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Newton's ring diameter change with liquid

Short answer

Refractive index μ = (D_air/D_liquid)² = (1.40/1.27)² = 1.215 ≈ 1.22.

In a Newton's ring experiment the diameter of the 10th ring changes from 1.40 cm to 1.27 cm when a liquid is introduced between the lens and the plate. Calculate the refractive index of the liquid.

In Newton's rings D_n² = 4nλR in air and D'_n² = 4nλR/μ with a liquid of index μ between lens and plate. Hence μ = D_n² / D'_n².

Given: D_10 (air) = 1.40 cm, D'_10 (liquid) = 1.27 cm.

  1. Ratio of squares:
μ = (1.40)² / (1.27)²
  = 1.9600 / 1.6129
  = 1.215

Answer: refractive index of the liquid μ ≈ 1.22.

✓ Answer checked by Claude Opus 5.5 (high effort)

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