---
title: "Newton's rings vs Haidinger fringes"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:newton-haidinger-bright-rings-transmitted"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/newton-haidinger-bright-rings-transmitted
site: BCT Engineering
---

# Newton's rings vs Haidinger fringes

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2071 Chaitra, 5 marks
- Video: [Lecture 15 (3rd Semester) - Haidinger & Fizeau Fringes](https://www.youtube.com/watch?v=Iyw26UKYrIM) — Physics for UnderGraduates
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/newton-haidinger-bright-rings-transmitted
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

What are Newton's rings? How are they different from Haidinger fringes? Derive an expression for the diameter of bright rings in transmitted light. How can you obtain central fringe dark in this system?

## Short answer

Newton's rings are concentric fringes of equal thickness in an air film between a lens and plate. Unlike Haidinger fringes (equal inclination, uniform film), they are localized on the film. In transmitted light, bright rings satisfy D_n² = 4nλR. Central dark is obtained by filling with a liquid of intermediate refractive index to introduce a π phase shift.

## Answer

**Newton's rings** are concentric circular fringes formed in the air film between a plano-convex lens of large radius R and a glass plate. They are fringes of equal thickness, localized in the film.

#### Difference from Haidinger fringes
| Newton's rings | Haidinger fringes |
|---|---|
| Fringes of **equal thickness** | Fringes of **equal inclination** |
| Formed by a film of varying thickness (air film between lens and plate) | Formed by a plane parallel film of uniform thickness |
| Localized on the film surface | Localized at infinity (seen with a telescope / lens) |
| Fringes crowd as the order increases (D² ∝ n) | Fringes are more widely spaced at the centre |
| Central ring dark (reflected light) | Centre bright or dark depending on t |

#### Bright rings in transmitted light
1. At radius r, the air film thickness is t = r²/(2R).
2. In transmitted light the two internal reflections each give a phase change of π, which cancel, so the path difference = 2t (normal incidence, air).
3. **Bright** ring: 2t = nλ → r² = nλR.
4. With D = 2r:

```text
D_n² = 4 n λ R      → D_n = 2√(nλR)
```

(n = 0 is the central bright spot.) The **dark** rings in transmitted light satisfy 2t = (2n+1)λ/2, D² = 2(2n+1)λR.

#### Central fringe dark in transmitted light
In transmitted light, the centre is normally bright (complementary to the reflected pattern). To make it dark, an extra phase difference of π must be introduced between the two interfering beams. This is done by filling the space between the lens and the plate with a liquid whose refractive index lies **between** those of the lens and the plate (e.g. lens of crown glass, μ = 1.5; plate of flint glass, μ = 1.7; liquid μ ≈ 1.6). The twice-reflected ray then suffers a phase change of π at the plate (liquid to denser flint glass) but none at the lens (liquid to rarer crown glass), so a net extra λ/2 appears. The pattern is reversed: the centre becomes dark in transmitted light (and bright in reflected light).

## Check yourself

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

1. Haidinger fringes are:
   A. equal thickness fringes from a wedge
   B. equal inclination fringes from a plane parallel film, localized at infinity
   C. circular fringes from an air wedge between lens and plate
   D. formed only in transmission
   Answer: B. Newton's rings are equal thickness; Haidinger fringes come from a uniform film and depend on inclination. (Key point: Newton vs Haidinger)

2. In transmitted light (air film, normal incidence), the bright ring condition is:
   A. 2t = (2n+1)λ/2
   B. 2t = nλ
   C. t = nλ
   D. 2t = nλ/2
   Answer: B. No net phase change, so bright when 2t = nλ, giving D_n^2 = 4nλR. (Key point: Transmitted bright ring)

3. Normally the centre in transmitted light is:
   A. dark
   B. bright
   C. coloured
   D. absent
   Answer: B. The transmitted pattern is complementary to the reflected pattern, whose centre is dark. (Key point: Centre in transmitted light)

4. How can the transmitted centre be made dark?
   A. Use a liquid with index between those of lens and plate
   B. Use a thicker lens
   C. Remove the plate
   D. Use white light
   Answer: A. That liquid makes both reflections the same type of phase change, adding a relative π and reversing the pattern. (Key point: Making the centre dark)

## Related in Engineering Physics

- [Newton's rings: why circular, transmitted](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-rings-circular-transmitted-theory) — Engineering Physics, chapter 4
- [Number of bright Newton's rings](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-ring-bright-ring-count) — Engineering Physics, chapter 4
- [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
- [Third bright fringe separation, two wavelengths](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adouble-slit-two-wavelengths-separation) — Engineering Physics, chapter 4
- [Newton's rings with two wavelengths](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-rings-two-wavelengths-diameter) — Engineering Physics, chapter 4, asked 2×
- [Order of Newton's ring with double diameter](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-ring-double-diameter-order) — Engineering Physics, chapter 4

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