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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

L4: Interference 62

4-5 THIN FILM INTERFERENCEInterference patterns can be observed whenever waves from two or more coherent sources cometogether. In Young's experiment the waves came from two separate sources but in thin filminterference, the waves come from one source. One wavefront is split into two parts which arerecombined after traversing different paths. Examples of thin film interference occur in oil slicks,soap bubbles and the thin layer of air trapped between two glass slabs. Here thin film means a layerof transparent material no thicker than several wavelengths of light.

Ray from onepoint on the source

Rays whichhave travelleddifferent opticalpaths

Figure 4.10

Thin filminterference

Figure 4.11

Interferencefringes in a soap

film

When light strikes one boundary of the film, some of it will be reflected and some will betransmitted through the film to the second boundary where another partial reflection will occur(figure 4.10). This process, partial reflection back and forth within the film and partial transmission,continues until the reflected portion of the light gets too weak to be noticed. The interference effectscome about when parts of the light which have travelled through different optical paths cometogether again. Usually that will happen when the light enters the eye.* Thus for example, light

* When light rays are brought to a focus either by the eye or a lens, there is no extra optical pathdifference introduced so the focussing has no effect on the conditions for the location of theinterference fringes.

L4: Interference 63

reflected back from the top surface of the film can interfere with light which has been reflected oncefrom the bottom surface and is refracted at the top surface.

The interference effect for monochromatic light, light or dark or somewhere in between, isdetermined by the amplitudes of the interfering waves and their phase difference. The conditions fora maximum or minimum in the irradiance are the same as before: a phase difference of m(2π) gives amaximum and a phase difference of (m + 12 )(2π) produces a minimum.

Change of phase at reflectionA new phenomenon reveals itself here. A straightforward interpretation of the conditions forinterference maxima and minima solely in terms of optical path difference gives the wrong answer!Two examples illustrate this point. In a very thin soap film it is possible to get a film thicknesswhich is much less than one wavelength. So the path difference between light reflected from the twosurfaces of the film is much less than a wavelength and the corresponding phase difference will bealmost zero. A zero phase difference should produce brightness, but the opposite is observed -when the film is very thin there is no reflection at all! The explanation is that whenever a light waveis reflected at a boundary where the refractive index increases, its phase jumps by π or half a cycle.In the case of the soap film, the light reflected from the first surface, air to soapy water, suffers aphase change, but light reflected at the water-air boundary has no phase change. You can observethis effect yourself in soap bubbles. Carefully watch the top of a bubble as the water drains away.As the film gets thinner you will see a changing pattern of coloured fringes. Just before the bubblebreaks, the thinnest part of the film looks black - indicating no net reflection.

Monochromatic light

Central dark fringe in reflected light

Figure 4.12. Newton's rings

The other example is a thin film interference pattern called Newton's rings which are formedusing a curved glass lens resting on a flat glass slab (figure 4.12). The thin film is the air betweenthe lens and the slab. The important feature is that where the optical path difference is zero, right inthe middle of the pattern where the lens actually touches the slab, there is darkness instead of abright fringe. The dark spot can be explained by saying that there is a phase change of π in the lightreflected at the boundary between air and glass.

L4: Interference 64

Analysis of thin film interferenceThe conditions for finding bright or dark fringes in a thin film clearly depend on the angle ofincidence of the light, but a useful approximation can be worked out assuming that the incident lightrays are normal to the surface, or almost so. In that case the optical path difference between parts ofan elementary wave reflected from the top and bottom surfaces of a film is just 2nb, where b is thethickness and n is the refractive index (figure 4.13).

n b

There is no extra optical path differencefrom here on when these rays areeventually brought together by a lens or an eye.

Figure 4.13. Calculating the optical path differenceFor near normal incidence, D = 2nb.

To work out the conditions for bright and dark fringes you have to include the effect of phasechanges at reflection. Each phase change of π has the same effect as the addition of an extra halfwavelength of optical path.No net phase change at reflectionIf there is no phase change at either boundary or a phase change at both boundaries (for example: afilm of water on glass), the conditions for maxima and minima arefor a bright fringe: 2nb = mλ ... (4.5a)

and for a dark fringe: 2nb =

m + l2 λ (m = 0, l, 2, 3, ...). ... (4.5b)

Phase change at one boundaryWhere there is a phase change at only one boundary (for example an air film trapped between twoglass plates or a soap bubble) the interference conditions depend on both the thickness and thephase change at reflection. The conditions are simply interchanged:

for a bright fringe: 2nb =

m + l2 λ ... (4.6a)

and for a dark fringe: 2nb = mλ (m = 0, l, 2, 3, ...). ...(4.6b)Notes• There is no point in trying to memorise these equations. It is better to work them out whenyou need them by combining the conditions expressed in terms of phase difference (equations 4.1aand 4.1b) with the phase changes at reflection and the relation between optical path and phasedifference.• It is important to remember that the value of wavelength to be used in these relations is thewavelength in vacuum (or air). If you need to know the value of the wavelength, λm, in the mediumwith refractive index n it can be calculated using the relation

λλm

= n .

L4: Interference 65

Example: fringe patterns in wedgesIf two flat glass plates are allowed to touch at one edge and are separated by a small object such as athin wire at the opposite edge, the space between the plates contains a wedge-shaped thin film of air(figure 4.14).

Monochromatic light

Glass

Glass

Wire

Figure 4.14. Interference fringes in a wedge of air

The vertical scale is greatly exaggerated.

