i
ε
i
sin
(
θ
i
)
)
{\displaystyle \omega ({\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}-\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})})=-2\omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin {(\theta _{r})}-(\omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})})}
η
1
=
μ
i
ε
i
=
μ
r
ε
r
η
2
=
μ
t
ε
t
{\displaystyle \eta _{1}=\mu _{i}\varepsilon _{i}=\mu _{r}\varepsilon _{r}{\eta _{2}=\mu }_{t}\varepsilon _{t}}
ω
(
η
2
sin
(
θ
t
)
−
η
1
sin
(
θ
i
)
)
=
−
ω
B
i
sin
(
θ
i
)
+
ω
B
r
sin
(
θ
r
)
−
ω
(
η
2
sin
(
θ
t
)
+
η
1
sin
(
θ
i
)
)
{\displaystyle \omega (\eta _{2}\sin {(\theta _{t})}-\eta _{1}\sin {(\theta _{i})})=-\omega B_{i}\sin {(\theta _{i})}+\omega B_{r}\sin {(\theta _{r})}-\omega (\eta _{2}\sin {(\theta _{t})}+\eta _{1}\sin {(\theta _{i})})}
ω
(
η
2
sin
(
θ
t
)
−
η
1
sin
(
θ
i
)
)
(
η
2
sin
(
θ
t
)
−
η
1
sin
(
θ
i
)
)
=
(
ω
+
ω
B
r
sin
(
θ
r
)
)
ω
B
i
sin
(
θ
i
)
{\displaystyle {\frac {\omega (\eta _{2}\sin {(\theta _{t})}-\eta _{1}\sin {(\theta _{i})})}{(\eta _{2}\sin {(\theta _{t})}-\eta _{1}\sin {(\theta _{i})})}}={\frac {(\omega +\omega B_{r}\sin {(\theta _{r})})}{\omega B_{i}\sin {(\theta _{i})}}}}
Γ
|
|
=
E
r
E
i
=
η
2
s
i
n
θ
t
−
η
1
s
i
n
θ
i
η
2
s
i
n
θ
t
+
η
1
s
i
n
θ
i
{\displaystyle \Gamma _{||}={\frac {E_{r}}{E_{i}}}={\frac {\eta _{2}sin\theta _{t}-\eta _{1}sin\theta _{i}}{\eta _{2}sin\theta _{t}+\eta _{1}sin\theta _{i}}}}
Demostración del ángulo de Brewster
editar
Es importante mencionar que se demostrará en distintos espacios se probará:
En primer lugar, es cuando la permitividad sea distinta en ambos medios. para demostrar se usan las ecuaciones 3.27 y 3.19 que relacionan la premitividad distinta
s
i
n
(
θ
B
)
=
ε
1
ε
1
+
ε
2
{\displaystyle sin\left(\theta _{B}\right)={\sqrt {\frac {\varepsilon _{1}}{\varepsilon _{1}+\varepsilon _{2}}}}}
Ec. 3.27
Γ
∥
=
η
2
sin
θ
t
−
η
1
sin
θ
i
η
2
sin
θ
t
+
η
1
sin
θ
i
{\displaystyle \Gamma _{\parallel }={\frac {\eta _{2}\sin \theta _{t}-\eta _{1}\sin \theta _{i}}{\eta _{2}\sin \theta _{t}+\eta _{1}\sin \theta _{i}}}}
Ec. 3.19
0
=
η
2
sin
θ
t
−
η
1
sin
θ
i
η
2
sin
θ
t
+
η
1
sin
θ
i
{\displaystyle 0={\frac {\eta _{2}\sin \theta _{t}-\eta _{1}\sin \theta _{i}}{\eta _{2}\sin \theta _{t}+\eta _{1}\sin \theta _{i}}}}
0
=
η
2
sin
θ
t
−
η
1
sin
θ
i
{\displaystyle 0=\eta _{2}\sin \theta _{t}-\eta _{1}\sin \theta _{i}}
μ
2
ε
2
sin
θ
t
=
μ
1
ε
1
sin
θ
i
{\displaystyle {\sqrt {\frac {\mu _{2}}{\varepsilon _{2}}}}\sin \theta _{t}={\sqrt {\frac {\mu _{1}}{\varepsilon _{1}}}}\sin \theta _{i}}
μ
2
ε
1
sin
θ
t
=
μ
1
ε
2
sin
θ
i
