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Interference original

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Interference original

  1. 1. UNIT-III CHAPTER-IINTERFERENCE
  2. 2. STRUCTURE:3.1.1. INTRODUCTION OBJECTIVES3.1.2. LIGHT AS A WAVE3.1.3. PHASE DIFFERENCE3.1.4. PATH DIFFERENCE3.1.5. RELATION BETWEEN PHASE DIFFERENCE AND PATH DIFFERENCE3.1.6. COHERENCE3.1.7. INTERFERENCE (DEFINITION) 1. PRINCIPLE OF SUPERPOSITION 2. TYPES OF INTERFERENCE(CONSTRUCTIVE & DESTRUCTIVE) 3. WHEN THE LIGHT WAVES HAVING DIFFERENT FREQUENCIES AND VARYING PHASE DIFFERENCE. 4. WHEN THE LIGHT WAVES HAVING EQUAL FREQUENCIES AND CONSTANT PHASE DIFFEENCE 5.CONDITIONS FOR SUSTAINED INTERFERENCE3.1.8. INTERFERENCE IN THINFILMS 1. STOKE’SPRINCIPLE 2. OPTICAL PATH OF LIGHT3.1.9. INTERFERENCE IN THINFILMS OF UNIFORM THICKNESS DUE TO REFLECTED AND TRANSMITTED LIGHTS3.1.10. INTERFERENCE IN THINFILMS OF NON UNIFORM THICKNESS 1. WEDGE METHOD (i) DETERMINATION OF THICKNESS OF PAPER (ii) DETERMINATIO OF FLATNESS OF GIVEN GLASS PLATES 2. NEWTON’S RINGS (i) DETERMINATION OF WAVELENGTH OF MONOCHROMATIC LIGHT (ii) DETERMINATION OF REFRACTIVE INDEX OF A GIVEN LIQUID3.1.11. MICHELSON INTERFERROMETER3.1.12. QUIZ3.1.13. SOLVED EXAMPLES3.1.14. PROBLEMS FOR PRACTICE3.1.15. SUMMARY
  3. 3. 3.1.1. INTRODUCTION Sunlight, as the rainbow shows us, is a composite of all colors of visiblespectrum. The colors reveal themselves in the rainbow because the incident wave lengthsare bent through different angles as they pass through raindrops that produce a bow.However, soap bubbles and oil slicks can also show striking colors produced not byrefraction but by constructive and destructive interference of light. The interfering wavescombine either to enhance or to suppress certain colors in the spectrum of sun light.Interference of light waves is thus a superposition phenomenon. This selective superposition of wavelengths has many applications. When lightencounters an ordinary glass surface, for example about 4% of incident light energy isreflected, thus weakening the transmitted beam by the amount. This unwanted loss of lightcan be a real problem in optical systems with many components. A thin transparent“interference film” deposited on glass surface, Can reduce the amount of reflected light bydestructive interference. The bluish cast of camera lens reveals the presence of such acoating. Interference coatings can also be used to enhance –rather than reduce- the abilityof a surface to reflect light. To understand interference, we must go beyond the restrictions of geometricaloptics and employ the full power of wave optics. In fact, as you will see, the existence ofinterference phenomena is perhaps our most convincing evidence that light is a wave-because interference cannot be explained other than with waves.OBJECTIVES:After going through this chapter we should able to: know 1) About light wave. 2) The relation between path difference and phase difference. 3) About interference. 4) About the interference phenomenon takes place in thin films. 5) Determination of thickness of a paper using wedge method. 6) Determination of wavelength of monochromatic light using Newton’s rings.3.1.2) LIGHT AS A WAVE:Light has dual nature. It has both particle nature and wave nature but it cannot act as awave and particle simultaneously. Light is also having the properties such as reflection,refraction, interference, diffraction, polarization. Light is also having two most famouseffects like photoelectric effect and Compton Effect. To explain the properties likeinterference diffraction and polarization we need to adopt wave nature for light.Light is a transverse wave. The first person to advance a convincing wave theory for light was DUTCH PHYSICISTCHRISTIAN HUYGENS, in 1678. Light wave should be represented as follows.Figure1
