Newton’s rings between two curved surfaces

o

0

A BC DP

Q

R1R2

x

Air film

L

t be the thickness of air film at point P is PQ.

t = PQ = PL - QL

0

A BC DP

Q

R1R2

x

Air film

L

PL = x2/2R1 AND QL = x2/2R2

21

2

2

2

1

2 11

222 RR

x

R

x

R

xt

The effective path difference between the two interferingRays in reflected light, for normal incidence is in case ofBright rings

2

122 nt

n = 0,1,2,3…..

2

1211

212

11

22

21

2

21

2

nRR

x

nRR

x

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

1for 2

1211

212

11

21

2

21

2

nRR

r

nRR

r

n

n

If Dn be the diameter of nth bright ring then

2n

nD

r

21

2

21

2

11

122

212

11

4

RR

nD

nRR

D

n

n

For dark ring of reflected light

nt 2similarly

nRR

x

21

2 11

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

1for 11

11

21

2,

21

2,

nRR

r

nRR

r

n

n

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

21

2

21

2

11

4'

11

4

'

RR

nD

nRR

D

n

n

o

P

L

Q

R1

R2

x

QLPLPQt

o

P

L

Q

R1

R2

x

21

2

2

2

1

2

2

2

1

2

11

222

2QL

2

RR

xt

R

x

R

xt

R

x

R

xPL

For nth bright ring

21

2

21

2

11

122

212

11

4

RR

nD

nRR

D

n

n

For nth dark ring

21

2

21

2

11

4'

11

4

'

RR

nD

nRR

D

n

n

S

A

B

C

N

P

E

T

AIR

AIR

t

i

r

r

i

r r

L L’

M

i

Q

O

P’

rU U’

Here interference pattern will not be perfectBecause intensities AT and CQ will not be the sameAnd 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.

a

at

att’

ar1

atr

atr2

atr3

atr4

atr5

atr6

atr7

atr8

atr2t’ atr4t’ atr6t’

atrt’atr3t’ atr5t’ atr7t’

1 2 3 4 5

rarer

denser

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 arear, 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

2r1

1

2r1

r'att

According to the principle of reversibility

tt’ = 1- r12 and r = - r1

12

2

)1(

)1(arar

r

rraA

So the resultant amplitude of 2,3,4,… is equal inMagnitude 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

at

n1

n2

n1> n2

taann

n2a

raann

nna

ii21

1t

ii21

21r

ai, ar, at are amplitudes of incident,reflected andtransmitted waves.

n2>n1 , ar is negative showing that reflection occurs at a denser medium a phase change comes

21

1

21

21

nn

n2t,

nn

nnr

r and t are reflection and transmissioncoefficients

n2

n1

a r’a

t’a

21

2

12

12

nn

n2't,r

nn

nn'r

2

2

21

21

221

21

rnn

nn

)nn(

nn41'tt1

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.

2

1

1

• 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.

ar

t

12

air(a)

film(f)

glass(g)

Thickness of nonreflecting thin film

g f> a

Two interfering beams will interfere constuctively if

2ftcosr = nRays will interfere destructively if

2ftcosr = (2n+1)/2For normal incidence <r=900

2ft = (2n+1)/2

So 2ft = /2 for min thickness, n = 0

f4t

If a film having thickness of /4f 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 thelower surface of the film. Such a film known as a non reflecting film.

ar

ar’tt’

1 2

na

nf

ng

a

atar’t

Amplitude of ray 1 = arAmplitude of ray 2 = ar’tt’

r, t

r’, t’

fa

a

fa

fa

nn

n2t,

nn

nnr

2' , 'f g f

f g a f

n n nr t

n n n n

ann

n2at,a

nn

nnar

fa

a

fa

fa

ann

nn

)nn(

nn2

ann

n2

nn

n2

nn

nn'tt'ar

gf

gf2

fa

fa

fa

f

fa

a

gf

gf

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

ann

nn

nn

nn4a

nn

nn

gf

gf2

fa

af

fa

fa

unity toqeualnearly very isnn

nn42

fa

af

gf2fgafa

gf2fgafa

gf

gf

fa

fa

nnnnnnn

nnnnnnn

ann

nna

nn

nn

gaf

ga2f

nnn

nnn

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

gf

g

f

nn

n

film thin ofindex refractiven

• 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

it

2221

21

sin2

1

•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?