Newton's rings: why circular, transmitted

Why the rings are circular

A plano-convex lens of large radius of curvature R rests on a flat glass plate. The air film between them has a thickness t that depends only on the distance r from the point of contact, because the lens surface is a sphere (symmetric about the axis through the contact point). All points at the same r have the same thickness, so the loci of constant path difference are circles centred on the contact point.

Theory for transmitted light

        light
          |
     _____|_____        lens (radius of curvature R)
    /     |     \
   ---------------      <- air film of thickness t
   ===============      glass plate
        P  O  P'

The ray transmitted straight through the film and the ray that is reflected twice inside the film (once at the plate, once at the lens surface) interfere. Both of these reflections take place at an air-to-glass boundary (first on the plate, then on the lens surface), and each gives a phase change of π. The two changes cancel, so there is no net phase change.

1. Path difference at normal incidence = 2μt (μ = 1 for air, so 2t).
2. Geometry: from the right triangle formed by the lens centre C, the foot of the perpendicular and the point on the lens surface, R² = (R − t)² + r² gives r² = 2Rt − t². Since t ≪ R, t² is neglected: t = r² / 2R.
3. Bright ring (maximum): 2t = nλ, so r²/R = nλ. r_n = √(nλR), D_n² = 4nλR (n = 0, 1, 2, ...; n = 0 is the bright centre).
4. Dark ring (minimum): 2t = (2n − 1)λ/2. r_n² = (2n − 1)λR / 2, D_n² = 2(2n − 1)λR (n = 1, 2, ...).

Conclusions

• The centre of the transmitted pattern is bright; it is the complement of the reflected pattern (dark centre).
• Diameter of a bright ring is proportional to √n, so D_n² ∝ n and the rings get closer as n increases.
• Contrast is low in transmitted light because the two interfering beams have very different intensities.
• Wavelength can be found from D²_(n+p) − D²_n = 4pλR.