Newton’s rings in reflected light
Δ  2 μ t cosr 
λ
2
Interference is maximum
2 t  (2 n  1 )
interference is minimum
t
r
2
bright fringe
2R
dark fringe
rn 
rn 
(2n  1)
2
2t  nλ
λR
2
nR
λ
Newton’s rings in transmitted light
No additional phase change of π
(or path difference of λ/2) in
transmitted rays
For bright rings
2  t cos r  n 
2r
2
 n  r 
nR
2R
For dark rings
2  t cos r  ( 2 n  1)

2
2r
2
2R
 ( 2 n  1)

2
 r
( 2 n  1)  R
2
CENTRE
IS BRIGHT
For n=0, r=0 where
t is zero
 condition is for maxima
Newton’s rings formed by two curved surfaces
Case I: Lower surface
concave
Case II: Lower surface
convex
Newton’s rings formed by two curved surfaces
Case I: Lower surface
concave
Newton’s rings formed by two curved surfaces
Case I: Lower surface
concave
t1 
r
2R 1
r
t1
O
T
2
Newton’s rings formed by two curved surfaces
Case I: Lower surface
concave
Newton’s rings formed by two curved surfaces
Case I: Lower surface
concave
r
t2
t2 
r
2
2R 2
Newton’s rings formed by two curved surfaces
Case I: Lower surface
concave
t1 
r
2
2R 1
t=t1-t2
r
2  t cos r  m 
t2 
2  ( t 1  t 2 ) cos r  m 
B right Fringe
D ark Fringe
 1
1 

r 


(2
m

1)

R
R
2
 1
2 
2
 1
1 
r 

  m
 R1 R 2 
2
m =0,1,2,...
r
2
2R 2
Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex
Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex
t1 
r
t1
r
2
2R 1
Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex
Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex
t1
r
t2
t2 
r
2
2R 2
Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex
  2  t co s r 

2
 2  ( t 1  t 2 ) co s r 
t1 

2
t2 
r
B right Fringe
D ark Fringe
 1
1 

r 

  (2 m  1)
2
 R1 R 2 
2
 1
1 
r 

  m
 R1 R 2 
2
m =0,1,2,...
r
2
2R 1
r
2
2R 2
Newton’s rings formed by two curved surfaces
Case 1: Lower surface concave
• Two curved surfaces of radii of curvature R1 and R2 in contact at point O.
• Thin air film of variable thickness enclosed between two surfaces.
• The dark and bright rings depending on the path difference
• The thickness of air film at P is
Case 1: Lower surface concave
From geometry
r2 = 2 R t
therefore
But PQ = t. the condition for dark rings in reflected light is given
by
2tcos r =m
For air (µ = 1) and normal incidence cos r =1 , then above equation
become
2t=m
Case 1: Lower surface concave
dark rings
For bright fringes the condition is
For air (µ = 1) and normal incidence
equation become
, then above
bright rings
Case II: Lower surface convex
But PQ = t. The condition for dark rings in reflected light is given by
For air (µ = 1) and normal incidence, then above equation become
Case II: Lower surface convex
dark rings
For bright fringes the condition is
For air (µ = 1) and normal incidence, then above equation become
bright rings
How can we make centre bright in reflected rays?
Two ways:
1. By using a liquid film with refractive index µliquid
with condition µconvex lens< µliquid < µplate.
Ex: crown glass=1.45, flint glass=1.63
Liquid with 1.45 < µ <1.63
Liquid film
2. By lifting convex lens upward with a distance λ/4.
Because 2t=nλ (dark)
→
2[t + (λ/4)] = nλ+ λ/2=(2n+1)λ/2 (bright)
Numerical: refractive index
•
•
•
In a Newton’s ring experiment the diameter of the 12th dark
ring changes from 1.40 cm to 1.27 cm when a liquid is
introduced between the lens and the plate. Calculate the
refractive index of the liquid.
=1.215
In Newton’s ring exp., the diameter of 4th and 12th dark rings
are 0.4 and 0.7 cm, what will be the diameter of 20th dark
ring.
D20=0.905cm
If the diameter of nth ring change from 0.3cm to 0.25 cm
after filling a liquid b/w the lens and plate, find out the
refractive index of liquid.
= 1.44
Numerical: Two curved surfaces
•
The convex surface of radius 40 cm of a plano-convex lens
rests on the concave spherical surface of radius 60 cm. If
the Newton’s rings are viewed with reflected light of
wavelength 6000 Å, calculate the radius of 4th dark ring.
D4 = 1.697 mm
•
Newton’s rings by reflection are formed between two
plano-convex lenses having equal radii of curvature being
100 cm each. Calculate the distance between 5th and 15th
dark rings for monochromatic light of wavelength 5400 Å
in use.
D15 - D5 = 1.701mm