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?

Step-by-step solution

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 × 420 = n - 1 630 ∴         n 630 - 420 = 630 ⇒       n 210 = 630 ⇒       n = 630 210 ⇒ n = 3 This is the