{\displaystyle {\sqrt {\mu _{2}\varepsilon _{1}}}\sin \theta _{t}={\sqrt {\mu _{1}\varepsilon _{2}}}\sin \theta _{i}}
μ
2
ε
1
sin
2
θ
t
=
μ
1
ε
2
sin
2
θ
i
{\displaystyle \mu _{2}\varepsilon _{1}\sin ^{2}\theta _{t}=\mu _{1}\varepsilon _{2}\sin ^{2}\theta _{i}}
μ
1
ε
2
μ
2
ε
1
sin
2
θ
i
=
sin
2
θ
t
{\displaystyle {\frac {\mu _{1}\varepsilon _{2}}{\mu _{2}\varepsilon _{1}}}\sin ^{2}\theta _{i}=\sin ^{2}\theta _{t}}
μ
1
ε
2
μ
2
ε
1
sin
2
θ
i
=
(
1
−
cos
2
θ
t
)
{\displaystyle {\frac {\mu _{1}\varepsilon _{2}}{\mu _{2}\varepsilon _{1}}}\sin ^{2}\theta _{i}=\left(1-\cos ^{2}\theta _{t}\right)}
μ
1
ε
2
μ
2
ε
1
sin
2
θ
i
=
1
−
μ
1
ε
1
μ
2
ε
2
cos
2
θ
i
{\displaystyle {\frac {\mu _{1}\varepsilon _{2}}{\mu _{2}\varepsilon _{1}}}\sin ^{2}\theta _{i}=1-{\frac {\mu _{1}\varepsilon _{1}}{\mu _{2}\varepsilon _{2}}}\cos ^{2}\theta _{i}}
(
μ
1
ε
2
μ
2
ε
1
−
μ
1
ε
1
μ
2
ε
2
)
sin
2
θ
i
=
1
−
μ
1
ε
1
μ
2
ε
2
{\displaystyle \left({\frac {\mu _{1}\varepsilon _{2}}{\mu _{2}\varepsilon _{1}}}-{\frac {\mu _{1}\varepsilon _{1}}{\mu _{2}\varepsilon _{2}}}\right)\sin ^{2}\theta _{i}=1-{\frac {\mu _{1}\varepsilon _{1}}{\mu _{2}\varepsilon _{2}}}}
μ
1
μ
2
(
ε
2
ε
1
−
ε
1
ε
2
)
sin
2
θ
i
=
1
−
μ
1
ε
1
μ
2
ε
2
{\displaystyle {\frac {\mu _{1}}{\mu _{2}}}\left({\frac {\varepsilon _{2}}{\varepsilon _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}\right)\sin ^{2}\theta _{i}=1-{\frac {\mu _{1}\varepsilon _{1}}{\mu _{2}\varepsilon _{2}}}}
sin
2
θ
i
=
μ
2
μ
1
−
ε
1
ε
2
ε
2
ε
1
−
ε
1
ε
2
{\displaystyle \sin ^{2}\theta _{i}={\frac {{\frac {\mu _{2}}{\mu _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}{{\frac {\varepsilon _{2}}{\varepsilon _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}}}
sin
θ
i
=
μ
2
μ
1
−
ε
1
ε
2
ε
2
ε
1
−
ε
1
ε
2
{\displaystyle \sin \theta _{i}={\sqrt {\frac {{\frac {\mu _{2}}{\mu _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}{{\frac {\varepsilon _{2}}{\varepsilon _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}}}}
Para el angulo de brewster se puede considerar que:
μ
2
=
μ
1
{\displaystyle \mu _{2}=\mu _{1}}
sin
θ
i
=
1
−
ε
1
ε
2
ε
2
ε
1
−
ε
1
ε
2
{\displaystyle \sin \theta _{i}={\sqrt {\frac {1-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}{{\frac {\varepsilon _{2}}{\varepsilon _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}}}}
sin
θ
i
=
ε
2
−
ε
1
ε
2
ε
2
2
−
ε
1
2
ε
1
ε
2
{\displaystyle \sin \theta _{i}={\sqrt {\frac {\frac {\varepsilon _{2}-\varepsilon _{1}}{\varepsilon _{2}}}{\frac {\varepsilon _{2}^{2}-\varepsilon _{1}^{2}}{\varepsilon _{1}\varepsilon _{2}}}}}}
sin
θ
i
=
ε
