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# chap36 - 36.1 y1 x a x a x y1 y1a x(1.35 10 3 m(7.50 10 4 m...

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36.1: 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 36.2: 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 36.3: 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 ° = × × - - 36.4: 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 36.5: 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 ± = ° 36.6: 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 36.7: 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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36.8: 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 36.9: 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 36.10: 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 36.11: 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, 0.0109 m. 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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