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Order of Newton's ring with double diameter

Short answer

In Newton’s rings, D_n² ∝ n. D' = 2 D₅₀ gives D'²/D₅₀² = 4, so n' = 4 × 50 = 200. The 200th dark ring has double the diameter.

Newton's rings are formed by sodium light viewed normally. What is the order of the dark ring which will have double the diameter of the 50th ring?

For dark rings in reflected light (Newton's rings), D_n² = 4nλR, so the square of the diameter is proportional to the order n.

Given: D' = 2 D₅₀ (same λ and R).

  1. D₅₀² = 4 × 50 × λR
  2. D'² = (2D₅₀)² = 4 D₅₀² = 4 × (4 × 50 λR)
  3. Also D'² = 4 n' λR, so n' = 4 × 50 = 200
n' / 50 = (D' / D₅₀)² = 2² = 4   →   n' = 200

Answer: the 200th dark ring has double the diameter of the 50th ring.

✓ Answer checked by Claude Opus 5.5 (high effort)

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