---
title: "Fringe width of air wedge numerical"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:wedge-fringe-width-wire-numerical"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/wedge-fringe-width-wire-numerical
site: BCT Engineering
---

# Fringe width of air wedge numerical

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2071 Chaitra, 5 marks
- Video: [Numerical problems based on wedge shaped film](https://www.youtube.com/watch?v=4q22-gBrMro) — Engineering Physics by Sanjiv
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/wedge-fringe-width-wire-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Two glass plates enclosing a wedge-shaped air film touching at one edge are separated by a wire of 0.03mm diameter at a distance of 15cm from the edge. Monochromatic light of wavelength 600 nm from a broad source falls normally on the film. Calculate the fringe width.

## Short answer

Fringe width β = λL/(2d) = (600×10⁻⁹ m × 0.15 m)/(2 × 3×10⁻⁵ m) = 1.5 mm.

## Answer

For an air wedge viewed by reflected light at normal incidence, the fringe width is β = λ/(2θ), where θ = d/L is the wedge angle (d = wire diameter, L = distance of the wire from the edge). So β = λL/(2d).

**Given:** d = 0.03 mm = 3×10⁻⁵ m, L = 15 cm = 0.15 m, λ = 600 nm = 6×10⁻⁷ m, μ = 1 (air).

1. Wedge angle:

```text
θ = d / L = 3×10⁻⁵ / 0.15 = 2×10⁻⁴ rad
```

2. Fringe width:

```text
β = λ / (2θ) = 6×10⁻⁷ / (2 × 2×10⁻⁴)
  = 6×10⁻⁷ / 4×10⁻⁴
  = 1.5×10⁻³ m
```

**Answer: fringe width β = 1.5 mm.**

## Check yourself

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

1. How is the wedge angle found from the wire?
   A. θ = L/d
   B. θ = d/L
   C. θ = d × L
   D. θ = λ/d
   Answer: B. The wire of diameter d at distance L from the edge gives θ = d/L. (Key point: Wedge angle)

2. For d = 0.03 mm and L = 15 cm, θ is:
   A. 2×10^-4 rad
   B. 2×10^-3 rad
   C. 5×10^3 rad
   D. 2×10^-6 rad
   Answer: A. 3×10^-5 / 0.15 = 2×10^-4 rad. (Key point: Numerical angle)

3. For an air wedge, fringe width is:
   A. β = 2λθ
   B. β = λ/(2θ)
   C. β = λθ/2
   D. β = λ/θ
   Answer: B. β = λ/(2μθ) with μ = 1. (Key point: Fringe width formula)

4. The fringe width for λ = 600 nm is:
   A. 1.5 mm
   B. 0.15 mm
   C. 15 mm
   D. 3 mm
   Answer: A. β = 6×10^-7 / (4×10^-4) = 1.5×10^-3 m. (Key point: Result)

## Related in Engineering Physics

- [Wedge film fringes and fringe width](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-film-fringes-fringe-width) — 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
- [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×
- [Double slit fringe separation numerical](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ayoung-double-slit-fringe-separation) — Engineering Physics, chapter 4

## Related in other subjects

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