 Solution

The upper-state density is clamped near threshold once the laser operates, so excess pumping mainly increases photon density and output power. For approximately pump-independent absorption efficiency, $r=P_p/P_{p,\mathrm{th}}$, and both $P_{\mathrm{out}}$ and $f_{\mathrm{ro}}^2$ should rise approximately linearly with $P_p-P_{p,\mathrm{th}}$.

The left ordinate is the square of the frequency, in ${\mathrm{kHz}}^2$, rather than frequency in kHz. Reading the original triangles gives a threshold near $12$–$13\mathrm{mW}$ and a slope about $1.0{\mathrm{kHz}}^2/\mathrm{mW}$. A representative straight-line fit to visually read marker centers gives $P_{p,\mathrm{th}}\simeq 12.2\mathrm{mW}$ and $A=d{f}_{\mathrm{ro}}^2/d{P}_p\simeq 1.02\times 10^6{\mathrm{Hz}}^2/\mathrm{mW}$. ThereforeThe displayed precision is limited by graph reading and scatter; a lifetime around $0.15\mu \mathrm{s}$ is the meaningful result. The output squares have a compatible threshold and an approximately constant slope above it.

The round-trip time isIf mirror transmission is the only appreciable passive loss, energy decays by $R_1{R}_2=e^{-t_{\mathrm{rt}}/{\tau }_c}$ per round trip. For $R_1=1$,a power reflectance of roughly 65%. The exact exponential is preferable to replacing $1-R_2$ by $t_{\mathrm{rt}}/{\tau }_c$, since the inferred round-trip loss is not tiny. If distributed fibre loss is appreciable, the lifetime instead fixes the total round-trip retention; without a separate loss measurement, the bare mirror reflectance is not uniquely determined and is at least this retention factor.