ch24-sol - Problem Solutions 24.1 ybright = ym +1 ym = = 8...

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Problem Solutions 24.1 () 1 9 2 3 1 632. 81 0 m5 . 00 m 1.58 10 m1 . 58 cm 0.200 10 m bright m m LL L yy y m m dd d λ λλ + Δ=− = + = × == × × = 24.2 (a) For a bright fringe of order m , the path difference is m δ = , where At the location of the third order bright fringe, and 0,1,2, m =… 3 m = 3 33 δλ 5 8 9 nm 1.77 10 nm 1.77 m μ = × = (b) For a dark fringe, the path difference is 1 2 m ⎛⎞ =+ ⎜⎟ ⎝⎠ , where m At the third dark fringe, 2 m = and 3 15 2 589 1.47 10 nm 1.47 m 22 = = × = 24.7 Note that, with the conditions given, the small angle approximation does not work well . That is, si n, tan , and θ are significantly different. The approach to be used is outlined below. (a) At the 2 m = maximum, n2 d θλ , or Ly sin y λθ + or 2 2 400 m ±²³³³²´ µ³³²´ ¶³³²´ 300 m 55.7 m 2 1000 m4 0 0 m ⎡⎤ ⎢⎥ + ⎣⎦
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(b) The next minimum encountered is the 2 m = minimum; and at that point, 1 5 sin 2 2 dm δ θλ ⎛⎞ == + = ⎜⎟ ⎝⎠ λ or () 11 555.7 m 5 si ns i n 27.7 = 22 3 0 0 m d θ −− ° Then, ( ) 1000 mtan 27. 75 2 y 4 m = so the car must travel an additional 124 m 24.8
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This note was uploaded on 10/13/2010 for the course PHYSICS 102 taught by Professor Abdulrazzaq during the Fall '08 term at West Virginia State University.

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ch24-sol - Problem Solutions 24.1 ybright = ym +1 ym = = 8...

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