- 2071 Chaitra · 5 marks
Newton's rings vs Haidinger fringes
Short answer
Newton's rings are concentric fringes of equal thickness in an air film between a lens and plate. Unlike Haidinger fringes (equal inclination, uniform film), they are localized on the film. In transmitted light, bright rings satisfy D_n² = 4nλR. Central dark is obtained by filling with a liquid of intermediate refractive index to introduce a π phase shift.
What are Newton's rings? How are they different from Haidinger fringes? Derive an expression for the diameter of bright rings in transmitted light. How can you obtain central fringe dark in this system?
Newton's rings are concentric circular fringes formed in the air film between a plano-convex lens of large radius R and a glass plate. They are fringes of equal thickness, localized in the film.
Difference from Haidinger fringes
| Newton's rings | Haidinger fringes |
|---|---|
| Fringes of equal thickness | Fringes of equal inclination |
| Formed by a film of varying thickness (air film between lens and plate) | Formed by a plane parallel film of uniform thickness |
| Localized on the film surface | Localized at infinity (seen with a telescope / lens) |
| Fringes crowd as the order increases (D² ∝ n) | Fringes are more widely spaced at the centre |
| Central ring dark (reflected light) | Centre bright or dark depending on t |
Bright rings in transmitted light
- At radius r, the air film thickness is t = r²/(2R).
- In transmitted light the two internal reflections each give a phase change of π, which cancel, so the path difference = 2t (normal incidence, air).
- Bright ring: 2t = nλ → r² = nλR.
- With D = 2r:
D_n² = 4 n λ R → D_n = 2√(nλR)
(n = 0 is the central bright spot.) The dark rings in transmitted light satisfy 2t = (2n+1)λ/2, D² = 2(2n+1)λR.
Central fringe dark in transmitted light
In transmitted light, the centre is normally bright (complementary to the reflected pattern). To make it dark, an extra phase difference of π must be introduced between the two interfering beams. This is done by filling the space between the lens and the plate with a liquid whose refractive index lies between those of the lens and the plate (e.g. lens of crown glass, μ = 1.5; plate of flint glass, μ = 1.7; liquid μ ≈ 1.6). The twice-reflected ray then suffers a phase change of π at the plate (liquid to denser flint glass) but none at the lens (liquid to rarer crown glass), so a net extra λ/2 appears. The pattern is reversed: the centre becomes dark in transmitted light (and bright in reflected light).
✓ Answer checked by Claude Opus 5.5 (high effort)
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