Interference 1.1 INTRODUCTION 1.2 SUPERPOSITION OF WAVES

1 Interference 1.1 INTROUCTION In this chapter we will iscuss the phenomena associate with the interference of light waves. At any point where two or more wave trains cross one another they are sai to interfere. In stuying the effects of interference we are intereste to know the physical effects of superimposing two or more wave trains. It is foun that the resultant amplitue an consequently, the intensity of light gets moifie when two light beams interfere. This moification of intensity obtaine by the superposition of two or more beams of light is calle interference. In orer to fin out resultant amplitue, when two waves interfere, we make use of the principle of superposition. The truth of the principle of superposition is base on the fact that after the waves have passe out of the region of crossing, they appear to have been entirely uninfluence by the other set of waves. Amplitue, frequency an all other characteristics of each wave are just as if they ha crosse an unisturbe space. The principle of superposition states that the resultant isplacement at any point an at any instant may be foun by aing the instantaneous isplacements that woul be prouce at the point by the iniviual wave trains if each were present alone. In the case of light wave, by isplacement we mean the magnitue of electric fiel or magnetic fiel intensity. 1. SUPERPOSITION OF WAVES 1..1 Superposition of Waves of Equal Phase an Frequency Let us assume that two sinusoial waves of the same frequency are travelling together in a meium. The waves have the same phase, without any phase angle ifference between them. Then the crest of one wave falls exactly on the crest of the other wave an so o the troughs. The resultant amplitue is got by aing the amplitues of each wave point by point. The resultant amplitue is the sum of the iniviual amplitues (Fig. 1.1). Amplitue 1 Amplitue Resultant amplitue Amplitue t Fig. 1.1 Superposition of waves of equal phase an frequency 1

ENGINEERING PHYSICS i.e., A = A + A +... The resultant intensity is the square of the sum of the amplitues I = (A 1 + A + A 3 +...) (1.1) 1.. Superposition of Waves of Constant Phase ifference Let us consier two waves that have the same frequency but have a certain constant phase angle ifference between them. The two waves have a certain ifferential phase angle φ. In this case the crest of one wave oes not exactly coincie with the crest of the other wave (Fig. 1.). The resultant amplitue an intensity can be obtaine by trigonometry. Amplitue 1 Resultant t Fig. 1. Superposition of two sine waves of constant phase ifference The two waves having the same frequency (ω = πf) an a constant phase ifference (φ) can be represente by the equations Y 1 = a sin ωt Y = b sin (ωt + φ) (1.) where φ is the constant phase ifference, a, b are the amplitues an ω is the angular frequency of the waves. The resultant amplitue Y is given by Y = Y 1 + Y = a sin ωt + b sin (ωt + φ) = a sin ωt + b (sin ωt cos φ + cos ωt sin φ) = a sin ωt + b sin ωt cos φ + b cos ωt sin φ = (a + b cos φ) sin ωt + b cos ωt sin φ (1.3) If R is the amplitue of the resultant wave an θ is the phase angle then Y = R sin (ωt + θ) = R {sin ωt cos θ + cos ωt sin θ} = R cos θ sin ωt + R sin θ cos ωt (1.4) Comparing Eqs. (1.3) an (1.4) R cos θ = a + b cos φ R sin θ = b sin φ R = a + b + ab cos φ θ = tan 1 bsin φ a+ bsin φ (1.5)

