Question

Problem 6: A thin layer of oil with index of refraction n. - 1.47 is floating above the water. The index of refraction of water is ry - 1.3. The index of refraction of air is na 1. A light with wavelength 1 - 575 nm goes in from the air to oil and water. Part (a) Express the wavelength of the light in the oil, 1o, in terms of land og Expression: Select from the variables below to write your expression. Note that all variables may not be required. 0, B2,,0, a, b, d, g, h, j, k, L, m, n, n, ny, P, S,

Answer 1

Part (a): The wavelength of light in the oil is

$\lambda _{o}=\frac{\lambda}{n_{o}} \\\\\therefore \lambda _{o}=\frac{575\ nm}{1.47}=391.1\ nm$

Part (b): The minimum path difference will be 2t_min. The condition for destructive interference is that the path difference should be odd multiple of half wavelength.

$2t_{min}=\frac{\lambda _{o}}{2}\ i.e.\ t_{min}=\frac{\lambda _{o}}{4}$

Part (c): The expression for t_min in terms of wavelength in air and refractive index of oil will be

$\ t_{min}=\frac{\lambda _{o}}{4}=\frac{\lambda}{4n_{o}}$

Part (d): The numerical value of t_min will be

$\ t_{min}=\frac{\lambda}{4n_{o}}=\frac{575}{4\times 1.47}=97.79\ nm$