Aberrations — chromatic, spherical (concepts): Worked examples

This page is the drill ground for the aberrations topic. The parent note built the ideas; here we grind through every kind of case a problem can throw at you. Before each answer you'll forecast — guess first, then check yourself.

Everything you need is already earned in the parent note. A handful of symbols recur, so let's re-anchor them in one line each:

Definition The symbols we keep using

$f$ = focal length: the distance from a thin lens to where parallel rays meet. Bigger $f$ = weaker bending. A converging lens has $f > 0$ ; a diverging lens has $f < 0$ .
$P = \frac{1}{f}$ = power: how strongly the lens bends. Measured in dioptres (D) when $f$ is in metres. Converging $\Rightarrow P > 0$ ; diverging $\Rightarrow P < 0$ .
$n$ = refractive index of the glass; $n_{C}, n_{D}, n_{F}$ are its values for red, yellow, blue light.
$K = \frac{1}{R _{1}} - \frac{1}{R _{2}}$ = shape (curvature) constant: fixed by the lens's two surface radii $R_{1}, R_{2}$ . It appears in the lensmaker's equation $\frac{1}{f} = (n - 1) K$ and does not change with colour.
$θ$ = the angle a ray makes with the axis as it hits a surface; small for central rays, large for edge rays.
$ω = \frac{n _{F} - n _{C}}{n _{D} - 1}$ = dispersive power: how much the glass smears colours per unit bending. Always positive. ( $C$ =red, $D$ =yellow, $F$ =blue Fraunhofer lines.)

The scenario matrix

Every aberration problem is one of these cells. The worked examples below are tagged with the cell they cover.

Cell	Case class	What makes it tricky	Example
A	Basic chromatic: find $f_{C} - f_{F}$	plug into $\frac{df}{f} = ω$	Ex 1
B	Sign of $df$ (red vs blue), converging and diverging	which colour is nearer? does sign flip?	Ex 2
C	Degenerate: $ω = 0$ (no dispersion)	zero-input — what survives?	Ex 3
D	Achromatic doublet design	opposite-sign powers, solve 2 eqns	Ex 4
E	Limiting: $ω_{1} = ω_{2}$ (same glass)	can you achromatise?	Ex 5
F	Spherical: aperture cube-law	blur $\propto h^{3}$ scaling	Ex 6
G	Spherical: circle of least confusion	where to put the screen	Ex 7
H	Parabola vs sphere (limiting shape)	zero SA on axis, real-world	Ex 8
I	Exam twist: mixed — is it C or S?	diagnose the aberration	Ex 9

Cells A–E are chromatic (colour matters). Cells F–H are spherical (single colour, ray height matters). Cell I forces you to tell them apart.

Cell A — Basic chromatic spread

Steps.

Compute the dispersive power $ω = \frac{n _{F} - n _{C}}{n _{D} - 1} = \frac{1.526 - 1.517}{1.520 - 1} = \frac{0.009}{0.520}$ . Why this step? $ω$ is exactly the fractional focal spread — the boxed result $\frac{f _{C} - f _{F}}{f} = ω$ from the parent note. Everything hangs off it.
$ω = 0.009/0.520 = 0.017308$ (dimensionless). Why? It's a ratio of $n$ -values, so it carries no units — it's a pure fraction.
Multiply by $f$ : $f_{C} - f_{F} = ω f = 0.017308 \times 20 cm = 0.3462 cm \approx 3.46 mm$ . Why? Rearranging $\frac{f _{C} - f _{F}}{f} = ω$ gives the actual distance.

Verify: Units check — $ω$ is unitless, so $ω f$ inherits cm from $f$ . Magnitude check — a fraction of ~1.7% of 20 cm is a few mm, matching the forecast. Red's focus ( $f_{C}$ ) is farther; blue's ( $f_{F}$ ) is nearer, so $f_{C} - f_{F} > 0$ . ✓

Cell B — Which colour focuses nearer (both lens signs)

Steps — Part (a) (converging).

