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Unformatted text preview: Chapter 36 9 The condition for a minimum of intensity in a singleslit diffraction pattern is a sin = m , where a is the slit width, is the wavelength, and m is an integer. To find the angular position of the first minimum to one side of the central maximum, set m = 1: 1 = sin 1 a = sin 1 589 10 9 m 1 : 00 10 3 m = 5 : 89 10 4 rad : If D is the distance from the slit to the screen, the distance on the screen from the center of the pattern to the minimum is y 1 = D tan 1 = (3 : 00 m) tan(5 : 89 10 4 rad) = 1 : 767 10 3 m. To find the second minimum, set m = 2: 2 = sin 1 2(589 10 9 m) 1 : 00 10 3 m = 1 : 178 10 3 rad : The distance from the pattern center to the minimum is y 2 = D tan 2 = (3 : 00 m) tan(1 : 178 10 3 rad) = 3 : 534 10 3 m. The separation of the two minima is y = y 2 y 1 = 3 : 534 mm 1 : 767 mm = 1 : 77 mm. 17 (a) The intensity for a singleslit diffraction pattern is given by I = I m sin 2 2 ; where = ( a= ) sin , a is the slit width and is the wavelength. The angle is measured from the forward direction. You want I = I m = 2, so sin 2 = 1 2 2 : (b) Evaluate sin 2 and 2 = 2 for = 1 : 39 rad and compare the results. To be sure that 1 : 39 rad is closer to the correct value for than any other value with three significant digits, you should also try 1 : 385 rad and 1 : 395 rad. (c) Since = ( a= ) sin , = sin 1 a : Now = = 1 : 39 = = 0 : 442, so = sin 1 : 442 a : Chapter 36 231 The angular separation of the two points of half intensity, one on either side of the center of the diffraction pattern, is = 2 = 2 sin 1 : 442 a : (d) For a= = 1 : 0, = 2 sin 1 (0 : 442 = 1 : 0) = 0 : 916 rad = 52 : 5 : (e) For a= = 5 : 0, = 2 sin 1 (0 : 442 = 5 : 0) = 0 : 177 rad = 10 : 1 : (f) For a= = 10, = 2 sin 1 (0 : 442 = 10) = 0 : 0884 rad = 5 : 06 : 21 (a) Use the Rayleigh criteria. To resolve two point sources, the central maximum of the diffraction pattern of one must lie at or beyond the first minimum of the diffraction pattern of the other. This means the angular separation of the sources must be at least R = 1 : 22 =d , where is the wavelength and d is the diameter of the aperture. For the headlights of this problem, R = 1 : 22(550 10 9 m) 5 : 10 3 m = 1 : 3 10 4 rad : (b) If D is the distance from the headlights to the eye when the headlights are just resolvable and ` is the separation of the headlights, then ` = D tan R D R , where the small angle approximation tan R R was made. This is valid if R is measured in radians. Thus D = `= R = (1 : 4 m) = (1 : 34 10 4 rad) = 1 : 10 4 m = 10 km....
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 Fall '98
 Heckman
 Physics, Diffraction

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