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06prob - Physics 9B-A Assignment#6 Cole[36Q10 Using...

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Physics 9B-A Assignment #6 Cole [36Q10] Using Rayleigh's criterion, the angle between two points that are at the limit of being resolved is θ λ = 1 22 . D , so the shorter wavelengths will give a smaller angle because the diffraction caused by the apperture of the telescope will be smaller. The shorter wavelengths will provided more detail so ultraviolet is better. [36P19] The intensity for the interference looks like: -4 -2 0 2 4 0.2 0.4 0.6 0.8 1 Intensity for d = 4a & d/¬ = 8 Note the m = 4 fringe is missing. a) If there are only 5 fringes the m = 3 fringe will be missing. The first diffraction minimum for the single slit light (∫ = 2π) must be happening where the third two source interference maximum is occuring ƒ = 3(2π). Taking the ratio of the phases: φ π λ θ = 2 d sin and β π λ θ = 2 a sin õ φ β π π = = ( ) = d a 3 2 2 3 b) The central maximum has a range of 4π and contains two fringes on either side of the central fringe. The first diffraction max to the side of the central runs from ∫ = 2π - 4π, so it must contain 2 fringes. Look at the first drawing, where it contains 3 fringes just as there are 3 fringes to either side of the central fringe. Example a system where d/a = 5 -2 2 4 6 8 10 12 0.2 0.4 0.6 0.8 1 Set #6, 11/12/04 Page 1 of 5 UC Davis
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[ 36.24] The diffraction minima occur at ∫ = n(2π) and the interference max's at ƒ = m (2π). As in 38.17 take the ratio φ β π π = = ( ) = d a m n 2 2 3 õ m = 3 n where n = 1, 2, 3 .... [38.31] The maxima are given by φ π λ θ π = = ( ) 2 2 d m sin õ sin θ λ = m d where m d λ 1 . There are two values of m. θ = × × sin ( . ) .
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