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A soap film of refractive index 1.4 forms a thin layer on a glass plate of refractive index 1.5. For nearly normal incidence of visible light, two successive maxima in reflected light occur at wavelengths 420 nm and 630 nm. What is the least possible thickness of the soap layer?

1. 420 nm
2. 450 nm
3. 630 nm
4. 1260 nm

Correct answer: 450 nm

Solution

Both surface reflections undergo a pi shift, so maxima satisfy 2nt = m*lambda. Successive maxima: m*630 = (m+1)*420 gives m = 2, so 2nt = 2*630 = 1260 nm and t = 1260/(2*1.4) = 450 nm.

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