---
title: "Wedge film fringes and fringe width"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:wedge-film-fringes-fringe-width"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/wedge-film-fringes-fringe-width
site: BCT Engineering
---

# Wedge film fringes and fringe width

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2074 Asoj, 5 marks
- Video: [How Interference fringes are produced in WEDGE SHAPED film with Animation & derivation, 👌, Physics](https://www.youtube.com/watch?v=EeJwn42EEqc) — SS Physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/wedge-film-fringes-fringe-width
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

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.

## 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).

## Answer

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
```text
       incident (normal)   reflected
           | |
   ========================= glass
     \                 .-- angle θ
      \_______________/
   edge --> x --->      glass
```
1. Light from a broad monochromatic source falls (almost) normally on the film of refractive index μ and wedge angle θ.
2. 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).
3. At the edge t = 0, so the path difference is λ/2 and the edge is a **dark** fringe.
4. **Dark**: 2μt = nλ; **Bright**: 2μt = (2n+1)λ/2.
5. Since t is constant along lines parallel to the edge, the fringes are straight, equally spaced bands parallel to the edge.

#### Fringe width
1. At distance *x* from the edge, t = x tan θ ≈ xθ (θ small).
2. For the *n*th dark fringe at x_n: 2μ x_n θ = nλ
3. For the (n+1)th: 2μ x_{n+1} θ = (n+1)λ
4. Subtract:

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What do the fringes of a wedge film look like?
   A. Concentric circles
   B. Straight, equally spaced bands parallel to the edge
   C. Random patches
   D. Radial lines
   Answer: B. Thickness is constant along lines parallel to the edge, giving straight equal-thickness fringes. (Key point: Fringe shape)

2. Why is the fringe at the edge (t = 0) dark?
   A. No light reaches it
   B. The λ/2 phase change in reflection gives path difference λ/2
   C. The film is thickest there
   D. The rays are not coherent
   Answer: B. At t = 0 the path difference is just the λ/2 from reflection, which is destructive. (Key point: Edge fringe)

3. How is the fringe width obtained?
   A. Add 2μ x_n θ = nλ and 2μ x_{n+1} θ = (n+1)λ
   B. Subtract the conditions for consecutive dark fringes
   C. Multiply the two conditions
   D. Divide by sin θ
   Answer: B. Subtracting gives 2μθ(x_{n+1} − x_n) = λ, so β = λ/(2μθ). (Key point: Fringe width derivation)

4. What happens to the fringe width as the wedge angle θ increases?
   A. It increases
   B. It decreases
   C. It stays the same
   D. It becomes zero immediately
   Answer: B. β = λ/(2μθ) is inversely proportional to θ. (Key point: Dependence on angle)

## Related in Engineering Physics

- [Wedge film fringes, thickness difference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-film-fringes-thickness-difference) — Engineering Physics, chapter 4
- [Thin film interference by reflected light](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Athin-film-reflected-interference-theory) — Engineering Physics, chapter 4, asked 2×
- [Fringe width of air wedge numerical](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-fringe-width-wire-numerical) — Engineering Physics, chapter 4
- [Wedge angle from fringe spacing](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-angle-fringe-spacing) — Engineering Physics, chapter 4
- [Reflected and transmitted interference complementary](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Athin-film-reflected-transmitted-complementary) — Engineering Physics, chapter 4, asked 2×