--- title: "Refractive index of liquid film by interference" subject: "Engineering Physics" chapter: 4 key: "semester_1/engineering_physics:coherent-sources-film-refractive-index-method" source: https://www.bctengineer.com/q/semester_1/engineering_physics/coherent-sources-film-refractive-index-method site: BCT Engineering --- # Refractive index of liquid film by interference Engineering Physics · Chapter 4: Physical Optics - Asked in: 2067 Asoj, 5 marks - Video: [Newton's Rings | To Determine Refractive Index of a Liquid | Interference](https://www.youtube.com/watch?v=nXHN9kNEix8) — JIWAN THAPA PHYSICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/coherent-sources-film-refractive-index-method - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4 ## Question What are coherent sources? Describe a method for determining the refractive index of a transparent liquid film using the interference phenomenon. ## Short answer Coherent sources have the same frequency and constant phase difference. Using Newton's rings, measure diameters of dark rings with air (D²(n+p)−D²(n) = 4pλR) and with the liquid (D'²(n+p)−D'²(n) = 4pλR/μ). Then μ = (D²(n+p)−D²(n))/(D'²(n+p)−D'²(n)). ## Answer #### Coherent sources Two sources are coherent when they emit waves of the same frequency (wavelength) with a constant phase difference between them over time. Only coherent sources give a stable, observable interference pattern. In practice, two coherent sources are obtained from one source by division of wavefront (Young's slits, Fresnel biprism) or division of amplitude (thin films, Newton's rings). #### Determination of the refractive index of a liquid by Newton's rings ```text microscope | ----+---- glass plate G (45 deg) | ____|____ \ lens / <- plano-convex lens, radius R ---------- <- film: air, then liquid ========== <- glass plate ``` **Procedure** 1. Place the plano-convex lens on the plate and observe the rings in reflected monochromatic light of wavelength λ with a travelling microscope. 2. With **air** in the film, measure the diameters of the *n*th and (n + p)th dark rings. Then D²_(n+p) − D²_n = 4pλR (air, μ = 1). 3. Put a little of the liquid between the lens and the plate (the film is now liquid of index μ). The rings contract. Measure the diameters D′ of the same *n*th and (n + p)th rings: D′²_(n+p) − D′²_n = 4pλR / μ. 4. Divide the two results: **μ = (D²_(n+p) − D²_n) / (D′²_(n+p) − D′²_n)** **Why it works:** the optical path in the film is 2μt, with t = r²/2R, so the ring condition becomes D² = 4nλR/μ and the diameters shrink by a factor √μ. R and λ cancel in the ratio, so neither needs to be known. **Precautions:** the lens and plate must be clean, monochromatic light must be used, and the microscope should be moved in one direction only to avoid backlash errors. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What defines two sources as coherent? A. Equal intensity only B. Same frequency with a constant phase difference C. Different wavelengths with random phase D. Same direction of travel only Answer: B. Coherence needs same frequency and a constant phase difference over time. (Key point: Coherent sources) 2. How are two coherent sources obtained in practice? A. Using two separate lamps B. Splitting light from one source by division of wavefront or amplitude C. Using two lasers of different colours D. Heating the source Answer: B. Splitting a single source's light (Young's slits, thin films) preserves a constant phase relation. (Key point: Obtaining coherence) 3. In the Newton's rings method, what gives μ of the liquid? A. Ratio of D^2 differences, liquid over air B. Difference of the two diameters C. Product of the two diameters D. Ratio of D^2 differences, air over liquid Answer: D. D^2(n+p) − D^2(n) = 4pλR for air and 4pλR/μ for liquid, so μ = air value / liquid value. (Key point: Formula for μ) 4. Why do the rings contract when liquid fills the film? A. The lens radius R decreases B. Optical path becomes 2μt, so D^2 = 4nλR/μ and diameters shrink by √μ C. Wavelength increases D. The plate absorbs light Answer: B. Larger μ means a smaller thickness gives the same optical path, so rings are smaller. (Key point: Why rings contract) ## Related in Engineering Physics - [Newton's rings, refractive index of liquid](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-rings-liquid-refractive-index) — Engineering Physics, chapter 4, asked 2× - [Thin film interference by reflected light](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Athin-film-reflected-interference-theory) — Engineering Physics, chapter 4, asked 2× - [Coherent sources and Newton's ring law](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-rings-coherent-sources-diameter-law) — Engineering Physics, chapter 4 - [Intensity distribution in interference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ainterference-intensity-resultant) — Engineering Physics, chapter 4, asked 2× - [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 - [Acceptance angle of fiber in a liquid](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Afiber-acceptance-angle-liquid-index-change) — Engineering Physics, chapter 6 ## 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 - [External vs internal noise; interference sources; need for modulation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aexternal-internal-noise-interference-needs) — Communication Systems, chapter 1