---
title: "Wedge angle from fringe spacing"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:wedge-angle-fringe-spacing"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/wedge-angle-fringe-spacing
site: BCT Engineering
---

# Wedge angle from fringe spacing

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2070 Chaitra, 5 marks
- Video: [Wedge Shaped Thin Film | Numerical Problems Part 1b | Interference | BSc & BTech Physics](https://www.youtube.com/watch?v=H7VFtexaj9Q) — JIWAN THAPA 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-angle-fringe-spacing
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Light of wavelength 6000 Å falls normally on a thin wedge shaped film of refractive index 1.4, forming fringes that are 2 mm apart. Find the angle of the wedge.

## Short answer

Using fringe width β = λ/(2μθ), θ = λ/(2μβ) = 6×10⁻⁷/(2×1.4×2×10⁻³) ≈ 1.07×10⁻⁴ rad (about 22 arc-seconds).

## Answer

For a wedge film of index μ viewed by normally reflected light, the fringe width is β = λ / (2μθ).

**Given:** λ = 6000 Å = 6×10⁻⁷ m, μ = 1.4, β = 2 mm = 2×10⁻³ m.

1. Rearrange: θ = λ / (2μβ)
2. Substitute:

```text
θ = 6×10⁻⁷ / (2 × 1.4 × 2×10⁻³)
  = 6×10⁻⁷ / 5.6×10⁻³
  = 1.07×10⁻⁴ rad
```

3. In degrees/seconds: θ = 1.07×10⁻⁴ × (180/π) = 6.1×10⁻³ degree ≈ 22 seconds of arc.

**Answer: wedge angle ≈ 1.07 × 10⁻⁴ rad (about 22 arc-seconds).**

## Check yourself

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

1. For a wedge film of index μ, fringe width is:
   A. β = λ/(2μθ)
   B. β = μλ/(2θ)
   C. β = 2μθ/λ
   D. β = λθ/(2μ)
   Answer: A. From 2μ x θ = nλ for consecutive fringes. (Key point: Fringe width formula)

2. Solving for the angle gives:
   A. θ = 2μβ/λ
   B. θ = λ/(2μβ)
   C. θ = λβ/(2μ)
   D. θ = μλ/β
   Answer: B. θ = λ/(2μβ). (Key point: Rearranged)

3. For λ = 6000 Å, μ = 1.4, β = 2 mm, θ is about:
   A. 1.07×10^-4 rad
   B. 1.07×10^-2 rad
   C. 4.3×10^-4 rad
   D. 1.07×10^-6 rad
   Answer: A. 6×10^-7 / (2×1.4×2×10^-3) = 1.07×10^-4 rad (about 22 arc-seconds). (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 film fringes, thickness difference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-film-fringes-thickness-difference) — Engineering Physics, chapter 4
- [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
- [Refractive index of liquid film by interference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acoherent-sources-film-refractive-index-method) — 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×

## Related in other subjects

- [Fringe measurement of flatness error](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Ainterference-fringes-flatness-numerical) — Metrology, chapter 8