Quantum Physics

A.26 Answers to Quick Quizzes, Odd-Numbered Conceptual Questions and Problems23. 3.8 mm25. n 2.0027. 20.0 cm29. (a) 40.0 cm beyond the lens, real, inverted, M 1.00(b) No image is formed. Parallel rays leave the lens.(c) 20.0 cm in front of the lens, virtual, upright, M 2.0031. (a) 13.3 cm in front of the lens, virtual, upright, M 1/3(b) 10.0 cm in front of the lens, virtual, upright, M 1/2(c) 6.67 cm in front of the lens, virtual, upright, M 2/333. (a) either 9.63 cm or 3.27 cm (b) 2.10 cm35. (a) 39.0 mm (b) 39.5 mm37. at distance 2 f in front of lens39. 40.0 cm41. 30.0 cm to the left of the second lens, M 3.0043. 7.47 cm in front of the second lens; 1.07 cm; virtual, upright45. from 0.224 m to 18.2 m47. real image, 5.71 cm in front of the mirror49. 38.6°51. 160 cm to the left of the lens, inverted, M 0.80053. q 10.7 cm55. 32.0 cm to the right of the second surface (real image)57. (a) 20.0 cm to the right of the second lens; M 6.00(b) inverted(c) 6.67 cm to the right of the second lens; M 2.00;inverted59. (a) 1.99(b) 10.0 cm to the left of the lens(c) inverted61. (a) 5.45 m to the left of the lens(b) 8.24 m to the left of the lens(c) 17.1 m to the left of the lens(d) by surrounding the lens with a medium having arefractive index greater than that of the lens material.63. (a) 263 cm (b) 79.0 cmChapter 24QUICK QUIZZES1. (c)2. (b)3. (b)4. The compact disc.CONCEPTUAL QUESTIONS1. You will not see an interference pattern from the automobileheadlights, for two reasons. The first is that the headlightsare not coherent sources and are therefore incapableof producing sustained interference. Also, theheadlights are so far apart in comparison to the wavelengthsemitted that, even if they were made into coherentsources, the interference maxima and minima wouldbe too closely spaced to be observable.3. The result of the double slit is to redistribute the energy arrivingat the screen. Although there is no energy at the locationof a dark fringe, there is four times as much energyat the location of a bright fringe as there would be withonly a single narrow slit. The total amount of energy arrivingat the screen is twice as much as with a single slit, as itmust be according to the law of conservation of energy.5. One of the materials has a higher index of refraction thanwater, and the other has a lower index. The material withthe higher index will appear black as it approaches zerothickness. There will be a 180° phase change for the lightreflected from the upper surface, but no such phasechange for the light reflected from the lower surface,because the index of refraction for water on the otherside is lower than that of the film. Thus, the two reflectionswill be out of phase and will interfere destructively.The material with index of refraction lower than waterwill have a phase change for the light reflected from boththe upper and the lower surface, so that the reflectionsfrom the zero-thickness film will be back in phase and thefilm will appear bright.7. For incidence normal to the film, the extra path lengthfollowed by the reflected ray is twice the thickness ofthe film. For destructive interference, this must be a distanceof half a wavelength of the light in the material ofthe film. For a film in air, no 180° phase change will occurin these reflections, so the thickness of the film must beone-quarter wavelength, which is the same as the conditionfor constructive interference of reflected light. Thismeans that the transmitted light is a minimum when thereflected light is a maximum, and vice versa.9. Since the light reflecting at the lower surface of the filmundergoes a 180° phase change, while light reflectingfrom the upper surface of the film does not undergo sucha change, the central spot (where the film has near zerothickness) will be dark. If the observed rings are not circular,the curved surface of the lens does not have a truespherical shape.11. For regional communication at the Earth’s surface, radiowaves are typically broadcast from currents oscillating intall vertical towers. These waves have vertical planes ofpolarization. Light originates from the vibrations of atomsor electronic transitions within atoms, which representoscillations in all possible directions. Thus, light generallyis not polarized.13. Yes. In order to do this, first measure the radar reflectivityof the metal of your airplane. Then choose a light,durable material that has approximately half the radarreflectivity of the metal in your plane. Measure its indexof refraction, and place onto the metal a coating equal inthickness to one-quarter of 3 cm, divided by that index.Sell the plane quick, and then you can sell the supposedenemy new radars operating at 1.5 cm, which the coatedmetal will reflect with extra-high efficiency.15. If you wish to perform an interference experiment, youneed monochromatic coherent light. To obtain it, youmust first pass light from an ordinary source through aprism or diffraction grating to disperse different colorsinto different directions. Using a single narrow slit, selecta single color and make that light diffract to cover bothslits for a Young’s experiment. The procedure is muchsimpler with a laser because its output is already monochromaticand coherent.17. Strictly speaking, the ribs do act as a diffraction grating,but the separation distance of the ribs is so much largerthan the wavelength of the x-rays that there are no observableeffects.19. As the edge of the Moon cuts across the light from thestar, edge diffraction effects occur. Thus, as the edge ofthe Moon moves relative to the star, the observed lightfrom the star proceeds through a series of maxima andminima.21. Larger. From Brewster’s law, n tan p , we see that theangle increases as n increases.