Newton’s rings between two
curved surfaces
o
C R2
A
B
R1
P
0
x
Q
L
D
Air film
t be the thickness of air film at point P is PQ.
t = PQ = PL - QL
B
C R2 A R1
0
PL = x2/2R1
x
AND
P
Q
L
D
Air film
QL = x2/2R2
x
x
x  1
1 
 

t




2 R1 2 R2
2  R1 R2 
2
2
2
The effective path difference between the two interfering
Rays in reflected light, for normal incidence is in case of
Bright rings
2t  2n  1
n = 0,1,2,3…..

2
 x2  1
1 

  2n  1
2   

2
R
R
2
  1
2 

1
1 

2


 x  
 2n  1

2
 R1 R2 
Assume that the nth bright ring passes through the point P.
If rn is the radius of nth bright ring, then x = rn
2

1
1 

  2n  1
rn 


R
R
2
2
 1
2

1
1 

  2n  1
 rn 


R
R
2
2
 1
for   1
If Dn be the diameter of nth bright ring then
Dn
rn 
2
 1
1 


  2n  1

4  R1 R2 
2
22n  1
2
 Dn 
 1
1 



 R1 R2 
2
Dn
For dark ring of reflected light
2 t  n 
similarly
2

1
1 
  n
x 


R
R
2 
 1
Assume that the nth dark ring passes through the point P.
If r’n is the radius of nth bright ring, then x = r’n
, 2
r n 
1
1 
  n


R
R
2 
 1

, 2
r n 
1
1 
  n


R
R
2 
 1
for   1
If D’n be the diameter of nth dark ring then
 1
1 

  n

4  R1 R2 
4n
2
 D 'n 
 1
1 



 R1 R2 
2
D 'n
R1
P
o
L
x
R2
Q
t  PQ  PL  QL
R1
P
o
L
x
R2
x2
PL 
2 R1
Q
x2
QL 
2 R2
x2
x2
x2  1
1 
 

t

t 
2 R1 2 R2
2  R1 R2 
For nth bright ring
 1
1 


  2n  1

4  R1 R2 
2
22n  1
2
 Dn 
 1
1 



 R1 R2 
2
Dn
For nth dark ring
 1
1 

  n

4  R1 R2 
4n
2
 D 'n 
 1
1 



 R1 R2 
2
D 'n
T
S
Q
P’
i
i
A
U
N
C
O i
AIR
U’
r
t
L
r r r
E
B
r
P
M

AIR
L’
Here interference pattern will not be perfect
Because intensities AT and CQ will not be the same
And their amplitude are different.
Amplitude depends on amount of light reflected
and transmitted through the films.
Intensity never vanishes completely and perfectly
dark fringes will not be observed.
But for multiple reflection intensity of minima
will be zero.
1
2
atrt’
ar1
a
5
4
3
atr3t’
atr5t’
atr7t’
rarer
atr3
atr7
atr5
atr
denser
at
atr2
att’
atr2t’
atr4
atr6
atr4t’
atr8
atr6t’
rarer
Amplitude of incident ray a
Reflection coefficient = r1 and r
Transmission coefficient from rarer to denser
medium = t
Transmission coefficient from denser to rarer
medium = t’
The amplitudes of the reflected rays are
ar, atrt’, atr3t’, atr5t’, atr7t’……
When ray 1 is reflected from the surface of
denser medium it undergoes a phase change 
Rays 2, 3, 4… are all in phase but out of phase
with ray 1 by 
Resultant amplitude of 2, 3, 4, 5 .. is
A = atrt’+atr3t’+atr5t’+atr7t’+…..
= att’r[1+r2+r4+r6+……]
= att’r
 1    att' r 
1  r 2  1  r 2 
According to the principle of reversibility
tt’ = 1- r12
A
and r = - r1
a (1  r
2
)r
(1  r
2
)
 ar   ar1
So the resultant amplitude of 2,3,4,… is equal in
Magnitude of the amplitude of ray 1 but out of
Phase with it.
One of the important applications of the thin
film interference is reducing the reflectivity of
lens surface.
ar n1 n > n
1
2
ai
at
n2
n1  n 2
ar 
a i  a ir
n1  n 2
2 n1
at 
ai  ait
n1  n 2
ai, ar, at are amplitudes of incident,reflected and
transmitted waves.
n2>n1 , ar is negative showing that reflection
occurs at a denser medium a phase change  comes
n1  n 2
2n1
r
,t 
n1  n 2
n1  n 2
r and t are reflection and transmission
coefficients
a
r’a
t’a
n2
n1
n 2  n1
2n 2
r' 
 r, t ' 
n 2  n1
n1  n 2
4n1n 2
1  tt'  1 
2
( n1  n 2 )
2
 n1  n 2 
2
  r
 
