# Sol-Ch36 - CHAPTER 36 E xercises 1(a 1/4 1/q = 1/0.05 thus...

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Exercises 1. (a) 1/4 + 1/q = 1/0.05, thus q = 0.0506 m = –q/p = –0.0127, thus y I = my O = 2.53 cm (b) m = 0.526, so y I = 10.5 cm 2. The image lies in the focal plane tan α = R/D = r/f, r = Rf/D (a) 9.06 mm; (b) 9.34 mm 3. (a) m = –q/p = –5, so p = 0.4 m; (b) 1/p+1/q = 1/f, so f = 0.33 m 4. (a) m = –q/p = 4, so p = 4 cm (b) 1/p+1/q = 1/f, so f = 5.33 cm 5. (a) m = –q/p = –1/3, so p = 18 cm; (b) 1/p+1/q = 1/f, so f = 4.5 cm 6. m = –q/p = –55.6, so p = 0.126 m 1/0.126+1/7 = 1/f, so f = 0.124 m 7. (a) 1/q = 1/5 – 1/200, so q = 5.13 cm (b) 1/q = 1/5 – 1/50, so q = 5.56 cm 8. q = +2p, 1/p + 1/2p = 1/15, so p = 22.5 cm 1/p – 1/2p = 1/15, so p = 7.5 cm 9. (a) q = 2p/3 = 8 cm; (b) 1/8+1/12 = 1/f, so f = 4.8 cm 10. (a) From q = +2.5p, find p = 49 cm; (b) From q = –2.5p, find p = 21 cm 11. (a) q = 0.4p, then 1/p + 1/0.4p = 1/f, so p = 70 cm; (b) q = –0.4p, then 1/p – 1/0.4p = 1/f, so p = –30 cm 12. (a) q = –p/5, then 1/p – 5/p = –1/20, so p = 80 cm (b) For real, erect and 150 %, q = –1.5 p. Then, 1/p – 2/3p = –1/20, so p = –6.67 cm CHAPTER 36

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13. 1/20 + 1/q 1 = 1/10, so q 1 = 20 cm, thus p 2 = –10 cm –1/10 + 1/q 2 = –1/15, so q 2 = 30 cm 14. 1/12+1/q 1 = 1/10, so q 1 = 60 cm, thus p 2 = –45 cm –1/45+1/q 2 = –1/20, so q 2 = 13.8 cm 15. 1/40+1/q 1 = 1/8, so q 1 = 10 cm, thus p 2 = 10 cm 1/10+1/q 2 = 1/12, so q 2 = –60 cm mT = m 1 m 2 = (–1/4)(+60/10) = –1.5 16. –1/f 1 +1/q 2 = 1/f 2 , q 2 = f eff = f 1 f 2 /(f 1 + f 2 ) 17. See Exercise 16 above: 14 = f 1 f 2 /(f 1 + f 2 ) Since f 1 = 10 cm, find f 2 = –35 cm 18. (a) M = 0.25/0.057 = 4.39; (b) 1/q = 1/6 – 1/5.7, so q = –114 cm 19. (a) –1/40 + 1/p = 1/4, thus p = 3.64 cm, m = –q/p:
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Sol-Ch36 - CHAPTER 36 E xercises 1(a 1/4 1/q = 1/0.05 thus...

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