solutions_chapter26

# The only visible wavelength in air for which there is

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values are shorter. The only visible wavelength in air for which there is destructive inter- ference is 550 nm. For constructive interference, and For and all other values are shorter. There are no visible wavelengths in air for which there is constructive interference. 26.57. Set Up: The two sheets of glass are sketched in Figure 26.57. The thickness of the air wedge at a distance x from the line of contact is Consider rays 1 and 2 that are reflected from the top and bottom surfaces, respectively, of the air film. Ray 1 has no phase change when it reflects and ray 2 has a phase change when it reflects, so the reflections introduce a net phase difference. The path difference is 2 t and the wavelength in the film is Figure 26.57 Solve: (a) Since there is a phase difference from the reflections, the condition for constructive interference is The positions of first enhancement correspond to and is a constant, so For For (b) The positions of next enhancement correspond to and The values of x are 3 times what there are in part (a). violet: 3.45 mm; green: 4.74 mm; orange: 5.16 mm. (c) so 26.58. Set Up: Figure 26.57 for Problem 26.57 applies, but now the wedge is jelly and Ray 1 has a phase shift upon reflection and ray 2 has no phase change. As in Problem 26.57, the reflections introduce a net phase difference. 180° 180° l 5 l air n . t foil 5 9.57 3 10 2 4 cm 5 9.57 m m. tan u 5 t foil 11.0 cm , tan u 5 l 4 x 5 400.0 3 10 2 9 m 4 1 1.15 3 10 2 3 m 2 5 8.70 3 10 2 5 . x tan u 5 3 l 4 . 2 t 5 3 l 2 . m 5 1 x 2 5 1 1.15 mm 2 1 600 nm 400 nm 2 5 1.72 mm. l 2 5 600 nm 1 orange 2 , 1.58 mm. x 2 5 1 1.15 mm 2 1 550 nm 400 nm 2 5 l 2 5 550 nm 1 green 2 , x 2 5 x 1 1 l 2 l 1 2 . l 1 5 400.0 nm. x 1 5 1.15 mm, x 1 l 1 5 x 2 l 2 . u x tan u 5 l 4 . 2 t 5 l 2 . m 5 0 2 t 5 A m 1 1 2 B l . 180° t n 5 1.55 n 5 1.55 x u 1 2 l 5 l air . 180° 180° t 5 x tan u . l air l air 5 275 nm m 5 1, l air 5 2 tn m 5 275 nm m . 2 t 5 m l air n l air l air 5 183 nm. m 5 1: l air 5 550 nm. m 5 0: l air 5 2 tn m 1 1 2 5 275 nm m 1 1 2 . 2 t 5 A m 1 1 2 B l air n t 5 l 4 5 l air 4 5 550 nm 4 3 1.432 4 5 96.0 nm. t 5 A m 1 1 2 B l 2 2 t 5 A m 1 1 2 B l . l 5 l air n . 180° 1 n 5 1.62 2 1 n 5 1.432 2 180° 1 n 5 1.432 2 26-12 Chapter 26

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Solve: Since the reflections introduce a net phase difference, the condition for destructive interference is so The separation between adjacent dark fringes is and 26.59. Set Up: Consider reflection from either side of the film. (a) At the front of the film, light in air reflects off the film and there is a phase shift. At the back of the film, light in the film reflects off the cornea and there is no phase shift. The reflections produce a net phase difference and the condition for constructive interference is where Minimum thickness is for and is given by (b) For For and all other values are smaller. No other visible wavelengths are reinforced. The condition for destructive interference is For and all other values are shorter. There are no visible wavelengths for which there is destructive interference.
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