Figure 21.1. New York City skyline at night. Photo by Skip Moen.
X = I: SHADOWS
As I walk home in the dark, the length of my shadow (l) appears to be increasing faster and faster as I walk at a constant speed v away from a street lamp almost right behind me. Is this in fact true? Let’s see. From Figure 21.2, by similar triangles,
where h is my height, x is my distance from the base of the lamp, and L is the height of the lamp. Therefore
Figure 21.2. Shadow geometry for a lamp at height L.
Using the fact that v = dx/dt, the rate of increase of my shadow length is
which is constant, so the apparent acceleration of shadow length must find its explanations in the realm of psychology and perception, not physics and mathematics . . .
Exercise: It is more frequently the case that the base of the lamp is some fixed distance y from me when I walk right past it. (I try to make sure that y ≠ 0.) Generalize the above argument to show that
where θ(t) is the angle the shadow makes with my direction of travel. Clearly this result approaches that of (21.3) as x increases.
Question: Will the shadow of my head move parallel to me as I walk?
X = α: DOUBLE IMAGES FROM PLATE GLASS DOORS AND WINDOWS
This is a nice little “inverse problem.” Sometimes we may see a double image of the moon or a distant lamp from a window as we walk along a road at night.
Figure 21.3. Geometry for multiple reflections in (slightly) nonparallel window surfaces for angle of incidence i (relative to the back surface CD). The (highly exaggerated) angle between the front and back surfaces is α radians.
Although this can occur if the window has plane parallel surfaces, it is a very tiny effect unless the surfaces are in fact not parallel. Suppose that the front and back faces of the window (AB and CD respectively in Figure 21.3) are inclined at a very small angle α, so the angle between the reflected rays I′ and II′ will be 2α. This is seen by calculating the deviation of a single reflected ray when the “mirror” is rotated through an angle α; using the fact that the angle of reflection is equal to the angle of incidence, it is 2α. In the (greatly exaggerated) Figure 21.3, a pair of parallel rays, I and II, from an assumed distant source are shown impinging on the back and front surfaces of the plate glass, respectively. If we imagine the back surface CD to be a mirror (which in effect it is), it will give a reflected image of the surface AB along A′B′, and an image of the ray I′ along I″. Then the ray II″ has effectively passed through a narrow prism with apex angle 2α, and the minimum angular deflection in the ray path caused by such a prism is 2(n − 1)α, which when added to the 2α above gives a total contribution of 2nα, n ≈ 1.5 being the refractive index of the glass. Thus if we can estimate the angular distance (3α) between the two reflections, the angle between the surfaces is one third of that amount.
Exercise: The minimum angular deflection D in the symmetric path of a ray through a triangular prism of apex angle γ can be shown to be
where n is the refractive index of the prism. Show that for the plate glass window this reduces to D = 2(n−1)α as stated, if α is small.
How large might the angle α be and how might we estimate it? Nowadays plate glass windows are made to a very high degree of precision, but in older houses there can be some small but significant angles between the surfaces. There are probably other distortions in the thickness of the glass, but these will be ignored here.
Suppose we are close enough to estimate the lateral distance d between the images on the glass, and that the distance between the eye and the window is L. From Figure 21.4, representing the approximate geometrical relationship between L, d, and the rays I′, II″, we see that p ≈ d cos i ≈ L tan 3α ≈ 3αL, so that the angle between the faces is
Suppose that L = 10 m, and d = 1 cm = 10−2 m for an angle of incidence i ≈ 30°. We now see that α ≈ 3 × 10−4 radians (or about one minute of arc). Over a distance of 30 cm (about a foot) then, and on the basis of these figures, the thickness of the glass changes by about 30 sin α ≈ 30α ≈ 10−2 cm or 0.1 mm, a tiny amount, to be sure! So, if the glass surfaces are very nearly parallel (as here), it follows that the angle α is extremely tiny indeed, and the angular distance between the two reflections is three times “extremely tiny,” that is, still very tiny! So the double images that can be observed from large distances are almost inevitably produced by nonparallel glass surfaces. If we and the light source are both close enough to parallel glass surfaces, however, then the images may then be quite well separated if the light is not at normal incidence to the window.
Figure 21.4. Detail for equation (21.5).
Figure 21.5. Generic ray paths for a wedge of angle α.
Let’s pursue this a little farther. From Figure 21.5 we see that the ray LABCO exits the upper glass surface at an angle i1 to the normal direction, where sin i1 = n sin (r + 2α). Also, sin i0 = n sin r. Using the fact that α is a very small angle, we may write that sin i1 ≈ n sin r + 2nα cos r. The difference between the angles i0 and i1 is small, so we can use differential notation to write
sin i1 − sin i0 ≈ δ(sin i) ≈ 2αn cos r.
Therefore
Note that k(0) = 2n. That is, for normal incidence, the angular separation of the images is approximately 2nα = 3α for n = 1.5 (a generic value for the refractive index of glass). This is the result obtained earlier in this section. A graph of the function k(i) is shown in Figure 21.5. Clearly, this indicates that Δi, the approximate angular separation of the images, becomes larger and larger the closer the ray’s angle of incidence is to the plane of the surface.
