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

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Ação laser<<strong>br</strong> />

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

245<<strong>br</strong> />

kql- [tg -1 (z2/z0) -tg -1 (z1/z0)] +( θm1+θm2)/2 = qπ (12.5)<<strong>br</strong> />

onde o fator ½ vem da consideração de meia volta, como já fizemos na<<strong>br</strong> />

seção 11.3. Entretanto, diferentemente daquela análise, a cavidade agora<<strong>br</strong> />

está preenchida com o meio ativo e neste caso o índice de refração, e<<strong>br</strong> />

consequentemente o vetor de propagação k, será alterado pela ressonância.<<strong>br</strong> />

Neste caso temos:<<strong>br</strong> />

ωqn<<strong>br</strong> />

⎛ χ′ ( ν)<<strong>br</strong> />

⎞<<strong>br</strong> />

⎜1+<<strong>br</strong> />

⎟ − 2<<strong>br</strong> />

c ⎝ 2n<<strong>br</strong> />

⎠<<strong>br</strong> />

l<<strong>br</strong> />

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

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

[ tg (z /z ) -tg (z /z ) ] + ( θ + θ )/2 = qπ<<strong>br</strong> />

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

Se considerarmos uma cavidade vazia (χ´ = 0) obtemos:<<strong>br</strong> />

ν<<strong>br</strong> />

q<<strong>br</strong> />

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

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

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

m1<<strong>br</strong> />

m2<<strong>br</strong> />

qc c ⎡ -1<<strong>br</strong> />

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

( θm1<<strong>br</strong> />

+ θm<<strong>br</strong> />

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

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

⎤<<strong>br</strong> />

⎢⎣<<strong>br</strong> />

tg (z 2/z<<strong>br</strong> />

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

-tg (z1/z<<strong>br</strong> />

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

−<<strong>br</strong> />

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

2πnl<<strong>br</strong> />

2 ⎥⎦<<strong>br</strong> />

que quando substituído na eq. (12.6) resulta em:<<strong>br</strong> />

(12.6)<<strong>br</strong> />

(12.7)<<strong>br</strong> />

⎛ χ′ ( ν)<<strong>br</strong> />

ν 1<<strong>br</strong> />

⎞<<strong>br</strong> />

⎜ + = ν 2 ⎟<<strong>br</strong> />

(12.8)<<strong>br</strong> />

q<<strong>br</strong> />

⎝ 2n<<strong>br</strong> />

⎠<<strong>br</strong> />

de onde vemos que a freqüência é modificada pela presença da<<strong>br</strong> />

ressonância atômica. Esta é uma equação transcendental e para simplificar<<strong>br</strong> />

sua solução vamos utilizar as equações (10.4) para escrever:<<strong>br</strong> />

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

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

ν 0 − ν)<<strong>br</strong> />

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

( ν 0 − ν)<<strong>br</strong> />

χ′ ( ν)<<strong>br</strong> />

= χ ′<<strong>br</strong> />

( ν)<<strong>br</strong> />

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

γ(<<strong>br</strong> />

ν)<<strong>br</strong> />

(12.9)<<strong>br</strong> />

Δν<<strong>br</strong> />

kΔν<<strong>br</strong> />

onde na última passagem utilizamos γ(ν) = kχ”(ν)/n 2 . Substituindo na eq.<<strong>br</strong> />

(12.8) e considerando que o ganho se estabiliza no valor de limiar,<<strong>br</strong> />

teremos:<<strong>br</strong> />

⎛ ( ν 0 − ν)<<strong>br</strong> />

γ t ( ν)<<strong>br</strong> />

⎞<<strong>br</strong> />

ν ⎜1+<<strong>br</strong> />

⎟ = ν<<strong>br</strong> />

(12.10)<<strong>br</strong> />

q<<strong>br</strong> />

⎝ kΔν<<strong>br</strong> />

⎠<<strong>br</strong> />

e considerando que ν será muito próximo de νq,<<strong>br</strong> />

c ⎛ 1<<strong>br</strong> />

ν ≈ ν<<strong>br</strong> />

⎞<<strong>br</strong> />

q − ( ν q − ν 0 ) ⎜α<<strong>br</strong> />

− ln( r1r2<<strong>br</strong> />

) ⎟<<strong>br</strong> />

2πnΔν<<strong>br</strong> />

⎝ l ⎠<<strong>br</strong> />

(12.11)

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