A thin film of soap solution (n =1.4) lies on the top of a glass plate (n = 1.5). When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 400 and 630 nm. The minimum thickness of the soap solution is

(a) 420 nm

(b) 500 nm

(c) 450 nm

(d) 490 nm

Solution:

Answer: (c)

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.

For Maximum intensity: 2μt=nλ

For n, nλ1=(n−1)λ2

or n×420 = (n−1)630

or n = 3

Maximum order, where they coincide

so, 2×1.4×t = 3×420

or t = 450 nm

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