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Unformatted text preview: 35.1: Measuring with a ruler from both 2 1 and S S to there different points in the antinodal line labeled m = 3, we find that the difference in path length is three times the wavelength of the wave, as measured from one crest to the next on the diagram. 35.2: a) At , λ 4 , 1 2 1 = r r S and this path difference stays the same all along the axis, y so , r r S m λ 4 , At . 4 1 2 2 = + = and the path difference below this point, along the negative yaxis, stays the same, so . 4 = m b) c) The maximum and minimum mvalues are determined by the largest integer less than or equal to . λ d d) If , 7 7 λ 2 1 7 + ≤ ≤ ⇒ = m d so there will be a total of 15 antinodes between the sources. (Another antinode cannot be squeezed in until the separation becomes six times the wavelength.) 35.3: a) For constructive interference the path diference is . . . , 2 , 1 , , λ ± ± = n m The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. 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 . . . , 2 , 1 , , λ ) ( 2 1 ± ± = + m m A path difference of 00 . 3 2 λ = ± / m is possible but a path difference as large as 00 . 9 2 / λ 3 = m is not possible. For a point a distance x from A and B x from 00 . 5 the path difference is m 00 . 1 gives m 00 . 3 ) m 00 . 5 ( m 00 . 4 gives m 00 . 3 ) m 00 . 5 ( ). m 00 . 5 ( = = = + = x x x x x x x x 35.4: a) The path difference is 120 m, so for destructive interference: . m 24 λ m 120 2 λ = ⇒ = b) The longest wavelength for constructive interference is . m 120 λ = 35.5: For constructive interference, we need λ ) m 00 . 9 ( λ 1 2 m x x m r r = ⇒ = , 1 0, 1, 2, 3, For m. 8.25 m, 7.00 m, 5.75 m, 4.50 m, 3.25 m, 2.00 m, 75 . ). m 25 . 1 ( m 5 . 4 Hz) 10 2(120 ) s m 10 00 . 3 ( m 5 . 4 2 m 5 . 4 2 λ m 5 . 4 6 8 = = ⇒ = × × = = = ⇒ m x m m f mc m x . 3 , 2 (Don’t confuse this m with the unit meters, also represented by an “m”). 35.6: a) The brightest wavelengths are when constructive interference occurs: nm. 408 5 nm 2040 λ and nm 510 4 nm 2040 λ , nm 680 3 nm 2040 λ λ λ 4 3 = = = = = = ⇒ = ⇒ = s m d m d b) The pathlength difference is the same, so the wavelengths are the same as part (a). 35.7: Destructive interference occurs for: . nm 453 5 . 4 nm 2040 λ and nm 3 58 5 . 3 nm 2040 λ 2 1 λ 4 3 = = = = ⇒ + = m d 35.8: a) For the number of antinodes we have: , 90 setting so, , 2317 . Hz) 10 (1.079 m) (12.0 ) s m 10 (3.00 λ sin 8 8 ° = = × × = = = θ θ m m df mc d m the maximum integer value is four. The angles are ° ± ° ± ° ± ° ± 9 . 67 and , . 44 , 6 . 27 , 4 . 13 for . 4 , 3 , 2 , 1 , ± ± ± ± = m b) The nodes are given by sin ). 2 1 ( 2317 ....
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.
 Fall '07
 Nearing
 Physics

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