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Lasers
Lasers 161
The two parallel mirrors of a semiconductor laser are formed by
cleaving facets along a set of well-defined crystal planes. The fact that
the mirror surfaces are directly related to atomic planes guarantees
that the cleaved surfaces at either end of the laser device are parallel.
The reflectivity of the mirrors is given by the Frensel equation (Eq.
6.11) and is equal to about 30%. So 70% of the photons are transmit-
ted to the outside world. The rest are reflected and continue to pro-
voke stimulated emission inside the structure.
The two mirrors form a resonant cavity around the gain region. The
length of the gain region is many times longer than the wavelength of
light. Only selected wavelengths of light can exist in such a cavity, ex-
actly the same condition that de Broglie cited for his proposal that
electrons have a wavelength. That is, the lightwave must retrace the
same path in amplitude and phase for each round-trip circuit in the
cavity. The round-trip distance, 2L, must therefore be an integral
number of wavelengths, p , where p is an integer. This is the condi-
tion for constructive interference to occur. All other wavelengths are
excluded because they lead to destructive interference.
The eligible wavelengths inside the cavity are separated from each
other by a constant increment of frequency of the lightwave:
c 2L
f = and =
n p
pc c
f = Therefore, f = (7.14)
2Ln 2Ln
where n is the index of refraction. (For example n InP = 3.4.)
Example 7.3
Find the mode index of laser emission in a cavity of GaInAsP at 1500
nm.
This is equivalent to finding the number of wavelengths that can fit
in a cavity. The mode index is equal to 1 when one round-trip in the
cavity equals one wavelength. Assume that the cavity length is 400
m.
The refractive index of GaInAsP at = 1500 nm is about 3.5. Note
that the wavelength inside the cavity is only 1500/3.5 = 429 nm.
2L 800 m
mode index = p = = = 1864
0.429
p
The cavity length in number of wavelengths = = 466.
2
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