Interference is maximum interference is minimum

1. Newton’s rings in reflected light • Interference is maximum • interferenceisminimum 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. 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 = 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