How Lasers Work — Question 10

Question 10

Stored inversion bounds the energy of a rapidly extracted laser pulse

A gain volume contains $N_i=5.00\times10^{15}$ excited atoms before a short extraction pulse. Lasing stops at $N_f=1.00\times10^{15}$. Each stimulated transition emits a photon of energy $E_L=2.00\times10^{-19}\,\mathrm J$. Assume negligible pumping and spontaneous decay during extraction. Of the extracted photons, fraction $\eta_o=.60$ leaves through the useful output port.

1. Find the maximum stimulated energy released from the atoms and useful output energy.

2. For a Gaussian output pulse of intensity FWHM $\Delta t=10.0\,\mathrm{ns}$, find its peak power using $U=P_{\mathrm{pk}}\Delta t\sqrt{\pi/(4\ln2)}$.

Question 10 - Solution

Stored inversion bounds the energy of a rapidly extracted laser pulse

(a) Method and calculation

$$N_\gamma=N_i-N_f=4.00\times10^{15}\Rightarrow\boxed{U_{\mathrm{stim}}=N_\gamma E_L=.800\,\mathrm{mJ}}.$$$$\boxed{U_{\mathrm{out}}=\eta_oU_{\mathrm{stim}}=.480\,\mathrm{mJ}}.$$

(b) Interpretation and further calculation

$$P_{\mathrm{pk}}=\frac{U_{\mathrm{out}}}{\Delta t}\sqrt{\frac{4\ln2}{\pi}}.$$$$\boxed{P_{\mathrm{pk}}=4.509\times10^4\,\mathrm W=45.09\,\mathrm{kW}}.$$

Physical check

The residual excited population cannot be counted as extracted photons. This energy bound assumes each depleted upper-state atom contributes one stimulated photon and that the stated output fraction accounts for subsequent losses.