--- title: "Newton's rings between two curved surfaces" subject: "Engineering Physics" chapter: 4 key: "semester_1/engineering_physics:newton-rings-two-curved-surfaces" source: https://www.bctengineer.com/q/semester_1/engineering_physics/newton-rings-two-curved-surfaces site: BCT Engineering --- # Newton's rings between two curved surfaces Engineering Physics · Chapter 4: Physical Optics - Asked in: 2071 Shrawan, 5 marks - Video: [Newton's Ring using two Curved surfaces || Optics - II ||](https://www.youtube.com/watch?v=T4DqliU7ZFw) — KC Physics Classes - 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-curved-surfaces - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4 ## Question Show that the diameters of the Newton's rings when two surfaces of radii R1 and R2 are placed in contact are related by the relation (1/R1) − (1/R2) = 4nλ/d²ₙ, where n is the integer number of the fringes. ## Short answer Film thickness t = (r²/2)(1/R1 − 1/R2). Dark ring condition 2t = nλ gives r²(1/R1 − 1/R2) = nλ. With r = d_n/2, 1/R1 − 1/R2 = 4nλ/d_n². ## Answer Two surfaces of radii R₁ and R₂ in contact (e.g. a convex lens of radius R₁ resting on a concave surface of larger radius R₂, curved the same way) enclose an air film whose thickness varies with the distance r from the point of contact. ```text convex R1 ____\__|__/____ ______\_|_/_____ concave R2 (R2 > R1) t at radius r ``` 1. For a spherical surface of radius R, the sag at distance r is r²/(2R). 2. Thickness of the film at radius r: t = r²/(2R₁) − r²/(2R₂) = (r²/2)(1/R₁ − 1/R₂) 3. In reflected light (normal incidence) a dark ring occurs when the effective path difference is (2n+1)λ/2. Including the λ/2 phase change at the lower surface: 2t + λ/2 = (2n+1)λ/2, so **2t = nλ**. 4. Substitute: r² (1/R₁ − 1/R₂) = nλ 5. Put r = d_n/2, r² = d_n²/4: ```text (d_n²/4)(1/R₁ − 1/R₂) = nλ 1/R₁ − 1/R₂ = 4nλ / d_n² ``` This is the required relation. If R₂ → ∞ (plane plate) it reduces to d_n² = 4nλR₁, the standard Newton's ring formula. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a convex surface R1 on a concave surface R2 (R2 > R1), the film thickness at radius r is: A. t = (r^2/2)(1/R1 + 1/R2) B. t = (r^2/2)(1/R1 − 1/R2) C. t = r^2 (R1 − R2) D. t = r^2/(2R1R2) Answer: B. t is the difference of the sags: r^2/2R1 − r^2/2R2. (Key point: Film thickness) 2. The sag of a spherical surface of radius R at distance r is approximately: A. r^2/(2R) B. r/(2R) C. 2R/r^2 D. r^2/R^2 Answer: A. From r^2 ≈ 2Rt. (Key point: Sag of a sphere) 3. With the λ/2 phase change, the dark ring condition becomes: A. 2t = (2n+1)λ/2 B. 2t = nλ C. t = nλ D. 2t = λ/2 Answer: B. 2t + λ/2 = (2n+1)λ/2 gives 2t = nλ. (Key point: Dark ring condition) 4. Substituting r = d_n/2 gives: A. 1/R1 − 1/R2 = 4nλ/d_n^2 B. 1/R1 + 1/R2 = 4nλ/d_n^2 C. 1/R1 − 1/R2 = nλ/d_n^2 D. R1 − R2 = 4nλ/d_n^2 Answer: A. (d_n^2/4)(1/R1 − 1/R2) = nλ, so 1/R1 − 1/R2 = 4nλ/d_n^2. 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