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6. Chapter Telescopes: Portals of Discovery This chapter focuses on telescopes and their uses. Note that, although these instruments are fundamental to modern astronomy, most of the material in this chapter is not prerequisite to later chapters. Thus, you can consider this chapter to be optional. As always, when you prepare to teach this chapter, be sure you are familiar with the relevant media resources (see the complete, section-by-section resource grid in Appendix 3 of this Instructor Guide) and the online quizzes and other study resources available on the MasteringAstronomy Website. Teaching Notes (By Section) Section 6.1Eyes and Cameras: Everyday Light Sensors As with all the chapters in Part II, we begin this chapter with a section on everyday light collection, discussing the human eye and cameras. Section 6.2Telescopes: Giant Eyes This section describes the general design of optical telescopes. Note our emphasis on two principal properties of telescopes: light-collecting area and angular resolution. Note that while we show Cassegrain, Newtonian, and Nasmyth/Coud foci in a figure, we do not expect students to learn the names, and we give the names only for reference. Although different observers tend to categorize observations differently, we have chosen to categorize observations as either imaging, spectroscopy, or timing. We believe that this categorization is pedagogically useful, because it most closely corresponds to the figures that students see in the book and in news reports: photographs (imaging), spectra (spectroscopy), and light curves (timing). Section 6.3Telescopes and the Atmosphere In this section, we turn to the atmospheric effects due to light pollution and turbulence (twinkling), leading to a discussion of how observing sites are chosen and of adaptive optics. This is where we point out that most wavelengths of light do not penetrate the atmosphere, and introduce the rationale for space telescopes. Section 6.4Telescopes and Technology This section covers telescopes designed to collect light of different wavelengths, and interferometry. 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 1 Answers/Discussion Points for Think About It/See It For Yourself Questions The Think About It and See It For Yourself questions are not numbered in the book, so we list them in the order in which they appear, keyed by section number. Section 6.1 (p. 182) The pupil will be wider in the eye exposed to light. Doctors dilate your pupils so that they can see through them into your eye. (p. 183) This question is intended simply to get students to see the correspondence between the ideas they are learning in this chapter and the cameras that they use at home. Section 6.2 (p. 184) You can find the angular resolution by measuring the distance at which the points blend together into one. Moving the holes farther apart makes this distance larger, but the angular resolution is still the same. (p. 190) CAT scans and MRIs use color-coded displays to show different types of tissue or different elements within tissue, allowing doctors effectively to see inside your body. Section 6.3 (p. 193) The coin appears to move because the water is moving and changing the path of the light as it comes through the water toward our eyes. Its basically the same reason that air turbulence causes twinkling. Section 6.4 (p. 199) If youve never tried this trick for seeing grazing incidence reflections of visible light, you and your students may all be surprised by how it works. Note that this demonstration of visible-light grazing incidence works only with fairly smooth and shiny surfaces. It is impressive if youve never noticed it before. Solutions to End-of-Chapter Problems (Chapter 6) 1. 2. The eye focuses light by bending (refracting) it so that parallel rays of incoming light all meet up on the same point in the back of our eye. A glass lens does the same thing, but unlike the eye it cannot adjust its shape to change its focus. The focal plane of a lens is the place where an image appears in focus. Cameras are better than eyes for astronomy because they are more reliable than people sketching what they see. In addition, they can use long exposure times to make images of faint objects. CCDs are better than film because they are more sensitive. Thus, we can get the same image with a shorter exposure time. They are also sensitive to a much broader range of brightness at one time, making it possible to take a picture of brighter and fainter objects at the same time. And finally, they automatically save their images digitally, 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 2 3. 4. 5. 6. 7. 8. 9. 10. which means we can import them into computers to manipulate them easily. Telescopes have two key properties: light-collecting area and angular resolution. One is light-collecting area. A telescope can collect a lot more light than the human eye can because it is larger. This is important because the objects astronomers want to look at are usually very faint. The other important property of telescopes is angular resolution: Telescopes can make out finer details than our eyes can. This is important because the objects we want to study appear small in our sky. The diffraction limit is a natural, physical limit on the finest resolution a given telescope can achieve at a particular wavelength. As the telescope gets larger, the diffraction limit decreases (leading to the possibility of finer resolution). However, as the wavelength increases, the diffraction limit increases so that