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2.2 Theoretical Analysis 33
vertically) are amplified per pass by

e 2gL (2.1)

but they are decayed per pass by
e −2αL R 1 R 2 . (2.2)

The lasingcondition is that the product of (2.1) and (2.2) is larger than 1,
namely
e
e 2gL −2αL R 1 R 2 ≥ 1. (2.3)
Therefore, the lasingthreshold gain g th is expressed as
1 1
g th = α + ln , (2.4)
2L R 1 R 2

where α =10 cm −1 ,L = 300 µm,R 1 = R 2 =0.32, then g th ≥ 49 cm −1
The light travels in the active region infinitesimally to establish a high
value of radiation density. The lasing light in the resonator generates the
standingwave fields. Since the integral multiple m of the half-wavelength
equals the cavity length L
λ
m = L. (2.5)
2n
When λ =1.3 µm,n =3.5and L = 300 µm, the integer m equals 1615. The
wavelength difference ∆λ when m increases by 1 becomes

λ 2
∆λ = − , (2.6)
2n eff L
where n eff is the effective refractive index of the LD. In terms of wave numbers,
the frequencies are separated by an interval of 1/∆λ. The calculated mode
interval of ∆λ equals 0.76 nm for λ =1.3 µm,n eff =3.7, and L = 300 µm.
The lasing-mode shift corresponds to the integral multiple of ∆λ.
λ 2 λ
Example 2.1. Derive the mode interval ∆λ = − from m = L, where
2n eff L 2n
λ is the wavelength, n eff is the effective refractive index, m is the integer, and
L is the cavity length of the LD.
Solution. Assumingthat refractive index n does not dependent on λ, (2.5)
becomes
2nL
λ = . (E2.1)
m
When m increases by 1, the wavelength becomes
2n
λ +∆λ = L. (E2.2)
m +1