By the end of this section, you will be able to:
The changing of a light ray’s direction (loosely called bending) when it passes through variations in matter is called refraction. The speed of light c not only affects refraction, it is one of the central concepts of Einstein’s theory of relativity. As the accuracy of the measurements of the speed of light were improved, c was found not to depend on the velocity of the source or the observer. However, the speed of light does vary in a precise manner with the material it traverses. These facts have far-reaching implications, as we will see in the chapter Special Relativity. It makes connections between space and time and alters our expectations that all observers measure the same time for the same event, for example. The speed of light is so important that its value in a vacuum is one of the most fundamental constants in nature as well as being one of the four fundamental SI units. Figure 1. Looking at the fish tank as shown, we can see the same fish in two different locations, because light changes directions when it passes from water to air. In this case, the light can reach the observer by two different paths, and so the fish seems to be in two different places. This bending of light is called refraction and is responsible for many optical phenomena. Why does light change direction when passing from one material (medium) to another? It is because light changes speed when going from one material to another. So before we study the law of refraction, it is useful to discuss the speed of light and how it varies in different media. Early attempts to measure the speed of light, such as those made by Galileo, determined that light moved extremely fast, perhaps instantaneously. The first real evidence that light traveled at a finite speed came from the Danish astronomer Ole Roemer in the late 17th century. Roemer had noted that the average orbital period of one of Jupiter’s moons, as measured from Earth, varied depending on whether Earth was moving toward or away from Jupiter. He correctly concluded that the apparent change in period was due to the change in distance between Earth and Jupiter and the time it took light to travel this distance. From his 1676 data, a value of the speed of light was calculated to be 2.26 × 108 m/s (only 25% different from today’s accepted value). In more recent times, physicists have measured the speed of light in numerous ways and with increasing accuracy. One particularly direct method, used in 1887 by the American physicist Albert Michelson (1852–1931), is illustrated in Figure 2. Light reflected from a rotating set of mirrors was reflected from a stationary mirror 35 km away and returned to the rotating mirrors. The time for the light to travel can be determined by how fast the mirrors must rotate for the light to be returned to the observer’s eye. Figure 2. A schematic of early apparatus used by Michelson and others to determine the speed of light. As the mirrors rotate, the reflected ray is only briefly directed at the stationary mirror. The returning ray will be reflected into the observer's eye only if the next mirror has rotated into the correct position just as the ray returns. By measuring the correct rotation rate, the time for the round trip can be measured and the speed of light calculated. Michelson’s calculated value of the speed of light was only 0.04% different from the value used today.The speed of light is now known to great precision. In fact, the speed of light in a vacuum c is so important that it is accepted as one of the basic physical quantities and has the fixed value c = 2.9972458 × 108 m/s ≈ 3.00 × 108 m/s, where the approximate value of 3.00 × 108 m/s is used whenever three-digit accuracy is sufficient. The speed of light through matter is less than it is in a vacuum, because light interacts with atoms in a material. The speed of light depends strongly on the type of material, since its interaction with different atoms, crystal lattices, and other substructures varies. We define the index of refraction n of a material to be n=cvn=\frac{c}{v}\\n=vc , where v is the observed speed of light in the material. Since the speed of light is always less than c in matter and equals c only in a vacuum, the index of refraction is always greater than or equal to one.
c = 2.9972458 × 108 m/s ≈ 3.00 × 108 m/s
n=cv\displaystyle{n}=\frac{c}{v}\\n=vc That is, n ≥ 1. Table 1 gives the indices of refraction for some representative substances. The values are listed for a particular wavelength of light, because they vary slightly with wavelength. (This can have important effects, such as colors produced by a prism.) Note that for gases, n is close to 1.0. This seems reasonable, since atoms in gases are widely separated and light travels at c in the vacuum between atoms. It is common to take n = 1 for gases unless great precision is needed. Although the speed of light v in a medium varies considerably from its value c in a vacuum, it is still a large speed.
Calculate the speed of light in zircon, a material used in jewelry to imitate diamond. The speed of light in a material, v, can be calculated from the index of refraction n of the material using the equation n=cvn=\frac{c}{v}\\n=vc . The equation for index of refraction states thatn=cvn=\frac{c}{v}\\n=vc . Rearranging this to determine v givesv=cnv=\frac{c}{n}\\v=nc .The index of refraction for zircon is given as 1.923 in Table 1, and c is given in the equation for speed of light. Entering these values in the last expression gives v=3.00×108 m/s1.923 =1.56×108 m/s\begin{array}{lll}v&=&\frac{3.00\times10^8\text{ m/s}}{1.923}\\\text{ }&=&1.56\times10^8\text{ m/s}\end{array}\\v ==1.9233.00×108 m/s1.56×108 m/s DiscussionThis speed is slightly larger than half the speed of light in a vacuum and is still high compared with speeds we normally experience. The only substance listed in Table 1 that has a greater index of refraction than zircon is diamond. We shall see later that the large index of refraction for zircon makes it sparkle more than glass, but less than diamond. Figure 3 shows how a ray of light changes direction when it passes from one medium to another. As before, the angles are measured relative to a perpendicular to the surface at the point where the light ray crosses it. (Some of the incident light will be reflected from the surface, but for now we will concentrate on the light that is transmitted.) The change in direction of the light ray depends on how the speed of light changes. The change in the speed of light is related to the indices of refraction of the media involved. In the situations shown in Figure 3, medium 2 has a greater index of refraction than medium 1. This means that the speed of light is less in medium 2 than in medium 1. Note that as shown in Figure 3a, the direction of the ray moves closer to the perpendicular when it slows down. Conversely, as shown in Figure 3b, the direction of the ray moves away from the perpendicular when it speeds up. The path is exactly reversible. In both cases, you can imagine what happens by thinking about pushing a lawn mower from a footpath onto grass, and vice versa. Going from the footpath to grass, the front wheels are slowed and pulled to the side as shown. This is the same change in direction as for light when it goes from a fast medium to a slow one. When going from the grass to the footpath, the front wheels can move faster and the mower changes direction as shown. This, too, is the same change in direction as for light going from slow to fast. Figure 3. The change in direction of a light ray depends on how the speed of light changes when it crosses from one medium to another. The speed of light is greater in medium 1 than in medium 2 in the situations shown here. (a) A ray of light moves closer to the perpendicular when it slows down. This is analogous to what happens