## Visibility

### Introduction

In other articles on meteorological optics, the scattering of light from a variety of atmospheric residents has been considered. The largest such residents are raindrops, millimetres in size, which are only temporary residents and produce the rainbow during their fall to earth. The rainbow is a result of refraction and reflection in spherical water drops, but requires wave theory to get everything right. The glory is produced by spheres of water, but smaller ones than raindrops, such as are found in clouds and fog. This is a very special direct backscattering, seldom observed, and for which the theory is difficult. Ice crystals of about the same size in high clouds create the halo, through refraction and reflection. Still smaller water drops and ice crystals are responsible for coronas, where the small-angle forward scattering is due to diffraction.

Now we consider scattering by still smaller particles, down to atomic sizes, which causes the blue sky and red sunsets. This affects light and illumination in nature, and provides many interesting observations. We will consider the general problem of atmospheric turbidity at the same time, and investigate the visibility of objects at a distance, an important practical consideration. Visibility involves considerations of visual acuity, the sensitivity of the eye, and loss of contrast from the addition of scattered light.

### Rayleigh Scattering

An oscillating electric dipole, p(t) = peiωt, radiates a power P = (ck4/3)|p|2, where k is the wave vector 2π/ω. For molecules or small material objects, the dipole is all that has to be considered, and it makes the case rather simple. The radiation is to the sides of the dipole, proportional to sin2θ, and there is no radiation in the direction of the axis of the dipole. The radiation from the dipole is polarized in a plane that contains the dipole.

The electric field of an electromagnetic wave exerts a force on electrons in the direction of the field. This creates an electric dipole oscillating at the frequency of the field. If the atom or small body is isotropic, that is, polarizes the same way in any direction, then the dipole is in the direction of the electric field of the wave. Since it radiates energy, it absorbs energy from the wave and scatters it (that is, radiates it in other directions). When the atoms or small bodies radiate independently, as when they are in random motion, the total radiation is the sum of the radiation from each radiating system, and is proportional to N, the number of such systems. This is called an incoherent superposition. The incoherent scattering by induced dipoles is known as Rayleigh scattering. It dominates for particles of sizes less than about 0.1 wavelength of light, or about 50 nm.

If a collimated beam of light passes through a gas, the intensity is given by I(x) = Ie-σx, where I is the intensity at x = 0. The quantity σ is called the extinction coefficient, with units of inverse length. For Rayleigh scattering, σ = (2k4/3πN)|n - 1|2, where n is the index of refraction of the gas, and N is the number of molecules per unit volume. The most remarkable thing about this formula is that it contains the molecular number density N. If there were no molecules, there would be no Rayleigh scattering! The second most remarkable thing is the strong dependence on wavelength through the factor k4, which means that short wavelengths, blue light, are scattered much more strongly than long wavelengths, red light.

Since scattered radiation is predominantly normal to the direction of the induced dipole, Rayleigh-scattered radiation is polarized. Skylight is polarized perpendicularly to a line from the point observed to the sun, and is a maximum at 90° from the sun. The polarization is not perfect, for several reasons (anisotropy, multiple scattering, aerosols) but is very distinct, and is easily detected with a polaroid filter. It is greatest in a deep-blue sky with clear air, and disappears for a white sky. The polarization can be detected by the eye through Haidinger's Brush (q.v.) by a trained observer. Insects use the polarization for navigation.

The atmosphere scatters blue light from the sunlight passing over our heads, leaving the red to color the sunset across the Atlantic. At one step Rayleigh scattering explains the blue of the sky and the red of the sunset and sunrise. It gives us the warm light of afternoon and the cool light of midday. When a scene is illuminated by the sun, the highlights are in the warm direct rays, while the shadows are illuminated by blue scattered light from the sky. When we look at distant mountains, they are veiled by the blue light scattered on the path between them and us, an effect called visual perspective. White snowcaps appear yellowish, since blue has been scattered from their light. The blue light against a dark background is more noticeable than the yellowish light against a bright background, because the contrast is greater.