longer wavelengths require larger telescopes to achieve the same resolution. The difference between a reflecting telescope and a refracting telescope lies in how the light is focused. In a refracting telescope, a lens is used to focus the light, whereas in a reflecting telescope we use mirrors. Today, we mainly use reflectors because they allow for larger mirrors and less loss of light. The three basic types of observing are imaging, spectroscopy, and timing. For imaging, the telescope functions like a camera, taking a picture of the object of interest. For spectroscopy, astronomers use a diffraction grating (or some other means) to spread the light into its component wavelengths and then measure the brightness of each wavelength. Finally, for timing observations, astronomers use multiple images or otherwise measure the incoming light at known times to measure how the brightness varies. An image in a nonvisible wavelength is just like an image in a visible wavelength, except that our eyes would never be able to see it without help. However, the colors in the image are not the actual colors of the object. They are simply used to aid us in studying the object. Spectral resolution measures how fine the details are that we can see in a spectrum. With better spectral resolution, we can make out features closer together in wavelength in the spectrum. High spectral resolution requires longer exposures (or brighter sources) than we need for images, because the light is spread out when we make a spectrum. The atmosphere has three negative effects on observations. First, it blocks light of most wavelengths from ever reaching the ground. Second, the atmosphere scatters human-generated light and makes it more difficult to make observations. And finally, the constant movement of air in the atmosphere (the source of turbulence when flying) causes stars to appear to twinkle, which effectively blurs telescopic images. Putting a telescope in space overcomes these problems because the telescope no longer has to look through the atmosphere. Adaptive optics is a technology that can essentially undo the blurring caused by twinkling. Figure 6.22 shows that the only wavelengths that make it all the way to the ground are the narrow range of visible light and the radio wavelengths. A few other wavelengths can be detected from high mountains or from aircraft, but most wavelengths require us to put telescopes in space, which is why space-based astronomy is so important. 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 3 11. 12. 13. 14. 15. 16. 17. 18. Radio astronomy uses dishes like satellite dishes to observe the objects. In some cases, the dishes work together to form images at a much higher resolution than any one of them could achieve alone. An example of a radio telescope is the Very Large Array in New Mexico. Infrared telescopes are very similar to optical telescopes, except that they use different detectors to sense the light. An example of an infrared telescope is the Spitzer Infrared Space Telescope. Most of the ultraviolet wavelengths act enough like the visible wavelengths that we can use the same sorts of telescopes, with different detectors. (At the shortest wavelengths, the ultraviolet waves actually act more like X rays so we use telescopes more like X-ray telescopes.) An example of an ultraviolet telescope is the Hubble Space Telescope. X-ray telescopes have to gently redirect the light because it tends to go straight through things in its path. X-ray mirrors are arranged so that the light just grazes them. An example of an X-ray telescope is the Chandra space telescope. Gamma-ray telescopes must be massive in order to intercept this high-energy light. An example of a gamma-ray telescope is Swift. Interferometry is the process of linking up multiple telescopes so that they combine their observations to make much higher resolution images than the individual telescopes can. This is a massive bonus since it is often impossible to make single telescopes as big as we would like. However, for a much smaller sum of money, a few cheaper telescopes can be used to get the same resolution (although not the same light-gathering power). The image was blurry, because the photographic film was not placed at the focal plane. This statement makes sense, because light is generally in focus only at the focal plane. By using a CCD, I can photograph the Andromeda Galaxy with a shorter exposure time than I would need with photographic film. This statement makes sense, because a CCD is more sensitive than photographic film and hence can record an image in less time. I have a reflecting telescope in which the secondary mirror is bigger than the primary mirror. This statement does not make sense. Remember that the secondary mirror is placed in front of the primary mirror in a reflecting telescope. Thus, if the secondary mirror was bigger than the primary mirror, it would block all light from reaching the primary mirror, rendering the telescope useless. The photograph shows what appear to be just two distinct stars, but each of those stars is actually a binary star system. This statement makes sense. If the angular resolution of the telescope isnt good enough to resolve the individual stars in each pair, then each pair will appear as a single star. I built my own 14-inch telescope that has a lower diffraction limit than most large professional telescopes. This statement does not make sense, because the diffraction limit depends only on the telescopes size: A small telescope cannot have a lower diffraction limit than a large telescope. Now that Ive bought a spectrograph, I can use my home for telescope spectroscopy as well as imaging. This statement makes sense, as long as you can hook up the 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4 spectrograph to your telescope. However, unless your telescope is unusually large, youll only be able to get spectra of relatively bright objects. 