Figure 21.6. The function k(i) in equation (21.6).
There are two additional factors that can modify the approximation (21.6), one being due to the divergence of the rays from a light source which is in practice not infinitely far away (and neither is the observer!). Suppose that the light source L is a distance RL from the glass, the distance of the observer O is RO, and the distance between these two points on the glass is d. Using simple geometrical optics for a single ray from the light source L to the observer’s eye at O (via the point P on the glass surface) we can show that (Figure 21.7)
The factor f is basically a measure of the angle of incidence for an individual surface-reflected ray as determined by the relative positions of source and observer (and is independent of the earlier geometric argument leading to equation (21.6)). It approaches one as the distance of the light from the glass (relative to the observer) becomes larger. Conversely, as this relative distance becomes smaller, f decreases toward zero. The factors k and f in equations (21.6) and (21.7), respectively, will tend to counteract one another as the angle of incidence approaches 90°.
Another geometric factor (cos ) comes from when the plane of incidence (containing the incident and reflected rays) is at an angle to the plane containing the wedge angle of the glass (i.e., the plane is rotated by an angle about the normal to the glass pane). This factor will reduce the angular distance between the two images unless = 90°. Taking all these factors into consideration, we have the more general approximation for δi, namely
Figure 21.7. The basic observer (O)-light source (L) geometry, neglecting the effects of wedge refraction for the additional factor f in equation (21.7).
X = r(t) : RAIN “SPARKS”
After it has been raining a while and there are several large puddles on the road, as you walk you may notice that the light’s reflection is surrounded by momentary “sparks” emanating from where the raindrops hit the surface. They appear to point radially outward. In his classic book Light and Colour in the Open Air, Marcel Minnaert [38] points out that “the explanation is simple.” Let’s see. When a drop hits the surface of the water it generates a set of concentric circular wave patterns (see Chapter 25 for more details about this). As the wave with center C expands (see Figure 21.8), a sequence of reflections occurs in an almost continuous fashion along the line CM (C1 and C2 are intermediate points of reflection)
. As the wave expands outward, so does the point of reflection, giving the appearance of rapidly moving sparks of light.
A related reflection phenomenon occurs when a street lamp is observed through the crown of a tree (Figure 21.9a). The light is reflected into the observer’s eyes by twigs and leaves, and the effect is enhanced if the tree is wet after rain, or even frost-covered. As indicated in Figure 21.9a, all the branches and twigs in a given plane may contribute reflections to the observer, but those parallel to the line EL will be greatly foreshortened compared with those perpendicular to that line. Those between these orientations will be foreshortened proportionately less. As such, the combined effect will be to create a “halo” effect—a set of approximately concentric circles (or arcs of such circles).
Figure 21.8. “Rain spark” geometry.
Figure 21.9a. “Rain circle” geometry when a source of light shines through wet leaves and twigs.
Figure 21.9b. Scratches in a stainless steel bowl exhibit circular and elliptical patterns similar to the “rain circle” phenomenon in Figure 21.9a. In this picture reflections from two kitchen lights produce the effect.
A very similar phenomenon occurs when you look at the sun through a train or airplane window. The many fine scratches give rise to a similar effect, and for very similar reasons; we only notice those scratches that are perpendicular to the plane of incidence of the light rays. As an example of the same type of thing arising in “kitchen optics,” Figure 21.9b shows a similar “rain circle” effect from scratches in a stainless steel mixing bowl. The “circles” are really more elliptical in shape because of the curvature of the bowl.
X = R: REFLECTIONS IN THE RAIN
On a rainy night, light from street lamps reflected by power lines and other cables may be noticed by the observant pedestrian. Let’s assume that you are that person. In Figure 21.10, what is the relationship between your position (at O say), that of the light and that of the reflection P, if any? Since any three noncollinear points define a plane (if we are permitted to assume that your eyes, the light, and its reflection are points here!), and you see the reflection, it follows that P lies on an ellipse, with you and the light source as foci. Why is this so? One fundamental property of an ellipse is that the lines joining the foci to any point on the ellipse make equal angles with the tangent line at that point (see Figures 21.11a and 21.11b), and hence with the normal line at that point. This is the law of reflection expressed as a property of the ellipse.
Figure 21.10. The reflection P of the lamp in a power line; the lamp and observer”s eye at O are at the foci of the ellipse.
Figure 21.11. (a) Basic geometry of the ellipse with foci A, B-, and point P lying on it, and tangent line at P. (b) Construction for the proof; Q′ lies on the ellipse. (c) Completion of the proof.
