Photons Emitted by Light Source

How many photons are emitted by a light source? To answer this question we need to evaluate the following simple formula at every wavelength in the spectral range we are interested in and add the values up:

(1)   \begin{equation*} N_{ph}= \frac{\text{Power of Light in }W/m^2}{\text{Energy of Average Photon in }J/photon} \end{equation*}

The astute reader will have realized that the units above are simply photons/m^2/s.  Written more formally:

(2)   \begin{equation*} N_{ph}= \int\limits_{\lambda_1}^{\lambda2} \frac{\Phi(\lambda)\cdot \lambda \cdot d\lambda}{hc} \text{  $photons/m^2/s$} \end{equation*}

where \lambda_1 and \lambda_2 limit the spectral range of interest, \Phi(\lambda) is the Spectral Power Distribution of the source (i.e. the power at every wavelength), \lambda is the wavelength of the light under evaluation and h and c are Planck’s constant and the speed of light respectively.

The spectral power distribution of some natural light sources like the sun or an incandescent light bulb are well understood and described by Planck and others as Lambertian Blackbody radiators.  Once stable they emit electromagnetic radiation of known spectrum and intensity as a function of wavelength \lambda and temperature T in degrees kelvin (K), properly termed Spectral Radiant Exitance in Watts/m^2/m:

(3)   \begin{equation*} M(\lambda,T)=\frac{2\pi h c^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)} \end{equation*}

with k Boltzmann’s constant.   We can express Spectral Radiant Exitance in units of photons instead of watts using (1).  The result is shown in the plot below with the number of photons emitted per square meter per second at the indicated wavelength in the visible range (which we will take to be 395-718nm, a typical passband for current digital camera Color Filter Arrays):

Spectral Radiant Exitance Photons

To obtain the total number of photons emitted by a blackbody radiator of known temperature in the visible range we integrate Spectal Radiant Exitance (3) over the visible range with equation (2):

(4)   \begin{equation*} N_{ph}= \int\limits_{\lambda_1}^{\lambda2} \frac{2\pi h c \cdot d\lambda}{\lambda^4(e^{\frac{hc}{\lambda kT}}-1)} \text{  $photons/m^2/s$} \end{equation*}

It is easily done numerically.  For instance the ‘daylight’ sun is considered to be a blackbody radiator of around temperature 5300K.  In units better suited to a photographic sensor context it therefore emits about 4.549 \times 10^{13} photons/s/micron^2 in the visible range as defined earlier.  All of these photons do not necessarily make it through the atmosphere which absorbs varying amounts of light energy depending on wavelength.  On the other hand a household incandescent bulb at around 3000K emits about 1.267 \times 10^{12} photons/s/micron^2 in the same range, most of which make it to our sensors.

Now that we know how to calculate how many photons are emitted by a blackbody light source of known temperature, the next step is relating this to Photography.  How many photons arrive from such a source at the Illuminance/Luminance/Exposure read out by a light meter?