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# hw21 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 21...

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PHYS-1200 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 λ θ m a = sin . Using the small angle approximation, we have λ 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 . 0 ) m 40 . 0 )( m 10 550 ( 4 4 so , 5 3 9 1 5 1 5 - - × × = - = - = - y y D a y y D a λ λ λ a = 2.5 ×10 -3 m = 2.5 mm b) m 10 5 . 2 m 10 550 3 9 1 - - × × = a λ θ θ 1 = 2.2 ×10 -4 rad = 0.013° 36.17. a) θ λ R d = = × × - - 122 122 550 10 50 10 9 3 . . . m m θ R = 1.34 ×10 -4 rad = 7.69 ×10 -3 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 θ θ 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, d λ θ 22 . 1 = . That means that longer wavelengths spread out farther. Therefore, the outer ring is produced by the longest visible wavelength. It is red.

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hw21 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 21...

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