1
(
ε
2
−
ε
1
)
ε
2
2
−
ε
1
2
{\displaystyle \sin \theta _{i}={\sqrt {\frac {\varepsilon _{1}\left(\varepsilon _{2}-\varepsilon _{1}\right)}{\varepsilon _{2}^{2}-\varepsilon _{1}^{2}}}}}
sin
θ
i
=
ε
1
(
ε
2
−
ε
1
)
(
ε
2
−
ε
1
)
(
ε
2
+
ε
1
)
{\displaystyle \sin \theta _{i}={\sqrt {\frac {\varepsilon _{1}\left(\varepsilon _{2}-\varepsilon _{1}\right)}{\left(\varepsilon _{2}-\varepsilon _{1}\right)\left(\varepsilon _{2}+\varepsilon _{1}\right)}}}}
En este punto se demostró de donde sale la ecuacion 3.27 cuando
μ
2
=
μ
1
{\displaystyle \mu _{2}=\mu _{1}}
s
i
n
(
θ
B
)
=
ε
1
ε
1
+
ε
2
{\displaystyle sin\left(\theta _{B}\right)={\sqrt {\frac {\varepsilon _{1}}{\varepsilon _{1}+\varepsilon _{2}}}}}
En segundo lugar, para el caso en el que la onda se propague en el espacio libre se usa la ecuacion 3.28 y la 3.24.
s
i
n
(
θ
B
)
=
ε
r
−
1
ε
r
2
−
1
{\displaystyle sin\left(\theta _{B}\right)={\frac {\sqrt {\varepsilon _{r}-1}}{\sqrt {\varepsilon _{r}^{2}-1}}}}
Ec. 3.28
Γ
∥
=
−
ε
r
sin
θ
i
+
ε
r
−
cos
2
θ
i
ε
r
sin
θ
i
+
ε
r
−
cos
2
θ
i
{\displaystyle \Gamma _{\parallel }={\frac {-\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}{\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}}}
Ec. 3.24
Polarización vertical
0
=
−
ε
r
sin
θ
i
+
ε
r
−
cos
2
θ
i
ε
r
sin
θ
i
+
ε
r
−
cos
2
θ
i
{\displaystyle 0={\frac {-\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}{\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}}}
0
=
−
ε
r
sin
θ
i
+
ε
r
−
cos
2
θ
i
{\displaystyle 0=-\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}
ε
r
sin
θ
i
=
ε
r
−
cos
2
θ
i
{\displaystyle \varepsilon _{r}\sin \theta _{i}={\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}
(
ε
r
sin
θ
i
)
2
=
ε
r
−
cos
2
θ
i
{\displaystyle \left(\varepsilon _{r}\sin \theta _{i}\right)^{2}=\varepsilon _{r}-\cos ^{2}\theta _{i}}
ε
r
2
sin
θ
i
2
=
ε
r
−
cos
2
θ
i
{\displaystyle \varepsilon _{r}^{2}\sin \theta _{i}^{2}=\varepsilon _{r}-\cos ^{2}\theta _{i}}
ε
r
2
sin
θ
i
2
=
ε
r
+
sin
2
θ
i
−
1
{\displaystyle \varepsilon _{r}^{2}\sin \theta _{i}^{2}=\varepsilon _{r}+\sin ^{2}\theta _{i}-1}
ε
r
2
sin
θ
i
2
−
sin
2
θ
i
=
ε
r
−
1
{\displaystyle \varepsilon _{r}^{2}\sin \theta _{i}^{2}-\sin ^{2}\theta _{i}=\varepsilon _{r}-1}
(
ε
r
2
−
1
)
sin
2
θ
i
=
ε
r
−
1
{\displaystyle \left(\varepsilon _{r}^{2}-1\right)\sin ^{2}\theta _{i}=\varepsilon _{r}-1}
sin
2
θ
i
=
ε
r
−
1
ε
r
2
−
1
{\displaystyle \sin ^{2}\theta _{i}={\frac {\varepsilon _{r}-1}{\varepsilon _{r}^{2}-1}}}
En este punto se demostró de donde sale la ecuacion 3.28 par el uso en espacio libre.
