UNIT-IIICHAPTER-I INTERFERENCE STRUCTURE: 3.1.1. INTRODUCTION OBJECTIVES 3.1.2. LIGHT AS A WAVE 3.1.3. PHASE DIFFERENCE 3.1.4. PATH DIFFERENCE 3.1.5. RELATION BETWEEN PHASE DIFFERENCE AND PATH DIFFERENCE 3.1.6. COHERENCE 3.1.7. 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 INTERFERENCE 3.1.8. INTERFERENCE IN THINFILMS 1. STOKE’SPRINCIPLE 2. OPTICAL PATH OF LIGHT 3.1.9. INTERFERENCE IN THINFILMS OF UNIFORM THICKNESS DUE TO REFLECTED AND TRANSMITTED LIGHTS 3.1.10. 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 LIQUID 3.1.11. MICHELSON INTERFERROMETER 3.1.12. QUIZ 3.1.13. SOLVED EXAMPLES 3.1.14. PROBLEMS FOR PRACTICE 3.1.15. SUMMARY 3.1.1. INTRODUCTION Sunlight, as the rainbow shows us, is a composite of all colors of visible spectrum. The colors reveal themselves in the rainbow because the incident wave lengths are 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 by refraction but by constructive and destructive interference of light. The interfering waves combine 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 light encounters an ordinary glass surface, for example about 4% of incident light energy is reflected, thus weakening the transmitted beam by the amount. This unwanted loss of light can 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 by destructive interference. The bluish cast of camera lens reveals the presence of such a coating. Interference coatings can also be used to enhance –rather than reduce- the ability of a surface to reflect light. To understand interference, we must go beyond the restrictions of geometrical optics and employ the full power of wave optics. In fact, as you will see, the existence of interference phenomena is perhaps our most convincing evidence that light is a wavebecause 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 a wave and particle simultaneously. Light is also having the properties such as reflection, refraction, interference, diffraction, polarization. Light is also having two most famous effects like photoelectric effect and Compton Effect. To explain the properties like interference 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 PHYSICIST CHRISTIAN HUYGENS, in 1678. Light wave should be represented as follows. Figure1 5) RELATION BETWEEN PHASE DIFFERENCE AND PATH DIFFERENCE: Figure 2 Phase Difference = λ Path Difference Phase Difference = 2 𝜆 2 𝜆 Path Difference Phase Difference =4 Path Difference .1. 'ωt' is Phase.3) PHASE DIFFERENCES: The angular separation between two points of wave is called Phase difference.Whose displacement is given b 𝑦 = 𝑎𝑠𝑖𝑛𝜔𝑡 Where ' a' is amplitude of wave.1. Fig. 𝑜𝑟 𝑦 = asin(𝜔𝑡 + 𝛿) 3. It is measured in radians or degrees. 