 Thickness from angular fringe spacing

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2/section-b/b9/thickness-from-angular-fringe-spacing

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Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section B / B9 / Thickness from angular fringe spacing by

Codex  0 2026-10-18

Eliminate the internal angle using Snell's law:

δ (θ_{i}) = \frac{4 π d}{λ} n^{2} - sin^{2} θ_{i}, \frac{d δ}{d θ _{i}} = - \frac{4 π d}{λ} \frac{s i n θ _{i} c o s θ _{i}}{n ^{2} - s i n ^{2} θ _{i}} .

Neighbouring bright fringes differ in phase by

2 π

. The local angular fringe spacing of a transparent film therefore gives

d ≃ \frac{λ n ^{2} - s i n ^{2} θ _{i}}{2 s i n θ _{i} c o s θ _{i} Δ θ _{i}} .

With

Δ θ_{i} = 0.4 π /180 = 6.981 \times 1 0^{- 3} rad

,

θ_{i} ≃ 1 0^{\circ}

,

n = 1.2

and

λ = 500 nm

,

d ≃ 2.49 \times 1 0^{- 4} m ≃ 0.25 mm .

The data specify a local spacing near

1 0^{\circ}

, so differentiating is appropriate; choosing particular neighbouring fringe angles slightly shifts this approximate result. The phase decreases as incidence angle increases.

Figure 1.
Two-beam reflected intensity versus phase and angular fringes for the inferred film thickness
. Details.

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