when a lawn mower goes from a footpath to grass. (b) A ray of light moves away from the perpendicular when it speeds up. This is analogous to what happens when a lawn mower goes from grass to footpath. The paths are exactly reversible.The amount that a light ray changes its direction depends both on the incident angle and the amount that the speed changes. For a ray at a given incident angle, a large change in speed causes a large change in direction, and thus a large change in angle. The exact mathematical relationship is the law of refraction, or “Snell’s Law,” which is stated in equation form as n1 sinθ1 = n2 sinθ2. Here n1 and n2 are the indices of refraction for medium 1 and 2, and θ1 and θ2 are the angles between the rays and the perpendicular in medium 1 and 2, as shown in Figure 3. The incoming ray is called the incident ray and the outgoing ray the refracted ray, and the associated angles the incident angle and the refracted angle. The law of refraction is also called Snell’s law after the Dutch mathematician Willebrord Snell (1591–1626), who discovered it in 1621. Snell’s experiments showed that the law of refraction was obeyed and that a characteristic index of refraction n could be assigned to a given medium. Snell was not aware that the speed of light varied in different media, but through experiments he was able to determine indices of refraction from the way light rays changed direction. A classic observation of refraction occurs when a pencil is placed in a glass half filled with water. Do this and observe the shape of the pencil when you look at the pencil sideways, that is, through air, glass, water. Explain your observations. Draw ray diagrams for the situation. Find the index of refraction for medium 2 in Figure 3a, assuming medium 1 is air and given the incident angle is 30.0º and the angle of refraction is 22.0º. The index of refraction for air is taken to be 1 in most cases (and up to four significant figures, it is 1.000). Thus n1=1.00 here. From the given information, θ1 = 30.0º and θ2 = 22.0º. With this information, the only unknown in Snell’s law is n2, so that it can be used to find this unknown. Snell’s law is n1 sinθ1 = n2 sinθ2. Rearranging to isolate n2 gives n2=n1sinθ1sinθ2\displaystyle{n}_2=n_1\frac{\sin\theta_1}{\sin\theta_2}\\n2=n1sinθ2sinθ1 Entering known values,n2=1.00sin30.0∘sin22.0∘=0.5000.375 =1.33\begin{array}{lll}{n}_2&=&1.00\frac{\sin30.0^{\circ}}{\sin22.0^{\circ}}=\frac{0.500}{0.375}\\\text{ }&=&1.33\end{array}\\n2 ==1.00sin22.0∘sin30.0∘=0.3750.5001.33 DiscussionThis is the index of refraction for water, and Snell could have determined it by measuring the angles and performing this calculation. He would then have found 1.33 to be the appropriate index of refraction for water in all other situations, such as when a ray passes from water to glass. Today we can verify that the index of refraction is related to the speed of light in a medium by measuring that speed directly.Suppose that in a situation like that in Example 2, light goes from air to diamond and that the incident angle is 30.0º. Calculate the angle of refraction θ2 in the diamond. Again the index of refraction for air is taken to be n1 = 1.00, and we are given θ1 = 30.0º. We can look up the index of refraction for diamond in Table 1, finding n2 = 2.419. The only unknown in Snell’s law is θ2, which we wish to determine. Solving Snell’s law for sin θ2 yields sinθ2=n1n2sinθ1\displaystyle\sin\theta_2=\frac{n_1}{n_2}\sin\theta_1\\sinθ2=n2n1sinθ1 . Entering known values,sinθ2=1.002.419sin30.0∘=(0.413)(0.500)=0.207\displaystyle\sin\theta_2=\frac{1.00}{2.419}\sin30.0^{\circ}=\left(0.413\right)\left(0.500\right)=0.207\\sinθ2=2.4191.00sin30.0∘=(0.413)(0.500)=0.207 . The angle is thus θ2 = sin−1 0.207 = 11.9º. For the same 30º angle of incidence, the angle of refraction in diamond is significantly smaller than in water (11.9º rather than 22º—see the preceding example). This means there is a larger change in direction in diamond. The cause of a large change in direction is a large change in the index of refraction (or speed). In general, the larger the change in speed, the greater the effect on the direction of the ray.
index of refraction: for a material, the ratio of the speed of light in vacuum to that in the material 1. 2.25 × 108 m/s in water; 2.04 × 108 m/s in glycerine 3. 1.490, polystyrene 5. 1.28 s 7. 1.03 ns 9. n = 1.46, fused quartz 13. (a) 0.898; (b) Can’t have n < 1.00 since this would imply a speed greater than c; (c) Refracted angle is too big relative to the angle of incidence. 15. (a) c5.00\frac{c}{5.00}\\5.00c ; (b) Speed of light too slow, since index is much greater than that of diamond; (c) Angle of refraction is unreasonable relative to the angle of incidence.CC licensed content, Shared previouslyPage 2By the end of this section, you will be able to:
Consider what happens when a ray of light strikes the surface between two materials, such as is shown in Figure 1a. Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than for the first, the ray bends away from the perpendicular. (Since n1 > n2, the angle of refraction is greater than the angle of incidence—that is, θ1 > θ2.) Now imagine what happens as the incident angle is increased. This causes θ2 to increase also. The largest the angle of refraction θ2 can be is 90º, as shown in Figure 1b.The critical angle θc for a combination of materials is defined to be the incident angle θ1 that produces an angle of refraction of 90º. That is, θc is the incident angle for which θ2 = 90º. If the incident angle θ1 is greater than the critical angle, as shown in Figure 1c, then all of the light is reflected back into medium 1, a condition called total internal reflection. The incident angle θ1 that produces an angle of refraction of 90º is called the critical angle, θc. Figure 1. (a) A ray of light crosses a boundary where the speed of light increases and the index of refraction decreases. That is, n2 < n1 . The ray bends away from the perpendicular. (b) The critical angle θc is the one for which the angle of refraction is. (c) Total internal reflection occurs when the incident angle is greater than the critical angle. Snell’s law states the relationship between angles and indices of refraction. It is given byn1 sin θ1 = n2 sin θ2. When the incident angle equals the critical angle (θ1 = θc), the angle of refraction is 90º (θ2 = 90º). Noting that sin 90º = 1, Snell’s law in this case becomesn1 sin θ1 = n2. The critical angle θc for a given combination of materials is thusθc=sin−1(n2n1)\theta_{c}=\sin^{-1}\left(\frac{n_2}{n_1}\right)\\θc=sin−1(n1n2) for n1 > n2. Total internal reflection occurs for any incident angle greater than the critical angle θc, and it can only occur when the second medium has an index of refraction less than the first. Note the above equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in the figure.What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air? The index of refraction for polystyrene is found to be 1.49 in Figure 2, and the index of refraction of air can be taken to be 1.00, as before. Thus, the condition that the second medium (air) has an index of refraction less than the first (plastic) is satisfied, and the equation θc=sin−1(n2n1)\theta_{c}=\sin^{-1}\left(\frac{n_2}{n_1}\right)\\θc=sin−1(n1n2) can be used to find the critical angle θc. Here, then, n2 = 1.00 and n1 = 1.49. The critical angle is given byθc=sin−1(n2n1)\theta_{c}=\sin^{-1}\left(\frac{n_2}{n_1}\right)\\θc=sin−1(n1n2) Substituting the identified values givesθc=sin−1(1.001.49) =sin−1(0.671) =42.2∘\begin{array}{lll}\theta_{c}&=&\sin^{-1}\left(\frac{1.00}{1.49}\right)\\\text{ }&=&\sin^{-1}\left(0.671\right)\\\text{ }&=&42.2^{\circ}\end{array}\\θc ===sin−1(1.491.00)sin−1(0.671)42.2∘ DiscussionThis means that any ray of light inside the plastic that strikes the surface at an angle greater than 42.2º will be totally