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anonymous
 5 years ago
A 250 nm thick film, of index n1 = 1.40, is on the surface of a glass plate, of index n2 = 1.55. A ray of
monochromatic light, of 500 nm wavelength, is incident normally upon the airfilm interface, and
undergoes reflections and transmissions. Consider points A, B, C, and D as being at a negligible
distance from their nearest interfaces, respectively. In Figure 35.2b, the phase difference in the wave
at C, with respect to the wave at A is closest to:
anonymous
 5 years ago
A 250 nm thick film, of index n1 = 1.40, is on the surface of a glass plate, of index n2 = 1.55. A ray of monochromatic light, of 500 nm wavelength, is incident normally upon the airfilm interface, and undergoes reflections and transmissions. Consider points A, B, C, and D as being at a negligible distance from their nearest interfaces, respectively. In Figure 35.2b, the phase difference in the wave at C, with respect to the wave at A is closest to:

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is the figure for the question: http://dl.dropbox.com/u/17638088/fig%2034.2b.PNG

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(already solved) I used\[\delta =(2\pi/\lambda)*d \sin \theta\] Also, point C is a reflection upon the glass, whose n is greater than the film, so the wave will undergo a phase change of pi, and thus pi must be subtracted from the answer.
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