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Unformatted text preview: nguyen (jmn727) homework 40 Turner (59070) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Consider the setup of a single slit experiment. The wavelength of the incident light is = 480 nm . The slit width and the distance between the slit and the screen is specified in the figure. y 1 6 . 1 m 550 m S 1 S 2 viewing screen Find the position y = y 1 of the first inten sity minimum. Use a small angle approxima tion sin = tan . Correct answer: 5 . 32364 mm. Explanation: Let : = 480 nm , L = 6 . 1 m , and a = 550 m . y 1 L a S 1 S 2 viewing screen a sin a bracketleftBig y L bracketrightBig For single slit diffraction, destructive in terference occurs when, a 2 sin = 2 , or sim ply when, a sin = . Thus, between the two end rays which correspond to the first minimum, the phase angle difference is 1 = 2 and the path length difference is 1 = . The small angle approximation gives us y 1 L = tan 1 1 sin 1 = 1 a , or y 1 = 1 a L = L a = sturt 5 . 32364 mm . 002 10.0 points Consider the setup of a single slit experiment. y 6 L a S 1 S 2 viewing screen 15 Find the height y 6 where the sixth mini mum occurs. Use a small angle approxima tion sin = tan . 1. y 6 = 15 2 L a 2. y 6 = 6 L a correct 3. y 6 = 11 2 L a 4. y 6 = 7 L a 5. y 6 = 5 L a 6. y 6 = 4 L a 7. y 6 = 17 2 L a 8. y 6 = 8 L a 9. y 6 = 9 2 L a 10. y 6 = 13 2 L a Explanation: Let : k 2 . nguyen (jmn727) homework 40 Turner (59070) 2 The first minimum is at = 2 , where = 2 = 2 , and = is the phase difference of the two rays for destructive interference....
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 Spring '08
 Turner
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