This is proved geometrically as follows. P is a point on the ellipse, with tangent line L, along which the point Q can move. When Q is coincident with P, the distance AQ + BQ is a minimum by virtue of the property that AQ′ + Q′B = constant. Now let A′ be the reflection of the point A in the line L, so that AC = CA′. Then AQ + QB = A′Q + QB which is a minimum when Q = P, and BPA is a straight line. This being the case, angles APC and CPA’ are equal, and hence so are the angles APC and QPB. The reflective property of the ellipse is established.
Exercise: Prove this result algebraically (you’ll also need a modicum of calculus).
Chapter 22
NIGHTTIME IN THE CITY—II
In a city at night, owing to artificial light sources, we may witness substantial modifications to light pillars, ice crystal halos, and rainbows when the weather conditions are right for producing them. The sources of light are much nearer than the sun (or moon) and thus the light from them is divergent, not parallel. This can give rise to some quite amazing three dimensional “surfaces” that differ significantly from their daytime (parallel light) counterparts. In order to appreciate this, some geometry associated with surfaces of revolution will be required. But before getting into that, let’s ask a much more basic question as an appetizer to the main course, a question that is surely on the minds of everyone in these dark times:
Question: How many light bulbs are there in Times Square? [6]
This is a question that has been radically changed by technology because there are so many types of “light bulbs.” Most of the illumination in Times Square comes from giant electronic billboards rather than from traditional light bulbs. These billboards emit their own light. In fact, each pixel of these billboards does so and could be counted as a separate “light bulb.” To answer the question, then, let’s estimate the number of these pixels. Each electronic billboard has about the resolution of a computer monitor or about 1000 × 1000 pixels = 1 million pixels. There are definitely more than one and fewer than 1000 billboards, so again, let’s use the geometric mean of 30. This gives about 30 million separate illuminated pixels.
Looking at a picture of Times Square, one can’t see that many traditional light bulbs. There are a few scrolling message signs on marquees. They are about 1 letter tall and 50 or so letters long. Each letter will have at least 6 × 8 ≈ 50 light bulbs, so let’s round up to 100. This gives 5000 light bulbs per scrolling message. There are five or ten of those, so that gives another 40,000 light bulbs. The static illuminated signs (traditional neon lights or plastic signs back-lit by fluorescent bulbs) will probably add another few thousand, but that’s just a rounding error.
Now we can start thinking about the effects of light on ice crystals in the night air—brrr!
X = αmax − αmin: LIGHT PILLARS FROM STREET LIGHTS
As plate-like ice crystals fall they can reflect and refract light. In the daytime sun pillars are produced by reflection from the faces of tilted crystals in the vicinity of the rising or setting sun. Specifically, the more common upper pillars result from the reflection of sunlight downward from the lower faces of such crystals, and lower pillars arise from sunlight being reflected upward from the upper faces of tilted crystals. Upper pillars can become brighter as the sun descends below the horizon, and can appear to extend very high up in the sky if the upper crystals have larger tilts than the lower ones (see Adam 2011 and references therein). Artificial sources of light can also produce light pillars as shown in Figure 22.1. The angular extent of the pillar is the maximum range of angles (α) subtended at the observer’s eye by light reflected from the crystals. This is determined by their distribution in the air between the observer and the light source (though some beyond the source may also contribute). The angular width of the pillar is similarly defined.
Figure 22.1. How light pillars are formed from ice crystals in the vicinity of a lamp. Each “ray” entering the observer’s eye makes a different angle α with the horizontal direction.
And now for some of the more esoteric artificial light phenomena, as promised earlier.
Question: What is a cone during the day and a cigar or an apple at night?
Answer: The locus of directions along which one sees an ice crystal halo or a rainbow.
Let’s unwrap this conundrum a little bit. When you look at a small portion of a rainbow, you are looking in a particular direction. Sunlight is scattered by raindrops from this direction into your eyes. For the primary (brighter) bow this direction is at about 42 degrees from the imaginary line joining the sun to the shadow of your head (the antisolar point). Scattered sunlight coming from all other directions at 42 degrees to this line will also reach your eye, resulting in the familiar circular arc we see as a rainbow. Thus we are looking along the surface of a cone with half-angle approximately 42 degrees. The same principle applies to the smaller 22-degree ice crystal halo commonly seen around the sun, or the moon at night; again, we are looking along the surface of a (smaller) cone when we observe this halo. We can think of the rainbow and halo arcs in another way: as the intersection of the imaginary cone with an imaginary plane perpendicular to its axis. And no matter how near or far away the raindrops or ice crystals may be
, they give rise to the same circular arcs, because we are looking in certain directions; furthermore, rainbows and halos are not objects, they are images! Put another way, you cannot “back up” to get more of a rainbow in your camera viewfinder. Of course, if you change your lens for a wide angle one, you can achieve that result! Note that the rainbow, an arc of a circle of angular radius 42° around the antisolar point, is the same as a circle of 138° around the sun. Thus a cone with ‘half-angle’ 138° opens up to the opposite direction; this establishes the notion that bows and halos are intimately connected in a geometrical sense as well as an optical one. This will be a valuable unifying concept in the next section.
X and the City: Modeling Aspects of Urban Life Page 18