solutions_chapter26

Solve y 1 5 l r a 5 1 546 3 10 2 9 m 21 1.75 m 2

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Unformatted text preview: 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 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 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 B1 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 . x m 1 1 5 A m 1 3 2 B l 2 tan u . x m 5 A m 1 1 2 B l 2 tan u t m 5 A m 1 1 2 B l 2 . t 5 x tan u . tan u 5 t x t x u 2 t 5 A m 1 1 2 B l . 26-6 Chapter 26 26.27. Set Up: is very small, so the approximate expression is accurate. is the distance from the center of the screen to the first dark fringe on either side of the central maximum. and Solve: Reflect: The central bright fringe is twice as wide as the other bright fringes. 26.28. Set Up: is very small, so the approximate expression is accurate. Solve: 26.29. Set Up: The angle that locates the first diffraction minimum on one side of the central maximum is given by The time between crests is the period T . and Solve: (a) (b) Africa-Antartica : a n d Australia-Antartica : and 26.30. Set Up: is very small, so the approximate expression is accurate. The distance between the two dark fringes on either side of the central maximum is Solve: 26.31. Set Up: is very small, so the approximate expression is accurate. The distance between the two dark fringes on either side of the central maximum is Solve: 26.32. Set Up: The dark fringes are located at angles that satisfy The largest value of is 1.00. Solve: (a) Solve for m that corresponds to The largest value m can have is 113. gives 226 dark fringes. (b) For and 26.33. Set Up: The wavelength of light in the liquid is The dark fringes are located at The first dark fringes on either side of the central bright fringe correspond to and If a feature is located by angle its distance from the center of the central maximum is In this problem is not small and Eq.(26.10) cannot be used. u y 5 R tan u . u , m 5 2 1. m 5 1 1 6 2, c . m 5 6 1, sin u 5 m l a , l 5 l air n . u 5 6 83.0°. sin u 5 6 113 1 585 3 10 2 9 m 0.0666 3 10 2 3 m 2 5 6 0.9926 m 5 6 113, 6 113 6 2, c , m 5 6 1, m 5 a l 5 0.0666 3 10 2 3 m 585 3 10 2 9 m 5 113.8. sin u 5 1: sin u 6 2, c . sin u 5 m l a , m 5 6 1, u 2 y 1 5 4.44 mm. y 1 5 l R a 5 1 475 3 10 2 9 m 21 3.50 m 2 0.750 3 10 2 3 m 5 2.22 mm. l 5 633 nm 1.333 5 475 nm....
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Solve y 1 5 l R a 5 1 546 3 10 2 9 m 21 1.75 m 2 0.437 3 10...

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