chapter 36 solutions - Chapter 36 12 th Edition 36.1. I...

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Unformatted text preview: Chapter 36 12 th Edition 36.1. I DENTIFY : Use tan y x = to calculate the angular position of the first minimum. The minima are located by Eq.(36.2): sin , m a = 1, 2, m = K First minimum means 1 m = and 1 sin / a = and 1 sin . a = Use this equation to calculate . S ET U P : The central maximum is sketched in Figure 36.1. E XECUTE : 1 1 tan y x = 1 1 tan y x = = 3 3 1.35 10 m 0.675 10 2.00 m-- = 3 1 0.675 10 rad - = Figure 36.1 3 3 1 sin (0.750 10 m)sin(0.675 10 rad) 506 nm a -- = = O = E VALUATE : 1 is small so the approximation used to obtain Eq.(36.3) is valid and this equation could have been used. 36.2. I DENTIFY : The angle is small, so m m y x a = . S ET U P : 1 10.2 mm y = E XECUTE : 7 5 1 3 1 (0.600 m)(5.46 10 m) 3.21 10 m. 10.2 10 m x x y a a y --- = = = E VALUATE : The diffraction pattern is observed at a distance of 60.0 cm from the slit. 36.3. I DENTIFY : The dark fringes are located at angles that satisfy sin , 1, 2, .... m m a = = S ET U P : The largest value of sin is 1.00. E XECUTE : (a) Solve for m that corresponds to sin 1 = : 3 9 0.0666 10 m 113.8 585 10 m a m -- = = = . The largest value m can have is 113. 1 m = , 2 s , , 113 s gives 226 dark fringes. (b) For 113 m = , 9 3 585 10 m sin 113 0.9926 0.0666 10 m -- = = and 83.0 = . E VALUATE : When the slit width a is decreased, there are fewer dark fringes. When a < there are no dark fringes and the central maximum completely fills the screen. 36.4. I DENTIFY and S ET U P : / a is very small, so the approximate expression m m y R a = is accurate. The distance between the two dark fringes on either side of the central maximum is 1 2 y . E XECUTE : 9 3 1 3 (633 10 m)(3.50 m) 2.95 10 m 2.95 mm 0.750 10 m R y a --- = = = = . 1 2 5.90 mm y = . E VALUATE : When a is decreased, the width 1 2 y of the central maximum increases. 36.5. I DENTIFY : The minima are located by sin m a = S ET U P : 12.0 cm a = . 40.0 cm x = . E XECUTE : The angle to the first minimum is = arcsin a = arcsin 9.00 cm 48.6 . 12.00 cm = So the distance from the central maximum to the first minimum is just 1 tan y x = = (40.0 cm)tan(48.6 ) 45.4 cm. = E VALUATE : 2 / a is greater than 1, so only the 1 m = minimum is seen. 36.6. I DENTIFY : The angle that locates the first diffraction minimum on one side of the central maximum is given by sin a = . The time between crests is the period T. 1 f T = and v f = . S ET U P : The time between crests is the period, so 1.0 h T = . E XECUTE : (a) 1 1 1 1.0 h 1.0 h f T- = = = . 1 800 km/h 800 km 1.0 h v f - = = = . (b) Africa-Antarctica: 800 km sin 4500 km = and 10.2 = ....
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This note was uploaded on 04/14/2008 for the course PHYS 2303 taught by Professor Hoffman during the Fall '07 term at University of Texas at Dallas, Richardson.

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chapter 36 solutions - Chapter 36 12 th Edition 36.1. I...

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