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Wedge film fringes, thickness difference

Short answer

Interference fringes in a thin wedge film arise from path difference 2μt plus a λ/2 phase change at the lower surface, giving straight, equally spaced fringes with a dark edge. The thickness difference between two points is Δt = Nλ/(2μ), where N is the number of fringes between them.

Explain how interference fringes are formed by a thin wedge shaped film, when examined by normally reflected light. How will you estimate the difference of film thickness between two points?

Formation of fringes

A thin wedge is an air (or other transparent) film between two plane glass plates inclined at a very small angle θ and touching at the edge. When it is lit by monochromatic light and viewed by normally reflected light, the ray reflected from the top surface of the film and the ray reflected from the bottom surface are coherent (they come from the same incident ray) and interfere.

      incident / reflected (normal)
        | |
   _____|_|______________  upper plate
   ~~~~~~~~~~~~~~~~~~~~~~~  film, thickness t = xθ
  edge  ----->  x  ------>  lower plate
   0                 t
  1. At a distance x from the edge the thickness is t = xθ.
  2. Path difference = 2μt cos r. For normal incidence, cos r = 1, so 2μt.
  3. Reflection at the lower surface (rarer to denser) gives an extra phase change of π, i.e. an extra λ/2.
  4. Bright fringe: 2μt = (2n + 1)λ/2, n = 0, 1, 2, ... Dark fringe: 2μt = nλ, n = 0, 1, 2, ...
  5. At the edge t = 0, so the edge is dark.
  6. Fringes are straight, parallel to the edge, and equally spaced.
  7. Fringe width: for two successive dark fringes, 2μ(x₂ − x₁)θ = λ, so β = λ / (2μθ).

Estimating the thickness difference between two points

Moving from one dark (or bright) fringe to the next changes the thickness by λ/2μ. If N fringes lie between the two points,

Δt = N λ / (2μ)

If the distance L between the points is known and the fringe width β is measured, N = L / β, so Δt = Lλ / (2μβ) = Lθ. For air, μ = 1 and Δt = Nλ/2.

Example: with λ = 589 nm and 10 fringes between two points in an air wedge, Δt = 10 × 589 nm / 2 ≈ 2.95 µm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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