INTERFERENCE 3 Clearly, R is maximum when φ = nπ an is minimum when φ = (n + 1)π where n = 0, 1,, 3,... When φ is an even multiple of π we say that waves are in phase an when φ is an o multiple of π, the waves are out of phase. When the amplitue of waves are equal to a say, then I = a (1 + cos φ) = 4a cos φ/ (1.6) A plot of I versus φ is shown in Fig. 1.3. Clearly, this reveals that the light istribution from the superposition of waves will consist of alternately bright an ark bans calle interference fringes. Such fringes can be observe visually if projecte on a screen or recore photo-electrically. In the above iscussion we have not consiere travelling waves (i.e., waves in which isplacement is also a function of istance). If λ is the wavelength, then the change of phase that occurs over a istance λ is π. Thus, if the ifference in phase between two waves arriving at a point is π, then ifference in the path travelle by these waves is λ. Let the phase ifference of two waves arriving at a point be δ an the corresponing path ifference be x. For a path ifference of λ, the phase ifference = π. Therefore, for a path ifference of x. Phase ifference = δ = π λ. x = π. path ifference λ an Path ifference = x = λ π phase ifference I = 4a cos I 4a a 5 4 3 0 3 4 5 6 7 Fig. 1.3 Intensity istribution for the interference fringes from two waves of same frequency an amplitue 1..3 Superposition of Waves of ifferent Frequencies So far we have assume that the waves have the same frequency. But light is never truly monochromatic. Many light sources emit quasimonochromatic light i.e., light emitte will be preominantly of one frequency but will still contain other ranges of frequencies. When waves of ifferent freqencies are superimpose, the result is more complicate. 1..4 Superposition of Waves of Ranom Phase ifferences When waves having ranom phase ifferences between them superimpose, no iscernible interference pattern is prouce. The resultant intensity is got by aing the square of the iniviual amplitues, N i.e., I = Ai = A1 + A Σ + A3 +... (1.7) i=1

4 ENGINEERING PHYSICS 1.3 YOUNG S OUBLE SLIT EXPERIMENT We have seen in the previous section that two waves with a constant phase ifference will prouce an interference pattern. Let us see how it can be realize in practice. Let us use two conventional light sources (like two soium lamps) illuminating two pin holes (Fig. 1.4). Then we will fin that no interference pattern is observe on the screen. This can be unerstoo from the following reasoning. In a conventional light source, light comes from a large number of inepenent atoms each atom emitting light for about 10 9 secons i.e., light emitte by an atom, is essentially a pulse lasting for only 10 9 secons. Even if the atoms were emitting uner similar conitions, waves from ifferent atoms woul iffer in their initial phases. Consequently light coming out from the holes S 1 an S will have a fixe phase relationship for a perio of about 10 9 sec. Hence, the interference pattern will keep on changing every billionth of a secon. The human eye can notice intensity changes which last at least for a tenth of a secon an hence we will observe a uniform intensity over the screen. However, if we have a camera whose time of shutter can be mae less than 10 9 sec, then the film will recor an interference pattern. We can summarize the above argument by noting that light beams from two inepenent sources o not have a fixe phase relationship over a prolonge time perio an hence, o not prouce any stationary interference pattern. Thomas Young in 180 evise an ingenious but simple metho to lock the phase relationship between two sources. The trick lies in the ivision of a wave front into two. These two split wave fronts act as if they emanate from two sources having a fixe phase relationship an therefore, when these two waves were allowe to interfere, a stationary interference pattern was prouce. In the actual experiment a light source illuminate a tiny pin hole S (Fig. 1.5). y S 1 S Screen Fig. 1.4 If two soium lamps illuminate two pin holes S 1 an S no interference pattern is observe on the screen z x y S S 1 S x Fig. 1.5 Young s arrangement to prouce interference pattern Light iverging from this pin hole fell on a barrier containing two rectangular apertures S 1 an S which were very close to each other an were locate equiistant from S. Spherical waves travelling from S 1 an S were coherent an on the screen beautiful interference fringes (Fig. 1.5) coul be obtaine. In the centre screen, where the light waves from two slits have travelle through equal istances an where the path ifference is zero, we have zeroth-orer maximum (Fig. 1.6). But maxima will also occur whenever the path ifference is one wavelength λ or an integral multiple of wavelength nλ. The integer n is calle the orer of interference.

