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
bright fringe is produced.
interference is minimum
A dark fringe is produced.

4. Newton’s Rings

5. Newton’s Rings

6. Newton’s Rings

7. The experimental set up consists of a monochromatic light from an extended source S rendered parallel by a lens L.
It is incident on a glass plate inclined at an angle 45o to the horizontal and is reflected normally down onto a plano-convex lens placed on a flat glass plate.
Part of the light incident on the system is reflected from the glass-to –air boundary, say from point D.
The remainder of the light is transmitted through the air film.
It is again reflected from the air-to-glass boundary say at point J.
The two rays reflected from the top and bottom of the air film are derived through the division of amplitude from the same incident ray CD and are therefore, coherent.
The rays 1 and 2 are close to each other and interfere to produce dark and bright fringes.
The condition of brightness or darkness depends on the path difference between the two reflected light rays.
This in turn depends on the thickness of the air film at the point of incidence.

8.  radius of the circular fringe
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

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

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

11. Mathematical Analysis of Newton’s RingsB C D A E O K
2R-t
Let the radius of the curved surface of the lens be R.
Consider a point K on the plate, such that the thickness of the air film at this popint be ‘t’ .

12. 1. CENTER IS DARK DARK CENTRAL SPOT
In the reflected system the central spot is dark.
The optical path difference in the reflected system is
At the point of contact ‘O’ of the lens and the glass plate , the thickness of the air film is negligibly small compared to the wave length of incident light. Therefore, t  0
Hence ,
The wave reflected from the lower surface of the air film suffers a phase change of π while the wave reflected from the upper surface of the film does not suffer such change.
Therefore, the superimposing waves are out of phase by
which is equivalent to a phase difference of 180o or π radians.
Thus the two interfering waves at the centre are opposite in phase → a dark spot.

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

14. This causes the rings to be 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.

15. 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
m+p m
Dm+p2-Dm2

16. 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 

17. Following the above relation the diameter of mth dark ring is
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

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

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

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