Newton's ring

Let s be a source of monochromatic light. light from s are made parallel be a lense L which are then allowed to fall on a glass plate inclined at 45degree with horizontal part of light will be trasnmitted and some part will be reflected down. The reflected lights then allowed to fall on the plane surface of a plano convex lense place on a base plate with convex surface touching base plate. Now, some part of light will be reflected from the lower(convex) surface of plano-convex lens and some parts will be reflected from upper surface of base plate. During this process, there will be interference in reflected lights because there is a thin film of air introduced in between plano-convex lense and base plate. The fringes will be circular and called as Newton's ring.
Here interference is because of thin film of air So,
for bright fringr
2mew t cosr = (2n+1)/2 * lamda-------------1
n = 0,1,2
For dark fringe
2 mew t cos r = n*lamda -------------------2
n=0,1,2
Diagram

Let a ring be observed at height(thickness) OD = t at a distance r from the point of contact.
Now, from fig. we can have
AD*DB = ED*OD
r*r = (EO-OD)*OD
or r^2 = (2R - t)* t
As t is very small, we can have
r^2 = 2Ry
t = r^2/2R

Now, condition for bright fringe will be
2mew * r^2/2R * cosr = (2n+1)/2 * lamda
if r is very small, cosr = 1
or r^2 = 2n+1/2* R*lamda/mew
r =???? --------------4mew = 1
n= 0,1,2.....

For DArk fringe
2mew * r^2/2R = n*lamda
r=??????
n = 0,1,2......