The Photon, the Quantum of Light — Question 4

Question 4

A flat power spectrum is not a flat photon-number spectrum

An ideal beam has constant power per unit wavelength $S_\lambda=P/(\lambda_2-\lambda_1)$ between $\lambda_1=400\,\mathrm{nm}$ and $\lambda_2=800\,\mathrm{nm}$, and zero elsewhere. Use $hc=1240\,\mathrm{eV\,nm}$ and $1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J$.

1. Integrate the photon rate over wavelength and find the mean photon energy.

2. Find the fraction of photons with wavelengths above $600\,\mathrm{nm}$.

Question 4 - Solution

A flat power spectrum is not a flat photon-number spectrum

(a) Method and calculation

$$\dot N=\int_{\lambda_1}^{\lambda_2}\frac{S_\lambda\lambda}{hc}\,d\lambda=\boxed{\frac{P(\lambda_1+\lambda_2)}{2hc}}.$$$$\boxed{\overline E=P/\dot N=2hc/(\lambda_1+\lambda_2)=2.06667\,\mathrm{eV}}.$$

(b) Interpretation and further calculation

$$F_{>600}=\frac{\int_{600}^{800}\lambda\,d\lambda}{\int_{400}^{800}\lambda\,d\lambda}.$$$$\boxed{F_{>600}=\frac{800^2-600^2}{800^2-400^2}=7/12}.$$

Physical check

The number spectrum contains an extra factor of wavelength because longer-wavelength photons carry less energy.