Athin film of acetone (n = 1.25) coats a thick glass plate (n = 1.50). white light is incident normal to the film. in the reflections, fully destructive interference occurs at 490 nm and fully constructive interference at 539 nm. calculate the thickness of the acetone film.

The answer is: the acetone film is 1078 nm thick.

Explanation:
We are in the case where


This means that the condition for constructive interference is


And the condition for destructive interference is


where
n = refractive index of acetone = 1.25
m = 0, 1, 2,...
= wavelength of costructive interference = 539 nm
= wavelength of distructive interference = 490 nm

Therefore, we need to find at which thickness L we have both constructive and destructive interference, which means to find for which m we get L₁ = L₂

For m = 0
L₁ = (0 · 539) / (2 · 1.25) = 0 nm
L₂ = (0 + 0.5) · [490 ÷ (2 · 1.25)] = 98 nm

For m = 1
L₁ = (1 · 539) / (2 · 1.25) = 215.6 nm
L₂ = (1 + 0.5) · [490 ÷ (2 · 1.25)] = 294 nm

For m = 2
L₁ = (2 · 539) / (2 · 1.25) = 431.2 nm
L₂ = (2 + 0.5) · [490 ÷ (2 · 1.25)] = 490 nm

For m = 3
L₁ = (3 · 539) / (2 · 1.25) = 646.8 nm
L₂ = (3 + 0.5) · [490 ÷ (2 · 1.25)] = 686 nm

For m = 4
L₁ = (4 · 539) / (2 · 1.25) = 862.4 nm
L₂ = (4 + 0.5) · [490 ÷ (2 · 1.25)] = 882 nm

For m = 5
L₁ = (5 · 539) / (2 · 1.25) = 1078 nm
L₂ = (5 + 0.5) · [490 ÷ (2 · 1.25)] = 1078 nm

Hence, the thickness of the acetone film is L = 1078 nm