Unusual atmospheric colors are produced when the path of the light to our eyes is shaded. The distant sky seen beneath a towering thunderhead appears orange, since the blue light has been scattered away. If the illumination comes from a smaller distance, the light may appear green, since the removal of less blue does not shift the color as far, and the eye is particularly sensitive to green. The wavelength dependence of Rayleigh scattering can produce a wide range of colors in nature.

We have assumed so far that the particles scatter light independently, and that the light that reaches our eye has made only a single encounter. In thicker smokes and hazes, multiple scattering is the rule. The light arriving at any scattering centre has perhaps already been scattered several times, and the light leaving will be scattered several times more before we see it. Multiple scattering always reduces the wavelength dependence of scattering. If there is little absorption, the multiply scattered light is the same color as the incident light. This explains the whiteness of clouds and of milk, as well as the whiteness of many other things. A cloud of smoke may appear blue near the edges, where it is less dense, because of Rayleigh scattering, while the thicker parts of the cloud are white.

Soot, on the other hand, is strongly absorbing, and the absorption increases for shorter wavelengths. Now, multiple scattering can result in strong attenuation, and a soot smoke appears black. Moreover, the transmitted light is a peculiar orange-red that is quite noticeable when the smoke from forest fires drifts overhead. This is a different color than that produced by Rayleigh scattering.

If we consider a small volume of a gas containing N molecules, then this number is constantly changing as molecules leave and arrive due to their incessant motion. The root-mean-square fluctuation in N, ΔN, is given by ΔN/N = 1/√N. For a volume containing a large number of molecules, this fluctuation is exceedingly small. However, if N is only 100, then ΔN/N is about 1/10, which represents a considerable fluctuation in the density of the gas on this scale. It is better to say that the polarization of these fluctuations causes Rayleigh scattering rather than the polarization of individual molecules. There isn't much difference for a thin gas, but there is for denser systems. At the critical point of a gas, the large fluctuations give rise to large scattering called critical opalescence.

For air at S.T.P., n - 1 is about 2.78 x 10-4, and N = 2.69 x 1019 cm-3. This gives an attenuation length σ-1 of 30 km for violet (410 nm), 77 km for green (520 nm) and 188 km for red (650 nm).

### Day Visibility

Although the word "visibility" has various colloquial uses, we want to give it a more precise definition, though even so its meaning may be subject to interpretation. Physically "seeing" an object is not enough. The object must be recognized for what it is. That is, it must be properly identified by the observer's visual sense in nature and position. Camouflage, which will be discussed below, interferes with recognition.

An important factor in seeing objects is visual acuity, the ability to perceive closely-spaced features. This can be defined most precisely in terms of spatial frequencies, but usually it is specified in terms of a limiting angle. Helmholtz assumed that the angle was 2' of arc, or 5.8 x 10-4 radians. The corresponding linear distance at 25 cm is 0.145 mm. At 100 m, it is 5.8 cm. At 5 km, it is 2.9 m. Smaller objects cannot be recognized, however clear the atmosphere may be. Telescopes and binoculars provide angular magnification, increasing the acuity by the magnification factor. 8X binoculars can resolve 36 cm, a little over a foot, at 5 km, and 7.25 mm at 100 m. Such instruments are aids to visual acuity.

Day visibility is limited by light scattering in the atmosphere, which reduces the contrast of an object seen at a distance. We usually consider a dark object against the sky. Sky brightness is due to scattering over an indefinitely long path. As distance increases, more and more scattered light obscures the object, reducing its contrast with the sky.

If B' is the brightness of an object (proportional to the light from the object that enters the observer's eye) and B is the brightness of the background, then the contrast C between the object and its background is the ratio C = (B' - B)/B = dB/B. It is found experimentally that the limiting contrast is independent of B. That is, if the background is made two times brighter, b = 2B, then for the same contrast, B' must also be doubled. It is not enough simply to keep the difference constant. This is expressed by Fechner's Law, S = k ln B, where S is the perceived sensation, B is the physical stimulus (in say, power units), and k is a constant. This is a natural property of our senses, so that they can handle a wide dynamic range of stimuli.

