UNIT I Interference - · PDF fileUNIT I Interference 1.1 Formation of Newton’s rings in...

$: UNIT I Interference - · PDF fileUNIT I Interference 1.1 Formation of Newton’s rings in reflected light. ... Newton’s ring experiment can be used to determine the refractive index$
Physics I Keystone Institute of Technology & Management, Surajgarh Unit I

By:- Manvendra Singh

UNIT I Interference 1.1 Formation of Newton’s rings in reflected light. Formation of Newton’s rings – When a plano-convex lens of large radius of curvature is placed with its convex surface in contact with a plane glass plate, an air-film of gradually increasing thickness from the point of contact is formed between the upper surface of the plate and the lower surface of the lens. If monochromatic light is allowed to fall normally on this film, then alternate bright and dark concentric rings with their centre dark are formed. These rings are known as Newton’s rings. The fringes are circular because the air film is symmetrical about the point of contact of the plano-convex lens with the plane glass plate. Newton’s rings are formed because of the interference (by division of amplitude) between the waves reflected from the top and bottom surfaces of an air-film formed between the plano-convex lens and the plate.

1.2 Refractive index of a liquid with the help of Newton’s rings

experiment with necessary formula. Newton’s rings in reflected light – We know that when monochromatic light

falls normally on a plano-convex lens resting on a plane glass plate, alternate bright and dark concentric rings with dark centre are formed due to waves reflected from the top and bottom surfaces of an air-film or any other medium of refractive index µ between the plano-convex lens and plane glass plate. For reflected system, the effective path difference is given by



2 cos 22 2

t r t

(since light is falling normally,cos 1r ) …1

At the point of contact 0t , therefore, effective path difference 2

This is the condition for minimum intensity. Hence, the centre of Newton’s rings is dark. For constructive interference (bright fringes/maxima)

22

t n

n = 0, 1, 2, 3…..

Or 2 2 12

t n

…2

For destructive interference (dark fringes/minima)

2 2 12 2

t n

n = 0, 1, 2, 3. ….

Or 2 t n …3

Diameters of Bright Rings –

From the above figure, , nOP R PN r and nON R t

Therefore, 2 2 2ON PN OP

Or 2 2 2

n nR t r R



Or 2 2 2 22n n nR t Rt r R

Or 2

2 nn

rt

R neglecting 2

nt , since nt is small …4

Determination of refractive index μ of a liquid - Newton’s ring experiment can be used to determine the refractive index μ of a liquid. The liquid whose refractive index is to be determined is placed between the plano-convex lens L and the glass plate P of the Newton’s ring set-up. In case liquid is rarer than glass, a phase change of π will occur at reflection from the lower surface of the liquid, but

if the liquid is denser than glass, phase change will occur at reflection from the upper surface of the film. Hence, in both the cases, path difference will be equal to λ/2.

therefore, effective path difference 2 cos2

t r

for normal incidence 0,cos 1r r

Or effective path difference 22

t

and from eq…4 2

2 nn

rt

R

for nth bright fringe, 22

t n

Or 2 1

22

nt

Or 2 2 1

2

nnr

R

Or 22 1

2n

nr R

…5

If nd is diameter of the thn ring, then 2n nd r

therefore, 2

2 2 1n

nd R

…6

If n pd is the diameter of th

n p ring,

then,

22 2 1

n p

n pd R

…7



Or 2 2 4pn p n

liquid

Rd d

…8

Since, 1 , for air

therefore, 2 2 4n p nair

d d p R …9

Or dividing eq…9 by eq…8

2 2

n p n

2 2

n p n

d d

d d

air

liquid

…10

By measuring diameters of thn and thn p rings for medium as air and liquid and

substituting the values in the eq…10, refractive index of the liquid can be determined.

since, liquid > 1, n liquidd < n air

d

therefore, when liquid is introduced between the lens and the plate, the diameters of the rings decrease, that is, rings are contracted.

