1. Newton’s rings in reflected light
Interference is maximum
interference is minimum
bright fringe
dark fringe

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
For dark rings
CENTRE IS BRIGHT
For n=0, r=0 where t is zero  condition is for maxima

3. Newton’s rings formed by two curved surfaces
Case I: Lower surface concave
Case II: Lower surface
convex

4. Newton’s rings formed by two curved surfaces
Case I: Lower surface concave

5. Newton’s rings formed by two curved surfaces
Case I: Lower surface concave r t1
O T

6. Newton’s rings formed by two curved surfaces
Case I: Lower surface concave

7. Newton’s rings formed by two curved surfaces
Case I: Lower surface concave r t2

8. Newton’s rings formed by two curved surfaces
Case I: Lower surface concave
t=t1-t2 r

9. Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex

10. Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex r t1

11. Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex

12. Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex t1 t2 r

13. Newton’s rings formed by two curved surfaces
Case II: Lower surface
convex r

14. 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

15. Case 1: Lower surface concave
From geometry
therefore
But PQ = t. the condition for dark rings in reflected light is given by
For air (µ = 1) and normal incidence cos r =1 , then above equation become
r2 = 2 R t
2tcos r =m
2t=m

16. 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

17. 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

18. 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

19. → 2[t + (λ/4)] = nλ+ λ/2=(2n+1)λ/2 (bright)
How can we make centre bright in reflected rays?
Two ways:
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)

20. 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

21. 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 = 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