  4. 4. Whose displacement is given bWhere a is amplitude of wave. ωt is Phase, Which gives the position and direction of wave at a time t3.1.3) PHASE DIFFERENCES: The angular separation between two points of wave is called Phase difference. It is measured in radians or degrees.3.1.4) PATH DIFFERENCE:The linear separation between two points of wave is called Path DifferenceIt is measured in mm or cm.Fig.3.1.5) RELATION BETWEEN PHASE DIFFERENCE AND PATH DIFFERENCE:Figure 2 Phase Difference = λ Path Difference Phase Difference = Path Difference Phase Difference = Path Difference
  5. 5. Therefore for 2 Phase Difference = λ (Path Difference) Phase Difference (δ)= What is the Path Difference(X) = δ = ( ) x path difference Phase Difference= ( ) x Path Difference Path Difference = ( ) x Phase Difference3.1.6) COHERENCE: Two light waves are said to be coherent if they are having 1) Same Frequency 2) Almost Same Amplitude 3) Moving in a medium with either zero or constant Phase Difference 4) S3.1.7) INTERFERENCE:Interference is the optical phenomenon in which brightness and darkness areproduced by the combination of two similar light waves.DEFINITION: When two light waves of same frequency having constant phase difference coincide inspace and time. There is modification in the intensity of light.The resultant intensity at any point depends upon amplitudes and phase relationshipsbetween two waves.This modification in the intensity is due to superposition of two light waves are calledInterference.And the pattern dark and bright fringes produced are called Interference pattern.
  6. 6. (i)PRINCIPLES OF SUPERPOSITION: When two or more waves reach a point simultaneously, the resultant displacementat that point is the algebraic sum of the displacements produced by the individual waves inabsence of others.Explanation: Let us consider the two waves of same frequency “ ” is If y1 is the displacement produced by one wave and y2 is the displacement produced by second wave δ is the phase difference between these two waves The resultant displacement produced by the superposition of these waves is (1) Let (2) (3)Where a1 and a2 are the amplitudes of these two waves Substitute equation (2) and (3) in (1), we get
  7. 7. (4) Let (5) (6) Sub eq(5) and eq(6) in (4) then (7) This is an equation for resultant displacement. Where A is resultant amplitude which can be calculated as follows. By squaring and Adding equation (5) and equation (6) = Square of the amplitude (A²) = Intensity of light (I)(2)TYPES OF INTERFERENCE: Interference is of two types. 1. Constructive interference / constructive superposition 2. Destructive interference / destructive superposition
  8. 8. Figure: Constructive and Destructive Interference (A)CONSTRUCTIVE INTERFERENCE: When crust of one light wave falls on the crust of another wave then the resultant intensity increases and this type of interference is called constructiveinterference.Here we get maximum intensity. The intensity becomes maximum ) when
  9. 9. , n=0, 1, 2, 3............... For δ=0, 2π, 4π.............we get bright band Path Difference = ) X Phase Difference Path Difference= ) x2nπ Path Difference= This is the condition for constructive interference.` For δ=0, 2π, 4 ……… s (B)DESTRUCTIVE INTERFERENCE: When crust of one light wave falls on the trough of another wave then theresultant intensity decreases. This type of interference is called destructive interference.Here we get minimum intensity.The intensity becomes minimum ) when , n=0, 1, 2, 3............... For .............we get dark band. Path Difference = )X Phase Difference Path Difference= Path Difference=
  10. 10. This is the condition for constructive interference.` For δ= , 3π, 5 ……… Variation of Intensity with phase:-3) When two light waves having different frequencies and varying phase difference: If two light waves have different frequencies and the phase difference betweenlight waves is not constant then we get unsustained interference pattern i.e., Uniformillumination is observed i.e., bright and dark bands cannot be seen.
