For the unnormalized <Zernike polynomial> with $n=4,m=0$, the three radial terms are
$$
\frac{4!}{2!2!}\rho^4-\frac{3!}{1!1!1!}\rho^2+\frac{2!}{2!0!0!}=6\rho^4-6\rho^2+1.
$$
Since $\cos0\phi=1$,
$$
\boxed{Z_4^0(\rho,\phi)=6\rho^4-6\rho^2+1.}
$$
Along a central chord with signed coordinate $s\in[-1,1]$, replace $\rho$ by $|s|$. The profile is even, equals one at the center and both edges, and has minima $-1/2$ at $s=\pm1/\sqrt2$. This <Zernike spherical mode> corresponds to \b[primary <spherical aberration>], balanced by defocus and piston terms rather than simply the raw quartic Seidel term.

\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-338-zernike-spherical-mode.png]
{title=The unnormalized Zernike spherical mode on a central chord}
{description=The even quartic has central and edge values one and two minima of minus one half. Its lower-order terms remove its projections onto piston and defocus.}
{height=360}

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 338 / 2 / a / vii / Solution

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Spherical aberration makes rays at different pupil radii focus at different axial positions. Its primary pupil dependence is quartic; the balanced <Zernike spherical mode> also contains quadratic and constant terms to remove defocus and piston.

Spherical aberration

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{title2=$Z_4^0=6\rho^4-6\rho^2+1$}

The unnormalized rotationally symmetric mode is $Z_4^0(\rho)=6\rho^4-6\rho^2+1$. It describes balanced primary <spherical aberration>. The lower-degree terms make it orthogonal to constant piston and quadratic defocus under the disk area measure: direct integration gives $\int_0^1 Z_4^0\rho\,d\rho=0$ and $\int_0^1 Z_4^0(2\rho^2-1)\rho\,d\rho=0$.

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 Incoming links: Zernike spherical mode