Figure 19. The parabolic path of a projectile that is fired from a hill in a horizontal direction. Point F is the focus of this parabola.
After canceling the terms α2z2 and α2z20, this reads
which is the same as our previous result, provided we take z0 = ℓ/α2. Note that a parabola of a given shape can be obtained from any cone, with any value of the angular parameter α (alpha), because the shape of any parabola (as opposed to its location and orientation) is entirely determined by a parameter ℓ with the units of length; we do not need to know separately any unit-free parameter like α or the eccentricity of an ellipse.
27. Tennis Ball Derivation of the Law of Refraction
Descartes attempted a derivation of the law of refraction, based on the assumption that a ray of light is bent in passing from one medium to another in the same way that the trajectory of a tennis ball is bent in penetrating a thin fabric. Suppose a tennis ball with speed vA strikes a thin fabric screen obliquely. It will lose some speed, so that after penetrating the screen its speed will be vB < vA, but we would not expect the ball’s passage through the screen to make any change in the component of the ball’s velocity along the screen. We can draw a right triangle whose sides are the components of the ball’s initial velocity perpendicular to the screen and parallel to the screen, and whose hypotenuse is vA. If the ball’s original trajectory makes an angle i to the perpendicular to the screen, then the component of its velocity along the direction parallel to the screen is vA sin i. (See Figure 20.) Likewise, if after penetrating the screen the ball’s trajectory makes an angle r with the perpendicular to the screen, then the component of its velocity along the direction parallel to the screen is vB sin r. Using Descartes’ assumption that the passage of the ball through the screen could change only the component of velocity perpendicular to the interface, not the parallel component, we have
vA sin i = vB sin r
and therefore
where n is the quantity:
Figure 20. Tennis ball velocities. The horizontal line marks a screen penetrated by a tennis ball with initial speed vA and final speed vB. The solid lines marked with arrows indicate the magnitude and direction of the velocities of the ball before and after it penetrates the screen. This figure is drawn with the ball’s path bent toward the perpendicular to the screen, as is the case for light rays entering a denser medium. It shows that in this case passage of the ball through the screen greatly reduces the component of its velocity along the screen, contrary to the assumption of Descartes.
Equation (1) is known as Snell’s law, and it is the correct law of refraction for light. Unfortunately, the analogy between light and tennis balls breaks down when we come to Eq. (2) for n. Since for tennis balls, vB is less than vA, Eq. (2) gives n < 1, while when light passes from air to glass or water, we have n > 1. Not only that; there is no reason to suppose that for tennis balls vB/vA is actually independent of angles i and r, so that Eq. (1) is not useful as it stands.
As shown by Fermat, when light passes from a medium where its speed is vA into one where its speed is vB, the index of refraction n actually equals vA/vB, not vB/vA. Descartes did not know that light travels at a finite speed, and offered a hand-waving argument to explain why n is greater than unity when A is air and B is water. For seventeenth-century applications, like Descartes’ theory of the rainbow, it didn’t matter, because n was assumed to be angle-independent, which is correct for light if not for tennis balls, and its value was taken from observations of refraction, not from measurements of the speed of light in various media.
28. Least-Time Derivation of the Law of Refraction
Hero of Alexandria presented a derivation of the law of reflection, that the angle of reflection equals the angle of incidence, from the assumption that the path of the light ray from an object to the mirror and then to the eye is as short as possible. He might just as well have assumed that the time is as short as possible, since the time taken for light to travel any distance is the distance divided by the speed of light, and in reflection the speed of light does not change. On the other hand in refraction a light ray passes through the boundary between media (such as air and glass) in which the speed of light is different, and we have to distinguish between a principle of least distance and one of least time. Just from the fact that a ray of light is bent when passing from one medium to another, we know that light being refracted does not take the path of least distance, which would be a straight line. Rather, as shown by Fermat, the correct law of refraction can be derived by assuming that light takes the path of least time.