reflected. This will make the inside surface of the clear plastic a perfect mirror for such rays without any need for the silvering used on common mirrors. Different combinations of materials have different critical angles, but any combination with n1 > n2 can produce total internal reflection. The same calculation as made here shows that the critical angle for a ray going from water to air is 48.6º, while that from diamond to air is 24.4º, and that from flint glass to crown glass is 66.3º. There is no total reflection for rays going in the other direction—for example, from air to water—since the condition that the second medium must have a smaller index of refraction is not satisfied. A number of interesting applications of total internal reflection follow. Figure 2. Light entering a thin fiber may strike the inside surface at large or grazing angles and is completely reflected if these angles exceed the critical angle. Such rays continue down the fiber, even following it around corners, since the angles of reflection and incidence remain large.Fiber optics is one application of total internal reflection that is in wide use. In communications, it is used to transmit telephone, internet, and cable TV signals. Fiber optics employs the transmission of light down fibers of plastic or glass. Because the fibers are thin, light entering one is likely to strike the inside surface at an angle greater than the critical angle and, thus, be totally reflected (See Figure 2.) The index of refraction outside the fiber must be smaller than inside, a condition that is easily satisfied by coating the outside of the fiber with a material having an appropriate refractive index. In fact, most fibers have a varying refractive index to allow more light to be guided along the fiber through total internal refraction. Rays are reflected around corners as shown, making the fibers into tiny light pipes. Figure 3. (a) An image is transmitted by a bundle of fibers that have fixed neighbors. (b) An endoscope is used to probe the body, both transmitting light to the interior and returning an image such as the one shown. (credit: Med_Chaos, Wikimedia Commons)Bundles of fibers can be used to transmit an image without a lens, as illustrated in Figure 3. The output of a device called an endoscope is shown in Figure 3b. Endoscopes are used to explore the body through various orifices or minor incisions. Light is transmitted down one fiber bundle to illuminate internal parts, and the reflected light is transmitted back out through another to be observed. Surgery can be performed, such as arthroscopic surgery on the knee joint, employing cutting tools attached to and observed with the endoscope. Samples can also be obtained, such as by lassoing an intestinal polyp for external examination. Figure 4. Fibers in bundles are clad by a material that has a lower index of refraction than the core to ensure total internal reflection, even when fibers are in contact with one another. This shows a single fiber with its cladding. Fiber optics has revolutionized surgical techniques and observations within the body. There are a host of medical diagnostic and therapeutic uses. The flexibility of the fiber optic bundle allows it to navigate around difficult and small regions in the body, such as the intestines, the heart, blood vessels, and joints. Transmission of an intense laser beam to burn away obstructing plaques in major arteries as well as delivering light to activate chemotherapy drugs are becoming commonplace. Optical fibers have in fact enabled microsurgery and remote surgery where the incisions are small and the surgeon’s fingers do not need to touch the diseased tissue. Fibers in bundles are surrounded by a cladding material that has a lower index of refraction than the core. (See Figure 4.) The cladding prevents light from being transmitted between fibers in a bundle. Without cladding, light could pass between fibers in contact, since their indices of refraction are identical. Since no light gets into the cladding (there is total internal reflection back into the core), none can be transmitted between clad fibers that are in contact with one another. The cladding prevents light from escaping out of the fiber; instead most of the light is propagated along the length of the fiber, minimizing the loss of signal and ensuring that a quality image is formed at the other end. The cladding and an additional protective layer make optical fibers flexible and durable.The cladding prevents light from being transmitted between fibers in a bundle. Special tiny lenses that can be attached to the ends of bundles of fibers are being designed and fabricated. Light emerging from a fiber bundle can be focused and a tiny spot can be imaged. In some cases the spot can be scanned, allowing quality imaging of a region inside the body. Special minute optical filters inserted at the end of the fiber bundle have the capacity to image tens of microns below the surface without cutting the surface—non-intrusive diagnostics. This is particularly useful for determining the extent of cancers in the stomach and bowel.Most telephone conversations and Internet communications are now carried by laser signals along optical fibers. Extensive optical fiber cables have been placed on the ocean floor and underground to enable optical communications. Optical fiber communication systems offer several advantages over electrical (copper) based systems, particularly for long distances. The fibers can be made so transparent that light can travel many kilometers before it becomes dim enough to require amplification—much superior to copper conductors. This property of optical fibers is called low loss. Lasers emit light with characteristics that allow far more conversations in one fiber than are possible with electric signals on a single conductor. This property of optical fibers is called high bandwidth. Optical signals in one fiber do not produce undesirable effects in other adjacent fibers. This property of optical fibers is called reduced crosstalk. We shall explore the unique characteristics of laser radiation in a later chapter. A light ray that strikes an object consisting of two mutually perpendicular reflecting surfaces is reflected back exactly parallel to the direction from which it came. This is true whenever the reflecting surfaces are perpendicular, and it is independent of the angle of incidence. Such an object, shown in Figure 5, is called a corner reflector, since the light bounces from its inside corner. Many inexpensive reflector buttons on bicycles, cars, and warning signs have corner reflectors designed to return light in the direction from which it originated. It was more expensive for astronauts to place one on the moon. Laser signals can be bounced from that corner reflector to measure the gradually increasing distance to the moon with great precision. Figure 5. (a) Astronauts placed a corner reflector on the moon to measure its gradually increasing orbital distance. (credit: NASA) (b) The bright spots on these bicycle safety reflectors are reflections of the flash of the camera that took this picture on a dark night. (credit: Julo, Wikimedia Commons) Corner reflectors are perfectly efficient when the conditions for total internal reflection are satisfied. With common materials, it is easy to obtain a critical angle that is less than 45º. One use of these perfect mirrors is in binoculars, as shown in Figure 6. Another use is in periscopes found in submarines. Figure 6. These binoculars employ corner reflectors with total internal reflection to get light to the observer’s eyes. Figure 7. Light cannot easily escape a diamond, because its critical angle with air is so small. Most reflections are total, and the facets are placed so that light can exit only in particular ways—thus concentrating the light and making the diamond