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