Exposure, Photometry, Radiometry, Sensors

How Many Photons on a Pixel at a Given Exposure

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

We can typically estimate the pixel’s effective area and the Spectral Power Distribution of the illuminant – so all we need to determine is what Exposure the relative irradiance corresponds to in order to obtain the answer.

Irradiance

The Spectral Power Distribution of an illuminant is its power per unit area at every small wavelength interval, I will give it the symbol M(\lambda) here, in units of Watts/m^2/nm for visible light applications.  The result over the wavelength range of interest (\lambda_1 to \lambda_2 ) is called Irradiance E_e, in W/m^2:

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

Since Watts (W) are joules per second and the Planck-Einstein relation informs us that the energy of one photon is given by \frac{hc}{\lambda}  joules per photon, Equation (1) can be expressed as a number of photons per second per meter squared as follows

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

where h and c are Planck’s constant and the speed of light respectively.  The quantity under the integral represents a number of photons per small wavelength interval (per second per unit area) – so it can be considered a Spectral Photon Distribution.

Illuminance

To determine what Exposure the equivalent quantities in Equation (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) V(\lambda) d\lambda, \text{ $lumens/m^2$ (or $lx$ )} \end{equation*}

where V(\lambda) is the photopic eye response function and K_m the luminous efficiency conversion constant, about 683.002 lm/W.   Note that in SI photometry lumens/m^2 are referred to as lux ( lx ).

The number of photons per second per square meter hitting the lens per Equation (2) results in illuminance E_v in lx per equation (3).  The latter 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 known illuminant to Exposure (H_v), simply divide (2) by (3) – check how the units play out:

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

lx–s, the units of Exposure, are in the denominator above because E_v \cdot t = H_v , with t exposure time in seconds.  The formula looks more complicated than it is but in practice it is easy to solve numerically.

Blackbodies

For instance, a common class of light sources like tungsten bulbs and stars can be considered Lambertian Blackbody Radiators, with Spectral Power Distribution defined by their absolute temperature T and termed Spectral Radiant Exitance M_e(\lambda,T), as discussed in an earlier article.

For a given photographic setup Exitance and Irradiance are related by a constant, which will however cancel out in Equation (4) since it appears both in the numerator and the denominator (see the Appendix in the article above).  We can therefore use either quantity interchangeably for this calculation.

If we assume a blackbody illuminant of approximate absolute temperature 5300 degrees Kelvin (refer to Figure 2) the number of visible-light* photons incident on a micron^2 per lx-s is about 11,170**.  I chose that temperature because that seems to be 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
Figure 1. Photons per lx-s per micron squared as a function of Lambertian Blackbody Radiator absolute temperature.

Note how the curve is relatively flat around ‘Daylight’ temperatures typical of landscape photography.  So far, so similar to Nakamura.****

Natural Daylight

The sun is considered to be a blackbody radiator out in space. However by the time sunlight reaches the surface of the earth it has had to go through the atmosphere, which filters it non-uniformly.  Illuminant D is a model for the relative spectral power distribution of sunlight on earth.  Below you can see the difference in Spectral Photon Distribution of D53, representative of such light with a correlated color temperature of about 5300K, compared to a blackbody source of the same temperature – and equi-energy Illuminant E:

Figure 2. Spectral Photon Distribution of a Blackbody Radiator at a temperature of 5300K, standard illuminant D at a correlated color temperature of 5300K, and equi-energy standard illuminant E.

Assuming the same measured illuminance, the number of photons in the ‘visible’* passband in natural daylight in typical landscape shooting conditions is estimated to be a few % less for D Illuminants than for a pure blackbody radiator at the same temperature, say about 11,000 photons/\mu m^2/lx-s  (+/- 50 or so).    You can peruse the Matlab script used for the calculations in the link at the bottom of the article***

As an approximate example of how this data can be used one could say that 11,000 visible-light photons per micron squared on a D610 sensor with square pixels of 5.9 micron pitch and 100% efficient microlenses would correspond to about 382,000 photons/lx-s/pixel.  The D610’s sensor starts to saturate at about 1.05 lx-s under daylight at its base ISO, 100, so one can easily derive the number of impinging photons in various settings from the earlier chart.

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 arbitrarily 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 loosely to the bandpass cutoffs at the infra-red and ultra-violet ends of typical monochrome and Bayer Color Filter Array 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 the given Spectral Exitance independent of wavelength.  If we make the same assumption in equation (4) above, we can pull out M(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.

*** The Matlab code used in the calculations can be downloaded by clicking here.  Note that some of the figures in the article were updated with the 2019 redefinition of SI base units but the attached code was not.

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

2 thoughts on “How Many Photons on a Pixel at a Given Exposure”

  1. Hello! My name is Bob Burridge. I would like to to produce an RGB image corresponding to the sum total of the number of photons contributing the corresponding pixels in several RGB images of the same size – a kind of stacking without averaging.

    Reply

    1. Hello Bob,

      so a sum of the various individual captures? By ‘corresponding to’ do you mean ‘proportional’? If so, ignoring color, just add up the captured raw data from the available images pixel by pixel in floating point and demosaic it to taste.

      If you do not have the raw data but just (s?)RGB images, things become messy because of non-linear corrections and tone curves applied during rendering that are not proportional to the number of incoming photons. At the very least you would have to undo the appropriate gamma and a guesstimated tone curve before performing the addition.

      Reply

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