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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 :

mstepffer

Answer:

The thickness of the film is [tex]T=2.2 \mu m[/tex]

Explanation:

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

[tex]n=\frac{m \lambda}{2T}[/tex]

where [tex]m[/tex] is the number of fringes, [tex]\lambda[/tex] is the wavelength of the incident light, and [tex]T[/tex] is the thickness of the film.

From this relation, we have that

[tex]T=\frac{m\lambda}{2n}=\frac{9}{2}*\frac{693nm}{1.4}=2227.5nm[/tex]

wich gives us that

[tex]T=2.2\mu m[/tex]

is the thickness of the film.

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