3.1. Which gives the position and direction of wave at a time 't' 3.4) PATH DIFFERENCE: The linear separation between two points of wave is called Path Difference It is measured in mm or cm. Therefore for 2 Phase Difference = λ (Path Difference) Phase Difference (δ)= What is the Path Difference(X) 𝛿 2 = 𝑥 𝜆 2 δ = ( 𝜆 ) x path difference Phase Difference= ( 𝜆 ) x Path Difference Path Difference = ( 𝝀 ) x Phase Difference 𝟐 2 3. The resultant intensity at any point depends upon amplitudes and phase relationships between two waves. This modification in the intensity is due to superposition of two light waves are called Interference.1. There is modification in the intensity of light.7) INTERFERENCE: Interference is the optical phenomenon in which brightness and darkness are produced by the combination of two similar light waves.1. . And the pattern dark and bright fringes produced are called Interference pattern.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) S 3. DEFINITION: When two light waves of same frequency having constant phase difference coincide in space and time. the resultant displacement at that point is the algebraic sum of the displacements produced by the individual waves in absence of others. we get 𝑦 = 𝑦1 + 𝑦2 y = a1 sin 𝜔𝑡 + 𝑎2 sin(𝜔𝑡 + 𝛿) y = a1 sin 𝜔𝑡 + 𝑎2 sin 𝜔𝑡 cos 𝛿 + 𝑎2 cos ωt sin 𝛿 . 𝜔 Explanation: Let us consider the two waves of same frequency “” is 2 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 𝑦 = 𝑦1 + 𝑦2 (1) Let y1 = a1 sin 𝜔𝑡 (2) y2 = a2 sin(𝜔𝑡 + 𝛿) (3) Where a1 and a2 are the amplitudes of these two waves Substitute equation (2) and (3) in (1).(i)PRINCIPLES OF SUPERPOSITION: When two or more waves reach a point simultaneously. By squaring and Adding equation (5) and equation (6) 𝐴2 𝑐𝑜𝑠 2 𝛿 = (𝑎1 + 𝑎2 𝑐𝑜𝑠𝛿)² 𝐴2 𝑠𝑖𝑛2 𝛿 = (𝑎2 𝑠𝑖𝑛𝛿)² 𝐴2 (𝑐𝑜𝑠 2 𝛿 + 𝑠𝑖𝑛2 𝛿) = (𝑎1 + 𝑎2 𝑐𝑜𝑠𝛿)² + (𝑎2 𝑠𝑖𝑛𝛿)² 𝐴2 = 𝑎1 ² + 𝑎2 ²𝑐𝑜𝑠²𝛿 + 2𝑎1 𝑎2 𝑐𝑜𝑠𝛿 + 𝑎2 ²𝑠𝑖𝑛²𝛿 𝐴2 = 𝑎1 ² + 𝑎2 ² + 2𝑎1 𝑎2 𝑐𝑜𝑠𝛿 Square of the amplitude (A²) = Intensity of light (I) 𝐼 = 𝑎1 ² + 𝑎2 ² + 2𝑎1 𝑎2 𝑐𝑜𝑠𝛿 𝐼𝑓 𝑎1 = 𝑎2 = 𝑎 𝐼 = 4𝑎2 𝑐𝑜𝑠 2 2 𝐴² = 4𝑎2 𝑐𝑜𝑠 2 2 𝐴 = 2𝑎𝑐𝑜𝑠 2 (2)TYPES OF INTERFERENCE: Interference is of two types. Destructive interference / destructive superposition 𝛿 𝛿 𝛿 . Constructive interference / constructive superposition 2. 1.𝑦 = (𝑎1 + 𝑎2 cos 𝛿) sin 𝜔𝑡 + (𝑎2 sin 𝛿) cos 𝜔𝑡 (4) Let (𝑎1 + 𝑎2 cos 𝛿) = 𝐴 cos 𝜃 (5) 𝑎2 sin 𝛿 = 𝐴 sin 𝜃 (6) Sub eq(5) and eq(6) in (4) then 𝑦 = 𝐴 sin 𝜔𝑡 cos 𝛿 + 𝐴 cos 𝜔𝑡 sin 𝛿 y = Asin(𝜔𝑡 + 𝛿) (7) This is an equation for resultant displacement. Where A is resultant amplitude which can be calculated as follows. 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 constructive interference. The intensity becomes maximum (𝐼 = 4𝑎2 𝑐𝑜𝑠 2 2 ) when 𝑐𝑜𝑠 2 2 = 1 𝑐𝑜𝑠 2 = ±1 𝑐𝑜𝑠 2 = 𝑐𝑜𝑠𝑛𝜋 𝛿 𝛿 𝛿 𝛿 . Here we get maximum intensity. . 1. 2π...........we get bright band Path Difference =(2) X Phase Difference Path Difference=(2) x2nπ Path Difference=𝑛𝜆 This is the condition for constructive interference.. 3𝜋..... n=0. For δ=0.... 𝛿 𝛿 𝜆 𝜆 The intensity becomes minimum (𝐼 = 4𝑎2 𝑐𝑜𝑠 2 2 ) when 𝑐𝑜𝑠 2 2 = 0 𝑐𝑜𝑠 2 = 0 𝑐𝑜𝑠 2 = cos(2𝑛 ± 1) 𝜋/2 𝛿 = (2𝑛 ± 1)𝜋 For . 