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Óptica Moderna Fundamentos e aplicações - Fotonica.ifsc.usp.br ...

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Interferência 135<<strong>br</strong> />

ρ′ τE<<strong>br</strong> />

0 exp{<<strong>br</strong> />

i(<<strong>br</strong> />

k 2AB<<strong>br</strong> />

+ k 2BC<<strong>br</strong> />

− ωt)<<strong>br</strong> />

}<<strong>br</strong> />

2<<strong>br</strong> />

em C: Refletido: ( ρ ) τE 0exp{<<strong>br</strong> />

i(<<strong>br</strong> />

k 2AB<<strong>br</strong> />

+ k 2BC<<strong>br</strong> />

− ωt)<<strong>br</strong> />

}<<strong>br</strong> />

ρ′ ττ′<<strong>br</strong> />

exp{<<strong>br</strong> />

i(<<strong>br</strong> />

k AB + k BC − ωt)<<strong>br</strong> />

}<<strong>br</strong> />

por:<<strong>br</strong> />

Incidente: (6.31a)<<strong>br</strong> />

′ (6.31b)<<strong>br</strong> />

Transmitido: (6.31c)<<strong>br</strong> />

E0 2<<strong>br</strong> />

2<<strong>br</strong> />

A frente de onda é constituída pelos campos em C e C’, dados<<strong>br</strong> />

E<<strong>br</strong> />

C<<strong>br</strong> />

= ρ′ ττ′<<strong>br</strong> />

E 0 exp<<strong>br</strong> />

= ρ′ ττ′<<strong>br</strong> />

E exp<<strong>br</strong> />

0<<strong>br</strong> />

{ i [ k 2 ( AB + BC)<<strong>br</strong> />

− ωt<<strong>br</strong> />

] }<<strong>br</strong> />

{ i ( 2k<<strong>br</strong> />

2 AB−<<strong>br</strong> />

ωt<<strong>br</strong> />

) }<<strong>br</strong> />

{ i(<<strong>br</strong> />

k AC′<<strong>br</strong> />

− ωt)<<strong>br</strong> />

}<<strong>br</strong> />

(6.32a)<<strong>br</strong> />

E C′<<strong>br</strong> />

= ρE<<strong>br</strong> />

0 exp<<strong>br</strong> />

(6.32b)<<strong>br</strong> />

1<<strong>br</strong> />

onde AB = BC . Por outro lado, vemos que AB = d / cos θ'<<strong>br</strong> />

' e<<strong>br</strong> />

AC′<<strong>br</strong> />

= ACsen<<strong>br</strong> />

θ , implicando que A C′<<strong>br</strong> />

= 2d tgθ′<<strong>br</strong> />

′ senθ.<<strong>br</strong> />

Definimos:<<strong>br</strong> />

φ = AC′<<strong>br</strong> />

= 2dk<<strong>br</strong> />

tgθ''sen<<strong>br</strong> />

θ<<strong>br</strong> />

1<<strong>br</strong> />

φ<<strong>br</strong> />

2<<strong>br</strong> />

k1 1<<strong>br</strong> />

=<<strong>br</strong> />

2 2<<strong>br</strong> />

2<<strong>br</strong> />

k AB = 2dk<<strong>br</strong> />

/ cos θ''<<strong>br</strong> />

(6.33a)<<strong>br</strong> />

(6.33b)<<strong>br</strong> />

Podemos ainda obter através das equações de Fresnel que<<strong>br</strong> />

2<<strong>br</strong> />

ρ=− ρ′ e ττ′ = 1−ρ<<strong>br</strong> />

. Desta forma o campo elétrico total na frente de<<strong>br</strong> />

onda será:<<strong>br</strong> />

E<<strong>br</strong> />

total<<strong>br</strong> />

= E +<<strong>br</strong> />

0<<strong>br</strong> />

1<<strong>br</strong> />

E 2 = E [ e i 1 e i 2<<strong>br</strong> />

0 ρ φ + ρ′ ττ′<<strong>br</strong> />

φ ] exp{<<strong>br</strong> />

− iωt<<strong>br</strong> />

}<<strong>br</strong> />

[ i ( φ −ωt)<<strong>br</strong> />

] [ 1 − ( 1−<<strong>br</strong> />

ρ 2 ) exp{<<strong>br</strong> />

i(<<strong>br</strong> />

φ − ) } ]<<strong>br</strong> />

= ρE<<strong>br</strong> />

exp<<strong>br</strong> />

φ (6.34)<<strong>br</strong> />

de forma que a intensidade será proporcional a:<<strong>br</strong> />

1<<strong>br</strong> />

2 2<<strong>br</strong> />

{ 1+<<strong>br</strong> />

( 1−<<strong>br</strong> />

ρ ) [ ( 1−<<strong>br</strong> />

ρ ) − 2cos(<<strong>br</strong> />

φ − ) ] }<<strong>br</strong> />

2 *<<strong>br</strong> />

2 2<<strong>br</strong> />

E total E total.<<strong>br</strong> />

E total = ρ E0<<strong>br</strong> />

2 φ1<<strong>br</strong> />

= (6.35)<<strong>br</strong> />

Se tivermos trabalhando com vidros teremos ρ ~ 0,2 ⇒ ρ 2 = 0,04<<strong>br</strong> />

1− ρ2<<strong>br</strong> />

= 0,<<strong>br</strong> />

96<<strong>br</strong> />

⇒ ( ) ~ 1. Então:<<strong>br</strong> />

A diferença de fases é:<<strong>br</strong> />

[ 1 − cos ( φ − ) ]<<strong>br</strong> />

2 2 2<<strong>br</strong> />

E total 2ρ<<strong>br</strong> />

E 0<<strong>br</strong> />

2 φ1<<strong>br</strong> />

2<<strong>br</strong> />

= (6.36)<<strong>br</strong> />

S. C. Zilio <strong>Óptica</strong> <strong>Moderna</strong> – <strong>Fundamentos</strong> e Aplicações<<strong>br</strong> />

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