Exposure, Radiometry

Photons Emitted by Light Source

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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 of interest and add the values up:

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

The Power of Light emitted in W/m^2 is called Spectral Exitance, with the symbol M_e(\lambda) when referred to  units of energy.  The energy of one photon at a given wavelength is

(2)   \begin{equation*} e_{ph}(\lambda) = \frac{hc}{\lambda}\text{ joules/photon} \end{equation*}

with \lambda the wavelength of light in meters and h and c Planck’s constant and the speed of light in the chosen medium respectively.  Since Watts are joules per second the units of (1) are therefore photons/m^2/s.  Writing it more formally:

(3)   \begin{equation*} M_{ph} = \int\limits_{\lambda_1}^{\lambda_2} \frac{M_e(\lambda)\cdot \lambda \cdot d\lambda}{hc} \text{ $\frac{photons}{m^2\cdot s}$} \end{equation*}

where wavelengths \lambda_1 and \lambda_2 limit the spectral range of interest, M_e(\lambda) is the Spectral Power Distribution emitted by the source (i.e. the radiated power per unit area of the source at every wavelength).

Blackbody Spectral Photon Distribution

The Spectral Exitance M_e(\lambda) of a class of natural light sources like the sun or an incandescent light bulb are well understood and can be grossly approximated  by Lambertian Blackbody radiators as described by Planck and Einstein-Bose.  Ignoring transmission losses, once stable they emit electromagnetic radiation of known spectrum and intensity as a function of wavelength \lambda in meters and absolute temperature T in degrees Kelvin.  Assuming perfect emittance

(4)   \begin{equation*} M_e(\lambda,T)=\frac{2\pi h c^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}\text{ $\frac{W}{m^2\cdot m}$} \end{equation*}

with k Boltzmann’s constant and the others as above.   To express Spectral Radiant Exitance in units of photons/s instead of Watts we divide by photon energy as instructed:

(5)   \begin{equation*} M_{ph}(\lambda,T)=\frac{2\pi c}{\lambda^4(e^{\frac{hc}{\lambda kT}}-1)} \text{ $\frac{photons}{m^2\cdot s \cdot m}$} \end{equation*}

The m^2 in the unit denominator refers to unit area on the source, while the m refers to the wavelength of light and it is in practice often converted to nm.  The result of Equation (5) is shown in the plot below for the visible range, with the number of photons corresponding to an area of a \mu m^2 per second per nm of bandwidth at the indicated wavelength:

Spectral Radiant Exitance in Photons/s
Figure 1. Photon flux from a Lambertian blackbody radiator at the given temperature

To obtain the total number of photons per unit area per second from a blackbody radiator of known temperature in a given wavelength range we integrate Spectral Exitance in Equation (5) over the range:

(6)   \begin{equation*} M_{ph}(\lambda,T) = \int\limits_{\lambda_1}^{\lambda_2} \frac{2\pi c \cdot d\lambda}{\lambda^4(e^{\frac{hc}{\lambda kT}}-1)} \text{ $\frac{photons}{m^2\cdot s}$} \end{equation*}

It is easily done numerically.  For instance once it reaches earth the ‘daylight’ sun is close to a blackbody radiator of around temperature 5300K.  In units better suited to digital imaging, a blackbody at that temperature emits about 4.5713 \times 10^{13} photons/s/\mu m^2 in the visible range, which here we will take to be 395-718nm, a typical passband for current digital camera Color Filter Arrays.  See the Appendix for why this choice of units makes sense in a photographic context.

All of these photons do not necessarily make it through the atmosphere which absorbs varying amounts of light energy depending on wavelength, see the next post for additional information on this.  On the other hand a household incandescent bulb at around 3000K emits about 1.2788 \times 10^{12} photons/s/\mu m^2 in the same range,  a good proportion of which should make it to our sensors after accounting for reflection and other losses.

Now that we know how to calculate how many photons to expect from a blackbody light source of known temperature, the next step is relating this to Photography.    What  Illuminance, Luminance and Exposure will a light meter read when so many photons hit it?

 

Appendix: Exitance to Radiance to Irradiance

Once it has been emitted and it is moving in free space, M is no longer referred to as Exitance – but it is called Irradiance instead, with symbol E.  It is effectively the same quantity in the same units of photons/m^2/s (or W/m^2) but the area now refers to the location in space where the relative photon flux \Phi is flowing through.

At a given small area A somewhere in space perpendicular to the source and light travel, Irradiance is by definition

(7)   \begin{equation*} E(\lambda) =\frac{\Phi(\lambda)}{A} = L(\lambda)\Omega \end{equation*}

with L(\lambda) Radiance in units of Watts or photons per m^2 per steradian (sr) per nm – and \Omega the solid angle of the source as seen from A in steradians.

