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Chapter 36

# Chapter 36 - 36.1 y1 x a x a y1 a x x y1(1.35 10 3 m(7.50...

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. : 361 m. 10 . 5 06 m . 2 00 m) 10 ( . 7 50 m) 10 ( . 1 35 λ λ 7 4 3 1 1 - - - × = × × = = = x a y a x y . : 362 m. 10 . 3 21 m 10 . 10 2 m) 10 ( . 5 46 m) ( . 0 600 λ λ 5 3 7 1 1 - - - × = × × = = y x a a x y . : 363 The angle to the first dark fringe is simply: = θ arctan a λ = arctan . . 0 15 m 10 . 0 24 m 10 633 3 9 ° = × × - - . : 364 m. 10 . 5 91 m 10 . 7 50 m) 10 ( . 6 33 m) ( . 2 3 50 λ 2 2 3 4 7 1 - - - × = × × = = = a x y D . : 365 The angle to the first minimum is θ = arcsin a λ = arcsin . . 48 6 cm . 12 00 cm . 9 00 ° = So the distance from the central maximum to the first minimum is just = = θ tan 1 x y cm. . 45 4 ) ( . 48 6 tan cm) ( . 40 0 ± = ° . : 366 a) According to Eq. . 36 2 a a m a m θ λ λ 1 ) 0 . 90 ( sin λ ) ( sin = = = ° = = Thus, mm. 10 . 5 80 nm 580 λ . 4 - × = = = a b) According to Eq. . 36 7 [ ] [ ] . 128 . 0 ) (sin ) (sin sin ) (sin ) (sin sin 2 4 4 2 0 = = = π π π π θ π θ π λ a λ a I I . : 367 The diffraction minima are located by . . . , 2 , 1 , sin ± ± = = m a λ m θ m . 1 00 m; . 0 2752 Hz) ( 1250 ) s m ( 344 λ = = = = a f v ; 6 . 55 , 3 ; 4 . 33 , 2 ; 0 . 16 , 1 ° ± = ± = ° ± = ± = ° ± = ± = θ m θ m m θ no solution for larger m

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. : 368 a) ) sin( max t ω kx E E - = Hz 10 . 5 73 m 10 . 5 24 s m 10 . 3 0 λ λ m 10 . 5 24 m 10 20 . 1 2 2 λ λ 2 14 7 8 7 1 7 × = × × = = = × = × = = = - - - c f c f k k π π π b) λ sin = θ a m 10 . 1 09 . 28 6 sin m 10 . 5 24 sin λ 6 7 - - × = ° × = = θ a c) . . ,. 3 , 2 , 1 ( λ sin = = m m a θ ) 74 2 2 sin 2 m 10 . 1 09 m 10 . 5 24 λ 2 6 7 ± = ± = ± = - - × × θ θ D . : 369 a θ λ sin = locates the first minimum m 920 . 0 ) 38 . 42 (sin m) 10 620 ( sin λ 38 . 42 and cm) ( . 40 0 cm) 5 . 36 ( tan , tan 9 μ θ a x y θ θ x y = ° × = = ° = = = = - θ . : 3610 a) m. 10 . 4 17 m 10 . 3 00 m) 10 ( . 5 00 m) ( . 2 50 λ λ 4 3 7 1 1 - - - × = × × = = = y x a a x y b) cm. . 4 2 m 10 . 4 17 m 10 . 3 00 m) 10 ( . 5 00 m) ( . 2 50 λ 2 3 5 1 = × = × × = = - - - y x a c) m. 10 . 4 17 m 10 . 3 00 m) 10 ( . 5 00 m) ( . 2 50 λ 7 3 10 1 - - - × = × × = = y x a . : 3611 a) m. 10 . 5 43 m 10 . 3 50 m) 10 ( . 6 33 m) ( . 3 00 λ 3 4 7 1 - - - × = × × = = a x y So the width of the brightest fringe is twice this distance to the first minimum, . m. 0 0109 b) The next dark fringe is at m . 0 0109 m 10 . 3 50 m) 10 . 6 33 ( m) ( . 2 3 00 λ 2 4 7 2 = × × = = - - a x y . So the width of the first bright fringe on the side of the central maximum is the distance from , y to 1 2 y which is m 10 43 . 5 3 - × .
36.12: . ) m 1520 ( m) (3.00 m) 10 20 . 6 ( m) 10 0 5 . 4 ( 2 λ 2 sin λ 2 1 7 4 y y π x y a a β - - - = × × = π θ π = a) . 760 . 0 2 ) m 10 00 . 1 ( ) m 1520 ( 2 : m 10 00 . 1 3 1 3 = × = × = - - - β y 0 2 0 2 0 822 . 0 760 . 0 ) 760 . 0 ( sin 2 ) 2 ( sin I I β β I I = = = b) . 28 . 2 2 ) m 10 00 . 3 ( ) m 1520 ( 2 : m 10 00 . 3 3 1 3 = × = × = - - - β y . 111 . 0 28 . 2 ) 28 . 2 ( sin 2 ) 2 ( sin 0 2 0 2 0 I I β β I I = = = c) . 80 . 3 2 ) m 10 00 . 5 ( ) m 1520 ( 2 : m 10 00 . 5 3 1 3 = × = × = - - - β y . 0259 . 0 80 . 3 ) 80 . 3 ( sin 2 ) 2 ( sin 0 2 0 2 0 I I β β I I = = = 36.13: a) m. 10 .75 6 m 10 2.40 m) 10 (5.40 m) (3.00 λ 3 4 7 1 - - - × = × × = = a x y b) . 2 λ λ 2 2 λ 2 sin λ 2 1 π ax x πa x y πa a β = = θ π = . m W 10 43 . 2 2 ) 2 ( sin ) m W 10 00 . 6 ( 2 ) 2 ( sin 2 6 2 2 6 2 0 - - × = × = = π π β β I I 36.14: a) . 0 0 sin 2 : 0 = = = o λ a π β θ b) At the second minimum from the center . 4 λ 2 λ 2 sin λ 2 π a πa θ πa β = = = c) 191 0 . 7 sin m 10 00 . 6 m) 10 50 . 1 ( 2 sin λ 2 7 4 = ° × × = = - - π θ πa β rad.

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Chapter 36 - 36.1 y1 x a x a y1 a x x y1(1.35 10 3 m(7.50...

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