Aberrations — chromatic, spherical (concepts): Visual walkthrough

Step 1 — What a lens does to a bundle of parallel rays

WHAT. Imagine light from something very far away — the Sun, a star. Its rays arrive as a bundle of parallel lines. A converging lens bends them all so they cross at a single point.

WHY start here. The whole story of chromatic aberration is about where that crossing point sits. So we must first name it. The distance from the lens to that crossing point is the focal length, written $f$ . The crossing point itself is the focus.

PICTURE. In the figure, straight blue lines come in from the left. After the lens they tilt and meet at one dot on the axis. The horizontal distance from lens to dot is labelled $f$ .

Step 2 — The rule that sets $f$ : the lensmaker's equation

WHAT. How strongly a lens bends light is fixed by two things: the shape of its glass surfaces, and how much the glass slows light down compared to air. The lensmaker's equation packages this:

$bending strength \frac{1}{f} = glass factor (n - 1) shape factor (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Let me name every piece the moment it appears:

$n$ = the refractive index — a plain number telling how much slower light travels in the glass than in air. Air is $n = 1$ ; glass is around $n = 1.5$ . See Dispersion and Refractive Index.
$(n - 1)$ = the excess over air. If the glass were the same as air ( $n = 1$ ) this is zero and the lens does nothing — which is exactly right.
$R_{1}, R_{2}$ = the radii of the two curved surfaces — how sharply each face is domed. These are pure geometry: once the glass is ground, they never change.

WHY combine them into one letter. Since $R_{1}$ and $R_{2}$ are frozen once the lens is made, we lump them into a single shape factor:

$K = \frac{1}{R _{1}} - \frac{1}{R _{2}} (a constant, decided by the grinder)$

Now the equation is beautifully simple:

$\frac{1}{f} = (n - 1) K$

PICTURE. The figure shows the same lens twice: on the left a fat lens (large $K$ , short $f$ ), on the right a thin one (small $K$ , long $f$ ). The only thing that will change $f$ after grinding is $n$ — and $n$ is about to betray us.

Step 3 — The hidden lie: $n$ is not one number

WHAT. White light is a mixture of colours. The glass does not slow every colour equally. It slows blue more than red. So there is not one $n$ , there is a whole family: $n_{red} < n_{yellow} < n_{blue}$ .

WHY this matters. Look back at Step 2: $f$ depends on $n$ . If $n$ is bigger for blue, then $(n - 1)$ is bigger for blue, then $1/ f$ is bigger for blue, then $f$ is smaller for blue. Blue's crossing point is nearer the lens. That splitting-by-colour is chromatic aberration.

PICTURE. The figure sends white light in; after the lens the rays fan into a little spectrum. The blue ray crosses the axis first (short $f$ ), the red ray crosses last (long $f$ ). The yellow sits between.

Step 4 — Turning "blue focuses closer" into a formula: why we differentiate

WHAT. We want to know: if $n$ nudges up by a tiny amount (red → blue), how much does $f$ shift? That is a rate-of-change question — "output change per input change" — and the tool built for exactly that question is the derivative $\frac{df}{d n}$ .

WHY the derivative and not just plugging in two numbers? We could compute $f_{red}$ and $f_{blue}$ separately and subtract. But the derivative gives us a clean, general formula that works for any small colour change, and reveals the sign (which way $f$ moves) in one line. The change $n_{F} - n_{C}$ is small (about $0.008$ ), so treating it as a tiny nudge $d n$ is honest.

The move, term by term. Start from Step 2 and let $n$ vary while $K$ stays frozen:

$\frac{1}{f} = (n - 1) K$

Take the derivative of both sides with respect to $n$ . On the left, $\frac{d}{d n} (\frac{1}{f}) = - \frac{1}{f ^{2}} \frac{df}{d n}$ (that minus sign is the whole story — it's where "closer" comes from). On the right, $\frac{d}{d n} [(n - 1) K] = K$ because $K$ is constant:

$- \frac{1}{f ^{2}} df = K d n$

Now replace $K$ using $K = \frac{1}{( n - 1 ) f}$ (just rearranged from Step 2), so the geometry vanishes and only measurable things remain:

$- \frac{1}{f ^{2}} df = \frac{1}{( n - 1 ) f} d n$

Multiply both sides by $- f$ (kills one power of $f$ , flips the sign):

$\frac{df}{f} = - \frac{d n}{n - 1}$

Reading each symbol:

$\frac{df}{f}$ = the fractional change in focal length (a percentage-like number).
$d n$ = the change in index between two colours.
the minus = up in $n$ means down in $f$ . Blue ( $+ d n$ ) → shorter $f$ . Exactly Step 3's picture, now in algebra.

PICTURE. The figure plots $f$ against $n$ : a downward-sloping curve. A tiny step right ( $+ d n$ ) produces a step down ( $- df$ ). The slope arrow is drawn in red to shout "negative."

