solutions_chapter26

# B the visible wavelength for which there is

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structive interference in the reflected light is 441 nm. (b) The visible wavelength for which there is destructive interference in the reflected light is 551 nm. This is the visible wavelength predominant in the transmitted light. Reflect: At a particular wavelength the sum of the intensities of the reflected and transmitted light equals the inten- sity of the incident light. l 0 5 367 nm. m 5 3: l 0 5 551 nm. m 5 2: l 0 5 1102 nm. m 5 1: l 0 5 2 tn m 5 1102 nm m . 2 t 5 m l 5 m l 0 n . l 0 5 315 nm. m 5 3: l 0 5 441 nm. m 5 2: l 0 5 735 nm. m 5 1: l 0 5 2200 nm. m 5 0: l 0 5 2 tn m 1 1 2 5 2 1 380 nm 21 1.45 2 m 1 1 2 5 1102 nm m 1 1 2 . 2 t 5 A m 1 1 2 B l 5 A m 1 1 2 B l 0 n . l 5 l 0 n . 2 t 5 m l . 2 t 5 A m 1 1 2 B l . a 5 l sin u 5 632.8 3 10 2 9 m sin 1.20° 5 30.2 m m. u 5 1.20°. tan u 5 y R 5 2.61 cm 125 cm 5 0.02088 y 5 R tan u m 5 6 1. sin u 5 m l a . v 5 f l 5 1 0.556 Hz 21 2.4 m 2 5 1.33 m / s l 5 2 1 13.51 m 2 12.32 m 2 5 2.4 m. r 2 2 r 1 5 l 2 r 1 5 " 1 12.0 m 2 2 1 1 2.8 m 2 2 5 12.32 m. r 2 5 " 1 12.0 m 2 2 1 1 6.2 m 2 2 5 13.51 m. 12.0m r 2 r 1 6.2m Shore Break water 9.0m 2.8m v 5 f l . f 5 10 18 s 5 0.556 Hz. r 2 2 r 1 5 l 2 . r 2 2 r 1 26-14 Chapter 26

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26.64. Set Up: If the glass plate is illuminated from above, the waves reflected at the top surface of the plate have a half-cycle reflection phase shift and the waves reflected at the bottom surface of the plate have no reflection phase shift. The condition for constructive interference is and the condition for destructive interference is Solve: (a) The visible wavelengths intensified in the reflected light are 594 nm and 424 nm. (b) The visible wavelength cancelled in the reflected light is 495 nm. 26.65. Set Up: The longest wavelength is for Solve: Reflect: for the longest wavelength in a particular order, and the longest wavelength for a given order decreases as the order m increases. 26.66. Set Up: The angular size of the object is Solve: Setting gives 26.67. Set Up: Rayleigh’s criterion says where s is the altitude and Solve: 26.68. Set Up: The diameter of Jupiter is The angular size of an object is Solve: Setting gives D 5 1.22 l u 5 1.22 1 550 3 10 2 9 m 3.39 3 10 2 9 rad 2 5 198 m. u res 5 u u 5 y s 5 1.38 3 10 8 m 4.3 1 9.461 3 10 15 m 2 5 3.39 3 10 2 9 rad. 1 light-year 5 9.461 3 10 15 m. 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 0 5 371 nm. m 5 4: l 0 5 495 nm. m 5 3: l 0 5 742 nm. m 5 2: l 0 5 1484 nm. m 5 1: l 0 5 2 tn m 5 1484 nm m . 2 t 5 m l 5 m l 0 n . l 0 5 330 nm. m 5 4: l 0 5 424 nm. m 5 3: l 0 5 594 nm. m 5 2: l 0 5 989 nm. m 5 1: 2968 nm. l 0 5 m 5 0: l 0 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 0 n . l 5 l 0 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.

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