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Newton’s Rings. Special case of interference in air film of variable thickness. Newton’s Rings. Newton’s Rings due to Reflected Light. Interference is maximum ∆ = n λ , bright fringe is produced. interference is minimum A dark fringe is produced. Newton’s Rings. Newton’s Rings.
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Newton’s Rings Special case of interference in air film of variable thickness
Newton’s Rings due to Reflected Light • Interference is maximum • ∆ = n λ, • bright fringeis produced. • interferenceisminimum • A darkfringeis produced.
RADII of DARK FRINGES radius of curvature of the lens thickness of the air film dark fringe be located R2 =rn2 + (R-t)2 As R >> t rn2 = 2Rt – t2 , since 2Rt >> t2 rn2 2Rt radius of the circular fringe
RADII of DARK FRINGES bright fringe at Q is dark fringe at Q is
DIAMETER OF THE RINGS Dn = 2rn rn2 2Rt Diameter of the nth dark ring Diameter of the nth bright ring
Mathematical Analysis of Newton’s Rings B 2R-t C R E D A t O K r
Each maximum and minimum → a locus of constant film thickness fringes of equal thickness. But the rings are unevenly spaced.
SPACING BETWEEN THE FRINGES The diameter of the dark rings is Therefore, Hence, the rings get closer and closer as the order of rings i.e. ‘n’ increases. This causes the rings to be unevenly spaced.
DETERMINATION OF WAVELENGTH OF LIGHT Diameter of the mth dark ring Dm2 = 4 mλR Diameter of the (m+p)th dark ring Dm+p2 = 4 (m+p) λR Dm+p2 - Dm2 = 4 pλR The slope is, Thus, R may be determined by using a spherometer λ is calculated. Dm2 Dm+p2-Dm2 m m+p m
Refractive index of a Liquid The liquid whose refractive index is to be determined is filled between the lens and glass plate. Now air film is replaced by liquid. The condition for interference then (for darkness) For normal incidence Therefore Liquid of refractive index
Liquid of refractive index Therefore Following the above relation the diameter of mth dark ring is Similarly diameter of (m+p)th ring is given by
On subtracting For air Therefore Liquid of refractive index
Newton’s rings in transmitted light The condition for maxima (brightness) is
And for minima (darkness) is For air (µ = 1) and normal incidence , for maxima (brightness) is for minima (darkness) is