 Solution

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 central spot is dark. This conclusion concerns the film reflection; other lens surfaces can add a weak background.

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 satisfiesThe 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 yieldsThe $m=0$ bright ring is the first bright ring around the dark contact point.

Curvature from ring diameters. Writing $D_m=2r_m$, subtract the squared diameters to eliminate the unknown initial order:Using SI units,

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 $2nt$, soHere $\lambda =500\mathrm{nm}$ is the illuminating vacuum wavelength, approximately its air value, rather than its shorter value inside the liquid. The ring labelled $m=2$ therefore has diameterFilling the gap contracts each ring radius by $1/\sqrt{n}$ at fixed order and curvature.