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Thickness of doubly refracting plate

Short answer

For a quarter-wave plate, (μe − μo) t = λ/4. Using μe = 1.54, μo = 1.53, λ = 5890 Å, t = λ / [4(μe − μo)] = 1.47 × 10⁻⁵ m (14.7 μm).

Calculate the thickness of a doubly refracting plate capable of producing a path difference of λ/4 between extraordinary and ordinary rays of wavelength 5890 Å. (Use μo = 1.53 and μe = 1.54)

A plate producing path difference Δ between the e-ray and o-ray has Δ = (μ_e − μ_o) t, where t is the plate thickness.

Given: Δ = λ/4, λ = 5890 Å = 5.890×10⁻⁷ m, μ_o = 1.53, μ_e = 1.54.

  1. Condition: (μ_e − μ_o) t = λ/4
  2. Solve for t:
t = λ / [4 (μ_e − μ_o)]
  = 5.890×10⁻⁷ / [4 × (1.54 − 1.53)]
  = 5.890×10⁻⁷ / 0.04
  = 1.4725×10⁻⁵ m

Answer: t ≈ 1.47 × 10⁻⁵ m (14.7 μm); this is a quarter-wave plate for 5890 Å.

✓ Answer checked by Claude Opus 5.5 (high effort)

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