---
title: "Newton's rings, refractive index of liquid"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:newton-rings-liquid-refractive-index"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/newton-rings-liquid-refractive-index
site: BCT Engineering
---

# Newton's rings, refractive index of liquid

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2070 Chaitra, 5 marks; 2068 Chaitra, 5 marks (asked 2 times)
- Video: [KTU Engineering Physics - Newtons Rings - Refractive Index of liquid](https://www.youtube.com/watch?v=6tS_2fgWkPM) — Neethu Balachandran
- 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-liquid-refractive-index
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

What are Newton's rings? How can you use these rings to determine the refractive index of a given liquid?

## Short answer

Newton's rings are circular interference fringes in a thin air film between a lens and plate. To find a liquid's refractive index μ, measure the diameter D of a dark ring in air, then with liquid D'. Using dark ring condition D² = 4nλR/μ, μ = (D_air/D_liquid)². More accurately, use the difference of squares of two rings: μ = (ΔD²_air)/(ΔD²_liquid).

## Answer

**Newton's rings** are circular, concentric interference fringes formed in the thin air film between a plano-convex lens of large radius of curvature *R* placed on a plane glass plate, when viewed in monochromatic light. They are fringes of equal thickness.

#### Theory
```text
     source -> glass plate at 45° -> lens
              ____|____
             /         \       air film
 ____________\_________/_____ plate
              t at radius r
```
1. For a ring of radius *r*, film thickness t = r²/(2R) (since r² = 2Rt − t² ≈ 2Rt).
2. Reflection at the lower surface adds λ/2, so for a dark ring 2μt = nλ.
3. Hence r² = nλR/μ and diameter **D² = 4nλR/μ**.

For air (μ = 1): D_air² = 4nλR. The centre is dark.

#### Determination of refractive index of a liquid
1. Set up the apparatus, view with a travelling microscope, and measure the diameter D_n of the *n*th dark ring in **air**.
2. Fill the space between the lens and plate with the liquid (using a container), and measure the diameter D'_n of the same *n*th ring.
3. From D_n² = 4nλR and D'_n² = 4nλR/μ:

```text
μ = D_n² / D'_n²
```

4. To reduce error, use a ring pair: D²_{n+p} − D²_n = 4pλR (air) and = 4pλR/μ (liquid), so

**μ = (D²_{n+p} − D²_n)_air / (D²_{n+p} − D²_n)_liquid**

This removes errors from the exact centre and the ring order. The liquid ring diameters are smaller than in air.

## Check yourself

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

1. Newton's rings are fringes of:
   A. equal inclination
   B. equal thickness
   C. equal wavelength
   D. equal polarization
   Answer: B. They form in an air film whose thickness depends on the radius, so each ring is a locus of equal thickness. (Key point: Nature of Newton's rings)

2. How does the dark ring diameter formula change with a liquid of index μ?
   A. D^2 = 4nλRμ
   B. D^2 = 4nλR/μ
   C. D^2 = 4nλR/μ^2
   D. It does not change
   Answer: B. The condition 2μt = nλ gives r^2 = nλR/μ, so D^2 = 4nλR/μ. (Key point: Effect of liquid)

3. Which procedure gives μ of the liquid?
   A. Measure the nth ring diameter in air and in the liquid, μ = D_air^2 / D_liquid^2
   B. Measure only the centre spot
   C. Measure D_liquid / D_air
   D. Measure the ring colour
   Answer: A. Since D_n^2 = 4nλR and D'_n^2 = 4nλR/μ, the ratio of squares gives μ. (Key point: Method)

4. Why use D^2(n+p) − D^2(n) for the determination?
   A. It removes errors from the exact centre and ring order
   B. It doubles the wavelength
   C. It makes the liquid index cancel
   D. It removes the need for a microscope
   Answer: A. The difference equals 4pλR (air) or 4pλR/μ (liquid), independent of the centre position and the order n. (Key point: Ring-pair method)

## Related in Engineering Physics

- [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
- [Newton's ring diameter change with liquid](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-ring-diameter-change-liquid) — Engineering Physics, chapter 4
- [Newton's rings vs Haidinger fringes](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-haidinger-bright-rings-transmitted) — 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×
- [Newton's rings between two curved surfaces](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-rings-two-curved-surfaces) — 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

## Related in other subjects

- [Refractive index, critical frequency and MUF](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Arefractive-index-fc-muf) — Propagation and Antenna, chapter 4