Sol-Ch38 - CHAPTER 38 E xercises 1. ( a) sin 1 = /a, then y...

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Exercises 1. (a) sin θ 1 = λ /a, then y 1 ≈ L λ /a = 2.04 cm. width 2y 1 = 4.08 cnl (b) sin θ 2 = 2 λ /a, so y 2 ≈ 2L λ /a = 4.08 cm then y 2 – y 1 = 2.04 cm 2. θ 1 = sin –1 ( λ /a); where λ = 6.8 cm. /width is 2 θ 1 . (a) 116 ° ; (b) 26.2 ° 3. ∆y 2 /∆y 1 = λ 2 / λ 1 , so ∆y 2 = (3)(436/589) = 2.22 cm 4. sin θ ≈ tan θ = y 1 /L = (∆y/2)L = 2 ×  10 –3 , a = λ /sin θ = 0.273 mm 5. ∆y = L(sin θ 2 – sin θ 1 ) = L λ /a. So a = L λ /∆y = 0.045 mm 6. (a) f = v/ λ = 340 ×  4/0.76 = 1790 Hz (b) sin θ 1 = λ /a = 1/4, leads to θ 1 = 14.5˚ 7. d/a = 4, so there are 7 fringes 8. sin θ 1 = 1.22 λ /a = 6.7 ×  10 –4 ; Width is 2 θ 1 . (a) 116 ° ; (b) 26.2 ° 9. θ c = 1.22 λ /a = 9.15 ×  10 –4 ; then s = d θ c = 1.46 cm 10. θ c = 1.22 λ /a = 1.83 ×  10 –2 ; then s = d θ c = 366 m 11. (a) θ c = 1.22 λ /a = 1.34 ×  10 –4 ; d = L θ c = 51.6 km; (b) θ c = 1.49 ×  10 –7 ; d = L θ c = 57.3 m 12. θ c = 1.22 λ /a = 1.14 ×  10 –6 ; s = d θ c = 0.205 m; 13. θ c = 1.22 λ /a = s/25, thus s = 50.8 µm 14. θ c = 1.22 λ /a = 4.47 × 10 –5 ; d = s/ θ c = 44.7 km; 15. (a) θ c = 1.22 λ /a = 1.2 ×  10 –7 ; s = d θ c = 1.2 ×  10 9 m; CHAPTER 38
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(b) θ c = 1.22 λ /a = 8.4 ×  10 –4 ; s = d θ c = 8.4 ×  10 –4 ; s = d θ c = 8.4 ×  10 –2 m 16. (a) θ c = 1.22 λ /a = 1.59 ×  10 –4 ; d = s/ θ c = 11.3 km;
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch38 - CHAPTER 38 E xercises 1. ( a) sin 1 = /a, then y...

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