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How Many Photons in a Photograph?

Inte The Light

Into The Light, Leica M6, Summicon 50mm, Tri-X, ~1/250 @ f5.6. But how many photons?

We photographers deal in photons, those baffling quanta of light. But just how many photons does it take to make a photo, to expose a single 35mm negative?

Let’s assume an arbitrary exposure at EV13, which is 1/1000 @ f5.6, ISO 400.  And let’s assume we are taking the worlds most boring photograph – a photo of an 18% grey card.  The luminance of the scene will be 20,480 lux or 1,903 foot candles.

But how does that relate to number of photons? You’ll all remember, of course, that the lux to PPF (μmol m-2 s-1) conversion factor for sunlight is 0.0185, and that 1 µmol = 6.02*1017 photons.

In this case the time can only be the exposure time of 0.001s. But what’s the area?

It can’t be the sensor area, because that would remain constant when we change the aperture/shutter speed combination (which changes the time factor). Based on that logic, aperture area seems a good candidate, but aperture area changes with focal length, and a constant numerical aperture across a focal length range will produce a similarly exposed negative for a given shutter speed (also evidenced by the fact EV values don’t change with focal length).

It must be some combination of aperture size and focal length, otherwise known as the f-stop, that we need to calculate the ‘area’. But how?

Having worked ourselves into a bit of a corner, let’s trying another angle – the definition of ISO’s sensor speed ratings.  This provides some insight.

From the gospel of Wikipedia: “ISO speed ratings of a digital camera are based on the properties of the sensor and the image processing done in the camera, and are expressed in terms of the luminous exposure H (in lux seconds) arriving at the sensor.”

Looking good. Let’s first find H:
H = q * L * t / N2

L is luminance of the scene (in candela per sqm, just to introduce another unit), t is the length of the exposure is seconds, q is a factor calculated from: “transmittance T of the lens, the vignetting factor v(θ), and the angle θ relative to the axis of the lens”, and N is aperture number.

t = 0.001, easy
N = 5.6, bosh
q = 0.65, based on ‘typical’ values
L = oh this is getting complicated again.

Luminance of the scene? “The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540*1012 hertz and that has a radiant intensity in that direction of 1⁄683 watt per steradian.” We’re going to need some approximation, some short cut here.

I think the problem here is candelas are used in the measurement of focussed light, the intensity of light, which would seem appropriate in this instance, but Lux is more useful for ambient light levels. To calculate this at different parts of the sensor would be possible, and we could then integrate over the surface…nope, not going to do that.

I found an online conversion tool that claimed to provide a ‘lux equivalent’ to ‘candela per sqm’ conversion.  Let’s go with that (this is probably the biggest hole in the working out, as I don’t understand how this conversion is done, or what ‘lux equivalents’ are):
20,480 lx ~= 6,512 candela per sqm

So now we have an H: 0.65 * 6512 * 0.001 / 5.62 = 0.135 lx s

But at this point, I’m going to abandon this route, as I fear I’ve been considering the ambient light levels, rather than the light that actually reaches the sensor.

Let’s work the other way around, by considering the ISO sensor speed calculation of S = 10/Hsos “where Hsos is the exposure that will lead to values of 118 in 8-bit pixels, which is 18 percent of the saturation value in images encoded as sRGB or with gamma = 2.2″ and S is the linear sensitivity number.  So here, for ISO 400:
Hsos = 10/400 = 0.025 lx s

Let’s do a check with a formula from another source. Kodak use this formula to determine the saturation ISO sensitivity of their sensors:
ISO = 15.4 * f2 / (L * t)

Where L is luminance, t is time, and f is f-stop number. That gives, in this case an L (cd/m^2) of:
L = 15.4 * 5.62 / (400 * 0.001) = 1207 cd/sqm

Do these numbers tie up?
1207 cd/sqm ~= 3793 lx

For our 1/000s exposure that’s 3.8 lx s. That’s 15.2 times the 18% saturation level, or virtually 4 stops of light (3.92) from midtone to saturation highlight.  Sounds approximately accurate, for digital, right?

So we’ve calculated a luminance of 0.025 lux seconds over the area of the sensor (0.024m x 0.036m). Taking account of the time aspect of the units, and using the very first conversion we came across, that’s:

0.0185 * 0.25 * 6.02*1017 * 0.024 * 0.036 = 2.4*1012

So our example photograph, is the result of 2.4 trillion photons!  Huzzah!

For a small compact camera at a high ISO, the photons could be measured in the billions, and for low sensitivity 10×8 film measured in the the quadrillions (left as an exercise for the reader).

Another thought: that’s approx 100,000 photons per pixel in a full frame 24MP digital full-frame image. Think about that next time you’re pushing that pixel in Photoshop – you’re manipulating the representation of 100,000 quanta of light energy.

NB: This is not only bad photometry and radiometry, it’s most likely bad physics and bad mathematics too.  It was just a fun exercise.  If anyone wants to take this apart, or suggest a wildly different figure, I’d be very interested to hear from you!  The resulting numbers seem a little low and I wouldn’t be surprised to learn I’m out my many orders of magnitude.

Sources:

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