P18_012 - ∆ φ ≈ 2 π ∆ x = 2 πD sin θ = 4 π...

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12. It is useful to study Sample Problem 18-3 before working this problem. We label the two point sources 1 and 2 and assume they are on the x axis (a distance D =2 λ apart). When we refer to the circle of large radius, we are assuming that a line drawn from source 1 to a point on the circle and a line drawn to it from source 2 are approximately parallel (and thus both at angle θ measured from the y axis). In terms of the theory developed in § 18-4, we Fnd that the phase di±erence at P (on the large circle of radius R ) for the two waves emitted from 1 and 2 is
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Unformatted text preview: ∆ φ ≈ 2 π ∆ x λ = 2 πD sin θ λ = 4 π sin θ . (a) ²or maximum signal, we set ∆ φ = 2 mπ ( m = 0 , ± 1 , ± 2 ,... ) to obtain sin θ = m/ 2. Thus we get a total of 8 possible values of θ between 0 and 2 π , given by θ = 0 , sin − 1 (1 / 2) = 30 ◦ , sin − 1 (1) = 90 ◦ and (using symmetry properties of the sine function) 150 ◦ , 180 ◦ , 210 ◦ , 270 ◦ , and 330 ◦ . (b) Since there must be a mininum in between two successive maxima, the total number of minima is also eight....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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