Question

A thin film with index of refraction n = 1.40 is placed in one arm of a Michelson interferometer, perpendicular to the optical path.

If this causes a shift of 9.0 bright fringes of the pattern produced by the light of wavelength 693 nm, what is the film thickness?

Answer 1

The thickness of the film is
`T=2.2 \mu m`

Step-by-step explanation:

The relation between the refraction index, and a wavelength is given by

`n=(m \lambda)/(2T)`

where
`m` is the number of fringes,
`\lambda` is the wavelength of the incident light, and
`T` is the thickness of the film.

From this relation, we have that

`T=(m\lambda)/(2n)=(9)/(2)*(693nm)/(1.4)=2227.5nm`

wich gives us that

`T=2.2\mu m`

is the thickness of the film.