Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems A.27PROBLEMS1. 1.58 cm3. (a) 2.6 mm (b) 2.62 mm5. (a) 36.2° (b) 5.08 cm (c) 5.08 10 14 Hz7. (a) 55.7 m (b) 124 m9. 75.0 m11. 11.3 m13. 148 m15. 91.9 nm17. 550 nm19. 0.500 cm21. (a) 238 nm (b) will increase (c) 328 nm23. 4.35 m25. 4.75 m27. No, the wavelengths intensified are 276 nm, 138 nm,92.0 nm, . . .29. 4.22 mm31. (a) 1.1 m (b) 1.7 mm33. 1.20 mm, 1.20 mm35. (a) 479 nm, 647 nm, 698 nm (b) 20.5°, 28.3°, 30.7°37. 5.91° in first order; 13.2° in second order; and26.5° in third order39. 44.5 cm41. 9.13 cm43. (a) 25.6° (b) 19.0°45. (a) 1.11 (b) 42.0°47. (a) 56.7° (b) 48.8°49. 31.2°53. 6.89 units55. (a) 413.7 nm, 409.7 nm (b) 8.6°57. 0.156 mm59. 2.50 mm61. Any positive integral multiple of 210 nm63. (a) 16.6 m (b) 8.28 m65. 127 m67. 0.350 mm69. 115 nmChapter 25QUICK QUIZZES1. (c)2. (a)CONCEPTUAL QUESTIONS1. The observer is not using the lens as a simple magnifier.For a lens to be used as a simple magnifier, the object distancemust be less than the focal length of the lens. Also,a simple magnifier produces a virtual image at the normalnear point of the eye, or at an image distance of aboutq 25 cm. With a large object distance and a relativelyshort image distance, the magnitude of the magnificationby the lens would be considerably less than one. Mostlikely, the lens in this example is part of a lens combinationbeing used as a telescope.3. The image formed on the retina by the lens and cornea isalready inverted.5. There will be an effect on the interference pattern—itwill be distorted. The high temperature of the flame willchange the index of refraction of air for the arm of theinterferometer in which the match is held. As the indexof refraction varies randomly, the wavelength of the lightin that region will also vary randomly. As a result, theeffective difference in length between the two arms willfluctuate, resulting in a wildly varying interferencepattern.7. Large lenses are difficult to manufacture and machinewith accuracy. Also, their large weight leads to sagging,which produces a distorted image. In reflecting telescopes,light does not pass through glass; hence, problemsassociated with chromatic aberrations are eliminated.Large-diameter reflecting telescopes are also technicallyeasier to construct. Some designs use a rotating pool ofmercury as the reflecting surface.9. In order for someone to see an object through a microscope,the wavelength of the light in the microscope mustbe smaller than the size of the object. An atom is muchsmaller than the wavelength of light in the visible spectrum,so an atom can never be seen with the use of visiblelight.11. farsighted; convergingPROBLEMS1. 30.0 cm beyond the lens, M 1/53. 177 m5. f/1.47. f/8.09. 40.0 cm11. 23.2 cm13. (a) 2.00 diopters (b) 17.6 cm15. 17.0 diopters17. (a) 5.8 cm (b) m 4.319. (a) 4.07 cm (b) m 7.1421. (a) M 1.22 (b) / 0 6.0823. 2.1 cm25. m 11527. f o 90 cm, f e 2.0 cm29. (b) fh/p (c) 1.07 mm31. (a) m 1.50 (b) m 1.9033. 492 km35. 0.40 rad37. 9.1 10 7 km39. 9.8 km41. No. A resolving power of 2.0 10 5 is needed, and thatavailable is only 1.8 10 5 .43. 50.4 m45. 4047. 98 fringe shifts49. (a) 2.67 diopters (b) 0.16 diopter too low51. (a) 44.6 diopters (b) 3.03 diopters53. (a) 1.0 10 3 lines (b) 3.3 10 2 lines55. m 10.757. (a) m 4.0 (b) m 3.0Chapter 26QUICK QUIZZES1. (a)2. No. From your perspective you’re at rest with respect tothe cabin, so you will measure yourself as having yournormal length, and will require a normal-sized cabin.3. (a), (e); (a), (e)4. (a) False (b) False (c) True (d) False5. (a)
A.28 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and ProblemsCONCEPTUAL QUESTIONS1. An ellipsoid. The dimension in the direction of motionwould be measured to be less than D.3. This scenario is not possible with light. Light waves are describedby the principles of special relativity. As you detectthe light wave ahead of you and moving away from you(which would be a pretty good trick—think about it!), itsvelocity relative to you is c. Thus, you will not be able tocatch up to the light wave.5. No. The principle of relativity implies that nothing cantravel faster than the speed of light in a vacuum, which is3.00 10 8 m/s.7. The light from the quasar moves at 3.00 10 8 m/s. Thespeed of light is independent of the motion of the sourceor the observer.9. For a wonderful fictional exploration of this question, geta “Mr. Tompkins” book by George Gamow. All of the relativityeffects would be obvious in our lives. Time dilationand length contraction would both occur. Driving homein a hurry, you would push on the gas pedal not to increaseyour speed very much, but to make the blocksshorter. Big Doppler shifts in wave frequencies wouldmake red lights look green as you approached and makecar horns and radios useless. High-speed transportationwould be both very expensive, requiring huge fuel purchases,as well as dangerous, since a speeding car couldknock down a building. When you got home, hungry forlunch, you would find that you had missed dinner; therewould