L 5 l n 2 t 5 1752 3 10 2 6 m t 5 l 4 n t 5 l 2 n m 5

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l 5 l 0 n . 2 t 5 17.52 3 10 2 6 m. t 5 l / 4 n . t 5 l / 2 n . m 5 2, t 5 m 1 505 nm 2 3 2.62 4 2 5 1 96.4 nm 2 m . 2 t 5 m l . 180° l 5 505 nm 2.62 . 180° 1 n 5 1.62 2 1 n 5 2.62 2 180° 1 n 5 2.62 2 1 n 5 1.00 2 Interference and Diffraction 26-5
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26.22. Set Up: The fringes are produced by interference between light reflected from the top and 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) is therefore The geometry of the air wedge is sketched in Figure 26.22. At a distance x from the point of contact of the two plates, the thickness of the air wedge is t . Figure 26.22 Solve: so and The distance along the plate between adjacent fringes is and The angle of the wedge is 26.23. Set Up: For the waves reflected at the top surface of the grease wedge, the waves are in material with refractive index 1.60 and reflect off material with refractive index 1.50. Therefore, there is no reflection phase shift. For the waves reflected at the bottom surface of the grease wedge, the waves are in material with refractive index 1.50 and reflect off material with refractive index 1.40, and there is no reflection phase shift. The condition for destructive interference therefore is Analysis similar to that in Example 26.5 gives and Successive dark fringes correspond to successive integer values of m and are spaced 0.833 mm apart, the same as the spacing between adjacent bright fringes in Example 26.5. 26.24. Set Up: Both reflections occur for waves in the plastic substrate reflecting from the reflective coating, so they both have the same phase shift upon reflection and the condition for destructive interference (cancellation) is where t is the depth of the pit. The minimum pit depth is for Solve: 26.25. Set Up: The condition for a dark fringe is Solve: Reflect: There are a finite number of dark fringes because can’t be larger than 1.00. This establishes a maximum value for m . 26.26. Set Up: so the approximate expression is accurate. Solve: y 1 5 l R a 5 1 546 3 10 2 9 m 21 1.75 m 2 0.437 3 10 2 3 m 5 2.19 3 10 2 3 m 5 2.19 mm y m 5 R m l a l V a m l a 5 sin u u 5 6 57.5°. m 5 6 5: u 5 6 42.4°. m 5 6 4: u 5 6 30.4°. m 5 6 3: u 5 6 19.7°. m 5 6 2: u 5 6 9.71°. m 5 6 1: sin u 5 m 1 632.8 3 10 2 9 m 0.00375 3 10 2 3 m 2 5 m 1 0.1687 2 . 6 2, c . m 5 6 1, sin u 5 m l a , t 5 l 4 5 l 0 4 n 5 790 nm 4 1 1.8 2 5 110 nm 5 0.11 m m. 2 t 5 l 2 . m 5 0. l 5 l 0 n . 2 t 5 A m 1 1 2 B l , x 5 A m 1 1 2 B l l 2 h 5 A m 1 1 2 B 1 0.833 m 2 . 2 xh l 5 A m 1 1 2 B l 2 t 5 A m 1 1 2 B l . 4.09 3 10 2 4 rad 5 0.0234° tan u 5 l 2 D x 5 546 3 10 2 9 m 2 1 0.0667 3 10 2 2 m 2 5 4.09 3 10 2 4 . 1.00 15.0 fringes / cm 5 0.0667 cm. D x 5 15.0 fringes / cm 5 1.00 D x D x 5 x m 1 1 2 x m 5 l 2 tan u .
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