To carry out this derivation, suppose that a ray of light travels from point PA in medium A in which the speed of light is vA to point PB in medium B in which the speed of light is vB. To make this easy to describe, suppose that the surface separating the two media is horizontal. Let the angle between the light rays in media A and B and the vertical direction be i and r, respectively. If points PA and PB are at vertical distances dA and dB from the boundary surface, then the horizontal distance of these points from the point where the rays intersect this surface are dA tan i and dB tanr, respectively, where tan denotes the tangent of an angle, the ratio of the opposite to the adjacent side in a right triangle. (See Figure 21.) Although these distances are not fixed in advance, their sum is the fixed horizontal distance L between points PA and PB:
L = dA tan i + dB tan r
To calculate the time t elapsed when the light goes from PA to PB, we note that the distances traveled in media A and B are dA/cos i and dB/cos r, respectively, where cos denotes the cosine of an angle, the ratio of the adjacent side to the hypotenuse in a right triangle. Time elapsed is distance divided by speed, so the total time elapsed here is
We need to find a general relation between angles i and r (independent of L, dA, and dB) that is satisfied by that angle i which makes time t a minimum, when r depends on i in such a way as to keep L fixed. For this purpose, consider δi, an infinitesimal variation δ (delta) of the angle of incidence i. The horizontal distance between PA and PB is fixed, so when i changes by amount δi, the angle of refraction r must also change, say by amount δr, given by the condition that L is unchanged. Also, at the minimum of t the graph of t versus i must be flat, for if t is increasing or decreasing at some i the minimum must be at some other value of i where t is smaller. This means that the change in t caused by a tiny change δi must vanish, at least to first order in δi. So to find the path of least time we can impose the condition that when we vary both i and r the changes δL and δt must both vanish at least to first order in δi and δr.
Figure 21. Path of a light ray during refraction. The horizontal line marks the interface between two transparent media A and B, in which light has different speeds vA and vB, and angles i and r are measured between the light ray and the dashed vertical line perpendicular to the interface. The solid line marked with arrows represents the path of a light ray that travels from point PA in medium A to point P on the interface between the media and then to point PB in medium B.
To implement this condition, we need standard formulas from differential calculus for the changes δ tan θ (theta) and δ(1/cosθ) when we make an infinitesimal change δθ in angle θ:
where R = 360°/2π = 57.293 . . . ° if θ is measured in degrees. (This angle is called a radian. If θ is measured in radians then R = 1.) Using these formulas, we find the changes in L and t when we make infinitesimal changes δi and δr in angles i and r:
The condition that δL = 0 tells us that
so that
For this to vanish, we must have
or in other words
with the index of refraction n given by the angle-independent ratio of velocities:
n = vA/vB
This is the correct law of refraction, with the correct formula for n.
29. The Theory of the Rainbow
Suppose that a ray of light arrives at a spherical raindrop at a point P, where it makes an angle i to the normal to the drop’s surface. If
there were no refraction, the light ray would continue straight through the drop. In this case, the line from the center C of the drop to the point Q of closest approach of the ray to the center would make a right angle with the light ray, so triangle PCQ would be a right triangle with hypotenuse equal to the radius R of the circle, and the angle at P equal to i. (See Figure 22a.) The impact parameter b is defined as the distance of closest approach of the unrefracted ray to the center, so it is the length of side CQ of the triangle, given by elementary trigonometry as
b = R sin i
We can equally well characterize individual light rays by their value of b/R, as was done by Descartes, or by the value of the angle of incidence i.
