Light of wavelength 630 nm is incident normally on a thin wedge-shaped film with index of refraction 1.50. There are ten bright and nine dark fringes over the length of the film. We have to find the amount through which the film thickness changes over this length.
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Solution: Click For PDF Version We assume that the wedge-shaped film with index of refraction 1.50 is surrounded by air. The light of wavelength 630 nm is incident normally on the wedge. Let the thickness of the wedge where a fringe is formed be d nm. The condition for constructive interference of waves reflected from the top of the film and the bottom of the film will be
The wavelength of light inside the film changes because the refractive index of the film is 1.50. We have
The first
fringe corresponds to
The tenth
fringe corresponds to
Therefore,
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The thickness of the film at the position of the first fringe will
therefore be
.
The thickness of the film at the position of the first fringe will
therefore be
