Question bank — Aberrations — chromatic, spherical (concepts)

False. Even an axial pencil of white light splits, because $n(\lambda )$ still differs by colour — see Dispersion and Refractive Index and figure 2. Paraxial-ness kills spherical aberration, not chromatic.

False. It is a monochromatic defect: with one colour, marginal rays (large $\theta$, large $h$) still bend too much from the $-{\theta }^3/6$ term and focus early (figure 3). Colour is irrelevant to it.

False. Blue has the larger $n$, and $f\propto 1/(n-1)$, so blue's $f$ is shorter — blue focuses nearer the lens.

False. The neglected $-{\theta }^3/6$ term bends marginal rays more toward the axis, so they cross sooner (nearer). Width does not mean reach.

False. It forces only two chosen colours (usually $F$ blue and $C$ red) to coincide; other colours retain a small residual called the secondary spectrum.

False. $\omega =(n_F-n_C)/(n_D-1)$ has $n_F>n_C$ and $n_D>1$, so $\omega >0$ always — that positivity is exactly why the doublet needs opposite-sign powers.

False. It is perfect only for on-axis parallel light; off-axis point sources still blur into coma. See Parabolic Mirrors and Telescopes.

False. It cures spherical aberration by removing marginal rays, but chromatic is a focal-length shift on the axis that a smaller hole does nothing to — you still get colour fringes.

False. With shape factor $K=\frac{1}{R_1}-\frac{1}{R_2}$, it still hides two approximations: $n$ as a single wavelength (breaks → chromatic) and $\sin \theta \approx \theta$ (breaks → spherical). See Lensmaker's Equation and Paraxial Approximation.

False. It comes from a ${\theta }^3\sim h^3$ term, so the transverse blur grows roughly as $h^3$ — halving $h$ cuts blur by about a factor of 8.

Only chromatic involves dispersion, i.e. $n(\lambda )$. Spherical happens with a single wavelength and is purely about ray height and the spherical shape.

Wrong tool: outer rays are the spherical villains. Chromatic is an on-axis focal-length difference between colours; you fix it with an Achromatic Doublet, not an aperture.

The chromatic spread depends on $dn$ (how much $n$ changes across colours, captured by $\omega$), not on $n$ itself. High-$n$ flint glass often disperses more, not less.

Equal-and-opposite $f$ would give zero net power ($P=1/f$, so the powers would cancel — no lens at all). The real condition is ${\omega }_1{P}_1+{\omega }_2{P}_2=0$ with net power surviving — the colour spreads cancel, the powers do not.

Backwards. The edges already turn light too hard; the cure is to flatten the effective curvature there — an aspheric/parabolic surface or best-form ("Coddington") bending, per Seidel Aberrations.

No. It lies between the marginal focus (nearer) and the paraxial focus (farther), where the converging cone of rays is narrowest — that's the best screen position (see figure 3).

The denominator is the yellow reference minus one, $n_D-1$: $\omega =(n_F-n_C)/(n_D-1)$. $D$ is yellow, $F$ blue, $C$ red — the Fraunhofer Lines.

The expansion $\sin \theta =\theta -{\theta }^3/6+\dots$ has no even-power term, so after the paraxial $\theta$ the first surviving correction is $-{\theta }^3/6$ — that is why it's called third-order (primary Seidel) aberration.

Because $\omega >0$ for both glasses, the balance ${\omega }_1{P}_1+{\omega }_2{P}_2=0$ can only hold if the powers $P_1$ and $P_2$ have opposite signs — and by convention $P>0$ means converging, $P<0$ means diverging.

The pinhole passes only a thin near-axial pencil (tiny $h$, tiny $\theta$), so the $h^3$ marginal error is negligible — the same reason stopping a lens down sharpens it.

A parabola is defined as the locus of equal optical path length to its focus, so all axial parallel rays arrive in phase at one point with zero ${\theta }^3$ error; a sphere only mimics this near its centre.

Reflection obeys ${\theta }_i={\theta }_r$ with no refractive index, so there is no $n(\lambda )$ and hence no colour-dependent focus. Mirrors are inherently free of chromatic aberration (a big reason large telescopes use mirrors).

Grinding controls the shape factor $K$ (the radii $R_1,R_2$), which fixes spherical-type errors, but colour spread comes from the material's $n(\lambda )$. You cannot polish dispersion away — you need a second, differently-dispersing glass.

It still splits colours. For a diverging lens $f<0$ (and $P<0$); since $f\propto 1/(n-1)$, blue's larger $n$ makes $|f|$ shorter, so blue's virtual focus sits nearer the lens on the incoming side. The displacement runs the opposite way to a converging lens — which is exactly why a diverging flint element can cancel a converging crown one in a doublet.

Exactly zero — the on-axis ray has $\theta =0$, so the ${\theta }^3$ term vanishes. Spherical aberration grows only as you move off-axis in height.

It vanishes: by the chromatic-shift relation $\frac{df}{f}=-\frac{dn}{n-1}=0$ (see the formula callout above), so all colours share one focus. No real glass is like this, but it shows chromatic aberration is entirely a dispersion effect.

Both the longitudinal and lateral spans collapse (blur $\sim h^3$), so the circle of least confusion shrinks toward the paraxial point — the image becomes ideal but dim.

No — paraxial kills spherical and single-colour kills chromatic, so both vanish. This is exactly the idealized regime in which the thin-lens formula is honest.

No — a small secondary spectrum remains for the uncorrected wavelengths, because $n(\lambda )$ isn't perfectly linear; correcting three colours (apochromat) reduces but never fully erases it.