2.𝛿 = 2𝑛𝜋 . This type of interference is called destructive interference.... n=0. 5𝜋. 4……… 𝑐𝑜𝑠𝛿 = 1 𝐼 = 𝑎1 ² + 𝑎2 ² + 2𝑎1 𝑎2 𝑐𝑜𝑠𝛿 𝐼 = (𝑎1 + 𝑎2 )² 𝐴² = (𝑎1 + 𝑎2 )² s𝐴 = (𝑎1 + 𝑎2 ) (B)DESTRUCTIVE INTERFERENCE: When crust of one light wave falls on the trough of another wave then the resultant intensity decreases.......... Here we get minimum intensity... 3. 1..... 3.. 𝛿 𝛿 𝛿 = 𝜋. 4π. 2π... 2..we get dark band.. Path Difference =(2)X Phase Difference Path Difference=(2 ) ∗ (2𝑛 ± 1)𝜋 Path Difference=(2𝑛 ± 1)𝜆/2 𝜆 𝜆 .... ` For δ=0........ ` For δ=. 3π. Uniform illumination is observed i..e. 5……… 𝑐𝑜𝑠𝛿 = −1 𝐼 = 𝑎1 ² + 𝑎2 ² − 2𝑎1 𝑎2 𝑐𝑜𝑠𝛿 𝐼 = (𝑎1− 𝑎2 )² 𝐴² = (𝑎1 − 𝑎2 )² 𝐴 = (𝑎1 − 𝑎2 ) 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 between light waves is not constant then we get unsustained interference pattern i.This is the condition for constructive interference. bright and dark bands cannot be seen.e. .. 4)When two light waves having equal frequencies and constant phase difference: If two light waves having equal frequencies and constant phase difference then we get sustained interference pattern. .Such a phase change cannot be observed for a light beam light beam which is initially passing through denser medium and reflects back into the same medium. 4) The amplitude must be the same. 5) They must travel in the same directions. they constant phase difference.1. 2) These two sources must be coherent i. 3) The frequency must be the same. a light beam which is initially passing through a rarer medium reflects back into the same rare medium from denser medium suffering sudden phase change of ∏ or path change of λ/2. Here the bright fringe becomes very bright and dark fringe becomes very dark. 1) STOKE'S PRINCIPLE: STATEMENT:According to stoke principle.9) INTERFERENCE IN THIN FILMS: To understand interference in thin films first of all we have to understand some basic concepts like stokes principle and optical path..e we get alternative bright and dark fringes. 5)Conditions for sustained or Good interference pattern : 1) We require two monochromatic light sources. To get sustained interference light waves has to be satisfying the following conditions. 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.e. 3. i. Optical path travelled by light beam =µ X Actual path travelled by light Thin films are of two types 1) Thin films which have uniform thickness 2) Thin films which have non uniform thickness Striking colors observed on thin films due to interference .