- 2074 Asoj · 5 marks
Wedge film fringes and fringe width
Short answer
A thin wedge-shaped film produces straight, equally spaced fringes parallel to the edge when viewed in normally reflected monochromatic light. At the edge, t=0 gives a dark fringe due to λ/2 phase change. Fringe width (distance between consecutive dark fringes) is β = λ/(2μθ); for an air wedge, β = λ/(2θ).
Explain how interference fringes are formed by a thin wedge shaped film examined by normally reflected light. Derive a relation for the fringe width on such a system of interference fringes.
A wedge-shaped film is a thin transparent film (such as an air film between two glass plates touching at one edge) whose thickness increases uniformly from zero at the edge. Fringes of equal thickness are formed, parallel to the edge.
Formation of fringes
incident (normal) reflected
| |
========================= glass
\ .-- angle θ
\_______________/
edge --> x ---> glass
- Light from a broad monochromatic source falls (almost) normally on the film of refractive index μ and wedge angle θ.
- Rays reflected from the top and bottom surfaces of the film interfere. At normal incidence, the path difference is 2μt (+ λ/2 for reflection at the denser surface).
- At the edge t = 0, so the path difference is λ/2 and the edge is a dark fringe.
- Dark: 2μt = nλ; Bright: 2μt = (2n+1)λ/2.
- Since t is constant along lines parallel to the edge, the fringes are straight, equally spaced bands parallel to the edge.
Fringe width
- At distance x from the edge, t = x tan θ ≈ xθ (θ small).
- For the nth dark fringe at x_n: 2μ x_n θ = nλ
- For the (n+1)th: 2μ x_{n+1} θ = (n+1)λ
- Subtract:
β = x_{n+1} − x_n = λ / (2μθ)
For an air wedge: β = λ/(2θ). If the wedge is formed by two plates separated by a wire of diameter d at a distance L from the edge, θ = d/L, so β = λL/(2d). The fringe width is constant, and it decreases as θ increases.
✓ Answer checked by Claude Opus 5.5 (high effort)
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