sparkle.Total internal reflection, coupled with a large index of refraction, explains why diamonds sparkle more than other materials. The critical angle for a diamond-to-air surface is only 24.4º, and so when light enters a diamond, it has trouble getting back out. (See Figure 7.) Although light freely enters the diamond, it can exit only if it makes an angle less than 24.4º. Facets on diamonds are specifically intended to make this unlikely, so that the light can exit only in certain places. Good diamonds are very clear, so that the light makes many internal reflections and is concentrated at the few places it can exit—hence the sparkle. (Zircon is a natural gemstone that has an exceptionally large index of refraction, but not as large as diamond, so it is not as highly prized. Cubic zirconia is manufactured and has an even higher index of refraction (≈2.17), but still less than that of diamond.) The colors you see emerging from a sparkling diamond are not due to the diamond’s color, which is usually nearly colorless. Those colors result from dispersion, the topic of Dispersion: The Rainbow and Prisms. Colored diamonds get their color from structural defects of the crystal lattice and the inclusion of minute quantities of graphite and other materials. The Argyle Mine in Western Australia produces around 90% of the world’s pink, red, champagne, and cognac diamonds, while around 50% of the world’s clear diamonds come from central and southern Africa. Explore bending of light between two media with different indices of refraction. See how changing from air to water to glass changes the bending angle. Play with prisms of different shapes and make rainbows. Click to download the simulation. Run using Java.
fiber optics: transmission of light down fibers of plastic or glass, applying the principle of total internal reflection corner reflector: an object consisting of two mutually perpendicular reflecting surfaces, so that the light that enters is reflected back exactly parallel to the direction from which it came zircon: natural gemstone with a large index of refraction 3. 66.3º 5. > 1.414 7. 1.50, benzene CC licensed content, Shared previouslyPage 3By the end of this section, you will be able to:
We see about six colors in a rainbow—red, orange, yellow, green, blue, and violet; sometimes indigo is listed, too. Those colors are associated with different wavelengths of light, as shown in Figure 2. When our eye receives pure-wavelength light, we tend to see only one of the six colors, depending on wavelength. The thousands of other hues we can sense in other situations are our eye’s response to various mixtures of wavelengths. White light, in particular, is a fairly uniform mixture of all visible wavelengths. Sunlight, considered to be white, actually appears to be a bit yellow because of its mixture of wavelengths, but it does contain all visible wavelengths. The sequence of colors in rainbows is the same sequence as the colors plotted versus wavelength in Figure 2. What this implies is that white light is spread out according to wavelength in a rainbow. Dispersion is defined as the spreading of white light into its full spectrum of wavelengths. More technically, dispersion occurs whenever there is a process that changes the direction of light in a manner that depends on wavelength. Dispersion, as a general phenomenon, can occur for any type of wave and always involves wavelength-dependent processes. Figure 2. Even though rainbows are associated with seven colors, the rainbow is a continuous distribution of colors according to wavelengths.Dispersion is defined to be the spreading of white light into its full spectrum of wavelengths. Refraction is responsible for dispersion in rainbows and many other situations. The angle of refraction depends on the index of refraction, as we saw in The Law of Refraction. We know that the index of refraction n depends on the medium. But for a given medium, n also depends on wavelength. (See Table 1. Note that, for a given medium, n increases as wavelength decreases and is greatest for violet light. Thus violet light is bent more than red light, as shown for a prism in Figure 3b, and the light is dispersed into the same sequence of wavelengths as seen in Figure 1 and Figure 2.
Any type of wave can exhibit dispersion. Sound waves, all types of electromagnetic waves, and water waves can be dispersed according to wavelength. Dispersion occurs whenever the speed of propagation depends on wavelength, thus separating and spreading out various wavelengths. Dispersion may require special circumstances and can result in spectacular displays such as in the production of a rainbow. This is also true for sound, since all frequencies ordinarily travel at the same speed. If you listen to sound through a long tube, such as a vacuum cleaner hose, you can easily hear it is dispersed by interaction with the tube. Dispersion, in fact, can reveal a great deal about what the wave has encountered that disperses its wavelengths. The dispersion of electromagnetic radiation from outer space, for example, has revealed much about what exists between the stars—the so-called empty space. Figure 4. Part of the light falling on this water drop enters and is reflected from the back of the drop. This light is refracted and dispersed both as it enters and as it leaves the drop. Rainbows are produced by a combination of refraction and reflection. You may have noticed that you see a rainbow only when you look away from the sun. Light enters a drop of water and is reflected from the back of the drop, as shown in Figure 4. The light is refracted both as it enters and as it leaves the drop. Since the index of refraction of water varies with wavelength, the light is dispersed, and a rainbow is observed, as shown in Figure 5a. (There is no dispersion caused by reflection at the back surface, since the law of reflection does not depend on wavelength.) The actual rainbow of colors seen by an observer depends on the myriad of rays being refracted and reflected toward the observer’s eyes from numerous drops of water. The effect is most spectacular when the background is dark, as in stormy weather, but can also be observed in waterfalls and lawn sprinklers. The arc of a rainbow comes from the need to be looking at a specific angle relative to the direction of the sun, as illustrated in Figure 5b. (If there are two reflections of light within the water drop, another “secondary” rainbow is produced. This rare event produces an arc that lies above the primary rainbow arc—see Figure 5c.) Figure 5. (a) Different colors emerge in different directions, and so you must look at different locations to see the various colors of a rainbow. (b) The arc of a rainbow results from the fact that a line between the observer and any point on the arc must make the correct angle with the parallel rays of sunlight to receive the refracted rays. (c) Double rainbow. (credit: Nicholas, Wikimedia Commons)Rainbows are produced by a combination of refraction and reflection. Dispersion may produce beautiful rainbows, but it can cause problems in optical systems. White light used to transmit messages in a fiber is dispersed, spreading out in time and eventually overlapping with other messages. Since a laser produces a nearly pure wavelength, its light experiences little dispersion, an advantage over white light for transmission of information. In contrast, dispersion of electromagnetic waves coming to us from outer space can be used to determine the amount of matter they pass through. As with many phenomena, dispersion can be useful or a nuisance, depending on the situation and our human goals.How does a lens form an image? See how light rays are refracted by a lens. Watch how the image changes when you adjust the focal length of the lens, move the object, move the lens, or move the screen. Click to run the simulation.
