1 / 20

Special case of interference in air film of variable thickness

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.

howard-walker
Download Presentation

Special case of interference in air film of variable thickness

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Newton’s Rings Special case of interference in air film of variable thickness

  2. Newton’s Rings

  3. Newton’s Rings due to Reflected Light • Interference is maximum • ∆ = n λ, • bright fringeis produced. • interferenceisminimum • A darkfringeis produced.

  4. Newton’s Rings

  5. Newton’s Rings

  6. Newton’s Rings

  7. 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

  8. RADII of DARK FRINGES bright fringe at Q is dark fringe at Q is

  9. DIAMETER OF THE RINGS Dn = 2rn rn2 2Rt Diameter of the nth dark ring Diameter of the nth bright ring

  10. Mathematical Analysis of Newton’s Rings B 2R-t C R E D A t O K r

  11. 1. CENTER IS DARK

  12. Each maximum and minimum → a locus of constant film thickness fringes of equal thickness. But  the rings are unevenly spaced.

  13. 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.

  14. 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

  15. 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 

  16. 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

  17. On subtracting For air Therefore  Liquid of refractive index 

  18. Newton’s rings in transmitted light The condition for maxima (brightness) is

  19. And for minima (darkness) is For air (µ = 1) and normal incidence , for maxima (brightness) is for minima (darkness) is

More Related