be a five-day delay in transit when you watch a liveTV program originating in Australia. Finally, we would notbe able to see the Milky Way, since the fireball of the BigBang would surround us at the distance of Rigel orDeneb.11. A photon transports energy. The relativistic equivalenceof mass and energy means that is enough to give itmomentum.13. Your assignment: measure the length of a rod as it slidespast you. Mark the position of its front end on the floorand have an assistant mark the position of the back end.Then measure the distance between the two marks. Thisdistance will represent the length of the rod only if thetwo marks were made simultaneously in your frame ofreference.PROBLEMS1. (a) t OB 1.67 10 3 s, t OA 2.04 10 3 s(b) t BO 2.50 10 3 s, t AO 2.04 10 3 s(c) t 90 s3. 5.0 s5. (a) 20 m (b) 19 m (c) 0.31c7. (a) 1.3 10 7 s (b) 38 m (c) 7.6 m9. (a) 2.2 s (b) 0.65 km11. 0.950c13. Yes, with 19 m to spare15. (a) 39.2 s (b) Accurate to one digit17. 3.3 10 5 m/s19. 0.285c21. 0.54c to the right23. 0.357c25. 0.998c toward the right27. (a) 54 min (b) 52 min29. c(√3/2)31. 0.786c33. 18.4 g/cm 335. 1.98 MeV37. 2.27 10 23 Hz, 1.32 fm for each photon39. (a) 3.10 10 5 m/s (b) 0.758c41. 1.42 MeV/c43. (a) 0.80c (b) 7.5 10 3 s (c) 1.4 10 12 m, 0.38c45. 0.37c in the x-direction47. (a) v/c 1 1.12 10 10 (b) 6.00 10 27 J(c) $2.17 10 2049. 0.80c51. (a) 0.946c (b) 0.160 ly (c) 0.114 yr (d) 7.50 10 22 J53. (a) 7.0 s (c) 1.1 10 4 muons59. 5.45 yr; Goslo is older.Chapter 27QUICK QUIZZES1. (b)2. (c)3. (c)4. (b)CONCEPTUAL QUESTIONS1. The shape of an object is normally determined by observingthe light reflecting from its surface. In a kiln, the objectwill be very hot and will be glowing red. The emittedradiation is far stronger than the reflected radiation, andthe thermal radiation emitted is only slightly dependenton the material from which the object is made. Unlike reflectedlight, the emitted light comes from all surfaceswith equal intensity, so contrast is lost and the shape ofthe object is harder to discern.3. The “blackness” of a blackbody refers to its ideal propertyof absorbing all radiation incident on it. If an observedroom temperature object in everyday life absorbs all radiation,we describe it as (visibly) black. The black appearance,however, is due to the fact that our eyes are sensitiveonly to visible light. If we could detect infrared light withour eyes, we would see the object emitting radiation. Ifthe temperature of the blackbody is raised, Wien’s lawtells us that the emitted radiation will move into the visiblerange of the spectrum. Thus, the blackbody could appearas red, white, or blue, depending on its temperature.5. All objects do radiate energy, but at room temperature thisenergy is primarily in the infrared region of the electromagneticspectrum, which our eyes cannot detect. (Pit vipershave sensory organs that are sensitive to infrared radiation;thus, they can seek out their warm-blooded prey in what wewould consider absolute darkness.7. Most metals have cutoff frequencies corresponding tophotons in or near the visible range of the electromagneticspectrum. AM radio wave photons have far too littleenergy to eject electrons from the metal.9. We can picture higher frequency light as a stream of photonsof higher energy. In a collision, one photon can giveall of its energy to a single electron. The kinetic energy ofsuch an electron is measured by the stopping potential.The reverse voltage (stopping voltage) required to stopthe current is proportional to the frequency of the incominglight. More intense light consists of more photonsstriking a unit area each second, but atoms are sosmall that one emitted electron never gets a “kick” frommore than one photon. Increasing the intensity of the
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27.3 X-Rays 881even when black card
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Problems 89917. When light of wavel
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28.2 Atomic Spectra 905l(nm) 400 50
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28.3 Th Bohr Theory of Hydrogen 909
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29.2 Binding Energy 943130120110100
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29.3 Radioactivity 947INTERACTIVE E
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Summary 965Photo Researchers, Inc./
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Problems 967CONCEPTUAL QUESTIONS1.
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Problems 96924. A building has beco
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Problems 97157. A by-product of som
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30.1 Nuclear Fission 975Applying Ph
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30.2 Nuclear Reactors 977Courtesy o
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30.2 Nuclear Reactors 979events in
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30.3 Nuclear Fusion 983VacuumCurren
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