rainbow: dispersion of sunlight into a continuous distribution of colors according to wavelength, produced by the refraction and reflection of sunlight by water droplets in the sky 2. 46.5º, red; 46.0º, violet 4. (a) 0.043º; (b) 1.33 m 6. 71.3º 8. 53.5º, red; 55.2º, violet CC licensed content, Shared previouslyPage 4By the end of this section, you will be able to:
The word lens derives from the Latin word for a lentil bean, the shape of which is similar to the convex lens in Figure 1. The convex lens shown has been shaped so that all light rays that enter it parallel to its axis cross one another at a single point on the opposite side of the lens. (The axis is defined to be a line normal to the lens at its center, as shown in Figure 1.) Such a lens is called a converging (or convex) lens for the converging effect it has on light rays. An expanded view of the path of one ray through the lens is shown, to illustrate how the ray changes direction both as it enters and as it leaves the lens. Since the index of refraction of the lens is greater than that of air, the ray moves towards the perpendicular as it enters and away from the perpendicular as it leaves. (This is in accordance with the law of refraction.) Due to the lens’s shape, light is thus bent toward the axis at both surfaces. The point at which the rays cross is defined to be the focal point F of the lens. The distance from the center of the lens to its focal point is defined to be the focal length f of the lens. Figure 2 shows how a converging lens, such as that in a magnifying glass, can converge the nearly parallel light rays from the sun to a small spot. Figure 1. Rays of light entering a converging lens parallel to its axis converge at its focal point F. (Ray 2 lies on the axis of the lens.) The distance from the center of the lens to the focal point is the lens’s focal length f. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces.The lens in which light rays that enter it parallel to its axis cross one another at a single point on the opposite side with a converging effect is called converging lens. The point at which the light rays cross is called the focal point F of the lens. The distance from the center of the lens to its focal point is called focal length f. Figure 2. Sunlight focused by a converging magnifying glass can burn paper. Light rays from the sun are nearly parallel and cross at the focal point of the lens. The more powerful the lens, the closer to the lens the rays will cross.The greater effect a lens has on light rays, the more powerful it is said to be. For example, a powerful converging lens will focus parallel light rays closer to itself and will have a smaller focal length than a weak lens. The light will also focus into a smaller and more intense spot for a more powerful lens. The power P of a lens is defined to be the inverse of its focal length. In equation form, this is P=1fP=\frac{1}{f}\\P=f1 .The power P of a lens is defined to be the inverse of its focal length. In equation form, this is P=1fP=\frac{1}{f}\\P=f1 , where f is the focal length of the lens, which must be given in meters (and not cm or mm). The power of a lens P has the unit diopters (D), provided that the focal length is given in meters. That is,1D=1m, or 1m−11\text{D}=\frac{1}{\text{m}}\text{, or }1\text{m}^{-1}\\1D=m1, or 1m−1 . (Note that this power (optical power, actually) is not the same as power in watts defined in the chapter Work, Energy, and Energy Resources. It is a concept related to the effect of optical devices on light.) Optometrists prescribe common spectacles and contact lenses in units of diopters.Suppose you take a magnifying glass out on a sunny day and you find that it concentrates sunlight to a small spot 8.00 cm away from the lens. What are the focal length and power of the lens? The situation here is the same as those shown in Figure 1 and Figure 2. The Sun is so far away that the Sun’s rays are nearly parallel when they reach Earth. The magnifying glass is a convex (or converging) lens, focusing the nearly parallel rays of sunlight. Thus the focal length of the lens is the distance from the lens to the spot, and its power is the inverse of this distance (in m). The focal length of the lens is the distance from the center of the lens to the spot, given to be 8.00 cm. Thus, f = 8.00 cm. To find the power of the lens, we must first convert the focal length to meters; then, we substitute this value into the equation for power. This givesP=1f=10.0800 m=12.5 DP=\frac{1}{f}=\frac{1}{0.0800\text{ m}}=12.5\text{ D}\\P=f1=0.0800 m1=12.5 D .DiscussionThis is a relatively powerful lens. The power of a lens in diopters should not be confused with the familiar concept of power in watts. It is an unfortunate fact that the word “power” is used for two completely different concepts. If you examine a prescription for eyeglasses, you will note lens powers given in diopters. If you examine the label on a motor, you will note energy consumption rate given as a power in watts. Figure 3. Rays of light entering a diverging lens parallel to its axis are diverged, and all appear to originate at its focal point F. The dashed lines are not rays—they indicate the directions from which the rays appear to come. The focal length f of a diverging lens is negative. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces.Figure 3 shows a concave lens and the effect it has on rays of light that enter it parallel to its axis (the path taken by ray 2 in the Figure is the axis of the lens). The concave lens is a diverging lens, because it causes the light rays to bend away (diverge) from its axis. In this case, the lens has been shaped so that all light rays entering it parallel to its axis appear to originate from the same point, F, defined to be the focal point of a diverging lens. The distance from the center of the lens to the focal point is again called the focal length f of the lens. Note that the focal length and power of a diverging lens are defined to be negative. For example, if the distance to F in Figure 3 is 5.00 cm, then the focal length is f = –5.00 cm and the power of the lens is P = –20 D. An expanded view of the path of one ray through the lens is shown in the Figure to illustrate how the shape of the lens, together with the law of refraction, causes the ray to follow its particular path and be diverged. A lens that causes the light rays to bend away from its axis is called a diverging lens. As noted in the initial discussion of the law of refraction in The Law of Refraction, the paths of light rays are exactly reversible. This means that the direction of the arrows could be reversed for all of the rays in Figure 1 and Figure 3. For example, if a point light source is placed at the focal point of a convex lens, as shown in Figure 4, parallel light rays emerge from the other side. Figure 4. A small light source, like a light bulb filament, placed at the focal point of a convex lens, results in parallel rays of light emerging from the other side. The paths are