Step 5 — Measuring the whole spread: dispersive power $ω$

WHAT. Step 4 handled an infinitesimal nudge. Now we ask for the total red-to-blue spread. We set $d n = n_{F} - n_{C}$ (blue index minus red index — the full colour gap) and use the middle colour for the denominator, $n - 1 \approx n_{D} - 1$ :

$\frac{f _{C} - f _{F}}{f} = \frac{n _{F} - n _{C}}{n _{D} - 1} \equiv ω$

The right-hand side gets its own name, the dispersive power $ω$ (see Dispersion and Refractive Index). Term by term:

numerator $n_{F} - n_{C}$ = how far apart blue and red are (the spread of the glass).
denominator $n_{D} - 1$ = the average bending strength of the glass.
$ω$ = spread ÷ strength = "how rainbow-y is this glass, relative to how much it bends at all."

WHY the ratio and not just $n_{F} - n_{C}$ . A strongly bending lens with a small colour gap can be worse or better than a weak lens with a big gap — what matters for the image is the gap relative to the bending. Dividing gives a pure number that compares glasses fairly.

PICTURE. Two glass samples side by side: crown glass (small $ω$ , tight colours) and flint glass (large $ω$ , fanned colours). Same incoming white ray, very different fans.

Step 6 — The cure in pictures: cancel a lie with its mirror image

WHAT. Put two lenses back-to-back: a converging crown ( $P_{1} > 0$ ) and a diverging flint ( $P_{2} < 0$ ). "Power" $P$ is just $1/ f$ — how strongly a lens bends. Powers add: total $P = P_{1} + P_{2}$ .

WHY this can work. Each lens smears colours by its own amount. From Step 5, the colour-smear of a single lens is $d P = P ω$ (the fractional spread $ω$ times its power). If we make the two smears equal and opposite, they cancel while the net bending survives.

Set the total colour-smear to zero:

$d P = d P_{1} + d P_{2} = P_{1} ω_{1} + P_{2} ω_{2} = 0$

Term by term:

$P_{1} ω_{1}$ = crown lens's colour error.
$P_{2} ω_{2}$ = flint lens's colour error.
setting the sum to zero = "the two rainbows destroy each other."

Rearranged:

$ω_{1} P_{1} + ω_{2} P_{2} = 0 \Rightarrow \frac{P _{1}}{P _{2}} = - \frac{ω _{2}}{ω _{1}}$

WHY one lens must diverge. Both dispersive powers $ω_{1}, ω_{2}$ are positive (a glass never un-spreads colours). For $ω_{1} P_{1} = - ω_{2} P_{2}$ to hold, $P_{1}$ and $P_{2}$ must have opposite signs — one converges, one diverges. This is the Achromatic Doublet.

PICTURE. Crown bends blue-below-red; flint (flipped) bends red-below-blue by the matching amount. After the pair the blue and red rays leave parallel again and reunite at one focus.

Recall Quick self-check

Why can't we just make the flint diverge a little? ::: The magnitudes are pinned: $ω_{1} P_{1} = - ω_{2} P_{2}$ fixes the exact ratio. Too little and colour survives; too much and it over-corrects (colours cross the other way).

The one-picture summary

WHAT it compresses. One frame with everything: white light hits a single lens → colours split ( $f_{blue} < f_{red}$ , from the minus sign in Step 4) → the spread measured by $ω$ (Step 5) → a second, diverging lens undoes it (Step 6). The chain of causes runs left to right.

Recall Feynman retelling — the whole walkthrough in plain words

A lens's job is to catch parallel sunlight and dump it into one dot; the reach to that dot is the focal length $f$ (Step 1). How hard the lens pulls is set by its shape and by how much the glass slows light — a number $n$ (Step 2). Here's the sneaky part: glass slows blue light more than red (Step 3). Slow-more means bend-more means a shorter reach, so blue lands nearer and red lands farther — the dot becomes a tiny smeared rainbow. To say exactly how much, we ask "how fast does the reach change when the colour-number nudges up?" — that's a derivative, and it comes out with a minus sign, which is math-speak for "up in $n$ , down in $f$ " (Step 4). Bundling the full red-to-blue nudge into one number gives the dispersive power $ω$ : how rainbow-y the glass is compared to how much it bends at all (Step 5). Finally, the trick: glue a diverging lens made of extra-rainbow-y glass behind the first. It spreads colours the opposite way by the same amount, the two rainbows kill each other, but the net converging still survives — and the dot is clean again (Step 6).

Recall

Recall Test yourself

Which term in the lensmaker's equation carries the colour dependence? ::: $(n - 1)$ , because $n = n (λ)$ . In $\frac{df}{f} = - \frac{d n}{n - 1}$ , what does the minus sign physically mean? ::: A bigger index (redder→bluer) gives a smaller focal length — blue focuses closer. Why is $ω$ a ratio rather than just $n_{F} - n_{C}$ ? ::: To compare the colour spread against the overall bending strength $(n_{D} - 1)$ — a pure, fair number. Why must an achromat contain a diverging lens? ::: Both $ω$ 's are positive, so $ω_{1} P_{1} + ω_{2} P_{2} = 0$ forces $P_{1}, P_{2}$ to have opposite signs.