sin
θ
i
=
ε
r
−
1
ε
r
2
−
1
{\displaystyle \sin \theta _{i}={\sqrt {\frac {\varepsilon _{r}-1}{\varepsilon _{r}^{2}-1}}}}
Demostración de la ecuación 3.20
editar
La demostración de la ecuación presentada en el libro Rappaport, la ecuación 3.20
E
r
E
i
=
η
2
sin
θ
t
−
η
1
∗
sin
θ
i
η
2
sin
θ
t
+
η
1
∗
sin
θ
i
{\displaystyle {\frac {E_{r}}{E_{i}}}={\frac {\eta _{2}\sin \theta _{t}-\eta _{1}*\sin \theta _{i}}{\eta _{2}\sin \theta _{t}+\eta _{1}*\sin \theta _{i}}}}
Partiendo de la ecuacion presentada anteriormente, empezaremos a hacer el analisis mediante la grafica b de la pagina 452 del libro de Matthew.
ε
i
=
ε
0
+
ε
r
{\displaystyle \varepsilon _{i}=\varepsilon _{0}+\varepsilon _{r}}
z
<
0
=
ε
1
,
G
1
,
μ
1
{\displaystyle z<0=\varepsilon _{1},G_{1},\mu _{1}}
z
>
0
=
ε
2
,
G
2
,
μ
2
{\displaystyle z>0=\varepsilon _{2},G_{2},\mu _{2}}
ε
i
=
H
i
{\displaystyle \varepsilon _{i}=H_{i}}
H
i
=
B
i
sin
(
θ
i
)
{\displaystyle H_{i}=B_{i}\ \sin {(\theta _{i})}}
B
i
=
ω
μ
i
ε
i
{\displaystyle B_{i}=\omega {\sqrt {\mu _{i}\varepsilon _{i}}}}
ε
r
=
H
r
{\displaystyle \varepsilon _{r}=H_{r}}
H
r
=
B
r
sin
(
θ
r
)
{\displaystyle H_{r}=B_{r}\ \sin {(\theta _{r})}}
B
r
=
ω
μ
r
ε
r
{\displaystyle B_{r}=\omega {\sqrt {\mu _{r}\varepsilon _{r}}}}
ε
t
=
H
t
{\displaystyle \varepsilon _{t}=H_{t}}
H
t
=
B
t
sin
(
θ
t
)
{\displaystyle H_{t}=B_{t}\ \sin {(\theta _{t})}}
B
t
=
ω
μ
t
ε
t
{\displaystyle B_{t}=\omega {\sqrt {\mu _{t}\varepsilon _{t}}}}
E
i
+
−
E
r
=
E
t
{\displaystyle E_{i}+-E_{r}=E_{t}}
E
i
=
E
r
+
E
t
{\displaystyle E_{i}=E_{r}+E_{t}}
E
r
=
E
i
−
E
t
{\displaystyle E_{r}=E_{i}\ -\ E_{t}}
E
t
=
E
i
−
E
r
{\displaystyle E_{t}=E_{i}\ -\ E_{r}}
E
i
=
E
i
−
E
t
+
E
i
−
E
r
{\displaystyle E_{i}=E_{i}\ -\ E_{t}+E_{i}\ -\ E_{r}}
E
r
+
E
t
=
E
i
−
E
t
+
E
i
−
E
r
{\displaystyle E_{r}\ +E_{t}=E_{i}\ -\ E_{t}+E_{i}\ -\ E_{r}}
B
r
sin
(
θ
r
)
+
B
t
sin
(
θ
t
)
=
B
i
sin
(
θ
i
)
−
B
t
sin
(
θ
t
)
+
B
i
sin
(
θ
i
)
−
B
r
sin
(
θ
r
)
{\displaystyle B_{r}\sin(\theta _{r})\ +\ B_{t}\sin(\theta _{t})\ =\ B_{i}\sin(\theta _{i})\ -\ B_{t}\sin(\theta _{t})\ +\ B_{i}\sin(\theta _{i})\ -\ B_{r}\sin(\theta _{r})\ }
0
=
−
B
r
sin
(
θ
r
)
+
B
t
sin
(
θ
t
)
+
B
i
sin
(
θ
i
)
−
B
t
sin
(
θ
t
)
+
B
i
sin
(
θ
i
)
−
B
r
sin
(
θ
r
)
{\displaystyle 0=-B_{r}\sin(\theta _{r})\ +\ B_{t}\sin(\theta _{t})\ +\ B_{i}\sin(\theta _{i})\ -\ B_{t}\sin(\theta _{t})\ +\ B_{i}\sin(\theta _{i})\ -\ B_{r}\sin(\theta _{r})}
θ
r
=
θ
i
