- 2073 Chaitra · 5 marks
Chromatic aberration and its minimization
Short answer
Chromatic aberration is the variation of focus with color due to dispersion. A single lens always has it because the change in focal length δf = −ω f is nonzero for any material. It can be minimized by an achromatic doublet (ω1/f1 + ω2/f2 = 0) or by two separated lenses of the same material with separation x = (f1+f2)/2.
What is chromatic aberration? Show that a single lens is always accompanied with such aberration. Discuss in brief how we can minimize chromatic aberration in the combination of lenses.
Chromatic aberration
The defect of a lens by which it focuses light of different colours at different points on the axis is called chromatic aberration. The refractive index is larger for violet than for red, so the violet focus F_V lies closer to the lens than the red focus F_R. The distance F_V F_R is the longitudinal chromatic aberration; the difference in image sizes is the lateral chromatic aberration.
white \
======> )-----F_V--F_R----->
light / <---> longitudinal chromatic
aberration
A single lens always has it
- Lens maker's formula: 1/f = (μ − 1)(1/R₁ − 1/R₂).
- Differentiate with respect to μ (R₁, R₂ fixed): −δf/f² = δμ (1/R₁ − 1/R₂) = δμ / [(μ − 1) f].
- So δf = −f δμ/(μ − 1).
- Between violet and red: f_V − f_R = −f (μ_V − μ_R)/(μ − 1) = −ω f, where ω = (μ_V − μ_R)/(μ_Y − 1) is the dispersive power of the lens material.
- Since μ_V > μ_R, ω is always positive and non-zero for any transparent material. Hence f_V ≠ f_R for every lens of any focal length, and the aberration cannot vanish for a single lens.
Minimizing chromatic aberration
1. Achromatic doublet (lenses in contact): a convex lens of crown glass and a concave lens of flint glass are cemented together. The condition is
ω₁/f₁ + ω₂/f₂ = 0
so f₁ and f₂ have opposite signs (one converging, one diverging), and the converging lens (crown, smaller ω) is the stronger one.
2. Two separated lenses of the same material: for lenses a distance x apart, the condition (ω₁/f₁) + (ω₂/f₂) = x(ω₁ + ω₂)/(f₁f₂) with ω₁ = ω₂ reduces to
x = (f₁ + f₂)/2
This is used in Huygens' eyepiece (f₁ = 3f, f₂ = f, x = 2f).
3. Using mirrors instead of lenses (reflecting telescopes) removes it completely, since reflection does not depend on colour.
✓ Answer checked by Claude Opus 5.5 (high effort)
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