  11. 11. 4)When two light waves having equal frequencies and constant phase difference: If two light waves having equal frequencies and constant phase difference then weget sustained interference pattern. i.e we get alternative bright and dark fringes. Here thebright fringe becomes very bright and dark fringe becomes very dark.To get sustained interference light waves has to be satisfying the following conditions.5)Conditions for sustained or Good interference pattern : 1) We require two monochromatic light sources. 2) These two sources must be coherent i.e., they constant phase difference. 3) The frequency must be the same. 4) The amplitude must be the same. 5) They must travel in the same directions. 6) The two sources must be the same 7) These two sources must be as near as possible and the screen must be as far from them as possible.3.1.9) INTERFERENCE IN THIN FILMS:To understand interference in thin films first of all we have to understand some basicconcepts like stokes principle and optical path.1) STOKES PRINCIPLE:STATEMENT:- According to stoke principle, a light beam which is initially passing through a rarermedium reflects back into the same rare medium from denser medium suffering suddenphase change of ∏ or path change of λ/2.Such a phase change cannot be observed for alight beam light beam which is initially passing through denser medium and reflects backinto the same medium.
  12. 12. 2) The second basic point is Optical Path:The optical path travel led by a light beam in a medium of refractive index μ is not equalto actual path travel led by the light beam.Optical path travelled by light beam =µ X Actual path travelled by lightThin films are of two types1) Thin films which have uniform thickness2) Thin films which have non uniform thicknessStriking colors observed on thin films due to interference
  13. 13. 3.1.10) INTERFERENCE IN THINFILMS OF UNIFORM THICKNESS DUE TO REFLECTED ANDTRANSMITTED LIGHTS:a) Interference due to reflected light :Let us consider a transparent thin film of uniform thickness t and refractive index as shownin figure.AB- is the incident ray on thin film.A part of light ray reflected along BR and transmitted along BC.BC reflected from lower surface of thin film along CD and DR1 is transmitted ray .Now theinterference takes place between reflected rays BR and DR1.These are reflected from upper and lower surfaces of thin film.The actual path difference between BR and DR1 is = (BC+CD) in film - BE in airThe optical path difference between BR and DR1 = (BC+CD) µ -BE air---------- (1) For BE From Snells law From ∆BED, ∆BFD, Therefore
  14. 14. BE= µBF--------------- (2) Sub eq’n (2) in eq’n (1) Path difference ∆= (BC+CD) µ - µBF = (BC+CD-BF) µ = (BC+CF) µ ------------ (3) ∆le BCQ, ∆le PCQ are congruent triangles. Therefore BC=PC BQ=PQ=t QC is common side and all angles are equal. Therefore ∆= (PC+CF) µ ∆=µ (PF) ------------- (4) From ∆le PBF PF= (PB) = (BQ+QP) = = 2t ---------- (5) Therefore eq’n (4) becomes ∆= µ (2t) ∆=2µt --------- (6)But according to stokes principles the light ray BR is undergoing additional path change of λ /2.Therefore the total path difference between BR and DR is equal to 2μtcosr± λ /2
  15. 15. Condition for bright band/bright fringesWe get bright fringes when path difference =n λCondition for dark band/dark fringeswe get dark fringes when the path difference3.4) NOTE: In thin film, due to transmitted light The conditions for bright is The conditions for dark isWe can say the interference pattern due to reflected and transmitted rays arecomplementary each other.3.1.11) INTERFERENCE IN THIN FILMS OF NON UNIFORM THICKNESS: 1) Wedge method 2) Newton’s Rings
  16. 16. (A)WEDGE METHOD: Let us consider two plane surfaces GH; GH1which is inclined at an angle gives wedge shapewhich encloses an air filmAB is the incident ray on GH. The interference takes place between two reflected rays BRand DR1.One is reflected from upper surface and other one is from lower surface of airfilm.The optical path difference between BR and DR1 is given byΔ = µ (BC+CD)-BE----------------> (1) For BE From Snells law From ∆BED, ∆BFD, Therefore BE= µBF--------------- (2)
  17. 17. Substitute equation (2) in equation (1) Δ =µ (BC+CD)-µBF Δ =µ (BC+CD-BF) Δ =µ (BC- BF+CD) Δ =µ (FC+CD) -----------------------------> (3) ΔCDQ, ΔCPQ are congruent triangles CD=CP Therefore Eq’n (3) becomes Δ =µ (FC+CP) Δ =µ (FP) -----------------------> (4) From ΔFPD FP=2t ------------------>(5) (from figure DP=2t) Sub (5) in (4) Δ=2µt --------------------------------- (6)From stoke principleThe ray BR undergoes reflection from denser medium; it suffers on additional path changeλ/2.Therefore the total path difference=Condition for bright fringe:We get bright fringes when path difference =nλ
  18. 18. Condition for dark fringe:We get dark fringe when path difference=3.4.2)TYPES OF FRINGES IN WEDGE SHAPED FILMS: In wedge shaped films ,at the edge of wedge the thickness of air film will be same .If wedraw a line at this edge all along this length we observed straight parallel alternative brightand dark fringes with equal Spacing.Fig3.4.3)Applications of Wedge method:1) Determination of thickness of a paper or diameter of a wire/hair.2) Verification of flatness of the given transparent surface.3.4.4)Determination of thickness of paper or thin filmFirst we measure the spacing between two consecutive dark/bright fringes which is knownas fringe width to determine the thickness of paper.Spacing between two consecutive dark fringesLet us assume that nth dark fringes is formed at a distance Xn and (n+1)th dark fringe at adistance Xn+1 from wedge.Now Xn+1 – Xn gives the values of fringe width .
