1006_PartUniversity Physics Solution

1006_PartUniversity Physics Solution - 35-6 Chapter 35...

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35-6 Chapter 35 35.22. IDENTIFY: The maximum intensity occurs at all the points of constructive interference. At these points, the path difference between waves from the two transmitters is an integral number of wavelengths. SET UP: For constructive interference, sin θ = m λ /d . EXECUTE: (a) First find the wavelength of the UHF waves: = c/f = (3.00 × 10 8 m/s)/(1575.42 MHz) = 0.1904 m For maximum intensity ( π d sin )/ = m π , so sin = m /d = m [(0.1904 m)/(5.18 m)] = 0.03676 m The maximum possible m would be for = 90&, or sin = 1, so m max = d/ = (5.18 m)/(0.1904 m) = 27.2 which must be –27 since m is an integer. The total number of maxima is 27 on either side of the central fringe, plus the central fringe, for a total of 27 + 27 + 1 = 55 bright fringes. (b) Using sin = m /d , where m = 0, –1, –2, and –3, we have s i n = m /d = m [(0.1904 m)/(5.18 m)] = 0.03676 m m = 0: sin = 0, which gives = 0& m = –1: sin = –(0.03676)(1), which gives = –2.11& m = –2: sin = –(0.03676)(2), which gives = –4.22& m = –3: sin = –(0.03676)(3), which gives = –6.33& (c) 2 0 sin cos d II π ⎛⎞ = ⎜⎟ ⎝⎠ = () 22 (5.18 m)sin(4.65 ) 2.00 W/m cos 0.1904 m ° = 1.28 W/m 2 . EVALUATE: Notice that sin increases in integer steps, but only increases in integer steps for small . 35.23. (a) IDENTIFY and SET UP: The minima are located at angles given by 1 sin . 2 dm =+ The first minimum corresponds to 0. m = Solve for . Then the distance on the screen is tan . yR = EXECUTE: 9 3 3 660 10 m sin 1.27 10 2 2(0.260 10 m) d × == = × × and 3 1.27 10 rad 3 (0.700 m)tan(1.27 10 rad) 0.889 mm. y = (b) IDENTIFY and SET UP: Eq.(35.15) given the intensity I as a function of the position y on the screen: 2 0 cos . dy R = Set 0 /2 = and solve for y . EXECUTE: 0 1 2 = says 2 1 cos 2 dy R = 1 cos 2 dy R = so rad 4 dy R ππ = 9 3 (660 10 m)(0.700 m) 0.444 mm 44 ( 0 . 2 6 0 1 0 m ) R y d × = × EVALUATE: 0 = at a point on the screen midway between where 0 = and 0. I = 35.24. IDENTIFY: Eq. (35.14): 2 0 cos sin . d = SET UP: The intensity goes to zero when the cosine±s argument becomes an odd integer multiple of 2 EXECUTE: sin ( 1/2) d m gives sin ( 1/2), = + which is Eq. (35.5). EVALUATE: Section 35.3 shows that the maximum-intensity directions from Eq.(35.14) agree with Eq.(35.4). 35.25. IDENTIFY: The intensity decreases as we move away from the central maximum. SET UP: The intensity is given by 2 0 cos dy R = . EXECUTE: First find the wavelength: = c/f = (3.00 × 10 8 m/s)/(12.5 MHz) = 24.00 m At the farthest the receiver can be placed, I = I 0 /4, which gives 2 0 0 cos 4 Id y I R = 2 1 cos 4 dy R = 1 cos 2 dy R = ±
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Interference 35-7 The solutions are π dy/ λ R = π /3 and 2 π /3. Using π /3, we get y = R/ 3 d = (24.00 m)(500 m)/[3(56.0 m)] = 71.4 m It must remain within 71.4 m of point C . EVALUATE: Using π dy/ R = 2 π /3 gives y = 142.8 m. But to reach this point, the receiver would have to go beyond 71.4 m from C , where the signal would be too weak, so this second point is not possible. 35.26. IDENTIFY: The phase difference φ and the path difference 12 rr are related by 2 () π = . The intensity is given by 2 0 cos 2 II ⎛⎞ = ⎜⎟ ⎝⎠ .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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1006_PartUniversity Physics Solution - 35-6 Chapter 35...

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