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442b - —.¢.— The answer is The interference pattern of...

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Unformatted text preview: —.¢.— The answer is @ The interference pattern of light and dark fringes is the result of interference between recombined rays reflectng off the glass-air and airvglass interfaces respec- tively at the top and bottom of the air wedge. Whether interference is constructive or destmctive depends on the path difference between the rays and the fact that the air~glass reflection is hard [the index of refraction of air is less than glass]: so reflection occurs with 130” phase change at that boundary, while the glass-air reflected rays do not undergo phase change. It follows, then, that the answer could be either 'a' or 'c’ only. because. in those choices, the area near where the two pieces of glass are in contact appears dark, where the air wedge is thinnest. This corresponds to the rays from the air-glass reflection directly superposed with the rays reflecting at the glass‘air boundary. Destructive interference is the result. Finally, the answer must he 'a’ instead of 'c'. The maximum path length difference for the reflected rays is twice the filament thickness. so. about 5 microns. or ten times the wavelength. In choice 'a' traveling up the wedge, the pattern cycles through approximately ten fringesI the path length difference for the reflected rays going back and forth between constructive and destructive interference. THERE IS A FORMULA. HEATiHG THIEHHEES OF AN AIR WEDGE, THE NUMBER DF DARK BANUS, AND THE WAVELENG‘H'I DF lLLUMIHATIDH ir=mLfIL BUT WINE IT THROUGH 15 ll'Jlf'l'El'il EE'ITER THAN MEMDRIIIHE A FDRWJLh. c zoos .J Int-HEEL ...
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