exactly the reverse of those shown in Figure 1. This technique is used in lighthouses and sometimes in traffic lights to produce a directional beam of light from a source that emits light in all directions. Figure 6. The light ray through the center of a thin lens is deflected by a negligible amount and is assumed to emerge parallel to its original path (shown as a shaded line).Ray tracing is the technique of determining or following (tracing) the paths that light rays take. For rays passing through matter, the law of refraction is used to trace the paths. Here we use ray tracing to help us understand the action of lenses in situations ranging from forming images on film to magnifying small print to correcting nearsightedness. While ray tracing for complicated lenses, such as those found in sophisticated cameras, may require computer techniques, there is a set of simple rules for tracing rays through thin lenses. A thin lens is defined to be one whose thickness allows rays to refract, as illustrated in Figure 1, but does not allow properties such as dispersion and aberrations. An ideal thin lens has two refracting surfaces but the lens is thin enough to assume that light rays bend only once. A thin symmetrical lens has two focal points, one on either side and both at the same distance from the lens. (See Figure 6.) Another important characteristic of a thin lens is that light rays through its center are deflected by a negligible amount, as seen in Figure 5. Figure 6. Thin lenses have the same focal length on either side. (a) Parallel light rays entering a converging lens from the right cross at its focal point on the left. (b) Parallel light rays entering a diverging lens from the right seem to come from the focal point on the right.A thin lens is defined to be one whose thickness allows rays to refract but does not allow properties such as dispersion and aberrations. Look through your eyeglasses (or those of a friend) backward and forward and comment on whether they act like thin lenses. Using paper, pencil, and a straight edge, ray tracing can accurately describe the operation of a lens. The rules for ray tracing for thin lenses are based on the illustrations already discussed:
The image formed in Figure 7 is a real image, meaning that it can be projected. That is, light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye, for example. Figure 8 shows how such an image would be projected onto film by a camera lens. This Figure also shows how a real image is projected onto the retina by the lens of an eye. Note that the image is there whether it is projected onto a screen or not. The image in which light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye is called a real image. Figure 8. Real images can be projected. (a) A real image of the person is projected onto film. (b) The converging nature of the multiple surfaces that make up the eye result in the projection of a real image on the retina.Several important distances appear in Figure 7. We define do to be the object distance, the distance of an object from the center of a lens. Image distance di is defined to be the distance of the image from the center of a lens. The height of the object and height of the image are given the symbols ho and hi, respectively. Images that appear upright relative to the object have heights that are positive and those that are inverted have negative heights. Using the rules of ray tracing and making a scale drawing with paper and pencil, like that in Figure 7, we can accurately describe the location and size of an image. But the real benefit of ray tracing is in visualizing how images are formed in a variety of situations. To obtain numerical information, we use a pair of equations that can be derived from a geometric analysis of ray tracing for thin lenses. The thin lens equations are 1do+1di=1f\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\do1+di1=f1 andhiho=dido=m\frac{h_{\text{i}}}{h_\text{o}}=\frac{d_{\text{i}}}{d_{\text{o}}}=m\\hohi=dodi=m (hiho)\left(\frac{h_{\text{i}}}{h_\text{o}}\right)\\(hohi) to be the magnification m. (The minus sign in the equation above will be discussed shortly.) The thin lens equations are broadly applicable to all situations involving thin lenses (and “thin” mirrors, as we will see later). We will explore many features of image formation in the following worked examples.The distance of the image from the center of the lens is called image distance.
1do+1di=1f\displaystyle\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\do1+di1=f1 hiho=dido=m\displaystyle\frac{h_{\text{i}}}{h_\text{o}}=\frac{d_{\text{i}}}{d_{\text{o}}}=m\\hohi=dodi=m A clear glass light bulb is placed 0.750 m from a convex lens having a 0.500 m focal length, as shown in Figure 9. Use ray tracing to get an approximate location for the image. Then use the thin lens equations to calculate both the location of the image and its magnification. Verify that ray tracing and the thin lens equations produce consistent results. Figure 9. A light bulb placed 0.750 m from a lens having a 0.500 m focal length produces a real image on a poster board as discussed in the example above. Ray tracing predicts the image location and size. Since the object is placed farther away from a converging lens than the focal length of the lens, this situation is analogous to those illustrated in Figure 7 and Figure 8. Ray tracing to scale should produce similar results for di. Numerical solutions for di and m can be obtained using the thin lens equations, noting that do = 0.750 m and f = 0.500 m. The ray tracing to scale in Figure 9 shows two rays from a point on the bulb’s filament crossing about 1.50 m on the far side of the lens. Thus the image distance di is about 1.50 m. Similarly, the image height based on ray tracing is greater than the object height by about a factor of 2, and the image is inverted. Thus m is about –2. The minus sign indicates that the image is inverted.The thin lens equations can be used to find di from the given information: 1do+1di=1f\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\do1+di1=f1 . Rearranging to isolate di gives1di=1f−1do\frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_\text{o}}\\di1=f1−do1 . Entering known quantities gives a value for1di\frac{1}{d_{\text{i}}}\\di1 :1di=10.500 m−10.750 m=0.667m\frac{1}{d_{\text{i}}}=\frac{1}{0.500\text{ m}}-\frac{1}{0.750\text{ m}}=\frac{0.667}{\text{m}}\\di1=0.500 m1−0.750 m1=m0.667 . This must be inverted to find di:di=m0.667=1.50 md_{\text{i}}=\frac{\text{m}}{0.667}=1.50\text{ m}\\di=0.667m=1.50 m . Note that another way to find di is to rearrange the equation:1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\di1=f1−do1 . This yields the equation for the image distance as:di=fdodo−fd_{\text{i}}=\frac{fd_{\text{o}}}{d_{\text{o}}-f}\\di=do−ffdo Note that there is no inverting here.The thin lens equations can be used to find the magnification m, since both di and do are known. Entering their values gives m=−dido=−1.50 m0.750 