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 equal to actual path travel led by the light beam. A part of light ray reflected along BR and transmitted along BC.BE in air The optical path difference between BR and DR1 = (BC+CD) µ -BE air---------- (1) For BE From Snell's law 𝜇 = 𝑠𝑖𝑛𝑖 𝑠𝑖𝑛𝑟 𝐵𝐸 From ∆BED.10) INTERFERENCE IN THINFILMS OF UNIFORM THICKNESS DUE TO REFLECTED AND TRANSMITTED LIGHTS: a) Interference due to reflected light : Let us consider a transparent thin film of uniform thickness t and refractive index as shown in figure. AB.is the incident ray on thin film.1. The actual path difference between BR and DR1 is = (BC+CD) in film .Now the interference takes place between reflected rays BR and DR1. BC reflected from lower surface of thin film along CD and DR1 is transmitted ray . These are reflected from upper and lower surfaces of thin film. 𝑠𝑖𝑛𝑖 = 𝐵𝐷 ∆BFD.𝑠𝑖𝑛𝑟 = 𝐵𝐷 Therefore 𝜇 = 𝐵𝐹 𝐵𝐸 𝐵𝐹 .3. ∆le PCQ are congruent triangles. Therefore the total path difference between BR and DR is equal to 2μtcosr ± λ /2 . Therefore BC=PC BQ=PQ=t QC is common side and all angles are equal.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. Therefore ∆= (PC+CF) µ ∆=µ (PF) ------------- (4) From ∆le PBF 𝑐𝑜𝑠𝑟 = 𝑃𝐹/𝑃𝐵 PF= (PB) 𝑐𝑜𝑠𝑟 = (BQ+QP) 𝑐𝑜𝑠𝑟 = (𝑡 + 𝑡)𝑐𝑜𝑠𝑟 = 2t 𝑐𝑜𝑠𝑟 ---------- (5) Therefore eq’n (4) becomes ∆= µ (2t) 𝑐𝑜𝑠𝑟 ∆=2µt 𝑐𝑜𝑠𝑟 --------- (6) But according to stokes principles the light ray BR is undergoing additional path change of λ /2. 11) INTERFERENCE IN THIN FILMS OF NON UNIFORM THICKNESS: 1) Wedge method 2) Newton’s Rings .Condition for bright band/bright fringes We get bright fringes when path difference =n λ 2𝜇𝑡𝑐𝑜𝑠𝑟 ± 𝜆 = 𝑛𝜆 2 𝜆 2 2𝜇𝑡𝑐𝑜𝑠𝑟 = (2𝑛 ± 1) Condition for dark band/dark fringes we get dark fringes when the path difference = (2𝑛 ± 1) 2𝜇𝑡𝑐𝑜𝑠𝑟 ± 𝜆 2 𝜆 𝜆 = (2𝑛 ± 1) 2 2 3.1. due to transmitted light 2𝜇𝑡𝑐𝑜𝑠𝑟 = 𝑛𝜆 The conditions for bright is 2𝜇𝑡𝑐𝑜𝑠𝑟 = 𝑛𝜆 The conditions for dark is 𝜆 2𝜇𝑡𝑐𝑜𝑠𝑟 = (2𝑛 ± 1) 2 We can say the interference pattern due to reflected and transmitted rays are complementary each other. 3.4) NOTE: In thin film. One is reflected from upper surface and other one is from lower surface of air film.𝑠𝑖𝑛𝑟 = 𝐵𝐷 Therefore 𝜇 = 𝐵𝐹 𝐵𝐸 𝐵𝐹 BE= µBF--------------- (2) . 𝑠𝑖𝑛𝑖 = 𝐵𝐷 ∆BFD.(A)WEDGE METHOD: Let us consider two plane surfaces GH. GH1which is inclined at an angle gives wedge shape which encloses an air film AB is the incident ray on GH. The interference takes place between two reflected rays BR and DR1. The optical path difference between BR and DR1 is given by Δ = µ (BC+CD)-BE----------------> (1) For BE From Snell's law 𝜇 = 𝑠𝑖𝑛𝑖 𝑠𝑖𝑛𝑟 𝐵𝐸 From ∆BED. Substitute equation (2) in equation (1) Δ =µ (BC+CD)-µBF Δ =µ (BC+CD-BF) Δ =µ (BC. ΔCPQ are congruent triangles CD=CP Therefore Eq’n (3) becomes Δ =µ (FC+CP) Δ =µ (FP) -----------------------> (4) From ΔFPD 𝑐𝑜𝑠(𝑟 + 𝛼) = 𝐷𝑃 𝐹𝑃 = 𝐷𝑃𝑐𝑜𝑠(𝑟 + 𝛼) 𝐹𝑃 FP=2t 𝑐𝑜𝑠(𝑟 + 𝛼) ------------------>(5) (from figure DP=2t) Sub (5) in (4) Δ=2µt 𝑐𝑜𝑠(𝑟 + 𝛼) --------------------------------.