Image saturation is a concept used to describe the purity or intensity of color in an image, such that an image with very little saturation approaches a black and white image. Pink, then, is said to be less saturated than red because in a pure red color, the red completely dominates other color components, and hues of grey are said to be de-saturated because all the color components contribute equally to the hue.

Although saturation is clearly something that can be manipulated during image-processing, several questions here also refer to saturation as being an attribute of camera sensors, lenses, and so on. If my understanding of camera sensors is correct, they typically capture the intensity of a specific color after it passes through a Bayer filter, so that the RAW output of the sensor is series of readings for red, green, and blue photosites. Variations exist here, but I think the end result ends up being about the same.

At this point, all sorts of processing kicks in, including demosaicing, which effectively assembles the separate red, green, and blue channels into pixels.

If this understanding is correct, is there any interpretation of "saturation" that makes sense as a hardware concept, or is it purely a product of how the RAW data is processed into an image (whether that happens in-camera or afterward)?

  • \$\begingroup\$ Are you asking if saturation is a artifact of digital processing? The idea of saturation of colors in a photo was around before digital photography. One example off the top of my head is that you usually got more saturated colors by slightly underexposing slide film. Also, depending on the circumstances, a polarizing filter could give more saturated colors than a straight photograph. \$\endgroup\$ Jan 27, 2012 at 18:45
  • \$\begingroup\$ Ah - that's a good one. So the polarizer can produce greens that are more green and reds that are more red and blues that are more blue by excluding scattered light, right? So that sounds like a non-image-processing way to affect saturation. \$\endgroup\$
    – D. Lambert
    Jan 27, 2012 at 18:56

2 Answers 2


From a technical standpoint, "saturation" is the extent of chromaticity for a certain hue...the hue's "colorfulness". Technically speaking, pink would be a less colorful magenta, but roughly the same hue, where as red would be a distinct and colorful hue on its own. You might think of light rose or salmon to be less colorful variations of red.

When it comes to things that can affect saturation in a photographic context, the story can be fairly complex. For the time being, lets eliminate post processing, and only factor in the physical aspects.

  • Pre-Camera Interference
    • Any obstructions between light and the camera can affect saturation...things that disperse and scatter light will usually produce a softer color than what it started out as...and extensive dispersion can eliminate most color entirely resulting in shades of gray. (i.e. fog, mist, clouds)
  • Optics in the Lens
    • The various materials used in a lens itself can affect saturation of different wavelengths of light.
    • These days, transmission of light through a lens is very high...above 90% in most cases. Even so, most lenses will absorb a certain amount of the light that passes through them, and will absorb different wavelengths to differing degrees. This can have a minor affect on saturation, but a measurable one.
    • Better optics, better multi-coating, fewer elements, etc. all contribute to improved transmission, saturation and resolution.
  • Sensor Design
    • The sensor itself is also a factor that affects saturation.
    • In a bayer sensor, you have an uneven distribution of red, green, and blue pixels, each of which have their own color filters above them that filter out wavelengths that don't fit their color (i.e. blue pixels filter out most green and all red light). The materials used for the filter will also filter out some amount of the desired wavelengths of light as well, as transmission is again less than 100%.
    • In a Foveon style sensor, each pixel has layers of photo-sensitive elements. Each subsequent layer is going to receive less total light, excluding the color from the preceding layer minus any transmission loss.
    • Some light in all sensors will simply be reflected or absorbed by non-photosensitive elements, affecting the saturation of those lost wavelengths.
  • Randomness
    • A variety of random factors can affect saturation, from natural randomness of photos, thermal factors, to issues of quantization and precision in conversion from analog to digital.

All these factors contribute to a loss of the original apparent saturation of light that we saw with our own eyes (which, while similar to cameras, have far fewer obstructions...our lenses are simple and highly transmissive, and our rods and cones are orders of magnitude more sensitive, all of which greatly reduce the amount of loss in accuracy before that light is processed by our brains.) The native, unprocessed RAW output from a modern sensor is pretty dull, lacking in both contrast and saturation (but usually appearing to have far more dynamic range than we usually work with in post). Most RAW processors apply an attenuated, multi-channel tone curve to RAW images by default (and only a couple open source options allow you to view the original native RAW image), producing the better initial results we see when we open up a photo in say Lightroom or Aperture. From that point on, its all a matter of mathematic adjustments that aim to recover the saturation that was lost.

