How Many Photons on a Pixel at a Given Exposure

How many photons impinge on a pixel illuminated by a known light source during exposure?  To answer this question in a photographic context we need to know the area of the pixel, the Spectral Power Distribution of the illuminant and the relative Exposure.

We know the pixel’s area and we know that the Spectral Power Distribution of a common class of light sources called blackbody radiators at temperature T is described by Spectral Radiant Exitance M(\lambda,T)  – so all we need to determine is what Exposure this irradiance corresponds to in order to obtain the answer.


The power of Spectral Radiant Exitance contained in light within a specified wavelength range is obtained by integrating M(\lambda,T)  over the range.  The result is called Irradiance E_e in W/m^2:

(1)   \begin{equation*} E_e = \int\limits_{\lambda_1}^{\lambda_2} M(\lambda,T) d\lambda \text{   $W/m^2$} \end{equation*}

As we have seen, this can be expressed in terms of photons as follows

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

where h and c are Planck’s constant and the speed of light respectively.


To determine what Exposure the equivalent quantities in (1) or (2) correspond to we simply convert the given Irradiance to photometric units by taking into consideration its visible wavelengths only, as described earlier.  The photometric version of Irradiance E_e is called Illuminance (E_v):

(3)   \begin{equation*} E_v = K_m \int\limits_{380}^{780} M(\lambda,T) V(\lambda) d\lambda \text{    $lumens/m^2$} \end{equation*}

where V(\lambda) is the photopic eye response function and K_m the luminous efficiency conversion constant 683 lm/W.  Note that lumens/m^2 are called lux (lx) and recall that Exposure H_v = E_v \cdot t lx-s, with t exposure time in seconds.

The number of photons per second N_{ph} per meter squared emitted from our blackbody source at the given temperature (2) results in illuminance E_v in lx (3).  This is what an incident light meter would show if directly exposed to it.

Irradiance divided by Illuminance

That’s all we need to tie the number of photons from a lambertian blackbody light source in the camera’s usable wavelength range to Exposure.  To obtain the number of photons incident on a square meter per lx-s of Exposure in those conditions we simply divide (2) by (3) – check how the units play out:

(4)   \begin{equation*} N_{ph}=\frac{ \int\limits_{\lambda_1}^{\lambda2} M(\lambda,T)\cdot \lambda \cdot d\lambda}{h c \cdot K_m \cdot \int\limits_{380}^{780} M(\lambda,T) V(\lambda) d\lambda}  \: \: \frac{photons}{\text{lx-s}\cdot m^2} \end{equation*}

This formula looks more complicated than it is because in practice it is easy to solve numerically.  For instance if we assume a ‘daylight’ illuminant of approximate temperature 5300K the number of visible-light* photons incident on a micron^2 per lx-s is about 11,260**.  I chose that temperature because that seems to be a decent average for the landscapes I shoot.

Below is a chart that shows the number of visible-light photons per micron^2 per lx-s that can be expected from a blackbody radiator at the indicated temperature with some approximate examples:

Photons per lxs per micron squared as a function of blackbody temp

Note how the curve is relatively flat around ‘Daylight’ temperatures typical of landscape photography.

As an example of how this data could be used one could say that 11,260 visible-light photons per micron squared on a D610 sensor with square pixels of 5.9 micron pitch would correspond to about 392,000 photons/lx-s/pixel.  The D610’s sensor saturates at about 1 lx-s at base ISO, so one can easily derive the number of impinging photons in various conditions from the above chart.

Incidentally, if the source is not a blackbody radiator but sunlight near the earth’s surface the figures change a bit.  Illuminant D53 is supposed to be representative of such light with a correlated color temperature of 5300K:

The number of photons in the ‘visible’* passband would be about 3% less for D53 than for the blackbody radiator at 5300K.

Now that we can calculate the number of photons incident on a pixel when exposed to a known light source the next step is determining how may photoelectrons are generated by the sensor under those conditions in order to estimate its Effective Quantum Efficiency.


*In these pages visible light is defined as the wavelength range 395-718nm, where the photopic eye response as measured by Sharpe et all (2005) is approximately 10 stops down from its peak at around 556nm.  These wavelengths correspond fairly well to the bandpass cutoffs at the infra-red and ultra-violet ends of typical monochrome and Bayer CFA digital camera sensors.  So ‘visible’ refers also to visibility by the sensing medium.

** A while back we arrived at similar results in a simplified way by assuming average light energy equal to that of the mean visible* wavelength, 556.5nm.  That approach presumed that light energy was the same throughout the wavelength range, equivalent to making M(\lambda,T) a constant.  If we make the same assumption with equation (4) above, we can pull out Spectral Radiant Exitance M(\lambda,T) from the integrals in the numerator and denominator, which will at that point cancel out.  (4) above would then reduce to equation (1) in the older post.

*** I understood most of the steps involved in this article only after reading Appendix A in Nakamura.