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In a Young's double-slit experiment with light of wavelength 5000 angstrom, a transparent film of thickness 2.5 x 10⁻³ cm is introduced in front of one slit. The fringe pattern shifts by an amount equal to 20 fringe widths. What is the refractive index of the film material?

1. 1.25
2. 1.356
3. 1.4
4. 1.5

Correct answer: 1.4

Solution

Introducing a film of thickness t and refractive index mu shifts the pattern by N = (mu - 1)*t/lambda fringes. Here N = 20, t = 2.5e-3 cm = 2.5e-5 m, lambda = 5000 angstrom = 5e-7 m. So (mu - 1) = N*lambda/t = 20 * 5e-7 / 2.5e-5 = 1e-5 / 2.5e-5 = 0.4, giving mu = 1.4.

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