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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.

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Answer:

442.75862 nm

Explanation:

m = Order

n = Refractive index

[tex]\lambda[/tex] = Wavelength

We have the relation of thickness and wavelength given by

[tex]2nt=m\lambda[/tex]

The consecutive spectral line is given by

[tex]2nt=(m+1)\lambda[/tex]

So,

[tex]\lambda=\dfrac{2nt}{m}\\\Rightarrow 642=\dfrac{2nt}{m}[/tex]

and

[tex]428=\dfrac{2nt}{m+1}[/tex]

Dividing the wavelengths we get

[tex]\dfrac{642}{428}=\dfrac{m+1}{m}\\\Rightarrow 1.5 m=m+1\\\Rightarrow m=\frac{1}{0.5}\\\Rightarrow m=2[/tex]

[tex]t=\dfrac{m\lambda}{2n}\\\Rightarrow t=\dfrac{2\lambda}{2n}\\\Rightarrow t=\dfrac{642}{1.45}\\\Rightarrow t=442.75862\ nm[/tex]

The film thickness of oil is 442.75862 nm

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