Special case of interference in air film of variable thickness

1 / 20

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

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Special case of interference in air film of variable thickness' - howard-walker

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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.

 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

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

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