---
title: "Number of bright Newton's rings"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:newton-ring-bright-ring-count"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/newton-ring-bright-ring-count
site: BCT Engineering
---

# Number of bright Newton's rings

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2069 Chaitra, 5 marks
- Video: [Solved Problems on Newton's Rings](https://www.youtube.com/watch?v=2aUUdOoq1ls) — Physics Jessy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/newton-ring-bright-ring-count
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

In a Newton's ring experiment, the radius of curvature of the lens is 5cm and the lens diameter is 20mm. (a) How many bright rings are produced? Assume that λ=589nm (b) How many bright rings would be produced if the arrangement were immersed in water (μ=1.33)?

## Short answer

Bright ring radius r_n² = (2n−1)λR/(2μ). For air (μ=1), n ≈ 3396; for water (μ=1.33), n ≈ 4516.

## Answer

In reflected light, the *n*th bright Newton's ring has radius r_n² = (2n − 1)λR / (2μ), where μ is the refractive index of the film between lens and plate. The number of bright rings is the largest *n* for which r_n does not exceed the lens radius.

**Given:** R = 5 cm = 0.05 m, lens diameter 20 mm so r = 1 cm = 0.01 m, λ = 589 nm = 5.89 × 10⁻⁷ m.

#### (a) In air (μ = 1)
```text
r² = (2n − 1) λR / 2
2n − 1 = 2r² / (λR)
       = 2 × (0.01)² / (5.89×10⁻⁷ × 0.05)
       = 2×10⁻⁴ / 2.945×10⁻⁸
       = 6791.2
n      = (6791.2 + 1) / 2 = 3396.1
```
Only whole rings count, so n = 3396.

**Answer (a): about 3396 bright rings.**

#### (b) In water (μ = 1.33)
The wavelength in the film is λ/μ, so r² = (2n − 1)λR / (2μ).
```text
2n − 1 = 2μ r² / (λR)
       = 1.33 × 6791.2
       = 9032.3
n      = (9032.3 + 1) / 2 = 4516.6
```
**Answer (b): about 4516 bright rings.**

The rings crowd closer together in water, so more of them fit inside the same lens aperture (ratio 4516/3396 ≈ 1.33 = μ).

**Note:** the centre is dark in reflected light, and the ring count is only a theoretical limit; in practice the rings become too closely packed to resolve long before this number. (If R = 5 m, as in the common textbook version of this problem, the same method gives 34 bright rings in air and 45 in water.)

## Check yourself

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

1. The nth bright ring radius in air is given by:
   A. r^2 = nλR
   B. r^2 = (2n − 1)λR/2
   C. r^2 = 2nλR
   D. r = nλR
   Answer: B. Bright rings in reflected light satisfy 2t = (2n − 1)λ/2, so r^2 = (2n−1)λR/2. (Key point: Bright ring radius)

2. The number of bright rings equals:
   A. The largest n with r_n not exceeding the lens radius
   B. The smallest n with r_n equal to zero
   C. The ratio R / λ
   D. The lens diameter in mm
   Answer: A. Rings only exist within the lens aperture, r = 1 cm here. (Key point: Counting rings)

3. In air, about how many bright rings form?
   A. 339
   B. 3396
   C. 6791
   D. 33960
   Answer: B. 2n − 1 = 6791.2, so n ≈ 3396. (Key point: Air result)

4. Why are more rings seen in water?
   A. The wavelength in the film is λ/μ, so r^2 = (2n−1)λR/(2μ) and rings crowd closer
   B. Water enlarges the lens
   C. Water removes the phase change
   D. Water reduces R
   Answer: A. The count scales by μ = 1.33, giving about 4516 rings. (Key point: Effect of water)

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