- 2069 Chaitra · 5 marks
Number of bright Newton's rings
Short answer
Bright ring radius r_n² = (2n−1)λR/(2μ). For air (μ=1), n ≈ 3396; for water (μ=1.33), n ≈ 4516.
In a Newton's ring experiment, the radius of curvature of the lens is 5cm and the lens diameter is 20mm. (a) How many bright rings are produced? Assume that λ=589nm (b) How many bright rings would be produced if the arrangement were immersed in water (μ=1.33)?
In reflected light, the nth bright Newton's ring has radius r_n² = (2n − 1)λR / (2μ), where μ is the refractive index of the film between lens and plate. The number of bright rings is the largest n for which r_n does not exceed the lens radius.
Given: R = 5 cm = 0.05 m, lens diameter 20 mm so r = 1 cm = 0.01 m, λ = 589 nm = 5.89 × 10⁻⁷ m.
(a) In air (μ = 1)
r² = (2n − 1) λR / 2
2n − 1 = 2r² / (λR)
= 2 × (0.01)² / (5.89×10⁻⁷ × 0.05)
= 2×10⁻⁴ / 2.945×10⁻⁸
= 6791.2
n = (6791.2 + 1) / 2 = 3396.1
Only whole rings count, so n = 3396.
Answer (a): about 3396 bright rings.
(b) In water (μ = 1.33)
The wavelength in the film is λ/μ, so r² = (2n − 1)λR / (2μ).
2n − 1 = 2μ r² / (λR)
= 1.33 × 6791.2
= 9032.3
n = (9032.3 + 1) / 2 = 4516.6
Answer (b): about 4516 bright rings.
The rings crowd closer together in water, so more of them fit inside the same lens aperture (ratio 4516/3396 ≈ 1.33 = μ).
Note: the centre is dark in reflected light, and the ring count is only a theoretical limit; in practice the rings become too closely packed to resolve long before this number. (If R = 5 m, as in the common textbook version of this problem, the same method gives 34 bright rings in air and 45 in water.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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