Recall $f \propto \frac{1}{n - 1}$ . Bigger $n$ ⇒ smaller $f$ ⇒ focus nearer the lens. Why this step? This is the entire sign story: more bending pulls the meeting point inward.
Since $n_{F} > n_{D} > n_{C}$ (blue > yellow > red), the focal lengths order as $f_{F} < f_{D} < f_{C}$ . Why? Reciprocal of $(n - 1)$ flips the inequality: largest $n$ gives smallest $f$ .
So blue focuses nearest, red farthest. Numerically, using $\frac{df}{f} = - \frac{d n}{n - 1}$ with $f_{D} = + 20$ cm (the yellow focal length is our reference $f$ ):
- Blue offset from yellow: $Δ f_{D F} = f_{D} \cdot \frac{n _{F} - n _{D}}{n _{D} - 1} = 20 \times \frac{0.006}{0.520} = 0.2308 cm$ (blue focus is $\approx 2.31$ mm nearer the lens).
- Red offset from yellow: $Δ f_{D C} = f_{D} \cdot \frac{n _{D} - n _{C}}{n _{D} - 1} = 20 \times \frac{0.003}{0.520} = 0.1154 cm$ (red focus is $\approx 1.15$ mm farther). Why? The formula gives the shift per colour step; each colour-gap is a different $d n$ . Here $f_{D}$ is positive, so a shorter $f_{F}$ literally sits closer to the lens on the real image side.

Steps — Part (b) (diverging, $f_{D} = - 20$ cm).

The relation $∣ f ∣ \propto \frac{1}{n - 1}$ still holds, so blue still has the smallest magnitude focal length: $∣ f_{F} ∣ < ∣ f_{D} ∣ < ∣ f_{C} ∣$ . Why? Dispersion acts on the glass, not on the lens sign — blue is always bent hardest.
But now $f < 0$ : the foci are virtual and lie on the same side as the object. Blue, with the smallest $∣ f ∣$ , has its virtual focus nearest the lens; red's virtual focus is farthest. The offset magnitudes are identical to Part (a): $∣Δ f_{D F} ∣ = 0.2308$ cm, $∣Δ f_{D C} ∣ = 0.1154$ cm — only their side of the lens flips. Why? $\frac{df}{f} = - \frac{d n}{n - 1}$ is a fractional rule; with $f = - 20$ cm, $df$ picks up the sign of $f$ , so the virtual foci reorder as mirror images of the converging case.

Verify: Part (a): $Δ f_{D F} + Δ f_{D C} = 0.2308 + 0.1154 = 0.3462$ cm $= f_{C} - f_{F}$ from Ex 1 ✓. Part (b): magnitudes match Part (a) exactly (0.2308 and 0.1154 cm) ✓, and the rule "blue bent hardest ⇒ blue focus nearest the lens" holds for both signs — it's a statement about bending, not about which side the image lands. ✓

Cell C — Degenerate input: no dispersion

Steps.

Dispersive power: $ω = \frac{n _{F} - n _{C}}{n _{D} - 1} = \frac{1.500 - 1.500}{1.500 - 1} = \frac{0}{0.5} = 0$ . Why this step? The numerator $n_{F} - n_{C}$ is the whole cause of colour spread. Kill it and $ω \to 0$ .
Longitudinal aberration: $f_{C} - f_{F} = ω f = 0 \times 20 = 0$ . Why? No dispersion ⇒ no focal spread ⇒ all colours meet at one point.
But the lens still works! Using the lensmaker's equation with the shape constant $K = \frac{1}{R _{1}} - \frac{1}{R _{2}}$ defined above, $\frac{1}{f} = (n - 1) K = (0.5) K$ is nonzero, so it still focuses. Why? Dispersion (colour) and refraction (bending) are separate: $ω = 0$ removes only the variation of $n$ with colour, not the bending itself — $K$ and $(n - 1)$ are both still nonzero.

Verify: A single-index medium is non-dispersive — this is the paraxial ideal that the parent note's "first lie" refers to. With $ω = 0$ there is literally no chromatic aberration to correct; no doublet is needed. ✓

Cell D — Design an achromatic doublet

Steps.

