We can use color-filtered b & w pictures as transparencies projected through color filters and superposed (registered) to produce color composites. To do this, imagine this setup: Pick any scene containing many features and classes of differing colors. First, replace the prints with positive transparencies (tonally analogous to prints). Work with three b & w transparencies, each representing its spectral band. Shine white light through each one mounted in its own lamp projector (total of three) on to a screen. Project the blue band b & w transparency through a blue filter, the green through green, and the red through red. Blue features on the ground are clear areas in the blue spectral band. When the blue band transparency is projected through the blue filter, the blue features will be blue on an observing screen, likewise the green band projects green objects through its filter as green, and red as red. Co-register (line up) the three projections by superimposing several distinctive patterns that are common within the photographed scene. The result will be a simulated natural color image. with any red, green, and blue objects or classes showing in the projected image as these colors respectively. Other colors present are additive mixes of two or more primaries (e.g., yellow is a mix of red and green; orange is a mix of more red and some green; white is an equal mix of all three primaries, and black is simply the absence of any colored light of any wavelength). The colors that result from combinations of blue, green, and red (the primaries) are indicated in this additive color diagram.
This way of combining individual colors to produce a composite is called color additive viewing. The filters and films used do not necessarily have to correspond by wavelength. For example, a green spectral band image can be projected through a blue filter and a red band through a green filter. We can modify the color assignments in making an image if one of the transparencies is the infrared band and is projected through a red filter; the result is one version of a what is called a false color (IR) composite. This procedure and the resulting colors that are produced are indicated in this diagram:
Note: the shading on this diagram is not as contrasty as intended (and the original figure has been lost). To clarify: 1) in the left square only the diamond is white (clear), the others are light gray; 2) in the center square the triangle and circle are white; 3) in the right diagram the cross and circle are clear. In the discussion in the next two paragraphs, one can consider the gray shades as very dark (in a transparency, would not allow light to pass through).
The setup is a variant of the one described in the first paragraph in which a true color image is formed. In the above figure, each of the four geometric designs in its large framing square is either clear (actually open) (shown as white inside its boundaries) and will pass all light impinging through it or is shaded gray inside which means no light passes through. Each is located in the same position in all three squares. Each square is labeled as "band filter" (corresponding to the band associated with the Multispectral Scanner on Landsat-1). Thus, the MSS Green band (bottom) shows the diamond as a positive bright figure in a black and white print images and the other shapes are dark. As applied to a real image corresponding to the MSS Green band output, this just means that objects in the scene (on the ground) that are green show up as light-toned in a print or largely clear in a transparency; blue-green and yellow-green objects (whose wavelength spreads are offset from those of a pure green object) are lighter shades of gray (less clear). The same argument applies to the Red and IR band squares, i.e., white refers to objects that give off Red and IR radiation respectively, and gray refers to objects that don't represent those wavelength bands.
The three band squares are now mounted each in its own light projector (for instance a 35 mm or a lantern slide type). Light passing through a clear design is projected on a screen surface; each projector is moved until designs with equivalent shapes are registered (superimposed). A color filter is placed in front of the lens of each projector: blue for square A, green for B, and red for C. As seen on the screen, the diamond will show as blue , the inverted triangle as green, and the + as red. The combination of green and red for the circle design (in B and C) produces yellow (in the color additive process).
To relate this to how it applies to multispectral photography, consider square A as representing a scene in which only green light is reflected from objects (not too fanciful - the Emerald City in the "Wizard of Oz" was all green). Square B represents a reflected red light only scene; square C represents a scene containing only objects that give off only infrared light. The screen receiving the projected and superimposed light would show a blue object, a green object, a red object, and a yellow object, corresponding to real world objects imaged by green, blue, and infrared filters; the projected objects would be separated in the superposition, corresponding to their initial separations in the square.
You have already seen false color images in the Overview. Let's see what a false color image looks like in the ground scene that follows. First we show the area, grassland and a field with natural shrub cover (left), as it appears in natural color in this aerial oblique view:
The companion photo below is a notably different color version (typical false color rendition) in which various kinds of vegetation display in several tones of red, pink, or yellow (the latter two may indicate a degree of stressed or unhealthy vegetation).
This type of projected color combination, yielding the false color IR composite, is ordinarily used to emphasize a property of healthy vegetation in which incident light in the range of 0.7 - 1.1 µm reflects strongly from the internal cells of plants, giving rise to bright tones in the color IR film.
When this film combination is used in military reconnaissance photography, weapons camouflage - simulating vegetation - is not red in a false-color scene, because the high near-IR reflectance from healthy vegetation is absent (synthetic fabric in clothing is not bright in the near-IR; if made usually from cotton or other organic, this too is at best a subdued bright). Or, in a more familiar mode, color IR photos (and sensor-derived images that include an IR band) show a football field with natural grass in bright red as compared with artificial turf (Astroturf) in a dark, non-red tone.
Lets re-examine the square frames example above by conducting another experiment. Instead of the clear areas passing light, let them reflect light back towards a film camera. The gray areas reflect no light (ignore the areas outside the designs). The objective of this experiment is to determine what gray level wil be recorded on visible or infrared-sensitive film when the clear designs are (1) assigned a color (including IR), and then (2) viewed through some highly sensitive filter (one that passes onlylight of its stated color).
So, going back to the diagram, if we look at the reflecting diamond design in A through a green filter (the Landsat MSS case), we ask what color(s) it can have that will pass through the filter, be recorded as very dark on a film negative, and then expressed as a white (or light gray, since the filter may also pass some adjacent wavelengths) diamond shape on a positive black and white print. The obvious answer is green. If the objects in A are any color other than green, the filter would not pass these wavelengths and the diamond would be very dark in the print. The same argument holds for B and C. C, for example, has two designs that reflect the light. When that light is passed by an infrared filter, if light tones for these designs in a b & w photo-print are the result, this would mean that both designs are brightly reflected in the IR and not in the visible range. Going back to the square A case: if the diamond were white in actual fact, not green, and a green filter were used, then only the green component of the visible spectrum would pass and the resulting print would be a low to moderate gray. (The fact that it is white would emerge from the lightness of tone in each square print [provided one was made with a blue filter].) If the diamond were black, no light would pass and the print would be dark.