Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 36 | xX y1a x 10 3 m x 10 4 m 7 y1 X --------------- --------------- x 10 7 m. a x m yt  0 600 m 546 x 1 m x w m. a y1 x 10 3m The angle to the first dark fringe is simply n X 633 x 10 9m 0 arctan I I arctan I o 1 I . 2xX 2 m x 10 m _ D 2y1 ----------------- -------------- x 10 3m. a x 104m The angle to the first minimum is 0 arcsin arcsin I cm I . aJ I cm J So the distance from the central maximum to the first minimum is just y1 x tan 0 cm tan cm. a According to Eq. mX rm i mX X sin 0 --- sin 1 -- a a a . Thus a X 580 nm x 10 4 mm. b According to Eq. I sin a sin 0 2 sin m sin . The diffraction minima are located by sin 0 m 2 a m 1 2 . X v f 344 m s 1250 Hz m a m m 1 0 m 2 0 m 3 0 no solution for larger m a E Emax sin kx - at 2 k -- 1 1 2n k 2n x 107m-1 x 10- 7m c x 108m s 14 f 1 c f -------- x 10 Hz 1 x 10-7m b a sin 0 1 1 a ------ sin 0 x 10-7m sin x 10-6m c a sin 0 m1 m 1 2 3 . sin 02 2 D 2 02 74 sin Q 1 a locates the first minimum y x tan Q tan Q y x cm cm and 0 a 1 sin Q 620 x 10 9 m sin m X x1 x1 m 10 7m -4 a y1 a 2--------- 3-----------L x 10 4m. 1 a y1 x 10-3m x1 m x 10-5m 2 b a 2------- x 10 2m cm. y1 x 10-3m . x1 m 10-10m . c a 2-------E------------L x 10 7m. y1 x 10 3m a y1 10 m x 10- 3m. So the width of the a x 10-4m brightest fringe is twice this distance to the first minimum m. b The next dark fringe is at y2 2 m 6-33 4x 0 mi m. a 10 m So the width of the first bright fringe on the side of the central maximum is the distance from y2 toy1 which is 10 3m. a 2Ki . . 2m y 2 r m z p sin 6 ------- - ------ y 1520 m 1 y. I lx xl0m m v AA a-3 fi 1520