INTERFERENCE 5 First-orer maximum First-orer minimum S 1 Zeroth-orer maximum S First-orer minimum First-orer maximum Fig. 1.6 Maxima an minima in Young s ouble slit experiment When the path ifference is a multiple of (n + 1/)λ we observe a ark fringe. In orer to calculate the position of the maxima, we procee as follows. Let be the istance between the slits an be the istance of the screen from the slits. P S 1 Y S O S Fig. 1.7 Path ifference in Young s ouble slit experiment Let P be the position of the maximum (Fig. 1.7). Then the path ifference between the two waves reaching P is sin θ = nλ or sin θ = n λ (n = 1,, 3,...) where λ is the wavelength of light use an θ is the angle as shown in Fig. 1.7. If Y is the istance of point P from O, the centre of the screen, then we have Y = tan θ For small angles of θ, Y = tan θ = sin θ Y = n λ or λ = Y n (1.8) Clearly, fringe with = Y n+1 Y n = β = λ (1.9)

6 ENGINEERING PHYSICS Hence, by measuring the istance between slits, the istance to the screen an the istance from the central fringe to some fringe on either sie, the wavelength of light proucing the interference pattern may be etermine. 1.4 COHERENCE An important concept associate with the iea of interference is coherence. Coherence means that two or more electromagnetic waves are in a fixe an preictable phase relationship to each other. In general the phase between two electromagnetic waves can vary from point to point (in space) or change from instant to instant (in time). There are thus two inepenent concepts of coherence namely temporal coherence an spatial coherence. Temporal Coherence : This type of coherence refers to the correlation between the fiel at a point an the fiel at the same point at a later time i.e. the relation between E (x, y, z, t 1 ) an E (x, y, z, t ). If the phase ifference between the two fiels is constant uring the perio normally covere by observations, the wave is sai to have temporal coherence. If the phase ifference changes many times an in an irregular way uring the shortest perio of observation, the wave is sai to be non coherent. Spatial Coherence : The waves at ifferent points in space are sai to be space coherent if they preserve a constant phase ifference over any time t. This is possible even when two beams are iniviually time incoherent, as long as any phase change in one of the beams is accompanie by a simultaneous equal phase change in the other beam (this is what happens in Young s ouble slit experiment). With the orinary light sources, this is possible only if the two beams have been prouce in the same part of the source. Time coherene is a characteristic of a single beam of light whereas space coherence concerns the relationship between two separate beams of light. Interference is a manifestation of coherence. Light waves come in the form of wave trains because light is prouce uring eexcitation of electrons in atoms. These wave trains are of finite length. Each wave train contains only a limite number of waves. The length of the wave train s is calle the coherence length. It is the prouct of the number of waves N containe in wave train an their wavelength λ i.e., s = Nλ. Since velocity is efine as the istance travelle per unit of time, it takes a wave train of length s, a certain length of time t, to pass a given point t = s/c where c is the velocity of light. The length of time t is calle the coherence time. The egree of temporal coherence can be measure using a Michelson s interferometer. It is clear from the above iscussion that the important conition for observing interference is that the two sources shoul be coherent. The observations of interference are facilitate by reucing the separation between the sources of light proucing interference. Further, in the Young s ouble slit experiment the istance between two sources an the screen shoul be large. The contrast between the bright an ark fringes is improve by making equal the amplitues of the light sources proucing interference. Further, the sources must be narrow an monochromatic. The concept of coherence is iscusse in greater etail in the chapter on lasers. 1.5 TYPES OF INTERFERENCE The phenomenon of interference is ivie into two classes epening on the moe of prouction of interference. These are (a) interference prouce by the ivision of wavefront an

