Sol5 - UAHuntsville Spring, 2010 PH 113: Homework Solutions...

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UAHuntsville Spring, 2010 PH 113: Homework Solutions No. 5 Solution 28: Graphical solutions for x=tan(x) are given below, from which the solution can be read. Of course, but changing the resolution of the plot one can obtain as many significant figures as desired.
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Solution 29: 1) a d = 30.0mm 0.15mm = 200 so that there are 401 fringes: 200 on either side, plus the central one. 2) Notice that there are many interference fringes within the first diffraction oscillation. This means that the diffraction pattern is changing very slowly, and we expect that at the third interference fringe the intensity will be almost the same as at the center. Let us now calculate: At the third interference fringe the angle θ can be found from π a λ sin θ = n n = 3 so that for the diffraction function β = d sin = d a × a sin = 3 d a = 3 200 The ratio of intensities is just I 3 I 0 = sinc 2 3 sinc 2 0 = sinc 2 3 = sinc 2 3 200 = 0.99926 Solution 30: T he fringe spacing between the m fringe and the m + 1 fringe is Δ y = . Δ y can be obtained from Figure P22.34. The separation between the
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Sol5 - UAHuntsville Spring, 2010 PH 113: Homework Solutions...

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