{\displaystyle \theta _{r}\ =\ \theta _{i}}
B
t
sin
(
θ
t
)
−
B
i
sin
(
θ
i
)
=
−
ω
μ
r
ε
r
sin
(
θ
r
)
−
ω
μ
t
ε
t
sin
(
θ
t
)
+
ω
μ
i
ε
i
sin
(
θ
i
)
−
ω
μ
t
ε
t
sin
(
θ
t
)
{\displaystyle B_{t}\sin(\theta _{t})\ -\ B_{i}\sin(\theta _{i})\ =\ -\ \omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin(\theta _{r})\ -\ \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})\ +\ \omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i})\ -\ \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})\ }
ω
μ
t
ε
t
sin
(
θ
t
)
−
ω
μ
i
ε
i
sin
(
θ
i
)
=
−
ω
μ
r
ε
r
sin
(
θ
r
)
−
ω
μ
t
ε
t
sin
(
θ
t
)
+
ω
μ
i
ε
i
sin
(
θ
i
)
−
ω
μ
r
ε
r
sin
(
θ
r
)
{\displaystyle \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})\ -\ \omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i})\ =-\ \omega {\sqrt {\mu _{r}\varepsilon _{r}}}\ \sin(\theta _{r})-\ \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i})\ -\ \omega {\sqrt {\mu _{r}\varepsilon _{r}}}\ \sin(\theta _{r})\ }
ω
(
μ
t
ε
t
sin
(
θ
t
)
−
ω
μ
i
ε
i
sin
(
θ
i
)
)
=
−
2
ω
μ
r
ε
r
sin
(
θ
r
)
−
ω
(
μ
t
ε
t
sin
(
θ
t
)
+
ω
μ
i
ε
i
sin
(
θ
i
)
)
{\displaystyle \omega ({\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})\ -\ \omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i}))\ =-2\ \omega {\sqrt {\mu _{r}\varepsilon _{r}}}\ \sin(\theta _{r})-\ \omega ({\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i}))\ }
η
1
=
μ
i
ε
i
{\displaystyle \eta _{1}={\sqrt {\mu _{i}\varepsilon _{i}}}}
η
2
=
μ
t
ε
t
{\displaystyle \eta _{2}={\sqrt {\mu _{t}\varepsilon _{t}}}}
ω
(
η
2
sin
(
θ
t
)
−
η
1
sin
(
θ
i
)
)
=
−
ω
i
sin
(
θ
i
)
+
ω
r
sin
(
θ
r
)
−
ω
(
η
2
sin
(
θ
t
)
+
η
1
sin
(
θ
i
)
)
{\displaystyle \omega (\eta _{2}\sin(\theta _{t})-\eta _{1}\sin(\theta _{i}))=-\omega _{i}\ \sin(\theta _{i})+\omega _{r}\sin(\theta _{r})-\ \omega (\eta _{2}\sin(\theta _{t})+\eta _{1}\sin(\theta _{i}))}
ω
sin
θ
i
(
η
2
sin
θ
t
−
η
1
sin
θ
i
)
/
η
1
sin
θ
i
+
sin
θ
t
=
ω
+
ω
B
r
sin
θ
r
{\displaystyle \omega \sin {\theta _{i}}(\eta _{2}\sin {\theta _{t}}-\eta _{1}\sin {\theta _{i}})\ /\ \eta _{1}\ \sin {\theta _{i}}\ +\sin {\theta _{t}}\ =\ \omega +\omega B_{r}\sin {\theta _{r}}}
η
2
sin
θ
t
−
η
1
sin
θ
i
)
/
(
η
1
sin
θ
i
+
η
2
sin
θ
t
)
=
B
r
sin
θ
r
/
B
i
sin
θ
i
{\displaystyle \eta _{2}\sin {\theta _{t}}-\eta _{1}\sin {\theta _{i}})/(\eta _{1}\sin {\theta _{i}}+\eta _{2}\sin {\theta _{t}})=B_{r}\sin {\theta _{r}}/B_{i}\sin {\theta _{i}}}
E
i
=
B
i
