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Soap film thickness from dark bands

Short answer

For consecutive dark bands, 2μt cos r = nλ₁ = (n+1)λ₂. Order n = λ₂/(λ₁−λ₂) = 60. With sin r = sin i/μ, cos r = 0.7989, thickness t = nλ₁/(2μ cos r) ≈ 1.72×10⁻³ cm.

White light is incident on a soap film at an angle sin⁻¹(4/5) and the reflected light on examination by a spectrometer shows dark bands. The consecutive dark bands correspond to wavelength 6.1×10⁻⁵ cm and 6.0×10⁻⁵ cm. If μ=1.33 for the film, calculate its thickness.

For a film viewed by reflected light, a dark band satisfies 2μt cos r = nλ. Two consecutive dark bands of wavelengths λ₁ > λ₂ correspond to orders n and (n+1).

Given: sin i = 4/5 = 0.8, μ = 1.33, λ₁ = 6.1×10⁻⁵ cm, λ₂ = 6.0×10⁻⁵ cm.

  1. Condition for the two bands: 2μt cos r = nλ₁ = (n+1)λ₂
  2. Order:
n = λ₂ / (λ₁ − λ₂) = 6.0×10⁻⁵ / 0.1×10⁻⁵ = 60
  1. Angle of refraction: sin r = sin i / μ = 0.8 / 1.33 = 0.6015, so cos r = √(1 − 0.3618) = 0.7989
  2. Thickness:
t = nλ₁ / (2μ cos r)
  = 60 × 6.1×10⁻⁵ / (2 × 1.33 × 0.7989)
  = 3.66×10⁻³ / 2.125
  = 1.72×10⁻³ cm

Answer: film thickness t ≈ 1.72 × 10⁻³ cm (≈ 17.2 μm = 1.72×10⁻⁵ m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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