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Special case of interference in air film of variable thickness - PowerPoint PPT Presentation


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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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Presentation Transcript
slide1

Newton’s Rings

Special case of interference in air film of variable thickness

slide3

Newton’s Rings due to Reflected Light

                  • Interference is maximum
  • ∆ = n λ,
  • bright fringeis produced.
  • interferenceisminimum
  • A darkfringeis produced.
slide8

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

slide9

RADII of DARK FRINGES

bright fringe at Q is

dark fringe at Q is

slide10

DIAMETER OF THE RINGS

Dn = 2rn rn2 2Rt

Diameter of the nth dark ring

Diameter of the nth bright ring

slide13

Each maximum and minimum

→ a locus of constant film thickness

fringes of equal thickness.

But  the rings are unevenly spaced.

slide14

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.

slide15

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

slide16

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 

slide17

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

slide18

On subtracting

For air

Therefore

Liquid of refractive index 

slide19

Newton’s rings in transmitted light

The condition for maxima (brightness) is

slide20

And for minima (darkness) is

For air (µ = 1) and normal incidence ,

for maxima (brightness) is

for minima (darkness) is