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

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Problem Solutions 24.1 ( ) ( ) ( ) 1 9 2 3 1 632.8 10 m 5.00 m 1.58 10 m 1.58 cm 0.200 10 m bright m m L L L y y y m m d d d λ λ λ + Δ = = + = × = = × × = 24.2 (a) For a bright fringe of order m , the path difference is m δ λ = , where At the location of the third order bright fringe, and 0,1,2, m = 3 m = ( ) 3 3 3 δ λ = = 589 nm 1.77 10 nm 1.77 m μ = × = (b) For a dark fringe, the path difference is 1 2 m δ λ = + , where m At the third dark fringe, 0,1,2, = 2 m = and ( ) 3 1 5 2 589 1.47 10 nm 1.47 m 2 2 δ λ μ = + = = × = nm 24.7 Note that, with the conditions given, the small angle approximation does not work well . That is, sin , tan , and θ θ θ are significantly different. The approach to be used is outlined below. (a) At the 2 m = maximum, sin 2 d δ θ λ = = , or 2 2 L y sin 2 2 y d d λ θ = = + or ( ) ( ) ( ) 2 2 400 m 300 m 55.7 m 2 1000 λ = = m 400 m +

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(b) The next minimum encountered is the 2 m = minimum; and at that point, 1 5 sin 2 2 d m δ θ λ = = + = λ or ( ) ( ) 1 1 5 55.7 m 5 sin sin 27.7 = 2 2
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ch24-sol - Problem Solutions 24.1 ybright = ym 1 ym = = 8...

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