Write the two conditions.
- Net power: $P_{1} + P_{2} = P = 2$ .
- Achromatic condition: $ω_{1} P_{1} + ω_{2} P_{2} = 0$ (from the parent's boxed result). Why this step? Two unknowns need two equations: one fixes the strength, one kills the colour error.
From the achromatic condition, $P_{1} = - \frac{ω _{2}}{ω _{1}} P_{2} = - \frac{0.030}{0.018} P_{2} = - \frac{5}{3} P_{2}$ . Why? Rearranging isolates $P_{1}$ ; the minus sign already tells us the two powers have opposite sign.
Substitute into $P_{1} + P_{2} = 2$ : $- \frac{5}{3} P_{2} + P_{2} = 2 \Rightarrow - \frac{2}{3} P_{2} = 2 \Rightarrow P_{2} = - 3 D$ (flint diverges). Then $P_{1} = 2 - (- 3) = + 5 D$ (crown converges). Why? Back-substitution finishes the linear system. The crown must over-converge so the flint can subtract.

Verify: Net: $P_{1} + P_{2} = 5 + (- 3) = 2$ D ✓. Achromatic: $ω_{1} P_{1} + ω_{2} P_{2} = 0.018 \times 5 + 0.030 \times (- 3) = 0.090 - 0.090 = 0$ ✓. Crown converging, flint diverging — matches "one must be negative." ✓

Cell E — Limiting case: identical glasses

Steps.

Achromatic condition with equal $ω$ : $ω P_{1} + ω P_{2} = ω (P_{1} + P_{2}) = 0$ . Why this step? Factoring out the shared $ω$ exposes what's forced.
Since $ω \neq = 0$ , we need $P_{1} + P_{2} = 0$ . Why? A nonzero factor can't make the product zero, so the bracket must vanish.
But $P_{1} + P_{2} = 0$ means net power $P = 0$ — a useless flat combination, not a $+ 2$ D lens. Why? You cannot cancel the colour error and keep power using one glass. The whole point of a doublet is two different $ω$ 's.

Verify: The design requires $ω_{1} \neq = ω_{2}$ . This is why real achromats pair crown + flint (a low- $ω$ and a high- $ω$ glass). Same-glass ⇒ only $P = 0$ satisfies both conditions. ✓

Cell F — Spherical aberration: the cube law

Steps.

Write the scaling law: $\frac{b}{b _{0}} = (\frac{h}{h _{0}})^{3}$ . Why this step? Spherical aberration comes from the neglected $- θ^{3} /6$ term in $sin θ$ (where $θ$ is the ray's angle to the axis, defined above). A ray at height $h$ hits the surface at an angle $θ \propto h$ , so blur $\propto θ^{3} \propto h^{3}$ . This is the cube in action.
(a) $h = 5$ mm ⇒ $(\frac{5}{10})^{3} = \frac{1}{8}$ , so $b = 0.80 \times \frac{1}{8} = 0.10 mm$ . Why? Halving $h$ cubes to one-eighth — dramatic, not gentle.
(b) $h = 2$ mm ⇒ $(\frac{2}{10})^{3} = \frac{8}{1000} = 0.008$ , so $b = 0.80 \times 0.008 = 0.0064 mm = 6.4 μ m$ . Why? Stopping to one-fifth aperture drops blur by $125 \times$ — this is why lenses are razor-sharp at $f /8$ .

Figure — Aberrations — chromatic, spherical (concepts)

Figure (s01): the black curve is blur $b$ against aperture radius $h$ ; it is flat near the axis and shoots up steeply at large $h$ — the signature of a cube. The three red dots mark our answers at $h = 10, 5, 2$ mm. Notice how the red dot at $h = 2$ hugs the floor: almost all the blur lives in the outermost rays.

Cell G — The circle of least confusion

Steps.

Longitudinal SA: $L_{p} - L_{m} = 100.0 - 98.0 = 2.0 mm$ . Why this step? This is the axial gap between "edge focus" and "centre focus" — the size of the smear.
Marginal rays cross nearer the lens (98 vs 100 mm), confirming the $- θ^{3} /6$ term bends them toward the axis, so they meet sooner. Why? Direction-of-error check — extra bending pulls the crossing inward.
Best screen at $\frac{3}{4}$ of the way from paraxial toward marginal: $L_{best} = L_{p} - \frac{3}{4} (L_{p} - L_{m}) = 100.0 - 0.75 \times 2.0 = 100.0 - 1.5 = 98.5 mm$ . Why? Neither focus gives the smallest spot — the tightest bundle (least confusion) lies between, closer to the marginal side.