(6) From stoke principle The ray BR undergoes reflection from denser medium. it suffers on additional path change λ/2. Therefore the total path difference= 2𝜇𝑡𝑐𝑜𝑠(𝑟 + 𝛼) ± 2 𝜆 Condition for bright fringe: We get bright fringes when path difference =nλ 2𝜇𝑡𝑐𝑜𝑠(𝑟 + 𝛼) ± 2 = 𝑛𝜆 𝜆 .BF+CD) Δ =µ (FC+CD) -----------------------------> (3) ΔCDQ. 2) Verification of flatness of the given transparent surface.If we draw a line at this edge all along this length we observed straight parallel alternative bright and dark fringes with equal Spacing.3)Applications of Wedge method: 1) Determination of thickness of a paper or diameter of a wire/hair.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 . Fig 3. Now Xn+1 – Xn gives the values of fringe width ''.2𝜇𝑡𝑐𝑜𝑠(𝑟 + 𝛼) = (2𝑛 ± 1) 2 𝜆 Condition for dark fringe: 𝜆 We get dark fringe when path difference= (2𝑛 ± 1) 2 2𝜇𝑡𝑐𝑜𝑠(𝑟 + 𝛼) ± 2 = (2𝑛 ± 1) 2 2𝜇𝑡𝑐𝑜𝑠(𝑟 + 𝛼) = 𝑛𝜆 3.4. 𝜆 𝜆 Spacing between two consecutive dark fringes Let us assume that nth dark fringes is formed at a distance Xn and (n+1)th dark fringe at a distance Xn+1 from wedge.4. 3.4. .4)Determination of thickness of paper or thin film First we measure the spacing between two consecutive dark/bright fringes which is known as fringe width to determine the thickness of paper. Condition for dark fringe is 2𝜇𝑡𝑐𝑜𝑠𝑟 = 𝑛𝜆 -------------.(d) Sub eq (d) in eq(c) 2𝑥𝑛 tan 𝛼 cos 𝛼 = 𝑛𝜆 2xn sin 𝛼 = 𝑛𝜆 .(a) For normal incidence i=0 r=0 For air µ =1 Eq--(a) becomes 2𝑡𝑐𝑜𝑠𝑟 = 𝑛𝜆 ------------------------->(b) For nth dark fringe 2𝑡𝑛 𝑐𝑜𝑠𝑟 = 𝑛𝜆 --------------------------(c) From figure (ii) 𝑡𝑛 𝑥𝑛 tan 𝛼 = 𝑥𝑛 tan 𝛼 = 𝑡𝑛 ---------------. (j) Thickness of paper: λ If the angle between two wedges is very small then AB=AC=l From ΔABC.(h) Now fringe width 𝛽 = 𝑥𝑛+1 − 𝑥𝑛 𝛽 = (𝑛+1)𝜆 2𝑠𝑖𝑛𝛼 − 2𝑠𝑖𝑛𝛼 𝜷 = 𝟐𝒔𝒊𝒏𝜶 -----------------(i) 𝝀 𝑛𝜆 As α is very small .𝑥𝑛 = For (n+1)th dark fringe 2𝑡𝑛+1 𝑐𝑜𝑠𝛼 = (𝑛 + 1)𝜆 -------------.(f) From fig(ii) 𝑡𝑎𝑛𝛼 = 𝑡𝑛+1 /𝑥𝑛+1 𝑛𝜆 − − − − − −(𝑒) 2𝑠𝑖𝑛𝛼 𝑡𝑛+1 = 𝑥𝑛+1 𝑡𝑎𝑛𝛼 ------------------------>(g) sub eq's(g) in (f) 2𝑥𝑛+1 𝑡𝑎𝑛𝛼𝑐𝑜𝑠𝛼 = (𝑛 + 1)𝜆 -----------------. 𝑠𝑖𝑛𝛼 = 𝛼 β = 2α -------------------. 𝑠𝑖𝑛𝛼 = 𝐵𝐶 𝐵𝐶 𝑡 = = 𝐴𝐵 𝐴𝐶 𝑙 . When monochromatic light is allowed to fall normally then we get fringes which are circular.This is equation (k) Sub eq’n (k) in eq’n(i). uniform in thickness and with the point of contact as the center. Where λ = wavelength incident light l = length of air film β = Fringe width (B)NEWTON'S RINGS: Plano-convex lens on flat black surface When a Plano convex lens with its convex surfaces is placed on glass plate air film is formed between two whose thickness increases gradually. This phenomenon was first described by Newton that why they are known as Newton’s rings. The thickness of air film at the point of contact is zero. These fringes are concentric circle. . fringe width becomes 𝛽 = 𝜆 2𝑠𝑖𝑛𝛼 𝜆𝑙 2𝑡 𝛽 = 𝑡 = 𝜆𝑙 2𝛽 This is an expression for thickness of paper. e one is reflected from upper surface of air film and another one reflected from surface of air film. Now interference take place between these two rays i. Thus we get alternative bright and dark fringes . α is so small and it is neglected.Experimental