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Unformatted text preview: PHYS1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 21. 36.06, 36.17, 36.28, 36.65; Q38.04, Q38.06 36.06. a) The minima of single slit diffraction are determined by m a = sin . Using the small angle approximation, we have m D y a m = . Here m y is the distance from the center of central maximum to the m th minimum. So we get ( 29 m 10 35 . ) m 40 . )( m 10 550 ( 4 4 so , 5 3 9 1 5 1 5 = = = y y D a y y D a a = 2.5 103 m = 2.5 mm b) m 10 5 . 2 m 10 550 3 9 1 = a 1 = 2.2 104 rad = 0.013 36.17. a) R d = =  122 122 550 10 50 10 9 3 . . . m m R = 1.34 104 rad = 7.69 103 deg = 27.7 arc seconds b) Let s be the distance between the headlights, and D the distance from the headlights to the eye. Then, 4 10 1.34 m .4 1 so , = = = R R s D D s D = 1.0 10 4 m = 10 km 36.28. a) The angular position of the first minimum of the diffraction pattern is given by,...
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This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Work, Diffraction

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