Suppose we have a completely black object, so that B' = 0, and its background is the sky. Then the contrast C = -1 when we are close to it, which is the maximum possible. The minus sign indicates that the object is less bright than the background. If the object is brighter than the background, then C is positive and can be as large as desired. As we recede, light is scattered into our eyes in the solid angle subtended by the object, and it becomes brighter. The contrast C decreases towards zero at the same time, since at a large distance the scattered light will be as bright as the background, since the light coming from the sky is just the light scattered in the atmosphere over a large distance. At some limiting contrast the object will no longer be distinguishable from the sky, and this will happen at some distance that we shall call the visibility V of the object.

Suppose that σ is the attenuation coefficient for light, so that the light isotropically scattered per unit cross sectional area in a distance dx is σdx. A collimated beam will be attenuated according to I(x) = Ie-σx, where I is the intensity at x = 0. In the diagram, consider a length dx of the cone between the black object S and the eye E. The amount of light scattered in this volume will be proportional to σ(S/r2)x2dx. Of this, the fraction that enters the eye through the pupil area A will be (A/4πx2)e-σx, since the light is scattered into a total solid angle 4π and is itself scattered on the way to the eye. Therefore, the total scattered light entering the eye in the whole cone will be B' = (SA/4πr2)∫(0,r)σe-σxdx = K(1 - e-σr), where K is a constant independent of σ.

We can find the background brightness B by simply extending the integral to infinity, and find that B = K. Therefore, B' = B(1 - e-σr). Now it is easy to find the contrast, C = (B' - B)/B = -e-σr. When r = 0, this gives C = -1, and when r = ∞, C = 0, so the limiting values are correct. If -D is the limiting value for C, defining the visibility, then D = e-σV, where r = V, the visibility. Solving for V, we find that V = (1/σ)ln(1/D). The inventor of this analysis, Koschmieder, used the value D = 0.02. In that case, V = 3.912/σ. This is a quantitative relation between the visibility and the attenuation coefficient, so that one can be determined from the other. This prediction is found to agree well enough with observations, and is generally used. The value of D is larger for objects subtending a small angle. For a 1° subtense, D is twice as large, and for 0.5°, D is four times as large. Because of the logarithmic relationship, the visibility does not vary rapidly with D.

We call B brightness here, but it is actually illuminance or something similar, since it is measured in lumens or lux, an energy-like quantity, and not in the subjective sensation of brightness, which is proportional to ln B. At a wavelength of 555 nm, energy is most effective in producing illumination, at a rate of 680 lumens per watt.

Poor visibility is said to exist when V is less than 5 km or 3 miles. This corresponds to an attenuation coefficient of 7.82 x 10-4 per metre, or an attenuation length of 1.278 km (distance for 1/e of the original intensity). From the values of σ for Rayleigh scattering given above, it is clear that Rayleigh scattering by pure air is no impediment to visibility. To estimate visibility by eye, visibility marks are established at various distances. These should be black objects subtending from 0.5° to 5° so that problems of visual acuity or large area are not encountered. The visibility is estimated from the distance to the furthest recognizable visibility mark. This method has the benefit that visibility is established by actual vision, not by some theoretically established related quantity. If there is some way of measuring the attenuation coefficient σ, its value can then be used to determine V in a way that is independent of the observer.

Visibility is affected by the stability of the air. A small lapse rate or an inversion inhibits turbulent mixing, trapping scattering particles at lower levels and decreasing visibility. A wind or a large lapse rate encourages mixing and the dissipation of scattering centres. Therefore, visibility is generally worse in the mornings, and improves by afternoon.

### Night Visibility

The visibility of lights by night is a completely different matter than the day visibility of a dark target. There is no question of the reduction of contrast by scattered light (except from the moon and artificial sources), but only of the sensitivity of the eye. The light from a source is weakened by scattering, and changed in color by Rayleigh scattering, the blue being preferentially eliminated. The illumination produced by a light of candlepower I at a distance r is given by E(r) = Ie-σr/r2. The sensitivity of the eye depends on its state of dark adaptation. The threshold of a dark-adapted (scotopic) eye is around 10-9 lux (lumens/m2), and for an eye adapted to normal light levels (photopic) the threshold is about 10-7 lux. These thresholds increase as the square root of background illumination. One source gives the threshold E = 3.5 x 10-7√B lux. Moonlight provides an illumination of about 0.17 lux, but the actual value varies greatly, of course. This gives a threshold of about 1.44 x 10-7 lux in moonlight. The direct sun provides 109,000 lux.

