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# ssm_ch36 - Chapter 36 9 The condition for a minimum of...

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Chapter36 9 The condition for a minimum of intensity in a single-slit diffraction pattern is a sin μ = , 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: μ 1 = sin ¡ 1 μ ¸ a = sin ¡ 1 μ 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 μ 1 = (3 : 00 m) tan(5 : 89 £ 10 ¡ 4 rad) = 1 : 767 £ 10 ¡ 3 m. To find the second minimum, set m = 2: μ 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 μ 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 single-slit diffraction pattern is given by I = I m sin 2 ® ® 2 ; where ® = ( ¼a=¸ ) sin μ , a is the slit width and ¸ is the wavelength. The angle μ 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 ® = ( ¼a=¸ ) sin μ , μ = sin ¡ 1 μ ®¸ ¼a : Now ®=¼ = 1 : 39 = 0 : 442, so μ = sin ¡ 1 μ 0 : 442 ¸ a : Chapter 36 231

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