 Solution

During the pulse, the photon density reaches its maximum when $\dot{n}=0$ with $n>0$, so $N^{\ast }=N_{\mathrm{th}}^{\ast }$. Evaluating i at this inversion givesThe ideal instantaneous output power is $P(t)=V_c\hslash {\omega }_{L}n(t)/{\tau }_c$. ThereforeFor $r\gg 1$, almost all the extractable energy is stored briefly as intracavity photons, and $P_{\max }\simeq E_{\max }/{\tau }_c$. The leading rise while the inversion remains near its initial value is $n\propto e^{(r-1)t/{\tau }_c}$, with short growth time ${\tau }_c/(r-1)$. Depletion then takes the inversion below threshold. At late times $N^{\ast }\simeq N_f^{\ast }$ andwhich approaches a tail with decay time ${\tau }_c$ at large $r$. The pulse is consequently asymmetric, with a rapid rise and a slower approximately exponential trailing edge. The negligible-seed build-up delay depends on the seed strength. The pumping and spontaneous population decay have been neglected over the short pulse; parasitic cavity loss multiplies useful power by the output fraction described in ii.