INTERFERENCE 7 (b) interference prouce by the ivision of amplitue. In the first case the incient wavefront is ivie into two parts by making use of the phenomenon of reflection, refraction or iffraction. The two parts of the wavefront travel unequal istances an reunite to prouce interference fringes. Young s ouble slit experiment is a classic examples for this. In Young s ouble slit experiment one uses two narrow slits to isolate beams from separate portions of the primary wavefront. In the secon case the amplitue of the incient light is ivie into two parts either by parallel reflection or refraction. These light waves with ivie amplitue reinforce after travelling ifferent istances an prouce interference. Newton s rings is an example for this type. 1.6 INTERFERENCE IN THIN FILMS The colours of thin films, soap bubbles an oil slicks can be explaine as ue to the phenomena of interference. In all these examples, the formation of interference pattern is by the ivision 1 B i E i A phase change r r No phase change C Fig. 1.8 Interference in plane parallel films (Reflection geometry) of amplitue. For example, if a plane wave falls on a thin film then the wave reflecte from the upper surface interferes with the wave reflecte from the lower surface. Such stuies have many practical applications as provie by the example of prouction of non-reflecting coatings. 1.6.1 Interference in Plane Parallel Films ue to Reflecte Light Let us consier a plane parallel film as shown in the Fig. 1.8. Let light be incient at A. Part of the light is reflecte towar B an the other part is refracte into the film towars C. This secon part is reflecte at C an emerges at, an is parallel to the first part. At normal incience, the path ifference between rays 1 an is twice the optical thickness of the film. Γ = µ At oblique incience the path ifference is given by Γ = µ(ac + C) AB = µ cos r AB = µ r cos µ tan r sin r [Q AB = A sin i = AE. sin i = tan r. sin i = tan r. µ sin r]

8 ENGINEERING PHYSICS R S T 1 i.e., Γ = µ tanrsinr cos r W = µ 1 sin r.cosr cos r W = µ where µ is the refractive inex of the meium between the surfaces. Since for air µ = 1, the path ifference between rays 1 an is given by Γ = cos r While calculating the path ifference, the phase change that might occur uring reflection has to be taken into account. Whenever light is reflecte from an interface beyon which the meium has lower inex of refraction, the reflecte wave unergoes no phase change. When the meium beyon the interface has a higher refractive inex there is phase change of π. The transmitte waves o not experience any phase change. Hence, the conition for maxima for the air film to appear bright is µ cos r + λ = nλ U V R S T U V or µ cos r = nλ λ = (n 1) λ where n = 1,, 3,... The film will appear ark in the reflecte light when µ cos r + λ = (n + 1) λ or µ cos r = nλ where n = 0, 1,,3... 1.6. Interference in Plane Parallel Films ue to Transmitte Light Figure 1.9 illustrates the geometry for observing interference in plane parallel films ue to transmitte light. We have two transmitte rays CT an EU which are erive from the same point source an hence, are in a position to interfere. The effective path ifference between these two rays is given by Γ = µ(c + E) CP But µ = sin i / sin r = CP CE CP QE / CE = QE CP = µ(qe) or Γ = µ(c + Q + QE) µ(qe) = µ(c + Q) = µ(i + Q) = µ(qi) = µ cos r In this case it shoul be note that, no phase change occurs when the rays are refracte unlike in the case of reflection. Hence, the conition for maxima is µ cos r = nλ an the conition for minima is µ cos r = (n 1) λ. Thus, the conitions of maxima an minima in transmitte light are just the reverse of the conition for reflecte light. 1.6.3 Interference in Wege Shape Film Let us consier two plane surfaces GH an G 1 H 1 incline at an angle α an enclosing a wege shape film (Fig. 1.10). The thickness of the film increases from G to H as shown in the figure. Let µ be the refractive inex of the material of the film. When this film is illuminate there is

INTERFERENCE 9 A I G i B F r Air H G 1 r r r C i P r i Q E ense meium H 1 T U Fig. 1.9 Interference in plane parallel films (Transmission geometry) R A R 1 i F 90 i H 1 G 1 B r E r+ r+ G C H r+ Q P X n Fig. 1.10 Interference in a wege shape film interference between two systems of rays, one reflecte from the front surface an the other obtaine by internal reflection at the back surface. The path ifference Γ is given by Γ = µ(bc + C) BF Γ = µ(be + EC + C) µbe L NM BF BE sin i BF Q sin i = ; sin r = ; µ = µ = B B sin r BE Γ = µ(ec + C) = µ(ec + CP) = µep = µ cos (r + α) ue to reflection an aitional phase ifference of λ/ is introuce. O QP

