A thin film of soap solution μ s = 1.4 lies on the top of a glass plate μ g = 1.5 . When visible light is incident almost normal to the plate, two wavelengths 420 nm and 630 nm are strongly reflected.

Question

A thin film of soap solution μ s = 1.4 lies on the top of a glass plate μ g = 1.5 . When visible light is incident almost normal to the plate, two wavelengths 420 nm and 630 nm are strongly reflected. What is the minimum thickness (in nm ) of the soap film?

TestHub · Accepted Answer

For reflection at the air - soap solution interface, the phase difference is $π$

For reflection at the interface of soap solution to glass also, there will be a phase difference of $π$

$∴$ The condition for the maximum intensity

$= 2 μ t = n λ$

For n, $n λ_{1} = (n - 1) λ_{2}$

$n \times 420 = (n - 1) 630$

$∴$ $n (630 - 420) = 630$

$\Rightarrow$ $n (210) = 630$

$\Rightarrow$ $n = \frac{630}{210}$

$\Rightarrow$ $n = 3$

This is the maximum order, where they coincide

$2 \times 1.4 \times t = 3 \times 420$

$\Rightarrow$ $t = \frac{3 \times 420}{2 \times 1.40} = 450 nm$

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