Because of refraction, the ray will actually enter the drop at angle r to the normal, given by the law of refraction:
where n 4/3 is the ratio of the speed of light in air to its speed in water. The ray will cross the drop, and strike its back side at point p'. Since the distances from the center C of the drop to P and to P' are both equal to the radius R of the drop, the triangle with vertices C, P, and P' is isosceles, so the angles between the light ray and the normals to the surface at P and at P' must be equal, and hence both equal to r. Some of the light will be reflected from the back surface, and by the law of reflection the angle between the reflected ray and the normal to the surface at P' will again be r. The reflected ray will cross the drop and strike its front surface at a point P", again making an angle r with the normal to the surface at P". Some of the light will then emerge from the drop, and by the law of refraction the angle between the emergent ray and the normal to the surface at P" will equal the original incident angle i. (See Figure 22b. This figure shows the path of the light ray through a plane parallel to the ray’s original direction that contains the center of the raindrop and the observer. Only rays that impinge on the drop’s surface where it intersects this plane have a chance of reaching the observer.)
Figure 22. The path of a ray of sunlight in a spherical drop of water. The ray is indicated by solid lines marked with arrows, and enters the drop at point P, where it makes an angle i with the normal to the surface. (a) Path of the ray if there were no refraction, with Q the point of the ray’s closest approach to the center C of the drop in this case. (b) The ray refracted on entering the drop at P, reflected from the back surface of the drop at P", and then refracted again on leaving the drop at P". Dashed lines run from the center C of the drop to points where the ray meets the surface of the drop.
In the course of all this bouncing around, the light ray will have been bent toward the center of the drop by an angle i – r twice on entering and leaving the drop, and by an angle 180° – 2r when reflected by the back surface of the drop, and hence by a total angle
2(i − r) + 180° − 2r = 180° − 4r + 2i
If the light ray bounced straight back from the drop (as is the case for i = r = 0) this angle would be 180°, and the initial and final light rays would be along the same line, so the actual angle φ (phi) between the initial and final light rays is
φ = 4r − 2i
We can express r in terms of i, as
where for any quantity x, the quantity arcsin x is the angle (usually taken between –90° and +90°) whose sine is x. The numerical calculation for n = 4/3 reported in Chapter 13 shows that φ rises from zero at i = 0 to a maximum value of about 42° and then drops to about 14° at i = 90°. The graph of j versus i is flat at its maximum, so light tends to emerge from the drop at a deflection angle φ close to 42°.
If we look up at a misty sky with the sun behind us, we see light reflected back primarily from directions in the sky where the angle between our line of sight and the sun’s rays is near 42°. These directions form a bow, usually running from the Earth’s surface up into the sky and then down again to the surface. Because n depends slightly on the color of light, so does the maximum value of the deflection angle φ, so this bow is spread out into different colors. This is the rainbow.
It is not difficult to derive an analytic formula that gives the maximum value of φ for any value of the index of refraction n. To find the maximum of φ, we use the fact that the maximum occurs at an incident angle i where the graph of φ versus i is flat, so that the variation δφ (delta phi) in φ produced by a small variation δi in i vanishes to first order in δi. To use this condition, we use a standard formula of calculus, which tells us that when we make a change δx in x, the change in arcsin x is
where if arcsin x is measured in degrees, then R = 360°/2π. Thus when the angle of incidence varies by an amount δi, the angle of deflection changes by
or, since δ sin i = cos i δi/R,
Hence the condition for a maximum of φ is that
Squaring both sides of the equation and using cos2i = 1 – sin2i (which follows from the theorem of Pythagoras), we can then solve for sin i, and find
At this angle, φ takes its maximum value:
For n = 4/3, the maximum value of φ is reached for b/R = sin i = 0.86, for which i = 59.4°, where r = 40.2°, and φmax = 42.0°.
30. Wave Theory Derivation of the Law of Refraction
The law of refraction, which as described in Technical Note 28 can be derived from an assumption that refracted light rays take the path of least time, can also be derived on the basis of the wave theory of light. According to Huygens, light is a disturbance in a medium, which may be some transparent material or space that is apparently empty. The front of the disturbance is a line, which moves forward in a direction at right angles to the front, at a speed characteristic of the medium.
