Answer:
The thickness of the film is 4.32 μm.
Explanation:
Given;
index of refraction of the thin film on one beam, n₂ = 1.5
number of bright fringes shift in the pattern produced by light, ΔN = 8
wavelength of the Michelson interferometer, λ = 540 nm
The thickness of the film will be calculated as;
[tex]\delta N = \frac{2L}{\lambda} (n_2 -n_1)[/tex]
where;
n₁ and n₂ are the index of refraction on the beam
L is the thickness of the film
[tex]\delta N = \frac{2L}{\lambda} (n_2 -n_1)\\\\L = \frac{\lambda}{2} (\frac{N}{n_2-n_1} )\\\\L = \frac{540*10^{-9}}{2} (\frac{8}{1.5-1} )\\\\L = 4.32*10^{-6} \ m\\\\L = 4.32 \mu m[/tex]
Therefore, the thickness of the film is 4.32 μm.