When monochromatic light is shone down on to this arrangement, interference fringes will beobserved in the reflected light. Since the existence of a bright or dark fringe depends on thethickness of the film at a particular place, fringes will be seen at various places across the air wedge.The analysis above shows that the spacing of the fringes is proportional to the wavelength. For agiven wavelength each fringe follows a line or contour of constant thickness in the air film. If youfollow across the fringe pattern, the thickness of the film will change by λ/2n as you go from onefringe to the next. If the medium in the wedge is air then n = 1.000, so the fringe spacingcorresponds to a change in thickness of λ/2. This gives a way of measuring the thickness of the thinobject used to prop the plates apart if you already know the wavelength: just count the total numberof fringes across the whole wedge and multiply by λ/2. The resolution in this measurement is abouthalf a wavelength, or better, depending on how well you can estimate fractions of a fringe.Alternatively, you could use this method and a wire of known diameter to find the wavelength. Localisation of the fringesAlthough a narrow light source (the single slit) is needed to produce coherence in Young'sexperiment, thin film fringes can be formed using extended light sources, even daylight from thesky. The difference is that in thin film interference every incident wavefront, no matter where itcomes from, is split into two wavefronts when it meets the first surface of the film. When the twowaves meet again they have a definite phase relationship so that interference is seen to occur. Thephase difference between the waves is locally constant and the fringes are said to be localised. Youcan see that when you look at thin film fringes - they appear to be located in (or just behind) thefilm. Coloured fringesIf a thin film is illuminated with white light the reflected light will contain a continuous range offringe patterns corresponding to the spectrum of wavelengths in the light. You do not, however, seethe same colours as the pure spectrum like a rainbow. Instead the colours are formed by subtractionfrom the white light. For example, at a place where the film thickness is just right for a dark fringein the green you will see white light minus green, which leaves the red end and the blue end of thespectrum; the resulting visual sensation is purple.

L4: Interference 66

Where does the energy go?There is a puzzle that needs to be answered: what happens to the energy of the light when thereflected light is removed by interference? The energy cannot be destroyed so it must go somewhereelse - it is transmitted through the film instead of being reflected. As in the case of Young'sexperiment, the energy is rearranged in space but it is never destroyed. If you are used to thinkingof energy as a kind of fluid, then that idea may be hard to understand. However experimentalevidence supports the wave theory, so the "fluid" model of energy needs to be abandoned. Energy isnot like matter, it does not have to flow continuously through space. Another way of resolving theproblem is to say that the principle of superposition (just adding things up) works for electric fieldsbut it does not apply to energy or wave intensity.4-6 APPLICATIONS OF THIN FILM INTERFERENCETesting for flatnessGiven a slab with a very accurately flat surface, thin film interference can be used to test the flatnessof the another surface. (At least one of the two objects needs to be transparent.) Interference fringesformed by the thin film of air between the surfaces gives a contour map of variations in the height ofthe surface being tested. The contour interval is equal to half a wavelength of the light in the gap.

Surface being tested

Optically flat surface

Thin filmof air

Incident light

Figure 4.15. Testing for flatness

Figure 4.16.Thin film contour

fringes

Blooming of lensesA common application of thin film interference is in anti-reflection coatings on lenses that are usedin cameras, microscopes and other optical instruments. A modern lens system may have as many asten glass surfaces each with a reflectivity of about 5%. Without some kind of treatment about halfthe light entering such a lens system would be reflected instead of going on to form the final image.Apart from the loss of brightness involved, multiple reflections in an optical system can also degradethe quality of an image.

The amount of light reflected from each surface can be greatly reduced using the technique ofblooming, that is the deposition of an anti-reflection coating. Interference in the reflected lightmeans that light is transmitted instead of being reflected. The choice of material for the coating isimportant. Clearly it must be transparent, but it should also result in approximately equalreflectivities at both surfaces, so that the reflected waves (at a chosen wavelength) can completely

L4: Interference 67

cancel each other. Cancellation is achieved exactly when the refractive index of the coating is equalto the geometric mean of the refractive indices of the air and the glass: n2 = n1n3 . See figure4.17. However it is not easy to find materials with exactly the right properties, so in practice acompromise is needed. Magnesium fluoride, which has a refractive index of 1.38, is often used.

The thickness of the coating is chosen to work best for light of a wavelength near the middleof the visible spectrum, for example a wavelength of 500 nm corresponding to yellow-green light. Inthat case the lens still reflects some light in the blue and red so it looks purple in reflected light. Therefractive index of the coating is between that of air and glass so there is a phase change at bothreflections. At the chosen wavelength we require 2n2b = (m + l2 ) λ for no reflection. Withm = 0, the film thickness is a quarter of a wavelength.

n 1 n 2 n 3

Front surface

Coating

Lens

Reflected rays interfere Transmitted light is brighter

Figure 4.17. Anti-reflection lens coating

THINGS TO DO Look for examples of interference in your environment. The colours in oil slicks are an example ofthin-film interference. Next time that you see one make a note of the colours and their sequence.Are they the same as the colours of the rainbow? Can you explain the differences or similarities?Other examples of thin-film interference may be found in soap bubbles, the feathers of some birdsand opals.

You can make a thin film using two sheets of transparency film like that used on overheadprojectors. Just place the sheets together and look at the reflected light. A dark background behindthe sheets will help. You should be able to see coloured contour fringes which map the thickness ofthe air between the sheets. To enhance the effect place the two sheets on a hard surface and byrubbing something like a handkerchief over them, try to squeeze the air out of the gap. What do yousee now? See what happens when you press your finger on one part of the top sheet. Does theangle at which you look make any difference? Does the angle of the incident light matter? Lookthrough the sheets and try to see the interference in the transmitted light; why is that harder to see?

Observe the colour of the light reflected from various camera lenses. Can you explain thecolour? Is the colour the same for all lenses?

1.1 INTRODUCTIONFig. 1.3 Intensity distribution for the interference fringes from two waves of same frequency and amplitude 1.2.3 Superposition of Waves of Different Frequencies 4

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