Question

A thin layer of oil with index of refraction no = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na = 1. A light with wavelength λ = 375 nm goes in from the air to oil and water.(a) Express the wavelength of the light in the oil, λo, in terms of λand no.(b) Express the minimum thickness of the film that will result in destructive interference, tmin, in terms of λo.(c) Express tmin in terms of λ and no.(d) Solve for the numerical value of tmin in nm.

Answer 1

`\lambda_0=(\lambda)/(n_0)`

`t=(\lambda_0)/(2)`

`t=(\lambda)/(2n_0)`

`1.27551* 10^(-4)\ mm`

Step-by-step explanation:

`n_0` = Oil index of refraction = 1.47

`\lambda` = 375 nm

We have the relation of wavelength and refractive index as

`(\lambda)/(\lambda_0)=n_0\\\Rightarrow \lambda_0=(\lambda)/(n_0)`

Thickness relation is

`t=(\lambda_0)/(2)`

From the above equations we have

`t=((\lambda)/(n_0))/(2)\\\Rightarrow t=(\lambda)/(2n_0)`

Thickness will be

`t=(375* 10^(-9))/(2* 1.47)\\\Rightarrow t=1.27551* 10^(-7)\ m=1.27551* 10^(-4)\ mm`

The thickness is
`1.27551* 10^(-4)\ mm`