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Unformatted text preview: The ratio of the minimum distance along the former line for destructive interference to that of constructive interference is a) &lt; 1 b) 1 c) &gt; 1 Note PAB form a right triangle. BP,AB are the sides and AP the hypotenuse. So 1 AP &gt; BP,AP = q 4 + ( BP ) 2 . For the minimum constructive interference value of BP,APBP = , while for destructive interference APBP = / 2. Thus for constructive interference BP/AP is less than for destructive interference (just draw the right triangles). Thus the ratio you want is &gt; 1, (c). Below is the full calculation. AP = q 2 . 00 2 + ( BP ) 2 . For destructive interfence the minimum BP is obtained from, q 2 . 00 2 + ( BP ) 2BP = / 2. Thus, 2 . 00 2 + ( BP ) 2 = ( BP ) 2 + BP + 2 / 4 or BP = 4 // 4 = 1 . 98 m. For constructive interference, q 2 . 00 2 + ( BP ) 2BP = , so BP = 2 // 2 = 0 . 363 m and the ratio is &gt; 1, (c). 2...
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This note was uploaded on 12/24/2011 for the course PHYS 2306 taught by Professor Ykim during the Fall '06 term at Virginia Tech.
 Fall '06
 YKim

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