19. If you lived on the Moon, youd never see stars twinkle. This statement makes sense, because twinkling is caused by the atmosphere and the Moon doesnt have an atmosphere. 20. New technologies will soon allow astronomers to use X-ray telescopes on the Earths surface. This statement does not make sense, because X rays do not reach the ground and there is no technology we can use for a telescope that will change this basic fact. 21. Thanks to adaptive optics, telescopes on the ground can now make ultraviolet images of the cosmos. This statement does not make sense, because ultraviolet light does not reach the ground and there is no technology we can use for a telescope that will change this basic fact. 22. Thanks to interferometry, a properly spaced set of 10-meter radio telescopes can achieve the angular resolution of a single, 100-kilometer radio telescope. This statement makes sense, because interferometry allows multiple small telescopes to achieve the angular resolution of a larger telescope. 23. b; 24. b; 25. b; 26. c; 27. c; 28. c; 29. a; 30. a; 31. a; 32. b. 33. Answers will vary. 34. a. Timing, since we are looking for variation. b. Spectroscopy, because we learn composition by identifying spectral lines. c. Spectroscopy, because we must compare the wavelengths of the objects spectral lines to the wavelengths of the same lines in the laboratory in order to determine the Doppler shift, from which we determine the speed. 35. You cannot gain any detail by blowing up a magazine or newspaper photograph, beyond the detail that is already there. In much the same way, additional magnification with an astronomical telescope cannot provide any more detail than the telescope is capable of obtaining as a result of its size and optical quality. 36. Answers will vary; the key point is to be sure students have considered the major factors that affect astronomical observations from the ground. 37. The five telescopes will be much more valuable if linked together by interferometry, because it will improve their angular resolution. Adaptive optics will do nothing at all for telescopes in space, since this technology is designed only to counteract blurring caused by Earths atmosphere. 38. Through a filter that transmits only red light, you would see the flags red stripes, but the rest would be black. Through a filter that transmits only blue light, you would see the blue background but not the red stripes. (Depending on the filter, you might see white portions of the flag through either filter, since white light contains all the colors.) 39. This is a project that should help students realize that brighter stars tend to twinkle more than dimmer ones and that stars twinkle more when nearer to the horizon then when higher overhead. 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 5 40. 41. 42. 43. This is a project requiring some research into how radio bands are assigned. This telescope review project requires some personal research. a. The 10-meter Keck telescope has twice the diameter of the 5-meter Hale telescope, so its light-collecting area is 22 = 4 times as much. b. A 100-meter telescope would have 10 times the diameter of the 10-meter Keck telescope, so its light-collecting area would be 102 = 100 times as much. We seek the angular separation of two stars with real separation s = 100 million km = 108 km at a distance d = 100 light-years. We begin by converting the distance from light-years to kilometers: km 100ly 1013 = 1015 ly Now we use the angular separation formula: = 8 s 10km 360 = 2 d 210 15 360(610) 6 Finally, we convert from degrees to arcseconds so that we can compare it to the 0.05-arcsecond resolution of the Hubble Space Telescope: 6060 = 11 The angular separation of the two stars in this binary system is below the angular resolution of the Hubble Space Telescope, so they will appear as a single point of light in Hubble Space Telescope images. a. Mathematical Insight 6.1 tells us that we can calculate the angular separation of two objects via the formula: physicalseparation angularseparation=206,265 distance where the physical separation and distance have to be measured in the same units. To use this formula for the Sun and Jupiter, we need the Sun-Jupiter distance. Appendix E says that this is 7.783 108 km. Before we can use the formula, well need to convert the distance, 10 light-years, to kilometers. Appendix A says that 1 light-year is 9.46 1012 km, so we convert: (610)0.02 6 44. 10lightyears 9.4610km 12 = 9.4610 13 1lightyear Calculating the angular separation: angularseparation206,265 = 7.78310km 8 9.4610km 13 = 1.70 From 10 light-years away, Jupiter would appear to be 1.70 arcseconds from the Sun. 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 6 b. Using the same formula with the same distance, this time with the Earth-Sun separation (1.496 108 km), we get: angular separation = 206,265 45. 