The Photon, the Quantum of Light — Question 9

Question 9

A required detection probability determines the mean photon budget

A weak pulse has Poisson-distributed incident photon number of mean $\mu$. Each photon is detected independently with probability $\eta=.200$. The detected count is Poisson with mean $\eta\mu$. Require at least one detection with probability .950. Use $hc=1240\,\mathrm{eV\,nm}$ and $1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J$.

1. Find the minimum incident mean photon number.

2. At wavelength 620 nm, find the required mean pulse energy and explain why a noninteger mean is admissible.

Question 9 - Solution

A required detection probability determines the mean photon budget

(a) Method and calculation

$$P(N_D\ge1)=1-P(N_D=0)=1-e^{-\eta\mu}.$$$$1-e^{-.200\mu}\ge.950\quad\Rightarrow\quad\boxed{\mu\ge5\ln20=14.9787}.$$

(b) Interpretation and further calculation

$$\overline U=\mu hc/\lambda=(5\ln20)(3.20\times10^{-19}\,\mathrm J).$$$$\boxed{\overline U_{\min}=4.79317\times10^{-18}\,\mathrm J}.$$

Physical check

Each individual count is an integer; the average over many pulses need not be.