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Condition for achromatism of two lenses

Short answer

For two thin lenses separated by distance x, achromatism requires the equivalent focal length to be independent of wavelength. Differentiating 1/F = 1/f1 + 1/f2 - x/(f1 f2) and using δ(1/f) = ω/f yields the condition ω1/f1 + ω2/f2 = x(ω1 + ω2)/(f1 f2).

Prove that the condition for achromatism for the combination of two lenses of focal length f1 and f2 having dispersive power ω1 and ω2 placed at a separate distance x is (ω1/f1) + (ω2/f2) = (x/f1f2)(ω1 + ω2).

Statement: two thin lenses of focal lengths f₁, f₂ and dispersive powers ω₁, ω₂, separated by a distance x, form an achromatic combination if ω₁/f₁ + ω₂/f₂ = x(ω₁ + ω₂)/(f₁f₂).

Proof

  1. The equivalent focal length F of the pair is given by 1/F = 1/f₁ + 1/f₂ − x/(f₁f₂). (1)
  2. For achromatism, F must be the same for all colours (red and violet), so δ(1/F) = 0 when the refractive indices change from μ_R to μ_V. x is fixed.
  3. For each lens, 1/f = (μ − 1)(1/R₁ − 1/R₂) = (μ − 1) a, where a depends only on the radii. So the change in 1/f for a change δμ is δ(1/f) = a δμ = (1/f) δμ/(μ − 1) = ω/f, with ω = δμ/(μ − 1) the dispersive power. Hence δ(1/f₁) = ω₁/f₁ and δ(1/f₂) = ω₂/f₂.
  4. Differentiate (1), treating δ(1/f₁), δ(1/f₂) as the variations: δ(1/F) = δ(1/f₁) + δ(1/f₂) − x [ (1/f₁) δ(1/f₂) + (1/f₂) δ(1/f₁) ]
  5. Substitute the results of step 3: δ(1/F) = ω₁/f₁ + ω₂/f₂ − x [ ω₂/(f₁f₂) + ω₁/(f₁f₂) ] = ω₁/f₁ + ω₂/f₂ − x (ω₁ + ω₂)/(f₁f₂)
  6. Setting δ(1/F) = 0:

ω₁/f₁ + ω₂/f₂ = x (ω₁ + ω₂) / (f₁ f₂)

Special cases

  • x = 0 (lenses in contact): ω₁/f₁ + ω₂/f₂ = 0, the achromatic doublet.
  • Same material (ω₁ = ω₂): x = (f₁ + f₂)/2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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