The answer is: the acetone film is 1078 nm thick.
Explanation:
We are in the case where
[tex] n_{air} \ \textless \ n_{acetone} \ \textless \ n_{glass} [/tex]
This means that the condition for constructive interference is
[tex]L_{1} = \frac{m \lambda_{c}}{2 n} [/tex]
And the condition for destructive interference is
[tex]L_{2} = (m + \frac{1}{2}) \frac{\lambda_{d}}{2 n} [/tex]
where
n = refractive index of acetone = 1.25
m = 0, 1, 2,...
[tex] \lambda_{c}[/tex] = wavelength of costructive interference = 539 nm
[tex] \lambda_{d}[/tex] = 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
