Interference-Newtons Ring Experiment

Interference: Newton's Ring Experiment

s

INTERFERENCE LECTURE BY DR.T.VISHWAM

� Newton's rings are an example of fringes of equalthickness

� Newton's rings are formed when a Plano-convexlens P of a large radius of curvature placed on asheet of plane glass AB is illuminated from the topwith monochromatic light

� The combination forms a thin circular air film ofvariable thickness in all directions around the pointof contactof thelensandtheglassplate

s

G

M

L

of contactof thelensandtheglassplate

�The locus of all points corresponding to specificthickness of the air film falls on a circle whosecentre id at O. Consequently, interference fringes areobserved in the form of a series of concentric ringswith their centre at O

�Newton originally observed the concentric circularfringes and hence they are called Newton's rings

P

A Bo

S: Source

L: Lens

G: Glass plate

P: Plano convex lens

AB: Plane glass plate

M: Microscope


Newton's ring exp setup

Newton's ring experimental set up


� Monochromatic light from extended source S is

rendered parallel by a lens L. It is incident on a glass

plate inclined at 45o to the horizontal, and is

reflected normally down onto a Plano convex lens

placed on a flat glass plate

� Part of the light incident on the system is reflected

from the glass to air boundary and remaining light is

transmitted through the air film. It is again reflected

form the air to glass boundary

� The two rays reflected from the top and bottom of

s

G

P

M

L

A Bo

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� The two rays reflected from the top and bottom of

the air film are derived through division of

amplitude form the same incident ray and they are

coherent

� The ray 1 and ray 2 are close to each other and

interfere to produce darkness or brightness

� The condition of brightness and darkness depends

on the path difference between the two reflected

rays, which in turn depends on the thickness of the

air film at the point of incidence

A


Condition for Bright and Dark rings

The optical path difference between the rays is given by

Δ = 2µt cos r –λ/2

Since µ=1 and cos r=1 for normal incidence of light

Δ = 2t –λ/2

Δ = mλIntensity maxima occur when the optical path difference Δ = mλ

Then

2t=(2m+1) λ/2 CONDITION FOR BRIGHT FRINGE

Intensity minima occur when the optical path difference Δ = (2m+1)λ/2

2t-λ/2=(2m+1) λ/2 Hence

2t= mλ CONDITION FOR DARK FRINGE


Radii of Dark fringes

R

N

R

M

Prm

t

Let R be the radius of curvature of the lens

Let a dark fringe be located at Q

Let the thickness of the film at Q be PQ=t

Let the radius of the circular fringe at Q be OQ= rm

By the Pythagoras theorem PM2= PN2+MN2

OQ

t

rm

By the Pythagoras theorem PM2= PN2+MN2

R2= rm2+ (R-t)2

rm2= 2Rt-t2

As R>>t, then 2Rt >> t2

therefore rm2 = 2Rt

The condition for darkness at Q is that

2t= mλ

rm2 = mλR

rm= (mλR)1/2


1

2

3

The radii of dark fringes can be found by inserting values m= 1,2, 3

then

r = 1 R

r = 2 R

r = 3 R

It means that the radii of dark rings are propotional to under root of

λ

λ

λ the

natural numbers

The above relation also implies that

m

The above relation also implies that

r = λ

thThe radius of the m dark ring is proportional to under root of wavelength

Ring Diameter thm m

m

m

Diameter of m dark ring D =2r

D = 2 2Rt

D = 2 m Rλ

SPACING BETWEEN FRINGES

m

It is seen that the diameter of the dark ring is given by

D = 2 m R

where m=1,2,3......

The diameters of dark rings are propotional to the square root of the nat

λ

ural numbers

Therefore , the diameter of the ring does not increase in the same proportion as the order of the ring

for ex, if m increases as 1 ,2,3,4......