m=−2.00\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{1.50\text{ m}}{0.750\text{ m}}=-2.00\\m=−dodi=−0.750 m1.50 m=−2.00 DiscussionNote that the minus sign causes the magnification to be negative when the image is inverted. Ray tracing and the use of the thin lens equations produce consistent results. The thin lens equations give the most precise results, being limited only by the accuracy of the given information. Ray tracing is limited by the accuracy with which you can draw, but it is highly useful both conceptually and visually. Real images, such as the one considered in the previous example, are formed by converging lenses whenever an object is farther from the lens than its focal length. This is true for movie projectors, cameras, and the eye. We shall refer to these as case 1 images. A case 1 image is formed when do > f and f is positive, as in Figure 10a. (A summary of the three cases or types of image formation appears at the end of this section.)A different type of image is formed when an object, such as a person's face, is held close to a convex lens. The image is upright and larger than the object, as seen in Figure 10b, and so the lens is called a magnifier. If you slowly pull the magnifier away from the face, you will see that the magnification steadily increases until the image begins to blur. Pulling the magnifier even farther away produces an inverted image as seen in Figure 10a. The distance at which the image blurs, and beyond which it inverts, is the focal length of the lens. To use a convex lens as a magnifier, the object must be closer to the converging lens than its focal length. This is called a case 2 image. A case 2 image is formed when do < f and f is positive. Figure 10. (a) When a converging lens is held farther away from the face than the lens’s focal length, an inverted image is formed. This is a case 1 image. Note that the image is in focus but the face is not, because the image is much closer to the camera taking this photograph than the face. (credit: DaMongMan, Flickr) (b) A magnified image of a face is produced by placing it closer to the converging lens than its focal length. This is a case 2 image. (credit: Casey Fleser, Flickr) Figure 11. Ray tracing predicts the image location and size for an object held closer to a converging lens than its focal length. Ray 1 enters parallel to the axis and exits through the focal point on the opposite side, while ray 2 passes through the center of the lens without changing path. The two rays continue to diverge on the other side of the lens, but both appear to come from a common point, locating the upright, magnified, virtual image. This is a case 2 image.Figure 11 uses ray tracing to show how an image is formed when an object is held closer to a converging lens than its focal length. Rays coming from a common point on the object continue to diverge after passing through the lens, but all appear to originate from a point at the location of the image. The image is on the same side of the lens as the object and is farther away from the lens than the object. This image, like all case 2 images, cannot be projected and, hence, is called a virtual image. Light rays only appear to originate at a virtual image; they do not actually pass through that location in space. A screen placed at the location of a virtual image will receive only diffuse light from the object, not focused rays from the lens. Additionally, a screen placed on the opposite side of the lens will receive rays that are still diverging, and so no image will be projected on it. We can see the magnified image with our eyes, because the lens of the eye converges the rays into a real image projected on our retina. Finally, we note that a virtual image is upright and larger than the object, meaning that the magnification is positive and greater than 1.An image that is on the same side of the lens as the object and cannot be projected on a screen is called a virtual image. Suppose the book page in Figure 11a is held 7.50 cm from a convex lens of focal length 10.0 cm, such as a typical magnifying glass might have. What magnification is produced? We are given that do = 7.50 cm and f = 10.0 cm, so we have a situation where the object is placed closer to the lens than its focal length. We therefore expect to get a case 2 virtual image with a positive magnification that is greater than 1. Ray tracing produces an image like that shown in Figure 11, but we will use the thin lens equations to get numerical solutions in this example. To find the magnification m, we try to use magnification equation, m=−didom=-\frac{d_{\text{i}}}{d_{\text{o}}}\\m=−dodi . We do not have a value for di, so that we must first find the location of the image using lens equation. (The procedure is the same as followed in the preceding example, where do and f were known.) Rearranging the magnification equation to isolate di gives1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\di1=f1−do1 . Entering known values, we obtain a value for1di\frac{1}{d_\text{i}}\\di1 :1di=110.0 cm−17.50 cm=−0.0333cm\frac{1}{d_\text{i}}=\frac{1}{10.0\text{ cm}}-\frac{1}{7.50\text{ cm}}=\frac{-0.0333}{\text{cm}}\\di1=10.0 cm1−7.50 cm1=cm−0.0333 . This must be inverted to find di:di=−cm0.0333=−30.0 cmd_{\text{i}}=-\frac{\text{cm}}{0.0333}=-30.0\text{ cm}\\di=−0.0333cm=−30.0 cm . Now the thin lens equation can be used to find the magnification m, since both di and do are known. Entering their values givesm=−dido=−−30.0 cm10.0 cm=3.00\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{-30.0\text{ cm}}{10.0\text{ cm}}=3.00\\m=−dodi=−10.0 cm−30.0 cm=3.00 DiscussionA number of results in this example are true of all case 2 images, as well as being consistent with Figure 11. Magnification is indeed positive (as predicted), meaning the image is upright. The magnification is also greater than 1, meaning that the image is larger than the object—in this case, by a factor of 3. Note that the image distance is negative. This means the image is on the same side of the lens as the object. Thus the image cannot be projected and is virtual. (Negative values of di occur for virtual images.) The image is farther from the lens than the object, since the image distance is greater in magnitude than the object distance. The location of the image is not obvious when you look through a magnifier. In fact, since the image is bigger than the object, you may think the image is closer than the object. But the image is farther away, a fact that is useful in correcting farsightedness, as we shall see in a later section. Figure 12. A car viewed through a concave or diverging lens looks upright. This is a case 3 image. (credit: Daniel Oines, Flickr)A third type of image is formed by a diverging or concave lens. Try looking through eyeglasses meant to correct nearsightedness. (See Figure 12.) You will see an image that is upright but smaller than the object. This means that the magnification is positive but less than 1. The ray diagram in Figure 13 shows that the image is on the same side of the