 n1  n 2 
These are the Stokes’ relations.
Non reflecting films
• Reflectivity is the fraction of incident light
reflected by a surface for normal incidence.
• Reflectivity depends upon the refractive index 
of the material. It is given by
  1


  1
2
• For glass  = 1.5.
• Reflectivity = 0.04
• 4% of incident light is reflected for normal
incidence. Remaining 96% is transmitted.
• The loss of energy due to reflection is one major reason of
clarity reduction. There is also a reduction in the intensity of
the images since less light is transmitted through the lenses.
• When films are coated on lens of prism surface the
reflectivity of these surfaces is appreciably reduced.
• Initially the coating were made by depositing several
monomolecular layers of an organic substance on glass
plates.
• Now it is done by either evaporating calcium or magnesium
fluoride on the surface in vacuum or by chemical treatment
of the surface with acids which leave a thin layer of silica
on the surface.
• No light is destroyed by non reflecting film, but there is
redistribution means decrease in reflection results increase
in transmission.
Thickness of nonreflecting thin film
1
a
r
2
air(a)
t
film(f)
glass(g)
g f> a
Two interfering beams will interfere constuctively if
2ftcosr = n
Rays will interfere destructively if
2ftcosr = (2n+1)/2
For normal incidence <r=900
2ft = (2n+1)/2
So 2ft = /2
for min
thickness, n = 0

t 
4 f
If a film having thickness of /4f and having
refractive index less than that of the glass is
coated on glass, then waves reflected from the
upper surface of the film destructively
interfere with the waves reflected from the
lower surface of the film. Such a film known as
a non reflecting film.
ar
r, t
1
a
2
ar’tt’
na
nf
r’, t’
at
ar’t
ng
Amplitude of ray 1 = ar
Amplitude of ray 2 = ar’tt’
na  nf
2n a
r
,t 
na  nf
na  nf
r'
n f  ng
n f  ng
,t ' 
2n f
na  n f
na  nf
2n a
ar 
a, at 
a
na  nf
na  nf
nf  ng
2n a
2n f
ar' tt' 


a
nf  ng na  nf na  nf
nf  ng
2n a n f


a
2
(n a  n f ) n f  n g
For complete destructive interference ray 1 and 2 must
Have the same amplitude, i.e.
nf  ng
na  nf
4n f n a
a

a
2
na  nf
na  nf  nf  ng
4n f n a
is verynearlyqeual to unity
2
n a  n f 
nf  ng
na  nf
a
a
na  nf
nf  ng
 nanf  nang  n  nf ng
2
f
 nanf  nang  n  nf ng
2
f
 n  na ng
2
f
 nf 
na ng
This equation gives the estimate of refractive index of
This film which should be coated on a surface to reduce
Its reflectivity. If na= 1 (for air) and ng = refractive index
of glass then
n f  refractiveindex of thinfilm
 ng
 nf  ng
• A soap film of refractive index  is illuminated with white
light incident at an angle i. the light refracted by it is
examined and two bright bands focused corresponding to
wave lengths 1 and 2. show that the thickness of the film
is

12 
1
t

1  2  2  2  sin 2 i






•White light falls normally upon a film of soapy water whose
thickness is 5 x 10-5 cm and refractive index is 1.33. what
wavelength in the Visible region will be reflected more
strongly?