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Unformatted text preview: UNIVERSITY OF CALIFORNIA College of Engineering Department of Materials Science & Engineering MSE 104 Spring Semester 2008 Professor R. Gronsky Instructions This is an OPEN-book (textbook only) exam. Calculators and drawing tools ARE permitted. Please PRINT your name in the box above and INITIAL all pages. Solutions MUST be written neatly and concisely in the spaces provided. Guidelines There are 5 equally-weighted questions. Average pace = 10 minutes per problem. Show ALL of your work for partial credit, as appropriate. Please read each question FULLY before commencing your answer. Solutions Midterm 02 Problem 1 (a) [10 points] Derive the specific formalism of Bragg's Law that describes the constructive interference condition resulting from monochromatic x-rays incidence on the family of crystalline lattice planes indicated in the sketch below. Label all appropriate variables directly on this sketch and express your answer in terms of those variables. State all assumptions. Bragg's Law models the phenomenon of diffraction as reflection from an array of semi-transparent mirrors. The fully labeled diagram above illustrates this model, which is described as follows. The incident wavefront of monochromatic radiation (this information is given) with wavelength identified in the figure is assumed to be fully coherent, that is, perfectly in phase, as it encounters the atomic planes from which it is diffracted. The wavefront remains coherent as long as it continues to travel the same path length, which, on the incident side, stops at the point of the dashed perpendicular line between the two beams incident upon the top and bottom planes. Because the process is modeled as a "reflection," the angle of incidence and angle of diffraction are identical, in this case labeled as . The wavefront on the diffraced side is shown to again be fully coherent, but with a phase shift that causes the lower beam to arrive exactly one wavelength behind the upper beam. This phase shift causes no loss of amplitude because both waves are shown to again be "in phase" along the diffracted wavefront, leading to constructive interference. Because the diffracted waves will also remain in phase as long as they travel the same path length, this condition persists on diffracted side back to the point of the perpendicular line between the two beams leaving the top and bottom planes in the diffracted direction. To complete the labeling, the distance between the two lattice planes is d . Now the derivation presents itself. The beam scattered from the lower plane travels a longer path than the reference beam scattered from the upper plane. This path difference is shown by the geometry of this symmetrical scattering condition to be equal to d sin on both the incident side and the diffraction side, for a total path difference of 2 d sin . The given figure shows that this path difference results in the waves arriving exactly one wavelength ( ) out of phase. So the specific formalism of Bragg's Law for this case is...
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- Spring '08