sin
θ
i
{\displaystyle E_{i}=B_{i}\sin {\theta _{i}}}
E
r
=
B
r
sin
θ
r
{\displaystyle E_{r}=B_{r}\sin {\theta _{r}}}
Realizando los cambios correspondientes obtendremos la siguiente ecuacion:
Γ
I
=
ε
r
ε
i
=
η
2
sin
(
θ
t
)
−
η
1
sin
(
θ
i
)
)
η
2
sin
(
θ
t
)
+
η
1
sin
(
θ
i
)
)
{\displaystyle \mathrm {\Gamma } _{I}={\frac {\varepsilon _{r}}{\varepsilon _{i}}}={\frac {\eta _{2}\sin {(\theta _{t})-\eta _{1}\sin {(\theta _{i}))}}}{\eta _{2}\sin {(\theta _{t})+\eta _{1}\sin {(\theta _{i}))}}}}}
Demostración ecuación 3.24
editar
η
1
=
μ
ϵ
0
∗
η
2
=
μ
ϵ
2
{\displaystyle \eta _{1}={\sqrt {\frac {\mu }{\epsilon _{0}}}}*\eta _{2}={\sqrt {{\frac {\mu }{\epsilon }}_{2}}}}
ϵ
0
μ
0
sin
90
−
θ
i
=
ϵ
2
μ
0
s
i
n
90
−
θ
t
{\displaystyle {\sqrt {\epsilon _{0}\mu _{0}}}\sin {90-\theta _{i}}={\sqrt {\epsilon _{2}\mu _{0}}}sin{90-\theta _{t}}}
μ
0
ϵ
0
μ
0
ϵ
2
sin
(
90
−
θ
i
)
=
sin
(
90
−
θ
t
)
{\displaystyle {\sqrt {\frac {\mu _{0}\epsilon _{0}}{\mu _{0}\epsilon _{2}}}}\sin {(90-\theta _{i})}=\sin {(90-\theta _{t})}}
ϵ
0
ϵ
2
sin
(
90
−
θ
i
)
=
sin
(
90
−
θ
t
)
{\displaystyle {\sqrt {\frac {\epsilon _{0}}{\epsilon _{2}}}}\sin {(90-\theta _{i})}=\sin {(90-\theta _{t})}}
Entidades trigonométricas
sin
(
90
−
θ
i
)
=
cos
θ
i
{\displaystyle \sin {(90-\theta _{i})}=\cos {\theta _{i}}}
cos
2
θ
i
=
1
−
sin
2
(
90
−
θ
i
)
{\displaystyle \cos ^{2}{\theta _{i}}=1-\sin ^{2}{(90-\theta _{i})}}
(
ϵ
0
ϵ
2
∗
cos
θ
i
=
cos
θ
t
)
2
{\displaystyle ({\sqrt {\frac {\epsilon _{0}}{\epsilon _{2}}}}*\cos {\theta _{i}}=\cos {\theta _{t}})^{2}}
ϵ
0
ϵ
2
cos
2
θ
i
=
cos
2
θ
t
{\displaystyle {\frac {\epsilon _{0}}{\epsilon _{2}}}\cos ^{2}{\theta _{i}}=\cos ^{2}{\theta _{t}}}
ϵ
0
ϵ
2
cos
2
θ
i
=
1
−
sin
2
(
θ
t
)
{\displaystyle {\frac {\epsilon _{0}}{\epsilon _{2}}}\cos ^{2}{\theta _{i}}=1-\sin ^{2}{(\theta _{t})}}
sin
2
(
θ
t
)
=
1
−
ϵ
0
ϵ
2
cos
2
θ
i
{\displaystyle \sin ^{2}{(\theta _{t})}=1-{\frac {\epsilon _{0}}{\epsilon _{2}}}\cos ^{2}{\theta _{i}}}
(
ϵ
2
ϵ
0
)
sin
2
(
θ
t
)
=
1
−
ϵ
0
ϵ
2
cos
2
θ
i
(
ϵ
2
ϵ
0
)
{\displaystyle ({\frac {\epsilon _{2}}{\epsilon _{0}}})\sin ^{2}{(\theta _{t})}=1-{\frac {\epsilon _{0}}{\epsilon _{2}}}\cos ^{2}{\theta _{i}}({\frac {\epsilon _{2}}{\epsilon _{0}}})}
Permitividad del vacío:
ϵ
r
=
ϵ
2
ϵ
0
{\displaystyle \epsilon _{r}={\frac {\epsilon _{2}}{\epsilon _{0}}}}
ϵ
r
sin
2
θ
t
=
ϵ
r
−
cos
2
θ
i
{\displaystyle \epsilon _{r}\sin ^{2}{\theta _{t}}=\epsilon _{r}-\cos ^{2}{\theta _{i}}}
(
ϵ
r
sin