arrangement lower L is a planoconvex lens with large radius of curvature . t=0 then eq--( 2) becomes . For normal incidence i=0. Here we get central ring as dark ring . monochromatic light is incident on L normally by using 45° arrangement of glass plate. For air film μ=1 Path difference from equation 1 becomes Path difference =2t+λ/2-----------. G is a plane glass plate L is placed with its convex surface on G. the reason is as follows : The path difference between two light rays due to reflected light equal to 2𝜇𝑡𝑐𝑜𝑠(𝑟 + 𝛼) ± 𝜆/2 Due to large radius of curvature of lens. r=0.(2) At point of contact of lens and glass plate. A part of light is reflected from curved surface of lens and another one is reflected from plane surface of glass plate. So central ring is dark due to reflected light.(4) Conditions for dark rings: We observe minimum intensity/ dark ring When Path difference = (2n+1) λ/2 𝜆 𝜆 2𝑡 + 2 = (2𝑛 + 1) 2 (from 2) 2t=n λ-----------------.(5) Calculation of Diameter of bright and Dark rings formed due to reflection of light : fig .The path difference = λ/2---------. 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----------.(3) This is a condition for minimum intensity. (Here d=t) From Figure t = thickness of air film D=Diameter of Newton's rings r=Radius of Newton’s ring =D/2 R = Radius of curvature of convex lens . Diameter of Dark Ring: From eq-5 2𝑡 = 𝑛𝜆 𝐷𝑑 ² = 4𝑅(2𝑡) 𝐷𝑏 ² = 4𝑅𝑛𝜆 𝐷𝑑 = √2𝑅𝜆𝑛 𝐷𝑏 𝛼 √𝑛 The diameter of the dark ring is proportional to square root of natural numbers.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. D²= 8Rt D²= 4R(2t) Condition for Diameter of bright ring From (4) 2𝑡 = (2𝑛 − 1) 𝜆 2 2 𝐷𝑏 = 4𝑅(2𝑡) 𝐷𝑏 ² = 4𝑅(2𝑛 − 1) 2 𝐷𝑏 = √2𝑅𝜆(2𝑛 − 1) 𝐷𝑏 𝛼√(2𝑛 − 1) 𝜆 Diameter of bright rings is proportional to square root of odd natural numbers. .t² is neglected.From properties of chords in circle ON * OD = OP *OQ t(2R . The condition for diameter of dark ring is given by 𝐷𝑑 ² = 4𝑅𝑛𝜆 If we consider mth dark ring . From this we can say that width decreases as order increases.Note: The distance between rings decreases when the order of rings increases. APPLICATIONS: 1) Determination of wavelength of monochromatic source 2) Determination of refractive index of liquid 1) Determination of wavelength of monochromatic source First Newton's Rings are formed by respective experimental arrangement. Reason: Condition for dark ring is 𝐷𝑑 = √2𝑅𝜆𝑛 D16= 8√Rλ D9= 6√Rλ D4= 4√Rλ D1= 2√Rλ D16 -D9=2√Rλ ------------->7 Fringes D9-D4 = 2√Rλ -------------->5 fringes D4-D1= 2√Rλ ---------------->3 Fringes Here the distance( 2√Rλ) remains constant but number of rings increases . (a) If we consider nth dark ring The diameter for nth dark ring is given by 𝐷𝑛 ² = 4𝑅𝑛𝜆---------------------. Again we measured the diameter of mth and nth rings whose diameters are D'²m and D'²n For liquid medium 𝜆 𝜇𝑙𝑖𝑞𝑢𝑖𝑑 𝐷1 𝑚 ² − 𝐷𝑛 1 = 4𝑅( 𝑚 − 𝑛) 2 . h is height of convex lens. The diameter of nth and mth rings are measured. (ii) DETERMINATION OF REFRACTIVE INDEX OF A LIQUID First the experiment is performed when there is an air film is formed between glass plate and Plano convex lens.equation(b) 𝐷𝑚 ² − 𝐷𝑛 2 = 4𝑅(𝑚 − 𝑛)𝜆 𝐷𝑚 ² − 𝐷𝑛 2 4𝑅( 𝑚 − 𝑛) 𝜆 = This is equation(c) R is radius of curvature of lens is measured by spherometer. For air film.