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 air-film 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:

1. anonymous

This is the figure for the question: http://dl.dropbox.com/u/17638088/fig%2034.2b.PNG

2. anonymous

(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.