Newton’s Rings
A
B
Plano convex lens
Air film
P
O
Q
Glass plate
Thickness of air film is zero at point of contact O.
Reflected light
L
Air film
G
Circular interference fringes are formed by reflected light
Microscope
45º
S
glass
L
Lens
Air film
Glass plate
Interference occurs between the light reflected
from the lower surface of the lens and the upper
surface of the glass plate G.
Since the convex side of the lens is a spherical
surface, the thickness of the air film will be constant over a circle (whose centre will be at O ) and
we will obtain concentric dark and bright rings.
Newton’s rings
It should be pointed out that in order to observe
the fringes the microscope has to be focused on the
Surface of the film.
Each ring will be locus of all such points where
thickness is same.
Condition for bright ring will be

2t cos r  (2n  1)
2
For air film ,  =1 and for near normal incidence
r is very small and hence cos r = 1
Thus,

2 t  (2n  1)
2
Where n = 0,1,2,3….
For dark rings,
2t cos r  n
Again for air film  = 1 and for small r we have
Condition for dark rings,
2t  n
O
R
R - tn
A
tn
tn
M
N
rn
B
R = radius of curvature of lens
t = thickness of air film at a
distance AB =rn
OA = R - t
From OAB
R2 = (R – t)2+rn2
rn2 = R2 - (R – t)2 = R2 - R2 –t2 +2Rt = 2Rt – t2
As R>>t , rn2 = 2Rt
 t = rn2/2R
So condition for bright rings
2t = (2n+1)/2
2r

or,
 (2n  1)
2R
2
(2n  1)R
2
 rn 
2

2n  1R
 rn 
2
2
n
 Diameter D n  2
n = 0,1,2,3,….
2n  1R
2
Similarly for dark rings,
2 t  n
2
n
r
2
 n
2R
 rn  nR
Diameter of dark rings,
D n  2rn  2 nR
n = 0,1,2,3,….
CENTER IS DARK
At n = 0, radius of dark ring = 0.
R
radius of bright ring =
2
Alternately dark and bright rings will be produced.
The spacing between second and third dark rings
is smaller than the spacing between the first and
second one.
Consider the diameter of dark rings
D1  2 1R  2 R
D 2  2 2R  2 2R
D3  2 3R  2 3R
D 4  2 4R  4 R
D 4  D1  2 R
Four fringes
D9  2 9R  6 R
D16  2 16R  8 R
D16  D9  2 R  8 fringes
Fringe width decreases with the order of the
Fringe and fringes get closer with increase in
their order.
Wavelength determination
Radius of the nth dark ring rn is given by
r  n R
2
n
2
n
D

 n R
4
2
 D n  4nR.....(1)
Similarly for (n+m)th dark band
D
2
n m
 4n  mR....(2)
(2) – (1)
D
2
n m
 D  4n  m R  4nR
2
n
 4mR

D
D
4mR
2
n m
2
n
Suppose diameter of 6th and 16th ring are
Determined then, m = 16-6 = 10
So
D D

4 10  R
2
16
2
6
Radius of curvature can be accurately measured
with the help of a spherometer and therefore
by measuring the diameter of dark or bright ring
you can experimentally determine the wavelength.
Newton’s rings with transmitted light
Transmitted light
Condition for bright fringes
2t cos r  n
Condition for dark fringes

2t cos r  (2n  1)
2
For air as thin film and near normal incidence
 = 1 and cosr = 1
So for bright fringes, 2t = n
2n  1
For dark fringes, 2 t 

2
2
r
But we know that t 
, r = radius of ring
2R
For bright rings
2
2r
 n  r  nR
2R
For dark rings
2r

(2n  1)R
 (2n  1)  r 
2R
2
2
2
If we put n = 0 then r = 0 for the bright ring
So for Newton's rings for transmitted rays the
central ring will be bright.
CENTRAL RING IS BRIGHT.
WAVELENGTH DETERMINATION
We know radius of the nth dark ring rn is
r  n R
2
n
2
n
D

 n R
4
2
 D n  4nR …….(1)
Similarly,
D
2
n m
 4(n  m)R
……..(2)
(2) – (1)
D
2
n m
 D  4(n  m)R  4nR
 4mR

2
n
……..(3)
D
D
4mR
2
n m
2
n
Suppose diameter of the 8th and 18th ring are
Determined then,
m = 18-8 =10 and
D D

4 10  R
2
18
2
8
Refractive index determination
air
Diameter of the dark rings
D  4mR
2
n
D
D
2
n m
2
n m
 4(n  m)R
 D  4(n  m)R  4nR
2
n
 4mR

Liquid of refractive index 
(for near normal incidence and g<
Condition for dark ring formation
2t n  n
'2
n
r
but t n 
2R
'2
n
r
nR
'2
 2
 n  r n 
2R

D
 
 2
'
n
2
 nR
4nR
'2
 
 Dn 
…..(4)



Similarly we can get
D
So, (5) – (4)
'2
n m
4(n  m)R …….(5)


4mR
D
D 

4mR
   '2
'2
D nm  D n
'2
n m
'2
n
…….(6)
This is the value of  if  is known.
 Can also be determine if  is unknown
We have from equation (3) and (6)
D
2
n m
 D  4(n  m)R  4nR
2
n
 4mR
D
'2
n m
……..(3)
4mR
D 

'2
n
…….(6)
Divide (3)/(6)

D
D
D
D
2
n m
'2
n m
2
n
'2
n