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Newton's rings between two curved surfaces

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.

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.

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.

        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:

(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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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