lens as the object and, hence, cannot be projected—it is a virtual image. Note that the image is closer to the lens than the object. This is a case 3 image, formed for any object by a negative focal length or diverging lens. Figure 13. Ray tracing predicts the image location and size for a concave or diverging lens. Ray 1 enters parallel to the axis and is bent so that it appears to originate from the focal point. Ray 2 passes through the center of the lens without changing path. The two rays appear to come from a common point, locating the upright image. This is a case 3 image, which is closer to the lens than the object and smaller in height.Suppose an object such as a book page is held 7.50 cm from a concave lens of focal length –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced? This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is thus the same, but the results are different in important ways. To find the magnification m, we must first find the image distance di using thin lens equation 1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\di1=f1−do1 , or its alternative rearrangementdi=fdodo−f{d_\text{i}}=\frac{fd_{\text{o}}}{d_{\text{o}}-f}\\di=do−ffdo .We are given that f = –10.0 cm and do = 7.50 cm. Entering these yields a value for 1di\frac{1}{d_{\text{i}}}\\di1 :1di=1−10.0 cm−17.50 cm=−0.2333cm\displaystyle\frac{1}{d_{\text{i}}}=\frac{1}{-10.0\text{ cm}}-\frac{1}{7.50\text{ cm}}=\frac{-0.2333}{\text{cm}}\\di1=−10.0 cm1−7.50 cm1=cm−0.2333 This must be inverted to find di:di=−cm0.2333=−4.29 cm\displaystyle{d_\text{i}}=-\frac{\text{cm}}{0.2333}=-4.29\text{ cm}\\di=−0.2333cm=−4.29 cm Ordi=(7.5)(−10)(7.5−(−10))=−7517.5=−4.29 cm\displaystyle{d}_{\text{i}}=\frac{\left(7.5\right)\left(-10\right)}{\left(7.5-\left(-10\right)\right)}=-\frac{75}{17.5}=-4.29\text{ cm}\\di=(7.5−(−10))(7.5)(−10)=−17.575=−4.29 cm Now the magnification equation can be used to find the magnification m, since both di and do are known. Entering their values givesm=−dido=−−4.29 cm7.50 cm=0.571\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{-4.29\text{ cm}}{7.50\text{ cm}}=0.571\\m=−dodi=−7.50 cm−4.29 cm=0.571 DiscussionA number of results in this example are true of all case 3 images, as well as being consistent with Figure 13. Magnification is positive (as predicted), meaning the image is upright. The magnification is also less than 1, meaning the image is smaller than the object—in this case, a little over half its size. The image distance is negative, meaning the image is on the same side of the lens as the object. (The image is virtual.) The image is closer to the lens than the object, since the image distance is smaller in magnitude than the object distance. The location of the image is not obvious when you look through a concave lens. In fact, since the image is smaller than the object, you may think it is farther away. But the image is closer than the object, a fact that is useful in correcting nearsightedness, as we shall see in a later section. Table 1 summarizes the three types of images formed by single thin lenses. These are referred to as case 1, 2, and 3 images. Convex (converging) lenses can form either real or virtual images (cases 1 and 2, respectively), whereas concave (diverging) lenses can form only virtual images (always case 3). Real images are always inverted, but they can be either larger or smaller than the object. For example, a slide projector forms an image larger than the slide, whereas a camera makes an image smaller than the object being photographed. Virtual images are always upright and cannot be projected. Virtual images are larger than the object only in case 2, where a convex lens is used. The virtual image produced by a concave lens is always smaller than the object—a case 3 image. We can see and photograph virtual images only by using an additional lens to form a real image.
Find several lenses and determine whether they are converging or diverging. In general those that are thicker near the edges are diverging and those that are thicker near the center are converging. On a bright sunny day take the converging lenses outside and try focusing the sunlight onto a piece of paper. Determine the focal lengths of the lenses. Be careful because the paper may start to burn, depending on the type of lens you have selected. Step 1. Examine the situation to determine that image formation by a lens is involved. Step 2. Determine whether ray tracing, the thin lens equations, or both are to be employed. A sketch is very useful even if ray tracing is not specifically required by the problem. Write symbols and values on the sketch. Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). Step 4. Make alist of what is given or can be inferred from the problem as stated (identify the knowns). It is helpful to determine whether the situation involves a case 1, 2, or 3 image. While these are just names for types of images, they have certain characteristics (given in Table 1) that can be of great use in solving problems. Step 5. If ray tracing is required, use the ray tracing rules listed near the beginning of this section. Step 6. Most quantitative problems require the use of the thin lens equations. These are solved in the usual manner by substituting knowns and solving for unknowns. Several worked examples serve as guides. Step 7. Check to see if the answer is reasonable: Does it make sense? If you have identified the type of image (case 1, 2, or 3), you should assess whether your answer is consistent with the type of image, magnification, and so on. We do not realize that light rays are coming from every part of the object, passing through every part of the lens, and all can be used to form the final image. We generally feel the entire lens, or mirror, is needed to form an image. Actually, half a lens will form the same, though a fainter, image.
1do+1di=1f\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{f}do1+di1=f1 andhiho=−dido=m\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=m\\hohi=−dodi=m (magnification).
diverging lens: a concave lens in which light rays that enter it parallel to its axis bend away (diverge) from its axis focal point: for a converging lens or mirror, the point at which converging light rays cross; for a diverging lens or mirror, the point from which diverging light rays appear to originate focal length: distance from the center of a lens or curved mirror to its focal point magnification: ratio of image height to object height power: inverse of focal length real image: image that can be projected virtual image: image that cannot be projected 2. 5.00 to 12.5 D 4. –0.222 m 6. (a) 3.43 m; (b) 0.800 by 1.20 m 7. (a) −1.35 m (on the object side of the lens); (b) +10.0; (c) 5.00 cm 8. 44.4 cm 10. (a) 6.60 cm; (b) –0.333 12. (a) +7.50 cm; (b) 13.3 D; (c) Much greater 14. (a) +6.67; (b) +20.0; (c) The magnification increases without limit (to infinity) as the object distance increases to the limit of the focal distance. 16. −0.933 mm CC licensed content, Shared previouslyPage 5 |