  19. 19. Condition for dark fringe is -------------- (a)For normal incidence i=0 r=0 For air µ =1 Eq--(a) becomes ------------------------->(b)For nth dark fringe --------------------------(c) From figure (ii) ---------------- (d) Sub eq (d) in eq(c)
  20. 20. For (n+1)th dark fringe -------------- (f)From fig(ii) ------------------------>(g)sub eqs(g) in (f) ------------------ (h)Now fringe width -----------------(i) As α is very small , -------------------- (j)Thickness of paper:If the angle between two wedges is very small then AB=AC=lFrom ΔABC,
  21. 21. This is equation (k) Sub eq’n (k) in eq’n(i), fringe width becomes This is an expression for thickness of paper.Whereλ = wavelength incident lightl = length of air filmβ = Fringe width(B)NEWTONS RINGS:Plano-convex lens on flat black surface When a Plano convex lens with its convex surfaces is placed on glass plate air film isformed between two whose thickness increases gradually. The thickness of air film at thepoint of contact is zero. When monochromatic light is allowed to fall normally then we getfringes which are circular. These fringes are concentric circle, uniform in thickness and withthe point of contact as the center. This phenomenon was first described by Newton thatwhy they are known as Newton’s rings.
  22. 22. Experimental arrangement L is a planoconvex lens with large radius of curvature . G is a plane glass plate L is placed with its convex surface on G. monochromatic light is incident on L normally by using 45° arrangement of glass plate. A part of light is reflected from curved surface of lens and another one is reflected from plane surface of glass plate. Now interference take place between these two rays i.e one is reflected from upper surface of air film and another one reflected fromlower surface of air film. Thus we get alternative bright and dark fringes .Here we get central ring as dark ring , the reason is as follows :The path difference between two light rays due to reflected light equal to Due to large radius of curvature of lens, α is so small and it is neglected. For normal incidence i=0, r=0. For air film μ=1 Path difference from equation 1 becomes Path difference =2t+λ/2------------ (2) At point of contact of lens and glass plate, t=0 then eq--( 2) becomes
  23. 23. The path difference = λ/2---------- (3) This is a condition for minimum intensity. So central ring is dark due to reflected light. Conditions for bright and dark rings: Conditions for bright rings: We observe maximum intensity/ bright ring when path difference=n λ 2t+λ/2=n λ (from eq---2) 2t= (2n-1) λ/2----------- (4) Conditions for dark rings: We observe minimum intensity/ dark ring When Path difference = (2n+1) λ/2 (from 2) 2t=n λ------------------ (5)Calculation of Diameter of bright and Dark rings formed due to reflection of light :fig
  24. 24. (Here d=t)From Figuret = thickness of air filmD=Diameter of Newtons ringsr=Radius of Newton’s ring =D/2R = Radius of curvature of convex lens
  25. 25. From properties of chords in circle ON * OD = OP *OQ t(2R - t) = r * r 2Rt-t ² = r ² 2Rt-t ²= (D/2)² =D²/4 D²= 8Rt-4t² As the thickness of air film is very small,t² is neglected. D²= 8Rt D²= 4R(2t)Condition for Diameter of bright ring From (4)Diameter of bright rings is proportional to square root of odd natural numbers.Diameter of Dark Ring: From eq-5The diameter of the dark ring is proportional to square root of natural numbers.
