Newton’s rings in reflected light

Δ



2 μ t cosr



λ 2



Interference is maximum

2t  

interference is minimum

t  r 2 2R bright fringe r n  2t  nλ λR 2 dark fringe r n  λ 2

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  2 r 2 2 R  n   r  n  R

For dark rings

2 r 2 2 R  2  t ( 2 n cos  1 ) r   2  ( 2 n r    1 ) 2 ( 2 n  1 )  R 2

For n=0, r=0 where t is zero



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 t 1  r 2 2R 1 r O T t 1

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 t 2 t 2  r 2 2R 2

Newton’s rings formed by two curved surfaces

Case I: Lower surface concave r t 1  r 2 2R 1 t=t 1 -t 2   1 t ) cos r 2 Bright Fringe r 2   1 R 1 Dark Fringe r 2   1 R 1  1 R 2  1 R 2       2 t 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 t 1  r 2 2R 1 r t 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 r t 1 t 2 t 2  r 2 2R 2

Newton’s rings formed by two curved surfaces

Case II: Lower surface convex   2 2 (t 1 t )cos r 2   2 r t 2 t 1  r 2  2R 1 2 r 2R 2 Bright Fringe r 2   1 R 1 Dark Fringe r 2   1 R 1  R  1 R 2 1 2       2

Newton’s rings formed by two curved surfaces Case 1: Lower surface concave

• • • • Two curved surfaces of radii of curvature R 1 and R 2 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

From geometry

Case 1: Lower surface concave

r 2

= 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 become 2t=m  cos r =1 , then above equation

Case 1: Lower surface concave

dark rings For bright fringes the condition is For air (µ = 1) and normal incidence , then above equation become 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 µ liquid with condition µ convex lens < µ liquid < µ plate .

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 4 th and 12 th dark rings are 0.4 and 0.7 cm, what will be the diameter of 20 th dark ring.

D 20 =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.

•

D 4 = 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.

D 15 - D 5 = 1.701mm