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Index difference for circular polarization

Short answer

For a quarter-wave plate, path difference (μ_e−μ_o)t = λ/4. Thus |μ_e−μ_o| = λ/(4t) = 600×10⁻⁹/(4×3×10⁻⁵) = 0.005.

A beam of plane polarized light is converted into circularly polarized light by passing it through a crystal slice of thickness 3×10⁻⁵ m. Calculate the difference in the refractive indices of the two rays inside the crystal. Wavelength of light is 600nm.

Plane polarized light becomes circularly polarized after passing through a crystal slice (quarter-wave plate) when the path difference between the e and o rays is λ/4, i.e. (μ_e − μ_o) t = λ/4 (lowest order, with vibration at 45° to the axis).

Given: t = 3×10⁻⁵ m, λ = 600 nm = 6×10⁻⁷ m.

  1. Condition: |μ_e − μ_o| t = λ/4
  2. Solve:
|μ_e − μ_o| = λ / (4t)
             = 6×10⁻⁷ / (4 × 3×10⁻⁵)
             = 6×10⁻⁷ / 1.2×10⁻⁴
             = 5×10⁻³

Answer: difference in refractive indices = 0.005.

✓ Answer checked by Claude Opus 5.5 (high effort)

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