A thin film of acetone (n = 1.25) coats a thick glass plate (n = 1.50). White li

Question

A thin 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 540 nm and fully constructive interference at 608 nm. Calculate the thickness of the acetone film.
the tolerance is +/-2%
A thin 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 540 nm and fully constructive interference at 608 nm. Calculate the thickness of the acetone film.
the tolerance is +/-2%

the tolerance is +/-2%

Explanation / Answer

given refractive index of acetone, n = 1.25
refractive index of glass plate , n' = 1.5

thickness of film = t
any incident light will go phase shift of pi/2 in air acetone interface and acetone glass interface because refractive index of air < n and n < n'
hence

path difference = 2nt

so from the given data
destructive interference ( condition for constructive interference as double phase shift of pi/2 is there)
lamdba = 540 nm
2nt = m*lambda = 540m

constructive interference
lambda = 608 nm
2nt = (2n - 1)lambda/2 = 608n - 304
hence
540m = 608n - 304
for n = 4, m = 3.9407
n = 5, m = 5.066

so, using n = 4
t = (608*4 - 304)/2*1.25
t = 851.2 nm

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