To Explain the World: The Discovery of Modern Science

by Steven Weinberg

  ϕ = nβ = (n − 1)θ

  So we see that ϕ is simply proportional to θ, and therefore, using our previous formula for f/r, we have

  This is independent of θ, so as promised all horizontal light rays entering the lens are focused to the same point on the centerline of the lens.

  If the radius of curvature r is very large, then the curvature of the front surface of the lens is very small, so that the lens is nearly the same as a flat plate of glass, with the bending of light on entering the lens being nearly canceled by its bending on leaving the lens. Similarly, whatever the shape of the lens, if the index of refraction n is close to 1, then the lens bends the light ray very little. In either case the focal length is very large, and we say that the lens is weak. A strong lens is one that has a moderate radius of curvature and an index of refraction appreciably different from 1, as for instance a lens made of glass, for which n 1.5.

  A similar result holds if the back surface of the lens is not plane, but a segment of a sphere of radius r’. In this case the focal length is

  This gives the same result as before if r’ is much larger than r, in which case the back surface is nearly flat.

  The concept of focal length can also be extended to concave lenses, like the lens that Galileo used as the eyepiece of his telescope. A concave lens can take rays of light that are converging and spread them out so that they are parallel, or even diverge. We can define the focal length of such a lens by considering converging rays of light that are made parallel by the lens; the focal length is the distance behind the lens of the point to which such rays would converge if not made parallel by the lens. Though its meaning is different, the focal length of a concave lens is given by a formula like the one we have derived for a convex lens.

  23. Telescopes

  As we saw in Technical Note 22, a thin convex lens will focus rays of light that strike it parallel to its central axis to a point F on this axis, at a distance behind the lens known as the focal length f of the lens. Parallel rays of light that strike the lens at a small angle γ (gamma) to the central axis will also be focused by the lens, but to a point that is a little off the central axis. To see how far off, we can imagine rotating the drawing of the ray path in Figure 16a around the lens by angle γ. The distance d of the focal point from the central axis of the lens will then be the same fraction of the circumference of a circle of radius f that γ is of 360°:

  and therefore

  (This works only for thin lenses; otherwise d also depends on the angle θ introduced in Technical Note 22.) If the rays of light from some distant object strike the lens in a range Δγ (delta gamma) of angles, they will be focused to a strip of height Δd, given by

  (As usual, this formula is simpler if Δγ is measured in radians, equal to 360° /2π, rather than in degrees; in this case it reads simply Δd = f Δγ.) This strip of focused light is known as a “virtual image.” (See Figure 17a).

  We cannot see the virtual image just by peering at it, because after reaching this image the rays of light diverge again. To be focused to a point on the retina of a relaxed human eye, rays of light must enter the lens of the eye in more or less parallel directions. Kepler’s telescope included a second convex lens, known as the eyepiece, to focus the diverging rays of light from the virtual image so that they left the telescope along parallel directions. By repeating the above analysis, but with the direction of the light rays reversed, we see that for the rays of light from a point on the light source to leave the telescope on parallel directions, the eyepiece must be placed at a distance f’ from the virtual image, where f’ is the focal length of the eyepiece. (See Figure 17b.) That is, the length L of the telescope must be the sum of the focal lengths

  L = f + fʹ

  The range Δγ' of directions of the light rays from different points on the source entering the eye is related to the size of the virtual image by

  Figure 17. Telescopes. (a) Formation of a virtual image. The two solid lines marked with arrows are rays of light that enter the lens on lines separated by a small angle Δγ. These lines (and others parallel to them) are focused to points at a distance f from the lens, with a vertical separation Δd proportional to Δγ. (b) The lenses in Kepler’s telescope. The lines marked with arrows indicate the paths of light rays that enter a weak convex lens from a distant object, on essentially parallel directions; are focused by the lens to a point at a distance f from the lens; diverge from this point; and are then bent by a strong convex lens so that they enter the eye on parallel directions.

  The apparent size of any object is proportional to the angle subtended by rays of light from the object, so the magnification produced by the telescope is the ratio of this angle when the rays enter the eye to the angle they would have spanned if there were no telescope:

  By taking the ratio of the two formulas we have derived for size Δd of the virtual image, we see that the magnification is

  To get a significant degree of magnification, we need the lens at the front of the telescope to be much weaker than the eyepiece, with f >> f’.

  This is not so easy. According to the formula for the focal length given in Technical Note 22, to have a strong glass eyepiece with short focal length f', it is necessary for it to have a small radius of curvature, which means either that it must be very small, or that it must not be thin (that is, with a thickness much less than the radius of curvature), in which case it does not focus the light well. We can instead make the lens at the front weak, with large focal length f, but in this case the length L = f + f’ f of the telescope must be large, which is awkward. It took some time for Galileo to refine his telescope to give it a magnification sufficient for astronomical purposes.

  Galileo used a somewhat different design in his telescope, with a concave eyepiece. As mentioned in Technical Note 22, if a concave lens is properly placed, converging rays of light that enter it will leave on parallel directions; the focal length is the distance behind the lens at which the rays would converge if not for the lens. In Galileo’s telescope there was a weak convex lens in front with focal length f, with a strong concave lens of focal length f’ behind it, at a distance f’ in front of the place where there would be a virtual image if not for the concave lens. The magnification of such a telescope is again the ratio f/f’, but its length is only f – f’ instead of f + f’.