10 ENGINEERING PHYSICS Hence, Γ = µ cos (r + α) + λ/ For constructive interference µ cos (r + α) + λ/ = nλ or µ cos (r + α ) = (n 1) λ/ where n = 1,, 3... For estructive interference µ cos (r + α) + λ = (n + 1) λ or µ cos (r + α) = nλ where n = 0, 1,, 3... Spacing between two consecutive bright bans is obtaine as follows. For n th maxima µ cos (r + α) = (n 1) λ Let this ban be obtaine at a istance X n from thin ege as shown in Fig. (1.10). For near normal incience, r = 0. Assuming, µ = 1, From the figure, = X n tan α X n tan α cos α = (n 1) λ For (n + 1) th maxima X n sin α = (n 1) λ X n+1 sin α = (n + 1) λ (X n+1 X n ) sin α = λ or fringe spacing, λ λ β = X n+1 X n = sin α α where α is small an measure in raians. 1.7 COLOURS OF THIN FILMS The iscussion of the interference ue to a parallel film an at a wege shoul now enable us to unerstan as to why films appear coloure. To summarize, the incient light is split up by reflection at the top an bottom of the film. The split rays are in a position to interfere an interference of these rays is responsible for colours. Since the interference conition is a function of thickness of the film, the wavelength an the angle of refraction, ifferent colours are observe at ifferent positions of the eye. The colours for which the conition of maxima will be satisfie will be seen an others will be absent. It shoul be note here that the conitions for maxima an minima in transmitte light are opposite to that of reflecte light. Hence, the colours that are absent in reflecte light will be present in transmitte light. The colours observe in transmitte an reflecte light are complimentary. 1.8 NEWTON S RINGS When a plano-convex lens with its convex surface is place on a plane glass plate, an air film of graually increasing thickness is forme between the two. If monochromatic light is allowe to fall normally an viewe as shown in the Fig. 1.11 then alternate ark an bright circular fringes are observe. The fringes are circular because the air film has a circular symmetry. Newton s rings are forme because of the interference between the waves reflecte from the top an bottom surfaces of the air film forme between the plates as shown in the Fig. 1.1.

INTERFERENCE 11 M B L 1 45 S L G Air Film Fig. 1.11 Experimental set up for viewing Newton s rings A 1 L B P C Fig. 1.1 Interference in Newton s rings setup The path ifference Γ between these rays (i.e., rays 1 an ) is µ cos r + λ i.e., Since r 0, µ = 1; Γ = + λ At the point of contact = 0, the path ifference is λ. Hence, the central spot is ark. The conition for bright fringe is + λ = nλ or = ( n 1 ) λ, an the conition for ark fringe is where n = 1,, 3... + λ = (n + 1) λ or = nλ where n = 0, 1,, 3... Now let us calculate the iameters of these fringes. Let LOL be the lens place on the glass plate AB (Fig. 1.13). The curve surface LOL is part of the spherical surface with the centre at C. Let R be the raius of curvature an r be the raius of Newton s ring corresponing to constant film thickness.

1 ENGINEERING PHYSICS C R L Q N P L A O B r Fig. 1.13 Calculation of iameter of Newton s Ring From the property of the circle. i.e., NP NQ = NO N i.e., r r = (R ) = R R i.e., r = R or = r /R Thus, for a bright fringe r n 1 = ( )λ or r = ( n λr R Replacing r by / where is the iameter we get Similarly, for a ark fringe n = λr n 1 r R or r = nλr n = 4nλR n = nλr Thus, the iameters of the rings are proportional to the square roots of the natural numbers. By measuring the iameter of the Newton s rings, it is possible to calculate the wavelength of light as follows. We have for the iameter of the n th ark fringe. n = 4nλR Similarly iameter for the (n + p) th ark fringe n + p = 4(n + p)λr n + p n = 4 pr λ or λ = n+ p n 4 pr λ can be calculate using this formula.