(b) Equation (a) .The diameter for mth dark ring is given by 𝐷𝑚 ² = 4𝑅𝑚𝜆 -----------------. R=l²/6h+h/2 Where l is the distance between two legs of spherometer. 𝜆 𝜇𝑎𝑖𝑟 = 𝐷𝑚 ² − 𝐷𝑛 2 4𝑅( 𝑚 − 𝑛) This is equation (d) Now the liquid whose refractive index is to be measured is powered in the container without disturbing the whole arrangement. Condition for bright spot 2𝑡 + 2 = 𝑛𝜆 2𝑡 = (2𝑛 − 1)𝜆/2 The mirror M1 is set at a distance x1 for one 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. two fringe patterns coincide practically. We move M2 again another indistinctness is observed . For nth fringe 𝜆 . 2) Determination of difference of nearly equal wavelengths 1)DETERMINATION OF WAVELENGTH OF MONOCHROMATIC LIGHT: Michelson interferometer is set for circular fringes set bright at center. Their wavelengths are nearly equal. As thickness between M2 and M2¹ is very small. Let it be zeroth fringe. Note Down this distance .Let the fringe distance be x. We move M 1 so that nth bright fringe of λ1 coincides with (n+1)th dark fringes of λ2.This is equation (e) From eq (d) /eq (e) then we get 𝜇𝑙𝑖𝑞𝑢𝑖𝑑 = 𝐷𝑚 ² − 𝐷𝑛 2 𝐷1 𝑚 ² − 𝐷𝑛 1 2 APPLICATIONS 1) Determination of wavelength of monochromatic light. So they form two separate fringe n . The difference in their wavelengths of D1 and D2 lines. At this position we observe indistinctness. By moving M1 we separate these two fringe pattern s. 13)QUIZ WHY RAINBOW FORMS: .2x=nλ1 𝑛 = 2𝑥 𝜆1 This is equation (1) For (n+1)th fringe 2x= (n+1)λ2 𝑛 + 1 = 2𝑥 𝜆2 This is equation (2) Equation (2) – equation (1) 𝑛 + 1 − 𝑛 = 2𝑥 𝜆1 2𝑥 𝜆2 − 𝜆1 − 𝜆2 = 𝜆1 − 𝜆2 = 𝜆1 𝜆2 2𝑥 𝜆²𝑎𝑣𝑔 2𝑥 This is difference between wavelengths. 3.1. Solution: Given that 𝐼1 = 𝑎1 2 =25 2 2 𝐼 𝑎2 10 Hence Now 𝑎1 3. What is the .WHY MORPHO’S BUTTERFLY CHANGES IT’S COLOR FROM BLUE TO BROWN .6647 wavelength out of phase. the resultant intensity is ∅ I=4𝑎2 𝑐𝑜𝑠 2 2 Or resultant amplitude is A=2acos2 ∅ 1 4 2.1351 =19.1. 3.14) SOLVED EXAMPLES: (1)Two coherent sources of intensity 10 w𝑚−2 and 25 w𝑚−2 interfere to from fringes.724 (2)Two sinusoidal waves of equal amplitudes are amplitude of the resultant? Solution: When two waves of equal amplitude but different phase superpose. .162 𝑎2 = 5 2 𝐼𝑚𝑎𝑥 (𝑎1+ 𝑎2) 𝐼𝑚𝑖𝑛 (𝑎1−𝑎2) 2 = =0. Find the ratio of maximum intensity to minimum intensity. Calculate the angle of wedge. 