= Solution
\b[Reflection and central spot.] At a boundary from lower to higher <refractive index>, the near-normal electric-field reflection coefficient is negative, $r=(n_1-n_2)/(n_1+n_2)$. This gives a $\pi$ <wave phase> reversal on reflection from the glass plate through the air film. Reflection at the lens's glass-to-air underside has no such reversal. These are consequences of the <Fresnel equations> for the reflected <electric field>.
At ideal contact the film thickness is zero, so the two reflected contributions have zero geometrical <optical path difference> but a relative $\pi$ phase reversal. They produce <destructive interference>, and the \b[central spot is dark]. This conclusion concerns the film reflection; other lens surfaces can add a weak background.
\b[Rings and their radii.] At radial distance $r$, let the air thickness be $t$. The two relevant rays reflect at the upper and lower film interfaces. The lower reflection accumulates an extra round-trip <optical path length> $2t$, plus the relative $\pi$ reflection shift. As $t$ increases away from contact, alternating constructive and destructive <thin-film interference> forms <Newton's rings>. At near-normal incidence, a bright ring satisfies
$$
\frac{2\pi}{\lambda}(2t)+\pi=2\pi(m+1),\qquad
2t=\left(m+\frac12\right)\lambda,\qquad m=0,1,2,\ldots.
$$
The extended source can illuminate these localized film fringes: each incident contribution interferes with its own two reflected paths; unrelated source points need not be mutually coherent.
For the spherical lens surface, geometry gives $r^2+(R-t)^2=R^2$, so $r^2=2Rt-t^2$. Since $t\ll R$, $t\simeq r^2/(2R)$. Combining the equations yields
$$
\boxed{r_m=\sqrt{R\lambda\left(m+\frac12\right)}.}
$$
The $m=0$ bright ring is the first bright ring around the dark contact point.
\b[Curvature from ring diameters.] Writing $D_m=2r_m$, subtract the squared diameters to eliminate the unknown initial order:
$$
D_{m+20}^2-D_m^2=4R\lambda(20).
$$
Using <SI> units,
$$
\boxed{R=\frac{(7.1\times10^{-3})^2-(3.3\times10^{-3})^2}
{80(500\times10^{-9})}=0.988\,\mathrm m\simeq0.99\,\mathrm m.}
$$
\b[Liquid-filled film.] With film <refractive index> $n=1.4$, still smaller than both glass indices, there remains exactly one relative reflection phase reversal. The round-trip <optical path length> is now $2nt$, so
$$
r_m^2=\frac{R\lambda}{n}\left(m+\frac12\right).
$$
Here $\lambda=500\,\mathrm{nm}$ is the illuminating vacuum wavelength, approximately its air value, rather than its shorter value inside the liquid. The ring labelled $m=2$ therefore has diameter
$$
\boxed{D_2=2\sqrt{\frac{(0.988)(500\times10^{-9})(2+1/2)}{1.4}}
=1.878\times10^{-3}\,\mathrm m\simeq1.9\,\mathrm{mm}.}
$$
Filling the gap contracts each ring radius by $1/\sqrt n$ at fixed order and curvature.
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