  24. Mountains on the Moon

  The bright and dark sides of the Moon are divided by a line known as the “terminator,” where the Sun’s rays are just tangent to the Moon’s surface. When Galileo turned his telescope on the Moon he noticed bright spots on the dark side of the Moon near the terminator, and interpreted them as reflections from mountains high enough to catch the Sun’s rays coming from the other side of the terminator. He could infer the height of these mountains by a geometrical construction similar to that used by al-Biruni to measure the size of the Earth. Draw a triangle whose vertices are the center C of the Moon, a mountaintop M on the dark side of the Moon that just catches a ray of sunlight, and the spot T on the terminator where this ray grazes the surface of the Moon. (See Figure 18.) This is a right triangle; line TM is tangent to the Moon’s surface at T, so this line must be perpendicular to line CT. The length of CT is just the radius r of the moon, while the length of TM is the distance d of the mountain from the terminator. If the mountain has height h, then the length of CM (the hypotenuse of the triangle) is r + h. According to the Pythagorean theorem, we then have

  (r + h)2 = r2 + d2

  and therefore

  d2 = (r + h)2 − r2 = 2rh + h2

  Since the height of any mountain on the Moon is much less than the size of the Moon, we can neglect h2 compared with 2rh. Dividing both sides of the equation by 2r2 then gives

  So by measuring the ratio of the apparent distance of a mountaintop from the terminator to the apparent radius of the Moon, Galileo could find the ratio of the mountain’s height to the Moon’s radius.

  Figure 18. Galileo’s measurement of the height of moun
tains on the Moon. The horizontal line marked with an arrow indicates a ray of light that grazes the Moon at the terminator T marking the boundary between the Moon’s bright and dark sides, and then strikes the top M of a mountain of height h at a distance d from the terminator.

  Galileo in Siderius Nuncius reported that he sometimes saw bright spots on the dark side of the Moon at an apparent distance from the terminator greater than 1/20 the apparent diameter of the Moon, so for these mountains d/r > 1/10, and therefore according to the above formula h/r > (1/10)2 /2 = 1/200. Galileo estimated the radius of the Moon to be 1,000 miles,* so these mountains would be at least 5 miles high. (For reasons that are not clear, Galileo gave a figure of 4 miles, but since he was trying only to set a lower bound on the mountain height, perhaps he was just being conservative.) Galileo thought that this was higher than any mountain on the Earth, but now we know that there are mountains on Earth that are almost 6 miles high, so Galileo’s observations indicated that the heights of mountains on the Moon are not very different from the heights of terrestrial mountains.

  25. Gravitational Acceleration

  Galileo showed that a falling body undergoes uniform acceleration—that is, its speed increases by the same amount in each equal interval of time. In modern terms, a body that falls from rest will after time t have a velocity v given by a quantity proportional to t:

  v = gt

  where g is a constant characterizing the gravitational field at the surface of the Earth. Although g varies somewhat from place to place on the Earth’s surface, it is never very different from 32 feet/second per second, or 9.8 meters/second per second.

  According to the mean speed theorem, the distance that such a body will fall from rest in time t is vmeant, where vmean is the average of gt and zero; in other words, vmean = gt/2. Hence the distance fallen is

  In particular, in the first second the body falls a distance g(1 second)2/2 = 16 feet. The time required to fall distance d is in general

  There is another, more modern way of looking at this result. The falling body has an energy equal to the sum of a kinetic energy term and a potential energy term. The kinetic energy is

  where m is the body’s mass. The potential energy is mg times the height (measured from any arbitrary altitude), so if the body is dropped from rest at an initial height h0 and falls distance d, then

  Epotential = mgh = mg(h0 − d)

  Hence with d = gt2/2, the total energy is a constant:

  E = Ekinetic + Epotential = mgh0

  We can turn this around, and derive the relation between velocity and distance fallen by assuming the conservation of energy. If we set E equal to the value mgh0 that it has at t = 0, when v = 0 and h = h0, then the conservation of energy gives at all times

  from which it follows that v2/2 = gd. Since v is the rate of increase of d, this is a differential equation that determines the relation between d and t. Of course, we know the solution of this equation: it is d = gt2/2, for which v = gt. So by using the conservation of energy we can get these results without knowing in advance that the acceleration is uniform.

