1. Newton’s Rings

2. AB P Q O Air film Plano convex lens Glass plate Thickness of air film is zero at point of contact O.

3. Reflected light Air film L G Circular interference fringes are formed by reflected light

4. S 45º Microscope glass Lens Air film Glass plate L

5. Interference occurs between the light reflected from the lower surface of the lens and the upper surface of the glass plate G. Since the convex side of the lens is a spherical surface, the thickness of the air film will be cons- tant over a circle (whose centre will be at O ) and we will obtain concentric dark and bright rings. Newton’s rings

6. It should be pointed out that in order to observe the fringes the microscope has to be focused on the Surface of the film. Each ring will be locus of all such points where thickness is same.

7. Condition for bright ring will be For air film,  =1 and for near normal incidence r is very small and hence cos r = 1 Thus, Where n = 0,1,2,3….

8. For dark rings, Again for air film  = 1 and for small r we have Condition for dark rings,

9. O R A B tntn rnrn R - t n tntn MN

10. R = radius of curvature of lens t = thickness of air film at a distance AB =r n OA = R - t From  OAB R 2 = (R – t) 2 +r n 2  r n 2 = R 2 - (R – t) 2 = R 2 - R 2 –t 2 +2Rt = 2Rt – t 2 As R>>t, r n 2 = 2Rt  t = r n 2 /2R

11. So condition for bright rings 2t = (2n+1) /2 n = 0,1,2,3,….

12. Similarly for dark rings, n = 0,1,2,3,…. Diameter of dark rings,

13. CENTER IS DARK At n = 0, radius of dark ring = 0. radius of bright ring = Alternately dark and bright rings will be produced. The spacing between second and third dark rings is smaller than the spacing between the first and second one.

14. Consider the diameter of dark rings Four fringes

15. Fringe width decreases with the order of the Fringe and fringes get closer with increase in their order.

16. Wavelength determination Radius of the nth dark ring r n is given by Similarly for (n+m)th dark band

17. (2) – (1)

18. Suppose diameter of 6 th and 16 th ring are Determined then, m = 16-6 = 10 So Radius of curvature can be accurately measured with the help of a spherometer and therefore by measuring the diameter of dark or bright ring you can experimentally determine the wavelength.

20. Newton’s rings with transmitted light

22. Transmitted light

23. Condition for bright fringes Condition for dark fringes For air as thin film and near normal incidence  = 1 and cosr = 1 So for bright fringes, 2t = n For dark fringes,

24. But we know that, r = radius of ring For bright rings For dark rings

25. If we put n = 0 then r = 0 for the bright ring So for Newton's rings for transmitted rays the central ring will be bright. CENTRAL RING IS BRIGHT.

26. WAVELENGTH DETERMINATION We know radius of the nth dark ring r n is …….(1)

27. Similarly, ……..(2) (2) – (1) ……..(3)

28. Suppose diameter of the 8 th and 18 th ring are Determined then, m = 18-8 =10 and

29. Refractive index determination air Diameter of the dark rings

30.  Liquid of refractive index 

31. (for near normal incidence and  g <  Condition for dark ring formation …..(4)

32. Similarly we can get So, This is the value of  if is known. …….(5) (5) – (4) …….(6)

33.  Can also be determine if is unknown We have from equation (3) and (6) ……..(3) …….(6)

34. Divide (3)/(6)