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Newton's rings with two wavelengths

Short answer

From nλ₁=(n+1)λ₂ get n=59. Diameter D=√(4nλ₁R)=√(4×59×6e−5×90)=√(1.2744)≈1.13 cm.

In a Newton's rings arrangement a source emitting two wavelengths λ1 = 6×10⁻⁵ cm and λ2 = 5.9×10⁻⁵ cm is used. It is found that the nth dark ring due to λ1 coincides with the (n+1)th dark ring due to λ2. Find the diameter of the nth dark ring. (radius of curvature of the lens R = 90 cm)

For dark rings in reflected light in Newton's rings, D² = 4nλR, where n is the ring order and R the radius of curvature.

Given: λ₁ = 6×10⁻⁵ cm, λ₂ = 5.9×10⁻⁵ cm, R = 90 cm. The nth dark ring of λ₁ coincides with the (n+1)th dark ring of λ₂.

  1. Equal diameters: 4nλ₁R = 4(n+1)λ₂R, so nλ₁ = (n+1)λ₂
  2. n(λ₁ − λ₂) = λ₂
n = λ₂ / (λ₁ − λ₂) = 5.9×10⁻⁵ / (0.1×10⁻⁵) = 59
  1. Diameter of the 59th dark ring (λ₁):
D² = 4 n λ₁ R = 4 × 59 × 6×10⁻⁵ × 90
   = 1.2744 cm²
D  = 1.129 cm

(Check with λ₂, n = 60: 4 × 60 × 5.9×10⁻⁵ × 90 = 1.2744 cm². Matches.)

Answer: diameter of the nth (59th) dark ring ≈ 1.13 cm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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