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chapter 35 solutions

# chapter 35 solutions - Solutions to Exercises 12th Ed 35.1...

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Solutions to Exercises 12 th Ed. 35.1. I DENTIFY : Compare the path difference to the wavelength. S ET U P : The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. E XECUTE : (a) For constructive interference the path difference is m λ , m = 0, ± 1, ± 2, . . . Thus only the path difference of zero is possible. This occurs midway between the two sources, 2.50 m from A. (b) For destructive interference the path difference is ( m + 1 2 ) λ , m = 0, ± 1, ± 2, .. . A path difference of ± λ / 2 = 3.00 m is possible but a path difference as large as 3 λ / 2 = 9.00 m is not possible. For a point a distance x from A and 5.00 - x from B the path difference is x - (5.00 m - x ). x - (5.00 m - x ) = + 3.00 m gives x = 4.00 m. x - (5.00 m - x ) = - 3.00 m gives x = 1.00 m . E VALUATE : The point of constructive interference is midway between the points of destructive interference. 35.8.I DENTIFY : The value of y 20 is much smaller than R and the approximate expression y m = R m λ d is accurate. S ET U P : y 20 = 10.6 × 10 - 3 m . E XECUTE : d = 20 R λ y 20 = (20)(1.20 m)(502 × 10 - 9 m) 10.6 × 10 - 3 m = 1.14 × 10 - 3 m = 1.14 mm E VALUATE : tan θ 20 = y 20 R so θ 20 = 0.51o and the approximation sin θ 20 tan θ 20 is very accurate. 35.9. I DENTIFY and S ET U P : The dark lines correspond to destructive interference and hence are located by Eq.(35.5): d sin θ = m + 1 2 λ so sin θ = m + 1 2 λ d , m = 0, ± 1, ± 2, K Solve for θ that locates the second and third dark lines. Use y = R tan θ to find the distance of each of the dark lines from the center of the screen. E XECUTE : 1st dark line is for m = 0 2nd dark line is for m = 1 and sin θ 1 = 3 λ 2 d = 3(500 × 10 - 9 m) 2(0.450 × 10 - 3 m) = 1.667 × 10 - 3 and θ 1 = 1.667 × 10 - 3 rad 3rd dark line is for m = 2 and sin θ 2 = 5 λ 2 d = 5(500 × 10 - 9 m) 2(0.450 × 10 - 3 m) = 2.778 × 10 - 3 and θ 2 = 2.778 × 10 - 3 rad (Note that θ 1 and θ 2 are small so that the approximation θ sin θ tan θ is valid.) The distance of each dark line from the center of the central bright band is given by y m = R tan θ , where R = 0.850 m is the distance to the screen. tan θ θ so y m = R θ m y 1 = R θ 1 = (0.750 m)(1.667 × 10 - 3 rad) = 1.25 × 10 - 3 m y 2 = R θ 2 = (0.750 m)(2.778 × 10 - 3 rad) = 2.08 × 10 - 3 m y = y 2 - y 1 = 2.08 × 10 - 3 m - 1.25 × 10 - 3 m = 0.83 mm

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E VALUATE : Since θ 1 and θ 2 are very small we could have used Eq.(35.6), generalized to destructive interference: y m = R m + 1 2 λ / d . 35.10. I DENTIFY : Since the dark fringes are eqully spaced, R ? y m , the angles are small and the dark bands are located by y m + 1 2 = R ( m + 1 2 ) λ d . S ET U P : The separation between adjacent dark bands is y = R λ d . E XECUTE : y = R λ d d = R λ y = (1.80 m) (4.50 × 10 - 7 m) 4.20 × 10 - 3 m = 1.93 × 10 - 4 m = 0.193 m. E VALUATE : When the separation between the slits decreases, the separation between dark fringes increases. 35.11. I DENTIFY and S ET U P : The positions of the bright fringes are given by Eq.(35.6): y m = R ( m λ / d ).
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