INTERFERENCE 13 Newton s rings set up coul also be use to etermine the refractive inex of a liqui. First the experiment is performe when there is air film between the lens an the glass plate. The iameters of the n th an (n + p) th fringes are etermine. Then we have n + p n = 4pλR Now the liqui whose refractive inex is to be etermine is poure into the container without isturbing the entire arrangement. Again the iameter of the n th an (n + p) th ark fringes are etermine. Again we have n + p n = 4 pλr µ from the above equations µ = n+ p n. + n p n REFERENCES 1. F.A. Jenkins an H.E. White, Funamentals of Optics, McGraw Hill Book Company, New York, 1985.. J.R. Meyer-Arent, Introuction to Classical an Moern Optics, Prentice Hall Pvt. Lt., New York, 1984. 3. A Ghatak, Optics, Tata McGraw Hill Publishing Co. Lt., New elhi, 1977. 4. R.K. Gaur & S.L. Gupta, Engineering Physics, hanpat Rai an Sons, 1987. 5. N. Subrahmanyan an Brijlal. A Text of Optics, Niraj Prakashan, 1968. SOLVE EXAMPLES 1. Two narrow an parallel slits 0.08 cm apart are illuminate by light of frequency 8 10 11 khz. It is esire to have a fringe with of 6 10 4 m. Where shoul the screen be place from the slits? = 0.08 cm = 0.08 10 m, β = 6 10 4 m frequency ν = 8 10 11 khz i.e., λ = c v = 3 10 8 10 10 8 11 3 m, =? From β = λ we have = β λ = 4 6 10 0. 08 10 8 10 3 10 8 14 = 18. m.. In Young s ouble slit experiment, a source of light of wavelength 400 Å is use to obtain interference fringes of with 0.64 10 m. What shoul be the wavelength of the light source to obtain fringes 0.46 10 m wie, if the istance between screen an the slits is reuce to half the initial value?

14 ENGINEERING PHYSICS In the first case λ = 400 Å = 400 10 10 m β = 0.64 10 m 0.64 10 400 10 = 10 In the secon case β = 0.46 10 m, λ =? 0.46 10 = λ / = λ iviing equation (i) by (ii) (i) (ii) 064. 10 046. 10 10 400 10 = λ 10 400 10 0. 46 λ = = 6037.5 Å. 064. 3. In Young s ouble slit experiment, the istance between the slits is 1 mm. The istance between the slit an the screen is 1 meter. The wavelength use in 5893 Å. Compare the intensity at a point istance 1 mm from the centre to that at its centre. Also fin the minimum istance from the centre of a point where the intensity is half of that at at the centre. Path ifference at a point on the screen istance y from the central point = Y. Here Y = 1 mm = 1 10 3 m = 1m = 1 mm = 1 10 3 m 3 3 1 10 1 10 Path ifference = = 1 10 6 m= 1 Phase ifference = 10 6 π π = 10 = 3.394 π raians λ 5893 10 Ratio of intensity with the central maximum = cos δ/ = cos (1.697π) = 0.337 When the intensity is half of the maximum, if δ is the phase ifference, we have cos δ/ = 0.5 or δ/ = 45 or δ = 90 = π/ Path ifference = = δ λ π = π λ λ = π 4 istance of the point on the screen from the centre = Y =.

INTERFERENCE 15 = λ 1 5893 10 4 = = 1. 473 10 m. 3 3 4 1 10 4 10 4. In a ouble slit experiment, fringes are prouce using light of wavelength 4800 Å. One slit is covere by a thin plate of glass of refractive inex 1.4 an the other slit by another plate of glass of the same thickness but of refractive inex 1.7. On oing so the central bright fringe shifts to the position originally occupie by the fifth bright fringe from the centre. Fin the thickness of the glass plate. We have nλ = (µ µ )t Here n = 5 µ µ = 0.3 λ = 4800 10 10 m 5 4800 10 t = = 8.0 10 8 m. 03. 5. A rop of oil of volume 0. cc is roppe on a surface of tank of water of area 1 m. The film spreas uniformly over the whole surface. White light which is incient normally is observe through a spectrometer. The spectrum is seen to contain one ark ban whose centre has a wavelength 5.5 10 5 cm in air. Fin the refractive inex of oil. 10 0. cm 5 The thickness of the film = = = 10 cm 100 100 The film appears ark by reflecte light for a wavelength λ given by the relation µ cos r = nλ For normal incience r = 0, cos r = 1 Further n = 1 an λ = 5.5 10 5 cm nλ 1 55. 10 µ = = tcos r 0 1 1375. 5 =. 6. A soap film 5 10 5 cm thick is viewe at an angle of 35 to the normal. Fin the wavelengths of light in the visible spectrum which will be absent from the reflecte light (µ = 1.33). Let i be the angle of incience an r the angle of refraction. Then µ = sin i sin 35 ; 1.33 = sin r sin r r = 5.55 cos r = 0.90 Applying the relation, µ cos r = nλ where = 5 10 5 cm (i) For the first orer n = 1 λ 1 = 1.33 5 10 5 0.90 = 1.0 10 5 cm Which lies in the infrare (invisible) region. 5 10