1

filmair in ring same theofdiameter

film liquidin ring a ofdiameter …11

1.3 Construction and working of Michelson’s interferometer and explain how circular and localized (straight) fringes are produced with it. Construction and working - Michelson’s interferometer – is a device that can be used to measure lengths or changes in length with great accuracy by means of interference fringes. The basic principle of this instrument was given by A.A. Michelson in 1881 according to which when a parallel beam of monochromatic light coming from an extended source is incident on a half silvered glass plate (also called as beam splitter), it is divided into two parts. One part is reflected wave and the other part is a refracted wave and both are coherent. In this experiment, coherent waves are produced by the method of division of amplitude. These waves proceed in the perpendicular directions and are incident normally on the two mirrors. After reflections from these mirrors, they superpose and produce interference fringes, which are observed with the help of a telescope as shown in the figure.



Form of fringes – In Michelson’s interferometer, the form of fringes depends on the separation d between M1 and M2

΄ and the shape of hypothetical air film formed between M1 and M2

΄, which is virtual image of M2. Circular fringes – are produced when the mirrors M1 and M2 are perpendicular to each other and thickness of air film between M1 and M2’ is not equal to zero, that is 0d . If 0d then, the whole pattern becomes dark.

Appearance of fringes in the Michelson’s interferometer as the mirrors are moved away from each other. Arrows on the far right figure indicate motion of the fringes.



If thickness of air film is d , the light waves reflected from the mirror M1 and M2

and reaching towards the telescope will coming parallel from M1’ and M2’ and will be equal to 2d . If these parallel waves make an angle θ with the normal, the path

difference between them will be 2 cosd .

We know that when a wave is reflected from a denser medium and another wave

are reflected from a rarer medium, path difference of 2

is created between them.

Hence, effective path difference between these waves will be 2 cos2

d

.

If 2 cos2

d n

n = 1, 2, 3…

Or 2 cos 2 12

d n

…1

Then a bright fringe will form due to constructive interference. Same condition will be at all points on the circle of inclination θ and bright fringe will appear circular.

If the effective path difference 2 cos 2 12

d n

n = 0, 1, 2…

Or 2 cosd n …2

Then a dark fringe will form due to destructive interference. Same condition will be at all points on the circle of inclination θ and dark fringe will appear circular. Hence, alternate bright and dark circular fringes are observed. Radii of circular fringes- It is clear that in the fringe system of Michelson’s interferometer, for given d , as n increases, cos increases and hence

decreases, that is, order of fringes increases towards center and decreases as



we move away from it. For central fringe 0 and order is n , then order of the

successive fringes from the central fringe are 1 , 2 , 3 ...n n n and so on.

Then from eq…2 2d n …3

If 1st, 2nd, 3rd… mth circular fringes subtend semi-angles 1 2 3, , ... m respectively

from the telescope, then

1

2

3

2 cos 1

2 cos 2

2 cos 3

2 cos m

d n

d n

d n

d n m

…4

Thus, if fringes are counted from the central fringe (assuming its order zero), then subtracting eq…4 from eq…3, we get

2 1 cos md m m = 1, 2, 3… …5

Or cos 12

m

m

d

…6

If radius of mth fringe is rm and final image of circular fringes is observed at a distance D (least distance of distinct vision), then

2 2cos 1

2m

m

D m

dr D

Or

12 2

1 12

m

mr D

d

…7

If angle m is very small, or 2m d , then using binomial approximation we get

m

mr D

d

…8

That is, near the central fringe, radius of fringes is directly proportional to square root of natural numbers. Localized fringes – are formed when mirrors are not orthogonal, that is, M1 and M2’ are not exactly parallel. A wedge shaped air film is formed between them giving rise to fringes of equal thickness. The path of the two waves reflected from mirrors M1 and M2’ and originating from a single wave, are no more parallel but intersect near M1 as shown in the figure below and so fringes are localized near



M1. The shapes of these fringes are curved with convex side towards thin edge of the wedge. As mirror M2’ is moved gradually the air film wedge varies successively and fringes change the shape and when mirrors M1 and M2’ intersect each other, fringes become straight as shown in the figure.