1

2

3

4

D = 2 R

D = 2 2 R 2(1.4) R

D = 2 3 R 2(1.7) R

D = 2

λ

λ λ

λ λ

=

=

4 R 2(2) R and so on...

Therefore, the rings get closer and closer as m increases

This is why the rings are not evenly spaced

λ λ=

Fringes of equal thicknessIt is observed that path difference between the reflected

rays arises due to the variation in the thickness ‘t’ of the air

film

Reflected light will be of minimum intensity for those

thickness for which the path difference is mλ and maximum

intensity for those thickness for which the path difference isintensity for those thickness for which the path difference is

(2m+1)λ/2.

Thus each maxima and minima is a locus of constant film

thickness.

Therefore the fringes are known as fringes of equal

thickness


Dark is central spot

Since the thickness at the center is zero (t=0)

Δ = 2t –λ/2

Δ= -λ/2

The wave reflected from the lower

surface of the air film suffers a phasesurface of the air film suffers a phase

change of π while the wave reflected

from the upper surface of the film does

not suffer such change

Thus the two interfering waves at the

centre are opposite in phase and produce

a dark spot


Determination of wave length of lightA Plano-convex lens of large radius of

curvature ( about 100 cm) and a flat glass

plate are cleaned. The lens is kept with its

convex surface on the glass plate and they

are held in position with the help of metal

ring arrangement

The system is held under a lower power

travelling microscope kept before a sodiumtravelling microscope kept before a sodium

vapor lamp. It is arranged that the yellow

light coming from the sodium falls on a

glass plate held at 45o light beam

The light is turned through 90o and is

incident normally on the lens-plate system

The microscope is adjusted till the circular

rings came in to focus


The centre of the cross wire is made to come into focus on the centre of the dark spot,

which is at the centre of the circular system

Now, turning the screw the microscope is moved on the carriage slowly towards one side,

say right side.

As the cross wires move in the field of view, dark ring is counted say it is 20th ring. Then

note down the reading and move slowly towards 19th ring take the reading. Like this note

down the value up to 1st ring and move slowly towards other side i.e., left side note down

the 1st dark ring reading and continue the same procedure up to 20th dark ring

The difference between the readings of theThe difference between the readings of the

right and left sides of the 5th dark ring gives the

diameter value

The procedure is repeated till the 20th ring is

reached and its reading is noted. From the

value of the diameters the square of the

diameters are calculated


The graph is plotted between Dm2 and the ring

number ‘m’. A straight line would be obtained

Dm2

m m+p

p

D (m+p)2 -Dm

2

(m+p)

(m+p)

2m

th

2

2 2m

we have

D 4

for the (m+p) ring,

D 4( )

D D 4

m R

m p R

p R

λ

λ

λ

=

= +

∴ − =

(m+p)

2 2mD D

4

The slope of the straight line gives the value of 4 R

pRλ

λ

−=

slope =

4RThe radius of curvature R of the lens may be determ ined

using spherometer and is computed w ith the help of the

above equations

λ

λ


Refractive index of the liquidT h e liq u id , w h o se re fra c t iv e in d e x is to b e d e te rm in e d , is f i l led in th e g a p b e tw e n th e le n s a n d p la n e g la ss p la te .N o w th e liq u id f i lm su b s t i tu tes th e a ir f i lmT h e c o n d it io n fo r in te rfe rn ce m a y b e w r i tte n a s

2µ tc o s r = mλ - D a rk n e ssw h e re µ is th e re f ra c t iv e in d e x o f th e f i lm s in c e 2µ t = mλ

22 2µ rr t = , = mλ 2 R 2 R

m λR2o r r =µ

4 mλR2 D =µ

4 mλR2 4 mλR2D =m µLthfo r th e (m + p ) r in g ,

4 (m + p )λR2 D =µ(m + p ) L

4 PλR2 2D - D =m µL(m + p )

2 2B u t w e k n o w th a t D - D = 4 PλRm(m + p ) a ir

2 2D - D m(m + p ) a ir µ =

D(m + p

2 2- D m L iq u id)


Thank you

References:

A text book of OPTICS by N.Subrahmanyam, Brijlal and M.N. Avadhanulu

Optics by Ajoy Ghatak


Interference-Newtons Ring Experiment

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