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

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<strong>Óptica</strong> de cristais<<strong>br</strong> />

de propagação sˆ . O sistema formado por esta equação tem três equações<<strong>br</strong> />

homogêneas, que só tem solução não trivial se o seu determinante for<<strong>br</strong> />

igual a zero, ou seja:<<strong>br</strong> />

n<<strong>br</strong> />

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

xx<<strong>br</strong> />

+ n<<strong>br</strong> />

n<<strong>br</strong> />

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

s<<strong>br</strong> />

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

y<<strong>br</strong> />

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

n s s<<strong>br</strong> />

z<<strong>br</strong> />

2 ( s −1)<<strong>br</strong> />

s<<strong>br</strong> />

x<<strong>br</strong> />

x<<strong>br</strong> />

x<<strong>br</strong> />

n<<strong>br</strong> />

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

yy<<strong>br</strong> />

n<<strong>br</strong> />

s<<strong>br</strong> />

+ n<<strong>br</strong> />

s<<strong>br</strong> />

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

x y<<strong>br</strong> />

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

y<<strong>br</strong> />

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

n s s<<strong>br</strong> />

2 ( s −1)<<strong>br</strong> />

z<<strong>br</strong> />

y<<strong>br</strong> />

+ n<<strong>br</strong> />

2 ( s −1)<<strong>br</strong> />

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

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

n<<strong>br</strong> />

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

zz<<strong>br</strong> />

n<<strong>br</strong> />

n<<strong>br</strong> />

s<<strong>br</strong> />

s<<strong>br</strong> />

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

x z<<strong>br</strong> />

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

sys<<strong>br</strong> />

z<<strong>br</strong> />

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

z<<strong>br</strong> />

275<<strong>br</strong> />

(14.22)<<strong>br</strong> />

Desse determinante resulta uma equação biquadrada, cujas raízes<<strong>br</strong> />

fornecem quatro valores para n. Só iremos considerar as raízes positivas,<<strong>br</strong> />

uma vez que n é positivo por definição. Se usarmos como sistema de<<strong>br</strong> />

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

referência os eixos dielétricos principais, que diagonalizam o tensor n t , a<<strong>br</strong> />

equação biquadrada terá uma forma mais simples. Usando novamente a<<strong>br</strong> />

eq. (14.8), temos:<<strong>br</strong> />

4 2<<strong>br</strong> />

An + Bn + C = 0<<strong>br</strong> />

(14.23)<<strong>br</strong> />

onde:<<strong>br</strong> />

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

A = n s + n s + n s<<strong>br</strong> />

(14.24a)<<strong>br</strong> />

x<<strong>br</strong> />

x<<strong>br</strong> />

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

( 1−<<strong>br</strong> />

s ) n n + ( 1−<<strong>br</strong> />

s ) n n + ( 1 s ) n n<<strong>br</strong> />

x<<strong>br</strong> />

y<<strong>br</strong> />

z<<strong>br</strong> />

y<<strong>br</strong> />

y<<strong>br</strong> />

y<<strong>br</strong> />

B = − (14.24b)<<strong>br</strong> />

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

x<<strong>br</strong> />

x<<strong>br</strong> />

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

y<<strong>br</strong> />

z<<strong>br</strong> />

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

z<<strong>br</strong> />

z<<strong>br</strong> />

z<<strong>br</strong> />

C = n n n<<strong>br</strong> />

(14.24c)<<strong>br</strong> />

Resolvendo a eq. (14.23) encontramos os dois valores possíveis para n.<<strong>br</strong> />

Para se obter as componentes do campo elétrico, referentes a cada valor<<strong>br</strong> />

de n, basta substitui-lo na eq. (14.21).<<strong>br</strong> />

z<<strong>br</strong> />

14.4 Superfície normal<<strong>br</strong> />

Usando as eqs. (14.14) e (14.19) podemos escrever a eq. (14.17)<<strong>br</strong> />

na seguinte forma:<<strong>br</strong> />

⎛ωµε<<strong>br</strong> />

⎜<<strong>br</strong> />

⎜<<strong>br</strong> />

⎜<<strong>br</strong> />

⎝<<strong>br</strong> />

x<<strong>br</strong> />

− k<<strong>br</strong> />

k<<strong>br</strong> />

k<<strong>br</strong> />

y<<strong>br</strong> />

z<<strong>br</strong> />

k<<strong>br</strong> />

k<<strong>br</strong> />

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

y<<strong>br</strong> />

x<<strong>br</strong> />

x<<strong>br</strong> />

− k<<strong>br</strong> />

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

z<<strong>br</strong> />

ωµε<<strong>br</strong> />

y<<strong>br</strong> />

k<<strong>br</strong> />

− k<<strong>br</strong> />

k<<strong>br</strong> />

x<<strong>br</strong> />

z<<strong>br</strong> />

k<<strong>br</strong> />

k<<strong>br</strong> />

y<<strong>br</strong> />

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

x<<strong>br</strong> />

y<<strong>br</strong> />

− k<<strong>br</strong> />

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

z<<strong>br</strong> />

ωµε<<strong>br</strong> />

z<<strong>br</strong> />

k<<strong>br</strong> />

k<<strong>br</strong> />

x<<strong>br</strong> />

y<<strong>br</strong> />

k<<strong>br</strong> />

k<<strong>br</strong> />

− k<<strong>br</strong> />

z<<strong>br</strong> />

z<<strong>br</strong> />

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

x<<strong>br</strong> />

x<<strong>br</strong> />

− k<<strong>br</strong> />

y<<strong>br</strong> />

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

y<<strong>br</strong> />

⎞⎛E<<strong>br</strong> />

x ⎞<<strong>br</strong> />

⎟⎜<<strong>br</strong> />

⎟<<strong>br</strong> />

⎟⎜<<strong>br</strong> />

E y ⎟ = 0<<strong>br</strong> />

⎟⎜<<strong>br</strong> />

E ⎟<<strong>br</strong> />

⎠⎝<<strong>br</strong> />

z ⎠<<strong>br</strong> />

(14.25)

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