Question

A film of oil that has an index of refraction of 1.45 rests on an optically flat piece of glass of index of refraction n = 1.60. When illuminated with white light at normal incidence, light of wavelengths 642 nm and 428 nm is predominant in the reflected light.
Determine the thickness of the oil film.

Answer 1

Answer:

442.75862 nm

Explanation:

m = Order

n = Refractive index

$\lambda$ = Wavelength

We have the relation of thickness and wavelength given by

$2nt=m\lambda$

The consecutive spectral line is given by

$2nt=(m+1)\lambda$

So,

$\lambda=\dfrac{2nt}{m}\\\Rightarrow 642=\dfrac{2nt}{m}$

and

$428=\dfrac{2nt}{m+1}$

Dividing the wavelengths we get

$\dfrac{642}{428}=\dfrac{m+1}{m}\\\Rightarrow 1.5 m=m+1\\\Rightarrow m=\frac{1}{0.5}\\\Rightarrow m=2$

$t=\dfrac{m\lambda}{2n}\\\Rightarrow t=\dfrac{2\lambda}{2n}\\\Rightarrow t=\dfrac{642}{1.45}\\\Rightarrow t=442.75862\ nm$

The film thickness of oil is 442.75862 nm