How Lasers Work — Question 6

Question 6

Passive photon lifetime follows the logarithm of the round-trip retention

A cavity has round-trip time $t_r=2.00\,\mathrm{ns}$. With gain switched off, each round trip retains fraction $\mathcal R=.90$ of its intracavity energy. Define photon lifetime by $U(t)=U_0e^{-t/\tau_c}$ at successive round-trip times.

1. Derive and calculate $\tau_c$.

2. Find the time for the energy to fall to one percent and the approximate number of round trips.

Question 6 - Solution

Passive photon lifetime follows the logarithm of the round-trip retention

(a) Method and calculation

$$U(mt_r)=U_0\mathcal R^m=U_0e^{m\ln\mathcal R}=U_0e^{-mt_r/\tau_c}.$$$$\boxed{\tau_c=-t_r/\ln\mathcal R=18.982\,\mathrm{ns}}.$$

(b) Interpretation and further calculation

$$.01=e^{-t/\tau_c}\Rightarrow\boxed{t=\tau_c\ln100=87.42\,\mathrm{ns}}.$$$$\boxed{t/t_r=43.71};\qquad\boxed{44\text{ complete round trips to fall below }1\%}.$$

Physical check

The continuous exponential matches the exact discrete decay at round-trip times. Replacing minus log R by one minus R is only a small-loss approximation.