core25 min studyLesson 207 in Physics

Spherical and chromatic aberration

Explain how aperture and wavelength produce different focus errors, then balance geometric blur against diffraction.

Before you begin

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The question for today

What you’ll understand

Distinguish spherical aberration from chromatic focus shifts.
Estimate focal-length changes from an index change.
Explain why stopping down cannot indefinitely improve a diffraction-limited image.

An ideal paraxial calculation sends every ray from an object point to a single image point. A real system departs from that prediction even when its surfaces are manufactured perfectly. Finite-aperture rays do not obey the same linearized geometry as near-axis rays, and different wavelengths generally experience different indices. These are different sources of aberration with different remedies. We will isolate an on-axis aperture error and a color-dependent power error before considering why simply making the opening smaller eventually introduces another limitation.

Spherical surfaces do not imply a spherical image point

Spherical aberration is an on-axis failure of finite-aperture rays to share the paraxial focus. A sphere approximates a suitably designed surface near its vertex, but higher-order shape terms differ. For many simple positive spherical elements, marginal rays focus closer to the element than paraxial rays, although the sign depends on design and conjugates. The effect occurs at one wavelength and does not require material dispersion, surface roughness, or an off-axis object.

A rotationally symmetric wavefront can be expanded in even powers of pupil radius. After the paraxial curvature and chosen defocus are accounted for, the leading spherical term is quartic. Its radial derivative produces a cubic ray-angle error. This scaling explains why reducing pupil radius can strongly reduce a geometric aberration. It also explains why one focus adjustment cannot make all pupil zones agree: a quadratic defocus term cannot cancel a quartic term over every radius.

W(r)=a_2r^2+a_4r^4+\cdots,\qquad \frac{dW}{dr}=2a_2r+4a_4r^3+\cdots

W is optical-path departure in metres at pupil radius r in metres. Coefficients a two and a four have units of inverse metres and inverse cubic metres. The derivative relates to a small wavefront-normal error within a specified optical geometry.↔ Scroll wide equations horizontally.

Chromatic error follows the material dispersion

For a fixed-shape thin lens in air, its paraxial power is a geometric curvature factor times $n(\lambda)-1$ . Ordinary transparent glasses often have higher visible index at shorter wavelength, giving greater positive power and a shorter focal length for blue than for red. This longitudinal chromatic aberration remains even for very small apertures because it changes the first-order imaging map itself. Closing the iris can mask some defocus blur without making the focal lengths equal.

An achromatic doublet combines positive and negative powers from materials with different dispersions so their leading color changes cancel at chosen wavelengths while net power stays useful. The correction is limited: secondary spectrum, other wavelengths, finite field, and monochromatic aberrations remain design questions. Mirrors avoid refractive chromatic power in their ideal geometry, but a reflecting instrument can still contain dispersive windows or eyepieces. “Achromatic” should describe a stated correction target rather than perfect imaging of every color.

P(\lambda)=\frac{1}{f(\lambda)}=[n(\lambda)-1]\left(\frac1{R_1}-\frac1{R_2}\right),\qquad \frac{\Delta f}{f}\simeq-\frac{\Delta n}{n-1}

Radii use the earlier signed convention. The fractional-change formula is first order for a small index change at fixed lens shape; it is a thin-lens air formula, not a general thick-lens dispersion model.↔ Scroll wide equations horizontally.

The best aperture balances different regimes

Stopping down rejects marginal rays and often reduces spherical aberration, coma, and defocus blur. Yet a finite circular opening also diffracts. For an ideal unobstructed circular pupil the first dark-ring radius at a paraxial image is approximately $1.22\lambda_0N$ in air. Increasing f-number by reducing pupil diameter therefore enlarges that diffraction scale. An optical system can move from aberration-dominated blur toward diffraction-dominated blur as the iris closes, rather than improving without bound.

A ray spot diagram addresses geometric ray crossings; a point-spread function includes interference across the pupil. Neither is interchangeable with a picture of manufacturing error. To diagnose a blur, vary one physical cause at a time: compare pupil zones at one wavelength for spherical error, then compare wavelengths at the same small aperture for chromatic error. A measured optimum also depends on detector sampling, signal level, and field position, so the lesson’s leading-order estimates should guide a controlled comparison rather than promise a universal best setting.

Work it through

A color-dependent focal length

A thin fixed-shape lens in air has focal length 100 mm where its index is 1.50. At a second wavelength its index is 1.51. Find its exact focal length in this thin-lens model and compare with the first-order shift estimate.

Let the fixed curvature factor be K. At the first wavelength, $K=(1/0.100)/(1.50-1)=20\,\mathrm{m^{-1}}$ . Geometry stays fixed when wavelength changes.
At the second wavelength the power is $(1.51-1)(20)=10.2\,\mathrm{m^{-1}}$ . Thus $f_2=1/10.2\,\mathrm m\simeq98.039\,\mathrm{mm}$ .
The exact shift is approximately −1.961 mm. The first-order fraction is $-0.01/0.50=-0.02$ , predicting a −2 mm shift, consistent with the exact value for this small index change.

The result

The higher-index wavelength focuses about 1.96 mm closer to the lens in this model. That is a chromatic power change, independent of which pupil zone a paraxial ray uses.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Phase agreement matters to a focus

Use two fixed-frequency components as a limited phase analogy. Reset, set Phase difference = 0° and Amplitude ratio = 1, and predict a summed amplitude of 2. For an independent comparison Reset, set Phase difference = 180° and Amplitude ratio = 1, and predict cancellation everywhere in this equal-frequency model.

SumFirst componentSecond component

Sum at x = 0: 0.866
Resultant amplitude: 1.732
Phase difference: 60°

Phase difference60°Amplitude ratio1

t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPupil phase errors can reduce coherent addition, but this lab only sums two same-direction waves. It contains no pupil, color dispersion, spherical surface, or diffraction point-spread calculation.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

A wavelength-dependent first-order focal position is longitudinal chromatic error. Zone-dependent focus at one wavelength instead diagnoses spherical aberration, while diffraction and throughput are separate effects.

Your turn

Try it yourself

At one wavelength an ideal circular pupil gives image-plane first-dark-ring radius $1.22\lambda_0N$ . Take $\lambda_0=0.50\,\mathrm{\mu m}$ . Calculate that radius at f/4 and f/8, and explain whether the larger f-number always means a sharper image.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

This radius is a diffraction estimate, so compare its trend with the geometric effect of rejecting marginal rays.

Compare with the solution

At f/4 the radius is $1.22(0.50)(4)=2.44\,\mathrm{\mu m}$ . At f/8 it is $4.88\,\mathrm{\mu m}$ , twice as large. Closing the pupil may nevertheless reduce geometric aberration enough to improve a previously poor image. Once diffraction dominates, further closure enlarges this diffraction scale. Thus aperture choice requires identifying the prevailing blur mechanism, not assuming one monotonic rule.

Keep the idea

Take it with you

Spherical aberration compares pupil zones at one wavelength.
Chromatic aberration changes power with wavelength through dispersion.
Stopping down reduces many geometric errors while increasing the diffraction scale.

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