Answer:
Diode Lasers
Consider a InGaAsP-InP laser diode which has an optical cavity of length 250
microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is
4. The optical gain bandwidth (as measured between half intensity points) will
normally depend on the pumping current (diode current) but for this problem
assume that it is 2 nm.
(a) What is the mode integer m of the peak radiation?
(b) What is the separation between the modes of the cavity? Please express your
answer as Δλ.
(c) How many modes are within the gain band of the laser?
(d) What is the reflection coefficient and reflectance at the ends of the optical
cavity (faces of the InGaAsP crystal)?
(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction
respectively. Estimate the x and y dimensions of the laser cavity. (Assume the
beam is a Gaussian beam with the waist located at the output. And the beam
waist size is approximately the x-y dimensions of the cavity.)
Solution:
(a) The wavelength λ of a cavity mode and length L are related by
n
mL
2
λ = , where m is the mode number, and n is the refractive index.
So the mode integer of the peak radiation is
1290
1055.1
10250422
6
6
= ×
××× == −
−
λ
nL
m .
(b) The mode spacing is given by nL
c f 2
=Δ . As
λ
c f = , λ
λ
Δ−=Δ 2
c f .
Therefore, we have nm
nL f
c
20.1
)10250(42
)1055.1(
2 || 6
2 2 26
= ×××
× ==Δ=Δ −
− λλ λ .
(c) Since the optical gain bandwidth is 2nm and the mode spacing is 1.2nm, the
bandwidth could fit in two possible modes.
For mode integer of 1290, nm
m
nL 39.1550
1290
10250422 6
= ××× ==
−
λ
Take m = 1291, nm
m
nL 18.1549
1291
10250422 6
= ××× ==
−
λ
Or take m = 1289, nm
m
nL 59.1551
1289
10250422 6
= ××× ==
−
λ .
Explanation:
