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# Sol-Ch37 - CHAPTER 37 E xercises 1 With sin = y/L d = m L/y...

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Exercises 1. With sin θ = y/L, d = m λ L/y = 0.17 mm 2. With sin θ = y/L, y = m λ L/d, find ∆y = 1.7 mm 3. (a) With sin θ = y/L, d = λ ω /2y = 0.141 mm; (b) d = λ L/y = 0.281 mm 4. ∆y = λ L/d = 3.07 mm 5. λ = yd/4L = 583 nm 6. (a) d = 3 λ L/y = 0.22 mm; (b) ∆y = λ L/d = 5.36 mm 7. dsin θ = 4.5 λ , so λ = 543 nm 8. 10(560) = 9.5 λ 2 , λ 2 = 589 nm 9. y = m λ L/d, so 480m 1 = 560m 2 , or 6m 1 = 7m 2 . Find m 1 = 7, and m 2 = 6. y = m 1 (480 nm)L/d = 1.68 cm 10. tan θ ≈ sin θ = 5.4 × 10 –3 ; d = 7 λ /sin θ = 0.763 mm. 11. sin θ 2 = sin( θ 1 + 10 ° ) = 2sin θ 1 . Expand sin( θ 1 + 10) to find tan θ 1 = sin10 ° /(2 – cos10 ° ), thus θ 1 = 9.71 ° d = λ /sin θ 1 = 17.8 cm 12. ∆y = λ L/d = 1/700, so d = 0.714 mm 13. δ = 3 λ /4 = dsin θ = dy/L, so y = 3 λ L/4d = 2.16 mm, upward 14. δ = λ = 2(D 2 + H 2 ) 1/2 – 2D; where D = 4 m and λ = 1.7 m. Find H = 2.74 m 15. dsin θ = λ /2 = (0.34 m)/2, then y = dtan θ = 1.38 m. CHAPTER 37

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16. (a) δ = (4 2 + d 2 ) 1/2 – 4 = λ /2, where λ = 3.58 m. Find d = 4.18m; (b) d = λ /2 = 1.79 m. 17. δ = (4 2 + d 2 ) 1/2 – 4 = λ /2, where λ = 0.68m. Find d = 1.7 m 18. Take speakers to be in phase. Δ = (4 2 + d 2 ) 1/2 – 4 = 0.472 m (a) δ = λ , so f = 340/ δ = 720 Hz (b) δ = λ /2, so f = 340/2 δ = 360 Hz 19. (a) δ = x – (d – x) = 2x – d = (m + 1/2) λ , thus x = 0.5d – (m + 0.5) λ /2; (b) δ = 2x – d = m λ , so x = 0.5d – m λ /2 20. Ф = 2π/ λ , so δ = 5 λ /2π. d sin θ = δ = 0.15/2π, thus θ = 1.7 ° . 21. (a) d(sin θ – sin α ); (b) θ = α ; (c) θ = 0, so δ = –dsin α = – λ /2, or α = sin – 1 ( λ /2d) 22. sin θ = y/f, so dy/f = (m + 1/2) λ , or y = (2m + 1) λ f/2d. 23. dsin θ = (m + 1/2) λ leads to m = 6, so there are 7 dark fringes. 24. Need to find the width of the reflected beam on the screen. d/L = y 2 /24L, y 2 = 24d; d/5L = y 1 /20L, y 1 = 4d, thus ∆y = 20d = 8 mm.
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