  26. 26. Note:The distance between rings decreases when the order of rings increases.Reason:Condition for dark ring is D16= 8√Rλ D9= 6√Rλ D4= 4√Rλ D1= 2√RλD16 -D9=2√Rλ ------------->7 FringesD9-D4 = 2√Rλ -------------->5 fringesD4-D1= 2√Rλ ---------------->3 FringesHere the distance( 2√Rλ) remains constant but number of rings increases .From this we can say that width decreases as order increases.APPLICATIONS: 1) Determination of wavelength of monochromatic source 2) Determination of refractive index of liquid1) Determination of wavelength of monochromatic sourceFirst Newtons Rings are formed by respective experimental arrangement.The condition for diameter of dark ring is given byIf we consider mth dark ring
  27. 27. The diameter for mth dark ring is given by ------------------ (a)If we consider nth dark ringThe diameter for nth dark ring is given by ---------------------- (b)Equation (a) - equation(b) This is equation(c)R is radius of curvature of lens is measured by spherometer. R=l²/6h+h/2Where l is the distance between two legs of spherometer. h is height of convex lens.(ii) DETERMINATION OF REFRACTIVE INDEX OF A LIQUIDFirst the experiment is performed when there is an air film is formed between glass plateand Plano convex lens.The diameter of nth and mth rings are measured.For air film, This is equation (d)Now the liquid whose refractive index is to be measured is powered in the containerwithout disturbing the whole arrangement.Again we measured the diameter of mth and nth rings whose diameters are D²m and D²nFor liquid medium
  28. 28. This is equation (e)From eq (d) /eq (e) then we getAPPLICATIONS 1) Determination of wavelength of monochromatic light. 2) Determination of difference of nearly equal wavelengths1)DETERMINATION OF WAVELENGTH OF MONOCHROMATIC LIGHT:Michelson interferometer is set for circular fringes set bright at center.Condition for bright spotThe mirror M1 is set at a distance x1 for one fringe.Let it be zeroth fringe.The mirror M1 is moved from x1 to x2 N fringes are crossed 2(X2~X1) =Nλ λ=2/N(X2~X1)2)DETERMINATION OF DIFFERENCE IN WAVELENGTH:There are two spectral lines D1 and D2 of sodium light.Their wavelengths are nearly equal.The difference in their wavelengths of D1 and D2 lines.So they form two separate fringe n .As thickness between M2 and M2¹ is very small, two fringe patterns coincide practically.By moving M1 we separate these two fringe pattern s.We move M 1 so that nth bright fringe of λ1 coincides with (n+1)th dark fringes of λ2.Note Down this distance , At this position we observe indistinctness.We move M2 again another indistinctness is observed .Let the fringe distance be x.For nth fringe
  29. 29. 2x=nλ1 This is equation (1)For (n+1)th fringe 2x= (n+1)λ2 This is equation (2)Equation (2) – equation (1) This is difference between wavelengths.3.1.13)QUIZWHY RAINBOW FORMS:
  30. 30. WHY MORPHO’S BUTTERFLY CHANGES IT’S COLOR FROM BLUE TO BROWN ..3.1.14) SOLVED EXAMPLES:(1)Two coherent sources of intensity 10 w and 25 w interfere to from fringes. Findthe ratio of maximum intensity to minimum intensity.Solution:Given that =Hence =Now = = =19.724(2)Two sinusoidal waves of equal amplitudes are wavelength out of phase. What is theamplitude of the resultant?Solution:When two waves of equal amplitude but different phase superpose, the resultant intensityis I=4 Or resultant amplitude is A=2acos
