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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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