  This is an elementary example of the conservation of energy, which makes the concept of energy useful in a wide variety of contexts. In particular, the conservation of energy shows the relevance of Galileo’s experiments with balls rolling down inclined planes to the problem of free fall, though this is not an argument used by Galileo. For a ball of mass m rolling down a plane, the kinetic energy is mv2/2, where v is now the velocity along the plane, and the potential energy is mgh, where h is again the height. In addition there is an energy of rotation of the ball, which takes the form

  where r is the radius of the ball, ν (nu) is the number of complete turns of the ball per second, and ζ (zeta) is a number that depends on the shape and mass-distribution of the ball. In the case that is probably relevant to Galileo’s experiments, that of a solid uniform ball, ζ has the value ζ = 2/5. (If the ball were hollow, we would have ζ = 2/3.) Now, when the ball makes one complete turn it travels a distance equal to its circumference 2πr, so in time t when it makes νt turns it travels distance d = 2πrνt, and therefore its velocity is d/t = 2πνr. Using this in the formula for the energy of rotation, we see that

  Dividing by m and 1 + ζ, the conservation of energy therefore requires that

  This is the same relation between speed and distance fallen d = h0 – h that holds for a body falling freely, except that g has been replaced with g/(1 + ζ). Aside from this change, the dependence of the velocity of the ball rolling down the inclined plane on the vertical distance traveled is the same as that for a body in free fall. Hence the study of balls rolling down inclined planes could be used to verify that freely falling bodies experience uniform acceleration; but unless the factor 1/(1 + ζ) is taken into account, it could not be used to measure the acceleration.

  By a complicated argument, Huygens was able to show that the time it takes a pendulum of length L to swing through a small angle from one side to the other is

  This equals π times the time required for a body to fall distance d = L/2, the result stated by Huygens.

  26. Parabolic Trajectories

  Suppose a projectile is shot horizontally with speed v. Neglecting air resistance, it will continue with this horizontal component of velocity, but it will accelerate downward. Hence after time t it will have moved a horizontal distance x = vt and a downward distance z proportional to the square of the time, conventionally written as z = gt2/2, with g = 32 feet/second per second, a constant measured after Galileo’s death by Huygens. With t = x/v, it follows that

  z = gx2/2v2

  This equation, giving one coordinate as proportional to the square of the other, defines a parabola.

  Note that if the projectile is fired from a gun at height h above the ground, then the horizontal distance x traveled when the projectile has fallen distance z = h and reached the ground is . Even without knowing v or g, Galileo could have verified that the path of the projectile is a parabola, by measuring the distance traveled d for various heights fallen h, and checking that d is proportional to the square root of h. It is not clear whether Galileo ever did this, but there is evidence that in 1608 he did a closely related experiment, mentioned briefly in Chapter 12. A ball is allowed to roll down an inclined plane from various initial heights H, then rolls along the horizontal tabletop on which the inclined plane sits, and finally shoots off into the air from the table edge. As shown in Technical Note 25, the velocity of the ball at the bottom of the inclined plane is

  where as usual g = 32 feet/second per second, and ζ (zeta) is the ratio of rotational to kinetic energy of the ball, a number depending on the distribution of mass within the rolling ball. For a solid ball of uniform density, ζ = 2/5. This is also the velocity of the ball when it shoots off horizontally into space from the edge of the tabletop, so the horizontal distance that the ball travels by the time it falls height h will be

  Galileo did not mention the correction for rotational motion represented by ζ, but he may have suspected that some such correction could reduce the horizontal distance traveled, because instead of comparing this distance with the value d = expected in the absence of ζ, he only checked that for fixed table height h, distance d was indeed proportional to , to an accuracy of a few percent. For one reason or another, Galileo never published the result of this experiment.

  For many purposes of astronomy and mathematics, it is convenient to define a parabola as the limiting case of an ellipse, when one focus moves very far from the other. The equation for an ellipse with major axis 2a and minor axis 2b is given in Technical Note 18 as:

  in which for convenience later on we have replaced the coordinates x and y used in Note 18 with z – z0 and x, with z0 a constant that can be chosen as we like. The center of this ellipse is at z = z0 and x = 0. As we saw in Note 18, there is a focus at z – z0 = –ae, x = 0, where e is the eccentricity, with e2 ≡ 1 – b2/a2, and the point of closest approach of the curve to this focus is at z – z0 =
–a and x = 0. It will be convenient to give this point of closest approach the coordinates z = 0 and x = 0 by choosing z0 = a and in which case the nearby focus is at z = z0 – ea = (1 – e)a. We want to let a and b become infinitely large, so that the other focus goes to infinity and the curve has no maximum x coordinate, but we want to keep the distance (1 – e)a of closest approach to the nearer focus finite, so we set

  1 − e = ℓ/a

  with ℓ held fixed as a goes to infinity. Since e approaches unity in this limit, the semiminor axis b is given by

  b2 = a2(1 − e2) = a2(1 − e)(1 + e) → 2a2(1 − e) = 2ℓa

  Using z0 = a and this formula for b2, the equation for the ellipse becomes

  The term a2/a2 on the left cancels the 1 on the right. Multiplying the remaining equation with a then gives

  For a much larger than x, y, or ℓ, the first term may be dropped, so this equation becomes

  This is the same as the equation we derived for the motion of a projectile fired horizontally, provided we take

  so the focus F of the parabola is at distance ℓ = v2/2g below the initial position of the projectile. (See Figure 19.)

  Parabolas can, like ellipses, be regarded as conic sections, but for parabolas the plane intersecting the cone is parallel to the cone’s surface. Taking the equation of a cone centered on the = α(z + z0), and the equation of a plane parallel to the cone as simply y = α(z – z0), with z0 arbitrary, the intersection of the cone and the plane satisfies