  31. 31. Where Ø is the phase difference.Given path difference =Hence phase difference Ø= Χ =Hence A=2a cos =2a =(3) A soap film of refractive index 4/3 and of thickness 1.5Χ cm is illuminated by whitelight incident at an angle of 60º. The light reflected by it is examined by a spectrometer inwhich is found a dark band corresponding to a wavelength of 5Χ Calculate the orderof interference of the dark band.Solution:For dark band 2µtcosr= λGiven =60ºHence µ=Or sin = = =0.6511i.e., 3.1.15) PROBLEMS FOR PRACTICE 1) Two coherent source of intensity ratio interfere. Prove that in the interference pattern 2) Two coherent sources whose intensity ratio is 81:1 produce interference fringes .Deduce the ratio of maximum intensity to minimum Intensity. 3) A parallel beam of light is incident on a thin glass plate such that the angle of refraction into the plate is 60°. Calculate the smallest thickness of the glass plate which will appear dark by reflection. 4) Fringes of equal thickness are observed in a thin glass wedge of refractive index, the fringe spacing is 1mm and wavelength of light is 5893 A°. Calculate the angle of wedge. 5) In Newtons Rings experiment, the diameter of the 4th and 12th dark rings is 0.400 cm and 0.700cm respectively. Find the diameter of 20th dark ring. 6) In Newtons Rings experiment, the diameter of the 10th rings from 1.40 cm to 1.27cm when a liquid is introduced between the lens and the plate. Calculate the
  32. 32. refractive index of the liquid. 7) Light containing two wavelengths λ1 and λ2 falls normally on a Plano convex lens of radius of curvature R resting on a glass plate .If nth dark ring due to λ1 coincides with (n+1)th dark ring due to λ2 prove that radius of nth dark ring of λ1 is 8) In a Michelson interferometer 200 fringes cross the field of view when the movable mirror is displaced through 0.0589 mm .Calculate the monochromatic light wavelength. 9) The movable mirror of Michelsons interferometer is moved through a distance of 0.02603 mm. find the number of fringes shifted across the cross wire of eye piece of the telescope ,if a wavelength of 5200 A° is used. 10) In an experiment with Michelsons interferometer scale readings for a pair of maximum indistinctness were found to be 0.6939 mm and 0.9884 mm .If the mean wavelength of the two components of D line are 5893 A°. Deduce the difference between wavelengths. 3.1.17) SUMMARY:We have seen that two independent sources of light cannot acts as coherent sources. Eventwo different parts of the same lamp cannot acts as coherent sources of light. Hence wechoose a ray which splits into two different parts such as reflected and transmitted to gettwo coherent waves. Due to the path difference between such split rays interferenceoccurs. The phenomena of interference are very useful in many areas such as holographyoptical switching calibration of instruments, displacements and measurements.  The displacement of wave light is y=a sin (wt+ δ)  Relation between path difference and phase difference  Path Difference= λ /2∏ * Phase Difference From principle of superposition  The intensity of resultant wave is given by I=a1²+a2²+2a1a2cosδ  Intensity I= A² When a1 =a2 =a  I=4a²cos²δ/2  A²=4a²cos²δ/2  A=2acosδ/2  Condition for maximum intensity /Bright fringe Path difference=nλ Phase difference =2nπ for n=0, 1,2....... I max= (a1+a2)²  Condition for minimum Intensity/dark fringe Path difference= (2n±1) λ / 2 Phase difference= (2n±1) π for n=0, 1, 2..... I min= (a1-a2)²
  33. 33. Interference in thin film for uniform thickness thin films: Condition for bright and dark fringes due to reflected light For Bright 2μtcosr= (2n ± 1) λ /2 For dark 2μtcosr=n λ Conditions for bright and dark fringes due to transmitted light For dark 2μtcosr=(2n ± 1) λ /2 For bright 2μtcosr=n λ Interference due to reflected and transmitted lights are complimentary to each other. B)FOR NONUNIFORM THICKNESS THIN FILMS: Wedge method: Bright fringe condition 2μtcos(r+α )=(2n ± 1) λ /2 Dark fringe condition 2μtcos(r+α )=nλ Fringe width β =λ /2sin α or λ /2α Thickness of thin film t= λl/2β Newtons Rings: Condition for bright ring 2t=(2n -1) λ /2 Condition for dark ring 2t=nλ Diameter of bright ring Db= √2R λ √ (2n-1 Diameter of dark ring Dd= √2R λ √n Wavelength of monochromatic light λ=D²m - D²n/4R(m - n) where R=l²/6h+h/2 Refractive index of a liquid µ liquid= D²m - D²n / D²m - D²n

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