𝑟 = 40. Calculate the smallest thickness of the glass plate which will appear dark by reflection. Given path difference =4 𝜆 Hence phase difference Ø= 𝜆 Χ4=2 Hence A=2a cos 4 1 𝑎 𝜋 2𝜋 𝜆 𝜋 =2a√2=√2 (3) A soap film of refractive index 4/3 and of thickness 1. Solution: For dark band 2µtcosr=𝑛λ Given 𝑖 =60º Hence µ=sin 𝑟 Or sin 𝑟= sin 𝑖 sin 60° 𝜇 sin 𝑖 = 1.27cm when a liquid is introduced between the lens and the plate.Calculate the order of interference of the dark band.40 cm to 1. Find the diameter of 20th dark ring. 5) In Newton's Rings experiment. the fringe spacing is 1mm and wavelength of light is 5893 A°.Where Ø is the phase difference.6511 i. 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 .6 3. 4) Fringes of equal thickness are observed in a thin glass wedge of refractive index.33 =0.5Χ 10−4 cm is illuminated by white light incident at an angle of 60º.Deduce the ratio of maximum intensity to minimum Intensity.1.700cm respectively.400 cm and 0.15) PROBLEMS FOR PRACTICE 1) Two coherent source of intensity ratio interfere. 6) In Newton's Rings experiment. Prove that in the interference pattern 2) Two coherent sources whose intensity ratio is 81:1 produce interference fringes . the diameter of the 10th rings from 1. The light reflected by it is examined by a spectrometer in which is found a dark band corresponding to a wavelength of 5Χ 10−5 𝑐𝑚. the diameter of the 4th and 12th dark rings is 0..e. Hence we choose a ray which splits into two different parts such as reflected and transmitted to get two coherent waves.If the mean wavelength of the two components of D line are 5893 A°.17) SUMMARY: We have seen that two independent sources of light cannot acts as coherent sources. 3..2. 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 .1.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.. 10) In an experiment with Michelson's interferometer scale readings for a pair of maximum indistinctness were found to be 0. Due to the path difference between such split rays interference occurs. 1.9884 mm ..02603 mm. 9) The movable mirror of Michelson's interferometer is moved through a distance of 0.6939 mm and 0...if a wavelength of 5200 A° is used. 2. Deduce the difference between wavelengths.refractive index of the liquid. find the number of fringes shifted across the cross wire of eye piece of the telescope .. displacements and measurements. Even two different parts of the same lamp cannot acts as coherent sources of light.. I min= (a1-a2)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. 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. The phenomena of interference are very useful in many areas such as holography optical switching calibration of instruments...0589 mm .Calculate the monochromatic light wavelength.. D²n / D'²m .D²n/4R(m .D'²n . 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 Newton's 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 .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.n) where R=l²/6h+h/2 Refractive index of a liquid µ liquid= D²m .