solutions_chapter26

# C now both rays have a phase change on reflection and

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(c) Now both rays have a phase change on reflection and the reflections don’t introduce any net phase shift. The expression for constructive interference in parts (a) and (b) now gives destructive interference and the expression in (a) and (b) for destructive interference now gives constructive interference. The only visible wavelength for which there will be destructive interference is 600 nm and there are no visible wavelengths for which there will be con- structive interference. Reflect: Changing the net phase shift due to the reflections can convert the interference for a particular thickness from constructive to destructive, and vice versa. 26.60. Set Up: The is related to the diameter of the lens opening by Rayleigh’s criterion for a circular aperture is Solve: (a) If the angular size of the object is set equal to then (b) and (c) Now instead of 50.0 mm. y 5 s u res 5 1 35.0 cm 21 1.59 3 10 2 4 2 5 0.0557 mm. u res 5 1.22 1 550 3 10 2 9 m 4.22 3 10 2 3 m 2 5 1.59 3 10 2 4 . D 5 135 mm 32 5 4.22 mm. f 5 135 mm 0 y r 0 5 0 m 0 y 5 1 0.167 21 0.151 mm 2 5 0.0252 mm. m 5 2 s r s 5 2 5.83 cm 35.0 cm 5 2 0.167. s r 5 sf s 2 f 5 1 35.0 cm 21 5.00 cm 2 35.0 cm 2 5.00 cm 5 5.83 cm. 1 s 1 1 s r 5 1 f 0.151 mm. y 5 s u res 5 1 35.0 cm 21 4.30 3 10 2 4 rad 2 5 0.0151 cm 5 u res , u 5 y s u res 5 1.22 1 550 3 10 2 9 m 1.56 3 10 2 3 m 2 5 4.30 3 10 2 4 rad. D 5 f f -number 5 50.0 mm 32 5 1.56 mm. u res 5 1.22 l D . f -number 5 f D . f -number 180° l air 5 299 nm m 5 1, l 5 2 tn m 5 299 nm m . 2 t 5 m l air n . l air 5 199 nm m 5 1, l air 5 598 nm. m 5 0, l air 5 2 nt m 1 1 2 5 2 1 1.45 21 103 nm 2 m 1 1 2 5 299 nm m 1 1 2 . t 5 l air 4 n 5 600 nm 4 1 1.45 2 5 103 nm. m 5 0, t 5 A m 1 1 2 B l air 2 n . l 5 l air n . 2 t 5 A m 1 1 2 B l , 180° 1 n 5 1.38 2 1 n 5 1.45 2 180° 1 n 5 1.45 2 1 n 5 1.00 2 n 5 525 3 10 2 9 m 2 1 0.633 3 10 2 3 m 21 1.875 3 10 2 4 2 5 2.21. tan u 5 0.015 3 10 2 3 m 8.00 3 10 2 2 m 5 1.875 3 10 2 4 . D x 5 6.33 mm 10 5 0.633 mm. n 5 l air 2 1 D x 2 tan u . D x 5 l air 2 n tan u D x x 5 m l air 2 n tan u . t 5 x tan u 2 t 5 m l air n . 180° Interference and Diffraction 26-13

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26.61. Set Up: At the point where no waves reach you the interference is destructive and the path difference is The frequency of the waves is The geometry of the situation is sketched in Figure 26.61. Figure 26.61 Solve: (a) and (b) Reflect: The nature of the interference at an observation point depends on the relation between the path difference for the two sources and the wavelength of the waves. Interference occurs for all types of waves, not just for light. 26.62. Set Up: The dark fringes are located by The first dark fringes are for is the distance from the center of the screen. From the center to one minimum is 2.61 cm. Solve: and 26.63. Set Up: For the waves reflected at the top surface of the oil film there is a half-cycle reflection phase shift. For the waves reflected at the bottom surface of the oil film there is no reflection phase shift. The condition for constructive interference is The condition for destructive interference is Wavelengths that are predominant in the transmitted light are those for which there is destructive interference in the reflected light. The range of visible wavelengths is approximately 400 nm to 700 nm. In the oil film, Solve: (a) The visible wavelength for which there is con- structive interference in the reflected light is 441 nm.

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