How Many Photons on a Pixel

How many visible photons hit a pixel on my sensor?  The answer depends on Exposure, Spectral power distribution of the arriving light and effective pixel area.  With a few simplifying assumptions it is not difficult to calculate that with a typical Daylight illuminant the number is roughly 11,760 photons per lx-s per \mu m^2.  Without the simplifying assumptions* it reduces to about 11,000.

Light from a source is reflected by the scene, travels through the air, traverses the lens and finally reaches the camera’s sensor.   If we assume equal energy per small wavelength interval Spectral Power Distribution at the sensor, the mean number of photons n_{ph} incident on it can be estimated by the following formula:

(1)   \begin{equation*} n_{ph}= \frac{\lambda_2 - \lambda_1}{E_{m}}\cdot \frac{1}{683 \cdot \int\limits_{380_{nm}}^{780_{nm}} V(\lambda) d\lambda} \: ,\:\frac{photons}{lx\text{-}s \cdot \: m^2} \end{equation*}

where \lambda_1 and \lambda_2 represent the wavelengths of the ultra-violet and infra-red bandpass limits of the sensor expressed in nanometers, E_{m} is the mean energy of photons in the range and V(\lambda) is the photopic eye response (Luminous Efficiency) function.  V(\lambda) is integrated over the 380-780nm wavelength range because that’s typically how incident light meters measure it.

The first term of Equation 1 represents the mean number of photons arriving per unit area per second at the sensor.  The denominator of the second term represents illuminance at the sensing plane in lx.   The constant Spectral Power Distribution (in \frac{W}{m^2\cdot nm}, not shown) in the numerator of the first term and denominator of the second are one and the same, so they cancel out.

As we will see the application of the formula is fairly simple because we know the photopic CIE “physiologically-relevant” luminous efficiency function V(\lambda)  – so all we need to figure out is the average photon energy in the range of interest to get our answer.

We will take visible light to be the range of wavelengths above a certain visibility threshold, namely 10 stops below peak (photopic) eye response.  This yields a range of visible wavelengths between 395nm and 718nm inclusive, the mean wavelength being 556.5nm.  This is fairly typical of the bandpass let through by the infra-red and ultra-violet filters in recent DSC sensors (including Bayer).

For photographic purposes we can simplify things further and assume the average energy of visible photons – from a well behaved ‘Daylight’ illuminant, reflected by a neutral subject, going through a neutral lens – to be that of the wavelength in the middle of the range.  We can therefore estimate the energy of the average photon in the visible band as follows:

(2)   \begin{equation*} E_{m}= \frac{h \cdot c}{\lambda_{avg}} \: joules/photon \end{equation*}

with h Planck’s constant, c the speed of light in vacuum (or air) and wavelength \lambda in meters .  At 556.5nm E_{m} the mean photon has an energy of about 3.569 \times 10^{-19} joules.

The integral of the photopic function is simply the sum of its value at one nm interval from 380-780nm, equal to 113.042 in the same units as the numerator.   With this information we can calculate the approximate number of impinging photons using (1)

(3)   \begin{equation*} \begin{split} n_{ph} &= \frac{718.5 - 394.5 }{3.569\times 10^{-19} \times 683.002 \times 113.042} \\ \\ &= 1.1760 \times 10^{16} \: \:\frac{photons}{lx\text{-}s \cdot \:m^2}\\ \\ &= 11,760 \: \:\frac{photons}{lx\text{-}s \cdot \: \mu m^2} \end{split} \end{equation*}

This is the approximate result we are after, a number of photons per unit area per lx-s, these last ones the units of Exposure.  We’ll deal with the figure that results without assuming constant Spectral Power Distribution and photon energy in a later post, but suffice it to say for now that for a Daylight standard ‘D’ illuminant at 5300 degrees K (D53) it works out to about 11,000.

Let’s work through an example using the more accurate estimate.  You and/or your camera decide on an appropriate Exposure to obtain the desired result from a specific capture in daylight and pull the trigger.  How many visible photons impinged on a pixel that time?  All we need is the Exposure and the area of the pixel to plug into (3).  If you have an incident light meter you can simply read Exposure off that.

Alternatively you can take a look at your raw data and work it out backwards from there as follows.  At base ISO my D610’s highest raw level is 15778.  It starts clipping highlights at an exposure H_{sat} of 1.05 lx-s** and it has square pixels of 5.9 micron pitch.  The green channel normally clips first with light reflected off a neutral subject in daylight.  Therefore the number of visible photons incident on one of its green pixels at full scale (raw level 15778) for the given Exposure under daylight approximated by illuminant D53 is roughly

n_{ph} \approx 11,000 \times 1.05 \times 5.9^2 \approx 400,000 \: photons

…at 10% of full scale (raw level 1578) about 40,000 photons, at 1% (raw level 158) 4,000, you get the idea.  Cut exposure in half, the number of photons is cut in half throughout the dynamic range.  Keep in mind that not all pixels are effectively square or have 100% effective active areas.

The next step is to figure out how many of these photons actually get converted to photoelectrons.

 

* For a more accurate treatment of this topic see this post.

** To figure out at what Exposure your digital camera saturates head to DxOmark.com.  They measure the Exposure at which modern digital cameras saturate (or clip) under a Daylight illuminant.  I assume they look at the green channel which has wavelengths around the middle of the range and should clip first.  They publish the data in terms of sensor sensitivity S_{sat}, what they call Measured ISO.   We can recover the saturation exposure that they actually measured using the formula

(4)   \begin{equation*} H_{sat}= \frac{78}{S_{sat}} \: lx-s \end{equation*}

At base ISO the majority of current DSCs saturates at around 1 lx-s +/- 20%.  For example Ssat for the D610 at base ISO is 74, which means that DxOmark actually measured that it saturates at an exposure of 1.05 lx-s.

2 thoughts on “How Many Photons on a Pixel”

  1. It is so informative and very helpful for my master’s thesis. Can you please tell me the reference for this study?

    1. Hello A.K.M.Masum,

      Wikipedia is your friend, though if you read the next few posts (click ‘Next Post’ at the bottom of each article), you will better understand where the theory comes from. If you just want to read one, read this one.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.