16 ENGINEERING PHYSICS (ii) For the secon orer n = λ = 1.33 5 10 5 0.90 = 6.0 10 5 cm which lies in the visible region. (iii) Similarly, taking n = 3, λ 3 = 4.0 10 5 cm which also lies in the visible region. (iv) If n = 4, λ 4 = 3.0 10 5 cm which lies in the ultraviolet (invisible region). Hence, absent wavelengths in the reflecte light are 6.0 10 5 an 4.0 10 5 cm. 7. Two glass plates enclose a wege shape air film, touching at one ege an separate by a wire of 0.05 mm iameter at a istance 15 cm from that ege. Calculate the fringe with. Monochromatic light of λ = 6000 Å from a broa source falls normally on the film. Fringe with β = λ α Clearly α = 005. mm 15 cm = 0.005 15 raian β = λ 6000 10 15 = = 0.09 cm. α 0. 005 8. An air wege of angle 0.01 raians is illuminate by monochromatic light of 6000 Å falling normally on it. At what istance from the ege of the wege, will the 10th fringe be observe by reflecte light. Here α = 0.01 raians n = 10 λ = 6000 10 10 m = ( n 1) λ where is the thickness of wege 9 But α = a x = α x α x = ( n 1) λ Here n = 10 x = ( 1) 10 n λ 19 6000 10 = m =.85 10 4α 4 001. 4 m. 9. A thin equiconvex lens of focal length 4 meters an refractive inex 1.50 rests on an is in contact with an optical flat. Using light of wavelength 5460 Å, Newton s rings are viewe normally by reflection. What is the iameter of the 5th bright ring? The iameter of the nth bright ring is given by n = ( n 1)λ R Here n = 5 λ = 5460 10 6 cm

INTERFERENCE 17 i.e., We have f = 400 cm µ = 1.50 1 1 1 = ( µ 1) + f R1 R Here R 1 = R = R 1 = ( µ 1) f R 1 400 = (1.50 1) R F HG I KJ R = 400 cm n = ( 5 1) 5460 10 400 = 0.67 cm. 10. In Newton s ring experiment, the iameters of the 4th an 1th ark rings are 0.400 cm an 0.700 cm respectively. Fin the iameter of the 0th ark ring. We have n+p n = 4pλR Here (n + p) = 1, n = 4, p = 1 4 = 8 1 4 0 4 = 4 3 λr... (i) = 4 16 λr... (ii) iviing (ii) by (i) 6 0 4 1 4 = 4 16 λr 4 8 λr = 0 (0.4) (0.7) (0.4) = 0 = 0.906 cm. 11. In a Newton s ring experiment the iameter of the 10th ring changes from 1.40 to 1.7 cm when a liqui is introuce between the lens an the plate. Calculate the refractive inex of the liqui. When the liqui is use the iameter of the 10th ring is given by ( 10 ) = 4 10 λ R (i) µ For air meium ( 10 ) = 4 10 λr (ii) iviing (i) by (ii) µ = 10 10 = 1.15. 1. In a Newton s ring experiment the iameter of the 5th ark ring was 0.3 cm an the iameter of the 5th ring was 0.8 cm. If the raius of the curvature of the plano-convex lens is 100 cms, fin the wavelength of the light use.

