solutions_chapter26

U 5 y s u res 5 1.22 l d 1.38 3 10 8 m s 5 yd 1.22 l

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Unformatted text preview: u 5 y s . u res 5 1.22 l D . 1.38 3 10 8 m. s 5 yD 1.22 l 5 1 65.0 m 21 4.00 3 10 2 3 m 2 1.22 1 550 3 10 2 9 m 2 5 3.87 3 10 5 m 5 387 km. y s 5 1.22 l D . y 5 65.0 m. u res 5 y s , D 5 4.00 mm. u res 5 1.22 l D . y 5 s u res 5 1 7.8 3 10 7 km 21 6.71 3 10 2 8 2 5 5.2 km u 5 u res u res 5 1.22 1 550 3 10 2 9 m 10.0 m 2 5 6.71 3 10 2 8 . u 5 y s . u res 5 1.22 l D . sin u 5 1 5 m l d l 5 d sin u m 5 1 2.50 3 10 2 6 m 21 1 2 4 5 625 nm. d 5 1.00 3 10 2 2 m 4000 5 2.50 3 10 2 6 m. sin u 5 1. m 5 4. d sin u 5 m l . l 5 371 nm. m 5 4: l 5 495 nm. m 5 3: l 5 742 nm. m 5 2: l 5 1484 nm. m 5 1: l 5 2 tn m 5 1484 nm m . 2 t 5 m l 5 m l n . l 5 330 nm. m 5 4: l 5 424 nm. m 5 3: l 5 594 nm. m 5 2: l 5 989 nm. m 5 1: 2968 nm. l 5 m 5 0: l 5 2 tn m 1 1 2 5 2 1 0.485 3 10 2 6 m 21 1.53 2 m 1 1 2 5 1484 nm m 1 1 2 . 2 t 5 A m 1 1 2 B l 5 A m 1 1 2 B l n . l 5 l n . 2 t 5 m l . 2 t 5 A m 1 1 2 B l Interference and Diffraction 26-15 26.69. Set Up: The bright fringes are located by First-order means The line density is The largest appears at the largest If the line is at then the line is to be at Solve: so so This gives and The line density is Reflect: and the second order pattern doesn’t include all of the visible spectrum. The first order is the only order that displays the entire visible spectrum. 26.70. Set Up: For destructive interference the net phase difference must be which is one-half a period, or In phase, A is or ahead of B . At points above the centerline, points are closer to A than to B and the signal from A gains phase relative to B because of the path difference. Destructive interference will occur when 1, At points at an angle below the centerline, the signal from B gains phase relative to A because of the phase difference. Destructive interference will occur when 1, Solve: Points above the centerline: Points below the centerline: 26.71. Set Up: The waves reflected from the surface of the water undergo a half-cycle phase shift and the waves that travel directly to the plane have no reflection and therefore no shift. The condition for constructive interference therefore is where is the path difference for the two sets of waves. Let y be the height of the airplane above the water. Let be the height of the cliffs. The two paths are sketched in Figure 26.71. The law of reflection says that the two angles labeled are equal. The diagram is not to scale, since is actually much larger than h or y . Figure 26.71 Solve: We first solve for the path difference where and and r 2 b 5 y sin u 5 y " D 2 1 1 h 1 y 2 2 h 1 y . r 2 a 5 h sin u 5 h " D 2 1 1 h 1 y 2 2 h 1 y . sin u 5 h 1 y " D 2 1 1 h 1 y 2 2 . sin 2 u 5 tan 2 u 1 1 tan 2 u 5 1 h 1 y 2 2 D 2 1 1 h 1 y 2 2 tan u 5 h 1 y D . D a 1 D b 5 h 1 y tan u D 5 D b 5 y tan u . D a 5 h tan u . r 2 5 r 2 a 1 r 2 b . r 2 2 r 1 , u u D a r 1 r 2 a r 2 b D b h h 2 y y Cliffs D 5 D a 1 D b 5 10.0 3 10 3 m u h 5 200 m r 2 2 r 1 r 2 2 r 1 5 A m 1 1 2 B l , u 5 55.2°. m 5 3: u 5 37.0°; m 5 2: u 5 22.5°; m 5 1: u 5...
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u 5 y s u res 5 1.22 l D 1.38 3 10 8 m s 5 yD 1.22 l 5 1...

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