---
title: "Newton's rings with two wavelengths"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:newton-rings-two-wavelengths-diameter"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/newton-rings-two-wavelengths-diameter
site: BCT Engineering
---

# Newton's rings with two wavelengths

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2073 Shrawan, 5 marks; 2067 Asar, 5 marks (asked 2 times)
- Video: [Newton's Rings  - Interference](https://www.youtube.com/watch?v=4pccfyupMq8) — PHYSICS Sujatha
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/newton-rings-two-wavelengths-diameter
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

In a Newton's rings arrangement a source emitting two wavelengths λ1 = 6×10⁻⁵ cm and λ2 = 5.9×10⁻⁵ cm is used. It is found that the nth dark ring due to λ1 coincides with the (n+1)th dark ring due to λ2. Find the diameter of the nth dark ring. (radius of curvature of the lens R = 90 cm)

## Short answer

From nλ₁=(n+1)λ₂ get n=59. Diameter D=√(4nλ₁R)=√(4×59×6e−5×90)=√(1.2744)≈1.13 cm.

## Answer

For dark rings in reflected light in Newton's rings, D² = 4nλR, where *n* is the ring order and *R* the radius of curvature.

**Given:** λ₁ = 6×10⁻⁵ cm, λ₂ = 5.9×10⁻⁵ cm, R = 90 cm. The *n*th dark ring of λ₁ coincides with the (n+1)th dark ring of λ₂.

1. Equal diameters: 4nλ₁R = 4(n+1)λ₂R, so nλ₁ = (n+1)λ₂
2. n(λ₁ − λ₂) = λ₂

```text
n = λ₂ / (λ₁ − λ₂) = 5.9×10⁻⁵ / (0.1×10⁻⁵) = 59
```

3. Diameter of the 59th dark ring (λ₁):

```text
D² = 4 n λ₁ R = 4 × 59 × 6×10⁻⁵ × 90
   = 1.2744 cm²
D  = 1.129 cm
```

(Check with λ₂, n = 60: 4 × 60 × 5.9×10⁻⁵ × 90 = 1.2744 cm². Matches.)

**Answer: diameter of the nth (59th) dark ring ≈ 1.13 cm.**

## Check yourself

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

1. Which formula gives the diameter of the nth dark ring in Newton's rings (air film, reflected light)?
   A. D^2 = 4nλR
   B. D^2 = 2nλR
   C. D = 4nλR
   D. D^2 = nλR/4
   Answer: A. For dark rings in reflected light, D^2 = 4nλR, with R the radius of curvature of the lens. (Key point: Dark ring diameter formula)

2. The nth dark ring of λ1 coincides with the (n+1)th dark ring of λ2. What condition follows?
   A. nλ2 = (n+1)λ1
   B. nλ1 = (n+1)λ2
   C. n + λ1 = (n+1) + λ2
   D. λ1 = λ2 + n
   Answer: B. Coinciding rings have equal diameters, so 4nλ1R = 4(n+1)λ2R, giving nλ1 = (n+1)λ2. (Key point: Coincidence condition)

3. With λ1 = 6×10^-5 cm and λ2 = 5.9×10^-5 cm, what is n?
   A. 60
   B. 59
   C. 6
   D. 590
   Answer: B. n = λ2/(λ1 − λ2) = 5.9×10^-5 / 0.1×10^-5 = 59. (Key point: Ring order)

4. What is the diameter of the 59th dark ring (R = 90 cm, λ1 = 6×10^-5 cm)?
   A. About 1.13 cm
   B. About 1.27 cm
   C. About 0.56 cm
   D. About 11.3 cm
   Answer: A. D^2 = 4×59×6×10^-5×90 = 1.2744 cm^2, so D ≈ 1.13 cm. (Key point: Final diameter)

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