The Photon, the Quantum of Light — Question 3

Question 3

Detector efficiency and dark counts must be removed before inferring power

A photon counter reports $R=5.20\times10^5\,\mathrm{s^{-1}}$, including dark rate $D=2.00\times10^4\,\mathrm{s^{-1}}$. Detection efficiency is $\eta=.250$ at $\lambda=620\,\mathrm{nm}$. Ignore dead time. Use $hc=1240\,\mathrm{eV\,nm}$ and $1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J$.

1. Infer the incident photon rate and optical power.

2. Find the factor by which power is underestimated if every reported count is treated as one incident photon.

Question 3 - Solution

Detector efficiency and dark counts must be removed before inferring power

(a) Method and calculation

$$R=\eta\dot N+D\quad\Rightarrow\quad\boxed{\dot N=(R-D)/\eta=2.00\times10^6\,\mathrm{s^{-1}}}.$$$$\boxed{P=\dot N hc/\lambda=6.40\times10^{-13}\,\mathrm W}.$$

(b) Interpretation and further calculation

$$P_{\mathrm{naive}}=Rhc/\lambda=1.664\times10^{-13}\,\mathrm W.$$$$\boxed{P/P_{\mathrm{naive}}=(R-D)/(\eta R)=50/13\simeq3.84615}.$$

Physical check

Dark counts add to the measured rate while inefficiency removes real detections; both corrections are necessary.