Newton’s rings between two curved surfaces

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

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

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…..

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

If Dn be the diameter of nth bright ring then

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

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

R1 P o L x R2 Q

o P L Q R1 R2 x

For nth bright ring

For nth dark ring

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

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.

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

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’……

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

According to the principle of reversibility tt’ = 1- r12 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.

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

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

r and t are reflection and transmission coefficients

r’a a n2 n1 t’a

These are the Stokes’ relations.

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.

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.

Thickness of nonreflecting thin film 1 2 r a air(a) film(f) t glass(g) g f> a

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 = /2 for min thickness, n = 0

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.

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’

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

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

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 1.33. what wavelength in the Visible region will be reflected more strongly?