18 ENGINEERING PHYSICS λ = n + p n 4pR Here 5 = 0.8 cm 5 = 0.3 cm P = 5 5 = 0 an R = 100 cm λ = ( 08. ) ( 03. ) 4 0 100 = 4.87 10 5 cm. QUESTIONS 1. What is interference of light waves? What are the conitions necessary for obtaining interference fringes?. Two inepenent non-coherent sources of light cannot prouce an interference pattern. Why? 3. efine spatial an temporal coherence. 4. escribe Young s ouble slit experiment an obtain an expression for fringe with. 5. Write a note on colours of thin films. 6. Show that colours exhibite by reflecte an transmitte systems are complementary. 7. Fin an expression for the with of the fringes obtaine in the case of air wege. How woul you use the result to fin the wavelength of a given monochromatic raiation? 8. What are Newton s rings? How are they forme? Why are they circular? 9. Explain why the centre of Newton s rings is ark in the reflecte system. 10. escribe how you woul use Newton s rings to etermine the wavelength of a monochromatic raiation an erive the relevant formula. 11. Obtain an expression for the raius of the n th ark ring in the case of Newton s rings. 1. Show that the raii of Newton s rings are in the ratio of the square roots of the natural numbers. PROBLEMS 1. Interference fringes are forme on a screen which is at a istance of 0.8 m. It is foun that the fourth bright fringe is situate at a istance of 0.00108 m from the central fringe. Calculate the istance between the two coherent sources. (given λ = 5896 Å). (Ans. 1.75 10 19 m). A parallel beam of light (λ = 5890 10 10 m) is incient on a thin glass plate (µ = 1.5) such that the angle of refraction into the plate is 60. Calculate the smallest thickness of plate which woul appear ark by reflection. (Ans. 3.96 10 7 m) 3. White light falls normally on a film of soapy water whose thickness is 5 10 5 cm an µ = 1.33. Which wavelength in the visible region will be reflecte most strongly? (Ans. 530 10 10 m) 4. White light is incient on a soap film at an angle of sin 4 4/5 an the reflecte light on examination by a spectroscope shows ark bans. Two consecutive bans corespon to wavelength 6.1 10 5 an 6.0 10 5 cm. If µ = 4/3, calculate its thickness. (Ans. 1.7 10 5 m) 5. If the angle of the air wege is 0.5 an the wavelengths of soium lines are 5896 an 5890 Å, fin the istance from the apex at which the maximum ue to two wavelengths first coincie when observe in reflecte light. (Ans. 6.63 cm) 6. A monochromatic light of wavelength 5893 10 10 m falls normally on an air wege. If the length of the wege is 0.05 m, calculate the istance at which the 1th ark an 1th bright fringes will form the line of contact of the glass plates forming the wege. (Given the thickness of the specimen = 154 10 6 m). (Ans. 9.61 10 4 m, 9.1 10 4 m)

INTERFERENCE 19 7. A square piece of cellophane film with refractive inex 1.5 has a wege shape section so that its thickness at two opposite sies is t 1 an t. If with a light of λ = 6000 Å, the number of fringes appearing in the film is 10, calculate the ifference t t 1. (Ans. 10 4 cm) 8. A Newton s ring arrangement is use with a source emitting two wavelengths λ 1 = 6 10 5 m an λ = 4.5 10 5 m. It is foun that nth ark ring ue to λ 1 coincies with (n + 1)th ark ring for λ. If raius of curvature of the lens is 90 cm fin the iameter of the nth ark ring. (Ans. 0.54 cm) 9. Light containing two wavelengths λ 1 an λ falls normally on a planoconvex lens of raius of curvature R resting on a glass plate. If the nth ark ring ue to λ 1, coincies with (n = 1)th ark ring ue to λ, prove that the raius of the nth ark ring of λ 1 is λλ 1 R λ λ b 1 g. 10. Newton s rings forme by soium light between a flat glass plate an a convex lens are viewe normally. What will be the orer of the ark ring which will have ouble the iameter of 40th ring? (Ans. 160)