 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2010/paper-1/section-a/2/solution

For the adjacent reflected minima of an air-film-air layer in thin-film interference, reflection at the front surface acquires a phase reversal, while reflection at the rear does not. Near normal incidence the reflected minima therefore obey

2 n t = mλ

for integral m. A single intermediate maximum means that the two stated minima are consecutive orders:

\frac{2 n t}{500 nm} - \frac{2 n t}{600 nm} = 1.

Thus

2 n t = 3000 nm

, with orders6 and5, and

t = \frac{3000}{2 ( 1.3 )} nm = 1.15 μ m .

The intermediate reflected maximum is at

λ = 3000/5.5 ≃ 545 nm

. This assumes negligible dispersion of the supplied refractive index and an air environment on both sides. ^[ref]

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