solutions_chapter26

# B the next three thicknesses that give constructive

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(b) The next three thicknesses that give constructive interference are for 3, 4: 472 nm, 708 nm and 944 nm. m 5 2, t 5 m 1 650 nm 2 3 1.38 4 2 5 1 236 nm 2 m . 2 t 5 m l . l 5 650 nm n . 180° 180° 1 n 5 1.51 2 1 n 5 1.38 2 180° 1 n 5 1.38 2 1 n 5 1.00 2 t 5 l 4 5 l 0 4 n 5 480 nm 4 1 1.49 2 5 80.5 nm. m 5 0. l 5 l 0 n . 2 t 5 A m 1 1 2 B l . t 5 m l 2 5 l 0 2 n 5 480 nm 2 1 1.33 2 5 180 nm l 5 l 0 n . m 5 1. 2 t 5 m l . t 5 l 4 5 l 0 4 n 5 650 nm 4 1 1.42 2 5 114 nm m 5 0. l 5 l 0 / n 2 t 5 A m 1 1 2 B l , u 5 6 51.6°. 2 4: m 5 3, u 5 6 34.1°. 2 3: m 5 2, u 5 6 19.6°. 2 2: m 5 1, u 5 6 6.4°. 2 1: m 5 0, sin u 5 A m 1 1 2 B l d 5 0.224 A m 1 1 2 B . 6 1, c . m 5 0, d sin u 5 A m 1 1 2 B l , m . 4. 0 sin u 0 u 5 6 63.6°. m 5 6 4: u 5 6 42.2°. m 5 6 3: u 5 6 26.6°. m 5 6 2: u 5 6 12.9°. m 5 6 1: u 5 0. m 5 0: sin u 5 m l d 5 m 1 1.23 m 5.50 m 2 5 0.224 m . 6 1, c . m 5 0, d sin u 5 m l , l 5 v f 5 340 m / s 277 Hz 5 1.23 m. 26-4 Chapter 26

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26.19. Set Up: Consider light reflected at the front and rear surfaces of the film. At the front surface of the film, light in air reflects from the film and there is a phase shift due to the reflection. At the back surface of the film, light in the film reflects from glass and there is no phase shift due to reflection. Therefore, there is a net phase difference produced by the reflections. The path difference for these two rays is 2 t , where t is the thickness of the film. The wavelength in the film is Solve: (a) Since the reflection produces a net phase difference, destructive interference of the reflected light occurs when The minimum thickness is 96.4 nm. (b) The next three thicknesses are for 3 and 4: 192 nm, 289 nm and 386 nm. Reflect: The minimum thickness is for Compare this to Problem 26.15, where the minimum thickness for destructive interference is 26.20. Set Up: The path length is The wavelength in the film is Solve: The number of waves is 26.21. Set Up: The fringes are produced by interference between light reflected from the top and from the bottom surfaces of the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the top surface of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the air wedge do have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) therefore is The geometry of the air wedge is sketched in Figure 26.21. Figure 26.21 Solve: so and The distance along the plate between adjacent fringes is The number of fringes per cm is Reflect: As the interference is destructive and there is a dark fringe at the line of contact between the two plates. t S 0 1.00 D x 5 1.00 0.0369 cm 5 27.1 fringes / cm. D x 5 x m 1 1 2 x m 5 l 2 1 8.89 3 10 2 4 2 5 656 3 10 2 9 m 2 1 8.89 3 10 2 4 2 5 3.69 3 10 2 4 m 5 0.369 mm. x m 1 1 5 A m 1 3 2 B l 2 1 8.89 3 10 2 4 2 . x m 5 A m 1 1 2 B l 2 1 8.89 3 10 2 4 2 t m 5 A m 1 1 2 B l 2 . t 5 1 8.89 3 10 2 4 2 x . tan u 5 t x tan u 5 0.0800 mm 90.0 mm 5 8.89 3 10 2 4 . u x 9.00cm t 0.0800mm Air 2 t 5 A m 1 1 2 B l . 2 t l 5 17.52 3 10 2 6 m 480 3 10 2 9 m 5 36.5. l 5 648 nm 1.35 5 480 nm.

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