1.4 How Michelson’s interferometer can be used to determine

wavelength of light. Determination of wavelength of monochromatic light – For this,

monochromatic light from source is allowed to fall on half silvered plate A and Michelson’s interferometer is adjusted for circular fringes. Then, mirror M1 is moved such that AM1 = BM2. The mirror M1 and M2 are made perfectly perpendicular to each other. Thus, concentric circular fringes are observed through telescope. Let the separation between real mirror M1 and virtual mirror M2’ is such that bright

fringe of thn order is formed at the center of the field of view and let reading of

micrometer screw is say 1x .

Then, path difference,

02 cos0d n

Or 2d n …1

Where d is separation between M1 and M2’.



Adding on both sides of the eq…1, we get

2 1d n

Or 2 12

d n

…2

From the above eq…2, it is observed that when d becomes2

d

, the thn

fringe at the center is replaced by 1th

n fringe. We can also say that if M1 is

moved by distance2

, one fringe is displaced in the telescope. Now the mirror M1

is gradually moved and number of fringes displaced is counted and reading of

micrometer screw is say 2x . If M1 is moved through distance 2 1x x x and the

number of fringes displaced is N . That is, by moving the mirror by2

, the number

of fringes displaced is one.

Therefore, on moving the mirror by distance 2 1x x x , the number of fringes

displaced will be

2x

N

Or wavelength 2x

N …3

Hence, by knowing the values of x and N experimentally, wavelength of

monochromatic light used can be calculated.

1.5. How Michelson’s interferometer can be used to determine

separation between two close wavelengths.

Determination of separation between two close wavelengths - For this,

light is allowed to fall on half silvered plate A and Michelson’s interferometer is

adjusted for circular fringes. Let two wavelengths 1 and 2 are very close to each

other. The two wavelengths form their separate fringe patterns, but because of very small difference in wavelengths, the two patterns overlap. As the mirror M1 is moved slowly, the two patterns separate out slowly and when the path

difference is such that the dark fringe due to 1 falls on the bright fringe due to 2 ,

the result is maximum indistinctness. When the path difference is such that,



bright fringe due to1 falls on the bright fringe due to

2 , or vice-versa, the result

is maximum distinctness.

Let the mirror M1 is moved through a distance 2 1x x x between two positions

1x and 2x of successive distinctness. In this position thn fringe due to

1 must

coincide with 1th

n fringe due to 2 . Therefore,

21

1

2 2

nnx

Or 1

2xn

…1

And 2

21

xn

…2

Subtracting eq...1 from eq…2, we get

2 1

1 11 2x

Or 1 2

1 2

1 2x

Or 1 21 2

2x

Or 2

1 22x

…3

Where 1 2 is geometric mean of the two wavelengths. Thus, by

measuring the distance x moved by the mirror M1, the difference between two

close wavelengths can be determined.

06. Compare the rings formed by Michelson’s interferometer and

Newton’s rings. 1. The fundamental difference between the two is that in Michelson’s interferometer rings originate as locus of equal inclination (also called as Haidinger’s fringes) whereas the Newton’s rings are locus of the air film of equal thickness (also called as Fizeau fringes).



2. In Michelson’s interferometer rings are located at infinity and are therefore viewed by a telescope whereas Newton’s rings are located in the plane of the film and hence viewed by traveling microscope. 3. The air film in Michelson’s interferometer is imaginary (hypothetical) whereas in Newton’s rings experiment it is real. 4. Center of circular rings in Michelson’s interferometer can be dark or bright whereas in Newton’s rings, in case of reflected light it is dark and in case of transmitted light it is bright. 5. In Michelson’s interferometer, order of the rings decrease when one moves outwards from the center whereas in Newton’s rings order of the rings increase when one moves away from the center. 6. In both, Michelson’s interferometer and Newton’s rings, the thickness of the rings decreases as radius of the rings increases, which is a common feature.

07. Write short note on anti-reflection coating.

Anti-reflection coating - Whenever a ray of light moves from one medium to

another, for example, when light enters a sheet of a glass after traveling through air, some portion of the light is reflected from the surface (known as interface) between the two media. The strength of the reflection depends on the refractive indices of the two media as well as the angle of the surface to the beam of light. When the light meets the interface at normal incidence (perpendicularly to the surface), the intensity of light reflected is given by the reflection coefficient or reflectance R .

