---
title: "Fringe measurement of flatness error"
subject: "Metrology"
chapter: 8
key: "semester_3/metrology:interference-fringes-flatness-numerical"
source: https://www.bctengineer.com/q/semester_3/metrology/interference-fringes-flatness-numerical
site: BCT Engineering
---

# Fringe measurement of flatness error

Metrology · Chapter 8: Interferometry

- Asked in: year not given, 6 marks
- Video: [Finding Flatness Error by using Interferometer](https://www.youtube.com/watch?v=tWOISJH-zjw) — Knowledge Capsules
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/metrology/interference-fringes-flatness-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/metrology/chapter-8

## Question

A surface 40 mm long is tested with an optical flat in sodium light ($\lambda = 0.5893\ \mu$m). Eight equally spaced straight fringes are seen along the 40 mm length. (a) Find the height of the air gap at the far end above the touching end, and the wedge angle. (b) On a second surface the fringes are curved, with maximum deviation equal to 0.4 of the fringe spacing. Find the flatness error. (c) A gauge block is wrung on a flat platen and the same flat is placed over both. The fringes over the block are displaced by 2.5 fringes relative to those over the platen. Find the difference between the block length and the height of the reference.

## Short answer

Each fringe represents λ/2 = 0.2947 µm. (a) Air gap height = 8×0.2947 = 2.357 µm; wedge angle = 2.357×10⁻³/40 = 5.9×10⁻⁵ rad. (b) Flatness error = 0.4×0.2947 = 0.118 µm. (c) Block length difference = 2.5×0.2947 = 0.737 µm.

## Answer

Each fringe represents a height change of $\lambda/2$:

$$
\frac{\lambda}{2} = \frac{0.5893}{2} = 0.2947\ \mu\text{m}
$$

#### (a) Wedge
$$
t = 8 \times 0.2947 = 2.357\ \mu\text{m}
$$

$$
\theta = \frac{t}{\text{length}} = \frac{2.357\times10^{-3}}{40} = 5.89\times10^{-5}\ \text{rad} \approx 12.2''
$$

The fringe spacing is $40/8 = 5$ mm. In general, spacing $= \lambda/(2\theta)$.

#### (b) Curved fringes
$$
\text{flatness error} = 0.4 \times 0.2947 = 0.118\ \mu\text{m}
$$

The surface is flat to within about $0.12\ \mu$m.

#### (c) Gauge block
Each fringe of displacement equals $\lambda/2$ of height:

$$
\Delta L = 2.5 \times 0.2947 = 0.737\ \mu\text{m}
$$

**Answer: (a) $t = 2.357\ \mu$m, $\theta = 5.9\times10^{-5}$ rad; (b) 0.118 $\mu$m; (c) 0.737 $\mu$m.**

## Check yourself

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

1. In sodium light (lambda = 0.5893 micrometre), one fringe corresponds to a height of:
   A. 0.5893 micrometre
   B. 0.2947 micrometre
   C. 1.1786 micrometre
   D. 0.1473 micrometre
   Answer: B. Each fringe is lambda/2 = 0.2947 micrometre. (Key point: Height per fringe)

2. Eight fringes across the 40 mm length give an air gap at the far end of:
   A. 2.357 micrometre
   B. 1.18 micrometre
   C. 0.295 micrometre
   D. 4.71 micrometre
   Answer: A. t = 8 x 0.2947 = 2.357 micrometre; angle = t/length = 5.89 x 10^-5 rad. (Key point: Wedge height)

3. Fringe curvature equal to 0.4 of the spacing gives a flatness error of:
   A. 0.4 micrometre
   B. 0.118 micrometre
   C. 0.295 micrometre
   D. 1.18 micrometre
   Answer: B. Error = 0.4 x lambda/2 = 0.4 x 0.2947 = 0.118 micrometre. (Key point: Curved fringe error)

4. A fringe displacement of 2.5 fringes between block and platen means a length difference of:
   A. 2.5 micrometre
   B. 0.295 micrometre
   C. 1.47 micrometre
   D. 0.737 micrometre
   Answer: D. Each fringe is lambda/2, so 2.5 x 0.2947 = 0.737 micrometre. (Key point: Fringe displacement)

## Related in Metrology

- [Optical flat and flatness interferometer](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Aflatness-interferometer-and-optical-flat) — Metrology, chapter 8
- [Straight edges and testing straightness](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Astraight-edge-and-flatness-check) — Metrology, chapter 6

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