Imagine a bucolic scene on a clear sunny day at the equator, sand warmed by the tropical sun with a typical irradiance () of about 1000 watts per square meter. As discussed earlier we could express this quantity as illuminance in lumens per square meter () – or as a certain number of photons per second () over an area of interest ().
(1)
How many per unit area can we expect on the camera’s image plane (irradiance )?
In answering this question we will discover the Camera Equation as a function of opening angles – and set the stage for the next article on lens pupils. By the way, all quantities in this article depend on wavelength, which will be assumed in the formulas to make them more readable.
In this article I bring together qualitatively the main concepts discussed in the series and argue that in many (most) cases a photographer’s job in order to obtain natural looking tones in their work during raw conversion is to get the illuminant and relative white balance right – and to step away from any slider found in menus with the word ‘color’ in them.
If you are an outdoor photographer trying to get balanced greens under an overcast sky – or a portrait photographer after good skin tones – dialing in the appropriate scene, illuminant and white balance puts the camera/converter manufacturer’s color science to work and gets you most of the way there safely. Of course the judicious photographer always knew to do that – hopefully now with a better appreciation as for why.
As we have seen in the previous post, knowing the characteristics of light at the scene is critical to be able to determine the color transform that will allow captured raw data to be naturally displayed from an output color space like ubiquitous sRGB.
White Point
The light source Spectral Power Distribution (SPD) corresponds to a unique White Point, namely a set of coordinates in the color space, obtained by multiplying wavelength-by-wavelength its SPD (the blue curve below) by the Color Matching Functions of a Standard Observer ()
Adding (integrating) the three resulting curves up we get three values that represent the illuminant’s coordinates in the color space. The White Point is then obtained by dividing these coordinates by the value to normalize it to 1.
The White Point is then seen to be independent of the intensity of the arriving light, as represents Luminance from the scene. For instance a Standard Daylight Illuminant with a Correlated Color Temperature of 5300k has a White Point of[1]
Building on a preceeding article of this series, once demosaiced raw data from a Bayer Color Filter Array sensor represents the captured image as a set of triplets, corresponding to the estimated light intensity at a given pixel under each of the three spectral filters part of the CFA. The filters are band-pass and named for the representative peak wavelength that they let through, typically red, green, blue or , , for short.
Since the resulting intensities are linearly independent they can form the basis of a 3D coordinate system, with each triplet representing a point within it. The system is bounded in the raw data by the extent of the Analog to Digital Converter, with all three channels spanning the same range, from Black Level with no light to clipping with maximum recordable light. Therefore it can be thought to represent a space in the form of a cube – or better, a parallelepiped – with the origin at [0,0,0] and the opposite vertex at the clipping value in Data Numbers, expressed as [1,1,1] if we normalize all data by it.
The job of the color transform is to project demosaiced raw data to a standard output color space designed for viewing. Such spaces have names like , or . The output space can also be shown in 3D as a parallelepiped with the origin at [0,0,0] with no light and the opposite vertex at [1,1,1] with maximum displayable light. Continue reading Linear Color Transforms→
In the last article we showed how a digital camera’s captured raw data is related to Color Science. In my next trick I will show that CIE 2012 2 deg XYZ Color Matching Functions , , displayed in Figure 1 are an exact linear transform of Stockman & Sharpe (2000) 2 deg Cone Fundamentals , , displayed in Figure 2
(1)
with CMFs and CFs in 3xN format, a 3×3 matrix and matrix multiplication. Et voilà:[1]
In the previous article we determined that the three values recorded in the raw data in the center of the image plane in units of Data Numbers per pixel – by a digital camera and lens as a function of absolute spectral radianceat the lens – can be estimated as follows:
(1)
with subscript indicating absolute-referred units and the three system Spectral Sensitivity Functions. In this series of articles is wavelength by wavelength multiplication (what happens to the spectrum of light as it progresses through the imaging system) and the integral just means the area under each of the three resulting curves (integration is what the pixels do during exposure). Together they represent an inner or dot product. All variables in front of the integral were previously described and can be considered constant for a given photographic setup. Continue reading Connecting Photographic Raw Data to Tristimulus Color Science→
In the previous article we (I) learned that the Spectral Sensitivity Functions of a given digital camera and lens are the result of the interaction of light from the scene with all of the spectrally varied components that make up the imaging system: mainly the lens, ultraviolet/infrared hot mirror, Color Filter Array and other filters, finally the photoelectric layer of the sensor, which is normally silicon in consumer kit.
In this one we will put the process on a more formal theoretical footing, setting the stage for the next few on the role of white balance.
Photography works because visible light from one or more sources reaches the scene and is reflected in the direction of the camera, which then captures a signal proportional to it. The journey of light can be described in integrated units of power all the way to the sensor, for instance so many watts per square meter. However ever since Newton we have known that such total power is in fact the result of the weighted sum of contributions by every frequency that makes up the light, what he called its spectrum.
Our ability to see and record color depends on knowing the distribution of the power contained within a subset of these frequencies and how it interacts with the various objects in its path. This article is about how a typical digital camera for photographers interacts with such a spectrum from the scene: we will dissect what is sometimes referred to as the system’s Spectral Response or Sensitivity.
We’ve seen how humans perceive color in daylight as a result of three types of photoreceptors in the retina called cones that absorb wavelengths of light from the scene with different sensitivities to the arriving spectrum.
A photographic digital imager attempts to mimic the workings of cones in the retina by usually having different color filters arranged in an array (CFA) on top of its photoreceptors, which we normally call pixels. In a Bayer CFA configuration there are three filters named for the predominant wavelengths that each lets through (red, green and blue) arranged in quartets such as shown below:
A CFA is just one way to copy the action of cones: Foveon for instance lets the sensing material itself perform the spectral separation. It is the quality of the combined spectral filtering part of the imaging system (lenses, UV/IR, CFA, sensing material etc.) that determines how accurately a digital camera is able to capture color information from the scene. So what are the characteristics of better systems and can perfection be achieved? In this article I will pick up the discussion where it was last left off and, ignoring noise for now, attempt to answer this question using CIE conventions, in the process gaining insight in the role of the compromise color matrix and developing a method to visualize its effects.[1]Continue reading The Perfect Color Filter Array→
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.
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)
The Power of Light emitted in is called Spectral Exitance, with the symbol when referred to units of energy. The energy of one photon at a given wavelength is
(2)
with the wavelength of light in meters and and 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 . Writing it more formally:
When first approaching photographic science a photographer is often confused by the unfamiliar units used. In high school we were taught energy and power in radiometric units like watts ($W$) – while in photography the same concepts are dealt with in photometric units like lumens ($lm$).
Once one realizes that both sets of units refer to the exact same physical process – energy transfer – but they are fine tuned for two slightly different purposes it becomes a lot easier to interpret the science behind photography through the theory one already knows.
It all boils down to one simple notion: lumens are watts as perceived by the Human Visual System.
When capturing a typical photograph, light from one or more sources is reflected from the scene, reaches the lens, goes through it and eventually hits the sensing plane.
In photography Exposureis the quantity of visible light per unit area incident on the image plane during the time that it is exposed to the scene. Exposure is intuitively proportional to Luminance from the scene $L$ and exposure time $t$. It is inversely proportional to lens f-number $N$ squared because it determines the relative size of the cone of light captured from the scene. You can read more about the theory in the article on angles and the Camera Equation.