1. Newton’s rings between two curved surfaces

2. t be the thickness of air film at point P is PQ.D R1 P
Air film Q L x
t be the thickness of air film at point P is PQ.
t = PQ = PL - QL

3. A B C D P Q R1 R2 x
Air film L
PL = x2/2R AND QL = x2/2R2

4. The effective path difference between the two interfering
Rays in reflected light, for normal incidence is in case of
Bright rings
n = 0,1,2,3…..

5. Assume that the nth bright ring passes through the point P.
If rn is the radius of nth bright ring, then x = rn

6. If Dn be the diameter of nth bright ring then

7. For dark ring of reflected light
similarly
Assume that the nth dark ring passes through the point P.
If r’n is the radius of nth bright ring, then x = r’n

8. If D’n be the diameter of nth dark ring then

9. R1 P o L x R2 Q

10. o P L Q R1 R2 x

11. For nth bright ring

12. For nth dark ring

13. T S Q P’ i N i
AIR U’ U A C O i r M t  r r r L E B L’
AIR r P

14. Here interference pattern will not be perfect
Because intensities AT and CQ will not be the same
And their amplitude are different.
Amplitude depends on amount of light reflected
and transmitted through the films.
Intensity never vanishes completely and perfectly
dark fringes will not be observed.
But for multiple reflection intensity of minima
will be zero.

15. 5 4 1 2 3
ar1
atrt’
atr7t’
atr5t’
atr3t’ a
rarer
atr3
atr7
atr5
atr
denser at
atr2
atr4
atr6
atr8
atr2t’
atr4t’
atr6t’
att’
rarer

16. Amplitude of incident ray a
Reflection coefficient = r1 and r
Transmission coefficient from rarer to denser
medium = t
Transmission coefficient from denser to rarer
medium = t’
The amplitudes of the reflected rays are
ar, atrt’, atr3t’, atr5t’, atr7t’……

17. When ray 1 is reflected from the surface of
denser medium it undergoes a phase change 
Rays 2, 3, 4… are all in phase but out of phase
with ray 1 by 
Resultant amplitude of 2, 3, 4, 5 .. is
A = atrt’+atr3t’+atr5t’+atr7t’+…..
= att’r[1+r2+r4+r6+……]
= att’r

18. According to the principle of reversibility
tt’ = 1- r and r = - r1
So the resultant amplitude of 2,3,4,… is equal in
Magnitude of the amplitude of ray 1 but out of
Phase with it.

19. One of the important applications of the thin film interference is reducing the reflectivity of lens surface. ai ar n1
n1> n2 at n2

20. ai, ar, at are amplitudes of incident,reflected and
transmitted waves.
n2>n1 , ar is negative showing that reflection
occurs at a denser medium a phase change  comes

21. r and t are reflection and transmission
coefficients

22. r’a a n2 n1
t’a

23. These are the Stokes’ relations.

24. Non reflecting films
Reflectivity is the fraction of incident light reflected by a surface for normal incidence.
Reflectivity depends upon the refractive index  of the material. It is given by
For glass  = 1.5.
Reflectivity = 0.04
4% of incident light is reflected for normal incidence. Remaining 96% is transmitted.

25. The loss of energy due to reflection is one major reason of clarity reduction. There is also a reduction in the intensity of the images since less light is transmitted through the lenses.
When films are coated on lens of prism surface the reflectivity of these surfaces is appreciably reduced.
Initially the coating were made by depositing several monomolecular layers of an organic substance on glass plates.
Now it is done by either evaporating calcium or magnesium fluoride on the surface in vacuum or by chemical treatment of the surface with acids which leave a thin layer of silica on the surface.
No light is destroyed by non reflecting film, but there is redistribution means decrease in reflection results increase in transmission.

26. Thickness of nonreflecting thin film1 2 r a
air(a)
film(f) t
glass(g)
g f> a

27. 2ftcosr = n 2ftcosr = (2n+1)/2 2ft = (2n+1)/2 thickness, n = 0
Two interfering beams will interfere constuctively if
2ftcosr = n
Rays will interfere destructively if
2ftcosr = (2n+1)/2
For normal incidence <r=900
2ft = (2n+1)/2
So 2ft = / for min
thickness, n = 0

28. If a film having thickness of /4f and having refractive index less than that of the glass is coated on glass, then waves reflected from the upper surface of the film destructively interfere with the waves reflected from the
lower surface of the film. Such a film known as a non reflecting film.

29. na nf ng ar 1 2 ar’tt’ a r, t r’, t’ ar’t at Amplitude of ray 1 = ar
Amplitude of ray 2 = ar’tt’

32. For complete destructive interference ray 1 and 2 must
Have the same amplitude, i.e.

34. This equation gives the estimate of refractive index of
This film which should be coated on a surface to reduce
Its reflectivity. If na= 1 (for air) and ng = refractive index
of glass then

35. A soap film of refractive index  is illuminated with white light incident at an angle i. the light refracted by it is examined and two bright bands focused corresponding to wave lengths 1 and 2. show that the thickness of the film is
White light falls normally upon a film of soapy water whose thickness is 5 x 10-5 cm and refractive index is what wavelength in the Visible region will be reflected more strongly?