It is inconvenient to solve the illumination formula for r, given E. This can be done by the SOLVE function of a calculator. Another way is to plot log E against log r, and read off the values of r that correspond to any E from the curve. As an example, I made a plot for σ = 7.8 x 10-4 m-1, which corresponds to a day visibility of 5 km. Log E ran from -6 to -14, and log r from 3 to 4.5. The same plot can be used for any candlepower by simply sliding reference levels up and down. For a 1 cp light, the photopic visibility was 1.6 km = 1 mi. The scotopic visibility for the same light was 4.8 km = 3 mi. This illustrates how the visibility depends on the dark adaptation of the eye. For a 100 cp light, the photopic visibility is 4.8 km, and the scotopic is 5.2 km. Note that the scotopic visibility for the 100 cp light is about equal to the day visibility in this case. This is only coincidence, of course. The attenuation is the main factor affecting the visibility, the inverse-square spreading having a smaller effect.

The candlepower of a light emitting L lumens is approximately I = L/4π. A 100-watt bulb has about 140 cp, and a 60-watt bulb has about 70 cp. It has been found that if the visibility of a 100 cp light is 111 yards, the day visibility is 55 yards. If the night visibility is 371 yd, the day visibility is 220 yd. If it is 1490 yd, the day visibility is 1100 yd. At 2-1/2 miles, the night and day visibilites are equal. If the lamp is visible at 9 miles by night, the day visibility will be 18 miles. By the use of lenses and mirrors, the candlepower in a certain direction (lumens per steradian) can be increased greatly.

Flashing lights are not noticed more readily than steady lights. The visibility of a flashing light depends on its time average candlepower, averaged over on and off periods. Red lights seem to be the easiest to detect, but there is little difference between lights of the same intensity but different colors. Point sources are much more easily detectable than extended sources.

### Tyndall Scattering

Very small particles, those smaller than about 0.1 μm, will display good Rayleigh scattering with its strong wavelength dependence. Cigarette smoke and internal-combustion engine exhausts are examples. They appears blue by reflected (scattered) light, and reddish by transmitted light. The strong wavelength dependence disappears for larger particles, say of 5 μm diameter, and light of all wavelengths is scattered equally, after passing through a region of less rapid wavelength dependence. This phenomenon is known as Tyndall scattering, typical of colloidal suspensions of particles larger than about 5 μm. Multiple scattering is also important in dense suspensions. Scattering by colloidal suspensions is an involved subject, with many interesting curiosities. Tyndall scattering renders a beam of light visible, as in the familiar case of searchlight beams, as well as for laser beams, visible in the dust motes that are always floating about.

For a particle of radius a larger than a wavelength, the scattering cross section is 2πa2, as shown in Jackson, p. 451. The scattering cross section is an area such that the part of the incident wave falling on it will be scattered. The extinction coefficient is then σ = 2πNa2, where N is the number of scattering centres per unit volume. The visibility is then V = (1/2πNa2)ln(1/D). If w is the mass of liquid water present per unit volume in the form of spherical drops of radius a, it is easy to show that V = (2aρ/3w)ln(1/D). If the number of droplets remains constant while w varies, then V will vary as w-2/3. This agrees pretty well with observation, where the exponent was found to be -0.6. Light scattered by a fog is white because the droplets are larger than a wavelength. These formulas can be used to estimate the visibility in fogs of known water content.

The reduction of visibility in a fog is due to the reduction of contrast between the object and the fog background, not to the blurring of edges, as is commonly thought. The edges are visible right up to the point of disappearance. The nature of the multiple scattering also shows that no color penetrates fog better than any other.

Scattering by larger particles is preferentially forward scattering at small angles. This explains why a thin mist is more easily seen when looking in the direction of the sun. This scattering makes a white corona around the source of light. The sun is usually surrounded by a bright corona.

