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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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- Spring '11