If 1 and 2 are refractive indices of the two media, then reflectance, R is given

by

2

2 1

2 1

R

…1

It is clear from the above eq…1 that reflection will not occur if 1 2

One of the practical applications of the interference phenomenon is the anti-reflection coating on the glass. The reflection from a lens or a prism can be decreased to a minimum by coating a thin transparent film of proper refractive index and proper thickness.



The idea behind anti-reflection coatings is that the creation of a double interface by means of a thin transparent film gives two reflected waves. If these waves are of nearly equal amplitude and out of phase, they partially or totally cancel. If the coating is of quarter wavelength thickness and has refractive index less than that of the glass then the two reflections are 180 degrees out of phase and complete destructive interference occurs and no reflected waves will emerge from the film.

The thickness of coating and refractive index is chosen in such a way that light waves reflected from the two layers have the same amplitude and out of phase so as to cancel one another.

If refractive index of coating be c , that of glass be g and that of air be 0 , then

the amplitude of reflected wave from the first surface (air to coating) is given by

2

0

1

0

c

c

R

…2

and the amplitude of reflected wave from the second surface (coating to glass) is given by

2

2

g c

g c

R

…3

The condition of equality of amplitude, that is, 1 2R R , at two reflections yield,

22

0

0

g cc

c g c

…4

Or 0

0

g cc

c g c

Or 2 2

0 0 0 0c g c g c c g c g c



Or 2

02 2c g

Or 0c g …5

Or c g since, 0 1 (for air) …6

That is, refractive index of coating should be equal approximately to the geometric mean of refractive indices of media on either side. The -phase condition gives the thickness of the coating film to be

2 2 12

ct n

…7

For minimum thickness, 1n

Or 22

ct

Or 4 c

t

…8

That is, optical thickness of the coating must be equal to one-quarter of a wavelength. Thus transparent coating satisfying eq…6 and eq…8 eliminates reflection completely. The best material known for this is MgF2 for which refractive index 1.38 . To have suitability at multi-wavelengths, for example,

white light, multi-layer coating is used. Each layer is optically quarter wave thick.

08. Write short note on interference filters.

Interference filters - are multilayer thin-film devices. They can be designed to function as an edge filter or band pass filter. In either case, wavelength selection is based on the property of destructive light interference. This is the same principle underlying the operation of a Fabry-Perot interferometer. Incident light is passed through two coated reflecting surfaces. The distance between the reflective coatings determines which wavelengths destructively interfere and which wavelengths are in phase and will ultimately pass through the coatings. If the reflected beams are in phase, the light is passed through two reflective surfaces. If, on the other hand, the multiple reflections are not in phase, destructive interference reduces the transmission of these wavelengths through the device to near zero. This principle strongly attenuates the transmitted intensity of light at wavelengths that are higher or lower than the wavelength of interest.



In many spectroscopic studies, it is required to have a narrow frequency band of light of width about 100Å or less, centered on a chosen wavelength of visible light. It can be obtained by an interference filter. In an interference filter, a thin transparent dielectric spacer like magnesium fluoride (MgF2) or cryolite is sandwiched between glass plates. Reflecting surfaces are coated by extremely thin semi-transparent layers of a good reflecting material like silver, deposited by vacuum evaporation method or a dielectric of desired characteristics. When a beam of light is incident normally on the filter, multiple reflections take place within the film.

Path difference between successive pair of emergent parallel rays is 2 t , for

normal incidence. With white light, the transmitted beam will be maximum for only those wavelengths which satisfies the condition 2 t n n = 1, 2, 3……

Or 2

nt

Where, is refractive index of the dielectric and t is its thickness.

If the effective thickness of the spacer is integral multiple of half of the desired wavelength, then other wavelengths will be attenuated by destructive interference and wavelength ,2 ... will be transmitted through the filter. If for a

particular thickness there are two maxima in the visible region, one of them can be eliminated by using colored glass filter. However, if the angle of incidence is and angle of refraction in the spacer is , then, the wavelength of light passing

through the filter can be obtained from 2 cost n



2

2

sin2 1t n

using

sin

sin

and 2 2sin cos 1

For 1n , 2

2

sin2 1t

Or 2

0 2

sin1

where

0 2 t

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