1.496 108 km 9.46 1013 km = 0.326 Earth would appear to be only 0.326 arcsecond from the Sun from 10 light-years away. c. Hubble has an angular resolution of 0.05 arcsecond. So both observations should, in theory, be possible. However, the Sun is much, much brighter than either planet. So the planetary light will probably be lost in the glare of the Sun, even though we should be able to resolve the planets. a. We use the formula from Mathematical Insight 6.2 to find the diffraction limit of the human eye, but instead of having a telescope diameter, we use the lens diameter of the eye, given as 0.8 centimeter: wavelength diffraction limit (arcseconds) = 2.5 10 5 lens diameter 500 10 9 m = 2.5 105 = 16 0.008 m (Note that the actual angular resolution of the human eye, about 1 arcminute, is not as good as the diffraction limit, which is a theoretical limit for a perfect optical system.) b. For a 10-meter telescope, the diffraction limit resolution is: wavelength diffraction limit (arcseconds) = 2.5 10 5 telescope diameter 500 10 9 m = 2.5 105 = 0.0125 10 m 46. The diffraction limit of the 10-meter telescope is smaller than the diffraction limit of the human eye by a factor of about 16/0.0125 = 1,280, or close to 1,300. For a 100-meter (104-centimeter) radio telescope observing radio waves with a wavelength of 21 centimeters, the diffraction limit is: wavelength diffraction limit (arcseconds) = 2.5 10 5 telescope diameter 21 cm = 2.5 105 4 = 525 10 cm This angular resolution of over 500 arcseconds is about 10,000 times poorer than the Hubble Space Telescopes 0.05-arcsecond resolution for visible light. In order to achieve significantly better angular resolution when observing 21-centimeter radio waves, a radio telescope must have an effective diameter much larger than 100 meters. Because it would be impractical to build such huge telescopes, radio 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 7 47. astronomers use the technique of interferometry to make many small radio telescopes achieve the angular resolution that a single very large one would achieve. We are told in Mathematical Insight 6.2 that the diffraction limit is given by the equation: wavelength of light diffraction limit = (2.5 10 5 arcseconds) diameter of telescope where the wavelength of light and the diameter of the telescope must be measured in the same units. In our case, the wavelength is 21 centimeters and the telescope is 0.5 meter across. Converting from meters to centimeters: 100 cm 0.5 m = 50 cm 1m We are ready to use the formula for the diffraction limit: diffraction limit = (2.5 10 5 arcseconds) 21 cm 50 cm = 1.05 105 The telescope would have a resolution of 1.05 105 arcseconds. We can convert this to degrees, to make it clearer how large this is: 1.05 105 arcseconds 1 arcminute 1 = 29.3 60 arcseconds 60 arcminutes 48. The angular resolution of your satellite dish is only 29.3roughly 60 times the 0.5 angular size of the full moon! Thus, your dish would not be very useful for making images; its also too small to gather much light, and the poor angular resolution means we couldnt pinpoint an objects location well enough for the dish to do useful spectroscopy either. a. The angular area of the advanced cameras field of view is about (0.06)2 = 0.0036 square degree. b. Given that the angular area of the entire sky is about 41,250 square degrees, taking pictures of the entire sky would require 41,250 square degrees = 11.5 million 0.0036 square degree 49. separate photographs by the advanced camera. If each of 11.5 million photographs required 1 hour to take, the total time required would be 11.5 million hours, which is the same as: 11.5 million hr 1 day 24 hr 1 yr = 1,310 yr 365 day That is, it would take more than a thousand years for the Hubble Space Telescope to photograph the entire sky, assuming 1 hour per photograph. Clearly, well need many more telescopes in space if we hope to get Hubble Space Telescope-quality photos of the entire sky. 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 8 50. We are told in Mathematical Insight 6.2 that the diffraction limit is given by the equation: wavelength of light diffraction limit = (2.5 10 5 arcseconds) diameter of telescope where the wavelength of light and the diameter of the telescope must be measured in the same units. We are told that the wavelength is 500 nanometers and that the diameter of the telescope is 300 meters. To make the units consistent, we will convert 500 nanometers into meters, recalling that a nanometer is one-billionth of a meter: 1m 500 nm = 5.00 10 7 m 9 1 10 nm Now, using the formula for angular resolution: diffractionlimit(2.510arcseconds) = 5 5.0010m 7 300m = 4.1710arcsecond 4 51. This telescope would have a diffraction limit of 4.17 104 arcsecond. This is a geometry problem. If the mirrors are 4 meters in diameter, they are 2 meters in radius. We want to know how far the photons at the outside, 2 meters from the center, must travel after being deflected 2 to meet with a photon that comes right down the center of the telescope. We can sketch the geometry as a right triangle. The photon thats coming down the telescopes central axis is the vertical leg of the triangle, while the photon that was deflected 2 out at the edge of the telescope is the hypotenuse: For right triangles, we know that: tan() = oppositeleg adjacentleg Solving for the opposite leg, which will be the length of the telescope in this case: oppositeleg(adjacentleg)tan() = Using = 88 and 2 meters for the adjacent leg (and remembering to set our calculators to degrees, not radians!), we get: oppositeleg2tan(88) = = 57m So Chandra should be around 57 meters long! In reality, Chandra is about 13 meters long. This is possible because it uses more than one set of mirrors to focus the X rays. 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley 9 ... View Full Document

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