Newton's rings with a liquid film: index, ring motion, central spot
In a Newton's rings experiment performed in reflected sodium light of wavelength 589.3 nm, a plano-convex lens of radius of curvature 800 cm rests on a flat glass plate. The student measures the diameter of the tenth dark ring as 1.37 cm with air in the gap. She then places a drop of a transparent liquid between the lens and the plate and measures the diameter of the same order dark ring as 0.98 cm. She reports the refractive index of the liquid as 1.96 and adds two remarks: (i) "the rings expand when the liquid is introduced, because the liquid has a higher refractive index than air"; (ii) "the centre of the pattern is bright because the two interfering beams travel equal optical paths there". The laboratory manual warns that diameters must be measured with a travelling microscope and that the order is counted from the central spot.
1. Obtain the refractive index of the liquid from the two diameters, and list the assumptions needed to justify the formula you use.
Read the worked answer
For reflected light with a liquid film of index mu, the dark rings satisfy 2mut = mlambda, with the film thickness t = r^2/(2R) for r much smaller than R. Hence r_m^2 = mlambda*R/mu, so d ∝ mu^(-1/2).
Therefore mu = (d_air/d_liquid)^2 = (1.37/0.98)^2 = (1.398)^2 = 1.955, i.e. about 1.96. The data are self-consistent with the quoted value.
Assumptions: both diameters refer to the same order m = 10; the contact is ideal (t → 0 at the centre); r << R so the parabolic approximation to the spherical surface holds; the liquid completely fills the gap and wets both surfaces; the refractive index of air is 1 to four figures. A useful check is that the ring order used must be counted from the centre, since a misidentified order shifts mu appreciably - the ratio method cancels m, so the result is robust to that error only when the same m is used in both measurements.
2. Evaluate remark (i): do the rings expand or contract when the liquid is introduced, and why?
Read the worked answer
The rings contract. The condition for a dark ring is that the round-trip optical path 2mut equals a whole number of wavelengths (with the extra pi phase change supplied by reflection at the lower, denser surface). For a fixed order m, mu*t must stay constant; raising mu from 1 to about 1.96 means the required thickness falls by a factor of about 2, so the ring moves to a smaller radius, and since r ∝ 1/sqrt(mu) every diameter shrinks by sqrt(1.96) = 1.40.
This matches the data: 1.37/0.98 = 1.40. The student has confused the size of the ring with the optical path; a larger refractive index packs more optical path into the same physical gap, so the interference condition is reached nearer the contact point. The remark is therefore qualitatively wrong even though the numerical value she quotes is right.
3. Evaluate remark (ii): what is actually seen at the centre in reflected light, and what changes in transmitted light?
Read the worked answer
At the centre the film thickness tends to zero, so the geometrical path difference tends to zero. But the beam reflected from the lower surface (glass or liquid against glass, i.e. a rarer-to-denser reflection) gains an extra phase of pi, whereas the beam reflected at the upper surface of the film does not (in the air case). The two beams therefore arrive in antiphase and the centre is dark, not bright. The same argument with the liquid film holds as long as the liquid index is below that of the glass; if the liquid index exceeded the glass index the extra pi shift would disappear and the centre would become bright - a useful diagnostic.
In transmitted light the pattern is complementary: the two transmitted beams differ by 2mut with no net extra pi shift, so at t → 0 they are in phase and the central spot is bright.