Quick QuizzesAnswers to Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsChapter 25 Optical Instruments Quick Quizzes 1. (c). The corrective lens for a farsighted eye is a converging lens, while that for a nearsighted eye is a diverging lens. Since a converging lens is required to form a real image of the Sun on the paper to start a fire, the campers should use the glasses of the farsighted person. 2. (a). We would like to reduce the minimum angular separation for two objects below the angle subtended by the two stars in the binary system. We can do that by reducing the wavelength of the light—this in essence makes the aperture larger, relative to the light wavelength, increasing the resolving power. Thus, we would choose a blue filter. 337338 CHAPTER 25 Answers to Conceptual Questions 2. The objective lens of the microscope must form a real image just inside the focal point of the eyepiece lens. In order for this to occur, the object must be located just outside the focal point of the objective lens. Since the focal length of the objective lens is typically quite short ()~1 cm , this means that the microscope can focus properly only on objects close to the end of the barrel and will be unable to focus on objects across the room. 4. For a lens to operate as a simple magnifier, the object should be located just inside the focal point of the lens. If the power of the lens is +20.0 diopters, it focal length is ()()1.00 m 1.00 m 20.0 0.050 0 m 5.00 cmf ==+==P The object should be placed slightly less than 5.00 cm in front of the lens. 6. The aperture of a camera is a close approximation to the iris of the eye. The retina of the eye corresponds to the film of the camera, and a close approximation to the cornea of the eye is the lens of the camera. 8. You want a real image formed at the location of the paper. To form such an image, the object distance must be greater than the focal length of the lens. 10. Under low ambient light conditions, a photoflash unit is used to insure that light entering the camera lens will deliver sufficient energy for a proper exposure to each area of the film. Thus, the most important criterion is the additional energy per unit area (product of intensity and the duration of the flash, assuming this duration is less than the shutter speed) provided by the flash unit. 12. The angular magnification produced by a simple magnifier is ()25 cmmf=. Note that this is proportional to the optical power of a lens, 1 f=P , where the focal length f is expressed in meters. Thus, if the power of the lens is doubled, the angular magnification will also double.Optical Instruments 339 Answers to Even Numbered Problems 2. 31 mm 4. 1.09 mm 6. (b) 1 100 s≈ 8. 2.2 mm farther from the film 10. For the right eye, ; for the left eye, 1.18 diopters=−P 0.820 diopters=−P . 12. (a) 33.3 cm (b) +3.00 diopters 14. (a) +50.8 diopters to +60.0 diopters (b) –0.800 diopters; diverging 16. (a) (b) +0.67 diopters 0.67 diopters−18. (a) (b) 2.0m =+ 1.0m=+ 20. (a) 4.17 cm in front of the lens (b) 6.00m=+ 22. (a) 0.400 cm (b) 1.25cm (c) 1000M=− 24. 0.806 mµ 26. 21.6 10 mi×28. (a) (b) 0.944 m 7.50m =30. (a) virtual image (b) 2q →∞ (c) 15.0 cm, 5.00 cmoeff==− 32. 0.77 m (≈30 inches) 34. 1.00 mrad 36. (a) (b) 43.6 m 42.29 10 rad−×38. 38 cm 40. (a) (b) 33.6 10 lines×31.8 10 lines×42. 31.31 10 fringe shifts×44. 39.6 mµ340 CHAPTER 25 46. 1.000 5 48. (a) -4.3 diopters (b) -4.0 diopters, 44 cm 50. (a) (b) 3.27 (c) 9.80 1.96 cm52. 32.0 10 radmθ−≤×54. (a) (b) 0.060 1 cm 2.00 cm− (c) 6.02 cm (d) 4.00m= 56. 5.07 mmOptical Instruments 341 Problem Solutions 25.1 Using the thin lens equation, the image distance is ()()150 cm 25.0 cm30.0 cm150 cm 25.0 cmpfqpf== =−− so the image is located 30.0 cm beyond the lens . The lateral magnification is 30.0 cm 1150 cm 5qMp=− =− = − 25.2 The f-number of a camera lens is defined as -number focal length diameterf=. Therefore, the diameter is 55 mm31 mm-number 1.8fDf=== 25.3 The thin lens equation, 111pqf+=, gives the image distance as ()()3100 m 52.0 mm52.0 mm100 m 52.0 10 mpfqpf−== =−−× From the magnitude of the lateral magnification, Mhh qp′==−, where the height of the image is , the height of the object (the building) must be 0.092 0 m 92.0 mmh′==()100 m92.0 mm 177 m52.0 mmphhq′=−= − = 25.4 The image distance is since the object is so far away. Therefore, the lateral magnification is qf≈Mh qp fp−≈−h′== , and the diameter of the Moon’s image is ()()68120 mm221.7410 m 3.84 10 moonfhMhp′== = × =×1.09mmmR342 CHAPTER 25 25.5 The exposure time is being reduced by a factor of 211256 s1132 s 8tt== Thus, to maintain correct exposure, the intensity of the light reaching the film should be increased by a factor of 8. This is done by increasing the area of the aperture by a factor of 8, so in terms of the diameter, ()222148 4DDππ= or 218=DD. The new f-number will be ()()1221-number4.0-number 1.4888ffffDD== = == or 1.4f 25.6 (a) The intensity is a measure of the rate at which energy is received by the film per unit area of the image, or 1imageIA∝ and x. Consider an object with horizontal and vertical dimensions yhh as shown at the right. If the vertical dimension intercepts angle θ, the vertical dimension of the image is yhqθ′=, or yhq′∝. Similarly for the horizontal dimension, xhq′∝, and the area of the image is 2image x yAhh′′= q∝. Assuming a very distant object, qf≈, so 2imageAf∝ and we conclude that 21If∝ . The intensity of the light reaching the film is also proportional to the area of the lens and hence, to the square of the diameter of that lens, or 2DI∝. Combining this with our earlier conclusion gives ()22DIf21fD∝= or ()21-numberIf∝ qhxhyh¢yh¢xqq » f (b) The total light energy hitting the film is proportional to the product of intensity and exposure time, It. Thus, to maintain correct exposure, this product must be kept constant, or giving 22 11It It=()()221122mber4.0 1s1100 s1.8 500mberItt tI = ≈  2211-nu-nuff==Optical Instruments 343 25.7 Since the exposure time is unchanged, the intensity of the light reaching the film should be