### Camouflage

It will be instructive to say a few words about camouflage, the art of concealing an object in open view. The aim of camouflage is to prevent recognition of the object by removing or altering clues to its nature. The word "camouflage," introduced by the French in World War I to denote scientific concealment, has strong military connotations, but we will use it as a general term. It is obviously a French word, and the verb is camoufler. However, the word did not begin as French, since it is taken from the Italian verb camuffare, "to disguise," which does not have the limited connotation of the French or English word. The military distinguishes cover, which is physical protection against gunfire, from concealment, which merely hides from view. Concealment is the realm of camouflage.

The principles of camouflage are to alter the form, shadows, texture, colors and silhouette of an object to hinder its recognition, and to make the object blend into the background. The most important techniques of camouflage are countershading and disruptive coloration. Nature shows many examples of camouflage in the animal and vegetable world. A cylinder is brighter above where the light falls, and darker in the shaded region below. Accordingly, a frog is darker above and lighter below to conceal its solidity. Even aircraft can be disguised in this fashion. Disruptive coloration was used in the First World War to camouflage ships by painting them in bold, bizarre angular patterns of varying colors, much unlike the horizontal lines and uniform color of a normal ship. Close up, this would fool nobody, but at sea in bad weather, with fog and foam, it was very effective at kilometer distances. It was given up, probably because it was not worth the effort, since it could not alter the distinctive silhouette of a ship on the horizon. Ships became fog grey in the Second World War, a color that hid them well enough in foul weather, and even against dull skies. Leaf and stick insects are excellent examples of blending into the background by imitating its colors, shapes and textures.

A soldier may camouflage himself with a camouflage suit. One type was reversible, with one side in two shades of green for spring and summer, and the other in two shades of brown for the autumn and winter. Of course, in snow a white suit is used. Modern camouflage suits seem more to be a fashion statement or military plumage than effective camouflage, often combining both green and brown, making the suit unsuitable for any season. Nevertheless, the random blotches of dull color help greatly in blending with the background. The helmet makes a distinctive silhouette often poked up against the sky by the unwary, which is broken up by sticking leafy branches in a helmet net. All exposed skin is darkened in blotches by anything available, from burnt cork to cosmetics. It is specially important to eliminate all reflective highlights, particularly metallic ones. The smallest motion can cause flashes that are easily noticed. At night, there should be nothing to reflect light.

Larger objects are camouflaged by draping nets over them. Burlap, or similar cloth, can be woven into the net to make an excellent distracting screen. Every effort is made to destroy familiar shapes and to imitate local colors. One principle of camouflage that is always emphasized is not to use too much. Camouflage need not cover completely to be good concealment. A soldier who puts more than a few sticks in his helmet net is likely to create a prominent shape, not break up a familiar outline. A camouflage net should contain just enough to shadow things below it, to break up outlines, and to blend into the surroundings. Too much, and you have made a recognizable roof. Vegetation used as camouflage must be changed often, since dead vegetation is soon easily distinguished from live vegetation, especially by infrared photography and similar means. Motion, also, can destroy concealment in an instant.

Bad camouflage can be used for "spoofing," to represent imaginary aircraft, tanks and guns to imply a greater force than actually is present. Then dead vegetation, and poles sticking out to be mistaken for gun barrels, are freely used, since the idea is to attract recognition, not hinder it. Butterflies have "eye spots" on their wings so that they may be mistaken for larger things. Inverse camouflage, such as the yellow and black of dangerous insects and reptiles, can be used to warn away attackers.

### References

R. A. R. Tricker, Introduction to Meteorological Optics (London: Mills and Boon, 1970). Chapter X.

M. Minnaert, The Nature of Light and Colour in the Open Air (New York: Dover, 1954). pp. 264-267.

F. A. Berry, E. Bollay and N. R. Beers, Handbook of Meteorology (New York: McGraw-Hill, 1945). pp. 242-251.

J. D. Jackson, Classical Electrodynamics, 2nd ed. (New York: Jophn Wiley & Sons, 1975. Chapter 9, especially pp. 418-427, (Rayleigh scattering).

Composed by J. B. Calvert
Created 3 August 2003
Last revised