How much saturation looks "good" or "realistic" is entirely a matter of artistic style and personal preference at that point. You can increase or decrease saturation at will, by small or very large amounts. Once a photo gets in the hands of the photographer, saturation takes on a very different meaning than it has when talking about color and saturation of a lens, or whether a sensor produces saturated images or not. Hardware-level color saturation really means less today than it may have in the past with film, as we have nearly unlimited control over our photos today that wasn't possible with color film and color photographic prints.

I thought I should add an edit, and note some caveats with the statement above about color and saturation at a hardware level not mattering. I had a fairly narrow vision of what that related to when I wrote the above, however its not entirely accurate. On a professional level, particularly for publications (newspaper, magazines, lots of various print and online editorials and the like), JPEG is still the most used format. A lot of professional sports, Olympics, photojournalist, political, etc. photographers tend to send their shots directly to their employers for immediate publication, as at-the-moment publication has become incredibly important these days. One of the things that made me think of this is the new Canon 1D X's ethernet port, which allows the camera to save images directly to a network attached resource.

With immediate delivery and publication, hardware level, in-camera image quality does have meaning. Rich, saturated, sharp, clear photos can be extremely important for up-to-the-moment publications. Given that such professional work is a huge driver of camera and lens sales, there are reasons why color quality, saturation...at a hardware level...are still very important.


Saturation is a mathematical concept of Color Theory, and it is abstracted to a higher plane than film, digital images, etc. While Hue, Saturation, and Value color models were developed for use in digital images, they are models designed to represent reality as perceived by the sensors in our eyes.

Just like a physical object can be described with mass, volume and temperature, a color can be partly described mathematically by its saturation.

So, just as a scale measures weight, and an NMR spectrometer measures chemical composition, you can think of a camera as an instrument that is designed to accurately measure the properties of the light coming into it. It can do that via a variety of methods (film, CCD, CMOS, etc), and the HSV it outputs is the mathematical "ideal" that is built upon the measurements from the camera 'instrument'

  • \$\begingroup\$ Technically speaking, a sensor doesn't output HSV, or for that matter RGB. Sensor output is generally a color-filtered luminance value...a single hue of varying saturation per pixel. Its only upon processing of sensor output in a RAW processor that those individual single-color luminance values called "pixels" get interpolated into the standard aggregates of RGB pixels that we call an image. \$\endgroup\$
    – jrista
    Jan 31, 2012 at 20:26
  • \$\begingroup\$ @jrista: the sensor doesn't even output a luminance value, the sensor outputs a voltage. That voltage is then interpreted by the camera hardware to represent a luminance value, and those values are interpreted to represent HSV. HSV is a mathematical model that lies at an abstraction far above the hardware implementation of the camera. \$\endgroup\$ Jan 31, 2012 at 20:49
  • \$\begingroup\$ In the case of JPEG, it would be a YCC variant, however in the case of RAW, the voltage is read and stored as a colored luminance value. HSV is a model for representing color, but I do not believe in any case that it is actually used in camera hardware. Most color space transformations that are applied by a RAW processor probably done in CIE 1931 XYZ, as thats exactly what it was designed for. The CIE 1931 color space is a luminance/chrominance color space modeled after human vision, with blue/yellow and magenta/green color planar axes and a luminance vertical axis. \$\endgroup\$
    – jrista
    Jan 31, 2012 at 22:27
  • \$\begingroup\$ HSV and HSB are derivatives of CIE's core XYZ color space, however their models are less accurate, and I would find it hard to believe they were used over the more accurate and precise CIE models. So, in the sense of using an abstracted model, you are correct...however I'm trying to point out that it's unlikely HSV is the actual model used...its a rather simplistic model that doesn't really take gamut into account. \$\endgroup\$
    – jrista
    Jan 31, 2012 at 22:31

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