Figure (s02): the solid black lines are steep marginal rays crossing the axis early at 98.0 mm; the dashed black lines are shallow paraxial rays crossing later at 100.0 mm. The red vertical line marks 98.5 mm — the circle of least confusion. Observe that the ray bundle is narrowest at the red line, not at either focus dot: that is where you put the screen.

Cell H — Parabola beats sphere (limiting shape)

Steps.

The parabola is defined as the locus of points equidistant (in optical path) from a focus and a flat wavefront. Why this step? "Equal path to the focus" is exactly the condition for all rays to arrive in phase and meet at one point — no $θ^{3}$ error, for any ray height $h$ .
A sphere only approximates a parabola near its vertex; expanding the sphere's sag $R - R^{2} - h^{2}$ shows sphere and parabola agree to order $h^{2}$ but differ at order $h^{4}$ . Why? The $h^{4}$ mismatch is precisely the spherical-aberration term — it grows with ray height $h$ .
Numerical taste: for $R = 200$ mm (radius of curvature) and $h = 40$ mm (ray height), the sag difference is $Δ z = \frac{h ^{4}}{8 R ^{3}} = \frac{4 0 ^{4}}{8 \times 20 0 ^{3}} = \frac{2 , 560 , 000}{64 , 000 , 000} = 0.04 mm .$ Why? This $0.04$ mm surface error is what a parabola removes and a sphere keeps — enough to blur a star.

Figure (s03): both curves start together at the vertex (bottom). The black curve is the sphere, the red curve is the parabola. They overlap near the centre but the black sphere pulls ahead at large $h$ ; the red double arrow marks the 0.04 mm sag gap at $h = 40$ mm. That tiny gap is the whole difference between a sharp star and a fuzzy one.

Cell I — Exam twist: diagnose the aberration

Steps.

Defect (i): coloured fringes, unchanged by aperture ⇒ chromatic aberration. It's a focal-length shift with wavelength, not a marginal-ray effect, so an iris can't fix it. Why this step? The diagnostic key: colour + aperture-independent ⇒ chromatic. Fix = achromatic doublet, not a smaller stop.
Defect (ii): monochromatic haze that shrinks with aperture ⇒ spherical aberration (a Seidel third-order effect). Fix = stop down / best-form bending / aspheric. Why? Single-colour + strongly aperture-dependent ⇒ spherical.
Going from $f /2.8$ to $f /5.6$ halves the aperture diameter, so the aperture radius $h \to h /2$ . By the cube law, the SA blur becomes $\frac{b _{new}}{b _{old}} = (\frac{h /2}{h})^{3} = (\frac{1}{2})^{3} = \frac{1}{8} = 0.125.$ Why? One full f-stop is a factor-of-two change in diameter; the $h^{3}$ law then cubes it to $1/8$ . So the haze drops to 12.5% of its wide-open value — an 8× improvement — while the colour fringes of defect (i) are untouched by this same move.

Verify: Reduction factor $= (1/2)^{3} = 0.125$ , i.e. blur drops to 12.5% ✓. The two defects respond to totally different cures — aperture kills spherical but not chromatic — exactly the parent note's key contrast. ✓

Recall Self-test — cover the answers

Which cell has $ω = 0$ and what survives? ::: Cell C (degenerate); refraction/power survives ( $K$ and $n - 1$ nonzero), chromatic aberration vanishes. In a doublet, which glass diverges — crown or flint? ::: The high- $ω$ flint diverges; the low- $ω$ crown converges (Ex 4: $P_{2} = - 3$ D flint, $P_{1} = + 5$ D crown). Does "blue focuses nearest" survive for a diverging lens? ::: Yes — blue is always bent hardest, so its (virtual) focus is nearest the lens; only the side flips (Ex 2b). Halving the aperture changes SA blur by what factor? ::: $(1/2)^{3} = 1/8$ (cube law, Ex 6). Does stopping down fix coloured fringes? ::: No — chromatic is aperture-independent; you need a doublet (Ex 9). Where is the circle of least confusion? ::: Between the foci, nearer the marginal focus (Ex 7: 98.5 mm).