For an ideal Lambertian source Radiance L(\lambda) is constant in all directions above it.  The relationship between Exitance and Radiance can be found by integrating over flux in all directions yielding**

(8)   \begin{equation*} L (\lambda)= \frac{M(\lambda)}{\pi} \end{equation*}

with \pi in units of sr.  The beauty of Radiance (and its photometric cousin Luminance) L is that it is Irradiance (or Illuminance) E into a solid angle \Omega, the latter representing a spherical cap normalized for distance.  Radiance L is preserved as it travels through space and through optics – other than of course for reflection and transmission losses.  The preservation of Radiance/ Luminance is related to the concepts of conservation of energy and Throughput (Etendue).

To estimate the Irradiance on a pixel (i.e. the number of photons per unit area per second impinging on it) given Radiance L, simply determine the projected solid angle \Omega of incoming light seen by the pixel and apply Equation (7).  In photography the solid angle is defined by the size of the Exit Pupil of the lens and its distance from the image plane.  In air it can be approximated as

(9)   \begin{equation*} \Omega \approx \frac{\pi}{4N^2}, sr \end{equation*}

for a circular aperture, with N the working f-number of the lens.  Combining  (7) and (9) we obtain the so-called ‘Camera Equation‘.  Combining all three of the Equations just above we see that in a digital imaging context Irradiance E_i on the image plane from a Lambertian Source is proportional to Exitance M and, ignoring reflections and losses along the way, the two are related by a simple constant:

(10)   \begin{equation*} E_i (\lambda) \approx \frac{M(\lambda)}{4N^2}. \end{equation*}

Since f-number N is dimensionless, M and E_i have the same units we started with.   If it surprises you that  the two are roughly of the same scale independently of the distance that separates them, just remember that the emitting and receiving areas are vastly different and related by lens magnification squared.

Reflection and transmission losses simply reduce the incoming Radiance/Luminance, as shown in the dedicated article above.  Once they are taken into account, by the time the original Irradiance from the source reaches a small area on the sensor there will only be a few to a few thousand photons per micrometer squared during a typical photographic Exposure.

 

* The Octave/Matlab code used to generate Figure 1 can be downloaded from here.

** See the Note at the bottom of the article on Angles and the Camera Equation.

 

6 thoughts on “Photons Emitted by Light Source”

  1. I really appreciate this whole series of photon related knowledge, which is really helpful for me to understand how much photons that a single camera pixel would capture.
    Besides, I have a small question about Equation (4). I think there is no “h” in the numerator. Since by taking Equation (3) to Equation (2), the h will be eliminated. Hope that I didn’t miss any important details.

    Reply

      1. Firstly, thank you for writing all these articles. I’m learning a lot from reading them.

        In equation (6) the per m² refers to per unit area of the emitting object but in the following paragraph you quote a number per square micrometre as being more appropriate for a photographic sensor. I think this is a mistake, or I’ve misunderstood something: you’d need to multiply by (R/D)² to get a value appropriate for the sensor, where R is the radius of the Sun and D is the distance to the Sun.

        Reply

        1. Hi Andrew,

          The starting point for Exitance is Planck’s Blackbody Radiation Law, which represents power per unit solid angle and per unit of area normal to propagation. So in the form shown above it is the density (intensity) of light per unit area per second.

          Jack

          Reply

          1. Hi Jack

            The Planck Law gives the power (per unit wavelength) per unit area of the *emitting* surface of a black body. Another way to see the issue : equation (6) couldn’t be for the power per unit area of the receiver because the distance to the receiver does not appear in it.

            Andrew

            Reply

        2. It appears I misunderstood your question Andrew. It’s a good one, perhaps I should add a paragraph about it above [Edit: I added it as an Appendix]. The reason why I bring it down to micrometers is that that is the scale of digital imaging pixels. And the reason why distance is not explicitly involved is that there are steradians hiding in there.

          Of course photons will spread out according to the relevant laws based on the physical setup of the illuminant, whether it be sunlight, moonlight or a light bulb. However in a photographic context, Radiant Exitance M from an ideal Lambertian source can be expresses as πL, with L Radiance from the source being conserved until it hits a very small pixel, ignoring reflection and transmission losses.

          Once one realizes that fact it becomes clear that the units of Spectral Exitance from the source and Irradiance on the sensor are the same up to a constant, which in photographic applications is of the order of the solid angle subtended by lens aperture at the pixel. You can find more detail on this in the article on camera angles:

          https://www.strollswithmydog.com/camera-equation-angles/#Camera

          Jack

          Reply

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