2
θ
t
=
ϵ
r
−
cos
2
θ
i
)
{\displaystyle {\sqrt {(\epsilon _{r}\sin ^{2}{\theta _{t}}=\epsilon _{r}-\cos ^{2}{\theta _{i}})}}}
ϵ
r
sin
θ
t
=
ϵ
r
−
cos
2
θ
i
{\displaystyle {\sqrt {\epsilon _{r}}}\sin {\theta _{t}}={\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}}
Γ
|
|
=
η
2
s
i
n
θ
t
−
η
1
s
i
n
θ
i
η
2
s
i
n
θ
t
+
η
1
s
i
n
θ
i
{\displaystyle \Gamma _{||}={\frac {\eta _{2}sin\theta _{t}-\eta _{1}sin\theta _{i}}{\eta _{2}sin\theta _{t}+\eta _{1}sin\theta _{i}}}}
Γ
|
|
=
ϵ
r
−
cos
2
θ
i
−
ϵ
r
sin
θ
i
ϵ
r
−
cos
2
θ
i
+
ϵ
r
sin
θ
i
{\displaystyle \Gamma _{||}={\frac {{\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}-\epsilon _{r}\sin {\theta _{i}}}{{\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}+\epsilon _{r}\sin {\theta _{i}}}}}
Γ
|
|
=
−
ϵ
r
sin
θ
i
+
ϵ
r
−
cos
2
θ
i
ϵ
r
−
cos
2
θ
i
+
ϵ
r
sin
θ
i
{\displaystyle \Gamma _{||}={\frac {-\epsilon _{r}\sin {\theta _{i}}+{\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}}{{\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}+\epsilon _{r}\sin {\theta _{i}}}}}
Se demuestra la ecuación 3.24
ESTE PROGRAMA REALIZA LA OPERACIÓN DE LA FUNCIÓN GAMA
import keyword ;
#keyword.iskeyword('pass')
import keyword
import math
from Lxtx2 import *
print ( 'Coeficiente de reflexión' )
print ( 'presione 1 para continuar' )
S = int ( input ())
if ( S == 1 ):
print ( 'ingrese Efectividad relativa:' )
Er = int ( input ())
print ( 'ingrese theta: ' )
T = int ( input ())
#print (Er, T)
Gamma ( Er , T )
(
import keyword ;
keyword . iskeyword ( 'pass' )
import keyword
import math
def Gamma ( Er , T ): \\
print ( & quot ; en grados es : & quot ;, math . radians ( T ))
n = ( math . sin ( math . radians ( T )) - math . sqrt ( Er - ( math . cos ( math . radians ( T )) ** 2 )))
d = ( math . sin ( math . radians ( T )) + math . sqrt ( Er - ( math . cos ( math . radians ( T )) ** 2 )))
divi = n / d
posi = abs ( divi )
print ( & quot ; Gama perpendicular es : & quot ;, posi )
(
import keyword ;
keyword . iskeyword ( 'pass' )
import keyword
import math
def Gamma ( Er , T ):
print ( & quot ; en grados es : & quot ;, math . radians ( T ))
n = ( - Er * math . sin ( math . radians ( T )) + math . sqrt ( Er - ( math . cos ( math . radians ( T )) ** 2 )))
d = ( Er * math . sin ( math . radians ( T )) + math . sqrt ( Er - ( math . cos ( math . radians ( T )) ** 2
divi = n / d
posi = abs ( divi )
print ( & quot ; Gama paralelo es : & quot ;, posi )
)