The Photon, the Quantum of Light — Question 2

Question 2

Pulse energy fixes photon number while duration fixes peak flux

A rectangular optical pulse has energy $U=3.20\,\mathrm{nJ}$, wavelength $620\,\mathrm{nm}$ and duration $\tau=8.00\,\mathrm{ns}$. Pulses repeat at $f_r=1.00\,\mathrm{MHz}$. Use $hc=1240\,\mathrm{eV\,nm}$ and $1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J$.

1. Find the mean photon count per pulse and photon rate during a pulse.

2. Find peak and average power and verify their duty-factor relation.

Question 2 - Solution

Pulse energy fixes photon number while duration fixes peak flux

(a) Method and calculation

$$E_\gamma=hc/\lambda=2.00\,\mathrm{eV}=3.20\times10^{-19}\,\mathrm J,\quad\boxed{N=U/E_\gamma=10^{10}}.$$$$\boxed{\dot N_{\mathrm{pulse}}=N/\tau=1.25\times10^{18}\,\mathrm{s^{-1}}}.$$

(b) Interpretation and further calculation

$$\boxed{P_{\mathrm{peak}}=U/\tau=0.400\,\mathrm W},\quad\boxed{P_{\mathrm{avg}}=Uf_r=3.20\,\mathrm{mW}}.$$$$d=\tau f_r=.00800,\qquad\boxed{P_{\mathrm{avg}}=dP_{\mathrm{peak}}}.$$

Physical check

The count is an ensemble mean determined by pulse energy; repetition rate changes average power without changing photon energy.