Demostración Ecuación 3.19
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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
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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 distintas 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. 3.27Γ ∥ = 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. 3.190 = 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. 3.28Γ ∥ = − 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 vertical0 = − 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
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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
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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 )
)