Figure 23. Refraction of a light wave. The horizontal line again marks the interface between two transparent media, in which light has different speeds. The crosshatched lines show a segment of a wave front at two different times—when the leading edge and when the trailing edge of the wave front just touch the interface. The solid lines marked with arrows show the paths taken by the leading and trailing edges of the wave front.
Consider a segment of the front of such a disturbance, which is of length L in medium 1, traveling toward an interface with medium 2. Let us suppose that the direction of motion of the disturbance, which is at right angles to this front, makes angle i with the perpendicular to this interface. When the leading edge of the disturbance strikes the interface at point A, the trailing edge B is still at a distance (along the direction the disturbance is traveling) equal to L tan i. (See Figure 23.) Hence the time required for the trailing edge to reach the interface at point D is L tan i/v1, where v1 is the velocity of the disturbance in medium 1. During this time the leading edge of the front will have traveled in medium 2 at an angle r to the perpendicular, reaching a point C at a distance v2 L tan i/v1 from A, where v2 is the velocity in medium 2. At this time the wave front, which is at right angles to the direction of motion in medium 2, extends from C to D, so that the triangle with vertices A, C, and D is a right triangle, with a 90° angle at C. The distance v2 L tan i/v1 from A to C is the side opposite angle r in this right triangle, while the hypotenuse is the line from A to D, which has length L/cos i. (Again, see Figure 23.) Hence
Recalling that tan i = sin i/cos i, we see that the factors of cos i and L cancel, so that
sin r = v2 sin i/v1
or in other words
which is the correct law of refraction.
It is not an accident that the wave theory, as worked out by Huygens, gives the same results for refraction as the least-time principle of Fermat. It can be shown that, even for waves passing through a heterogeneous medium in which the speed of light changes gradually in various directions, not just suddenly at a plane interface, the wave theory of Huygens will always give a light path that takes the shortest time to travel between any two points.
31. Measuring the Speed of Light
Suppose we observe some periodic process occurring at some distance from us. For definiteness we will consider a moon going around a distant planet, but the analysis below would apply to any process that repeats periodic
ally. Suppose that the moon reaches the same stage in its orbit at two consecutive times t1 and t2; for instance these might be times that the moon consecutively emerges from behind the planet. If the intrinsic orbital period of the moon is T, then t2 – t1 = T. This is the period we observe, provided the distance between us and the planet is fixed. But if this distance is changing, then the period we observe will be shifted from T, by an amount that depends on the speed of light.
Suppose that the distances between us and the planet at two successive times when the moon is at the same stage in its orbit are d1 and d2. We then observe these stages in the orbit at times
tʹ1 = t1 + d1/c tʹ2 = t2 + d2/c
where c is the speed of light. (We are assuming here that the distance between the planet and its moon may be neglected.) If the distance between us and this planet is changing at rate v, either because it is moving or because we are or both, then d2 – d1 = vT, and so the observed period is
(This derivation depends on the assumption that v should change very little in a time T, which is typically true in the solar system, but v may be changing appreciably over longer time scales.) When the distant planet is moving toward or away from us, in which case v is respectively negative or positive, its moon’s apparent period will be decreased or increased, respectively. We can measure T by observing the planet at a time when v = 0, and then measure the speed of light by observing the period again at a time when v has some known nonzero value.
This is the basis of the determination of the speed of light by Huygens, based on Rømer’s observation of the changing apparent orbital period of Jupiter’s moon Io. But with the speed of light known, the same calculation can tell us the relative velocity v of a distant object. In particular, the light waves from a specific line in the spectrum of a distant galaxy will oscillate with some characteristic period T, related to its frequency ν (nu) and wavelength λ (lambda) by T = 1/ν = λ/c. This intrinsic period is known from observations of spectra in laboratories on Earth. Since the early twentieth century the spectral lines observed in very distant galaxies have been found to have longer wavelengths, and hence longer periods, from which we can infer that these galaxies are moving away from us.
To Explain the World: The Discovery of Modern Science Page 34