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1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film.

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Presentation on theme: "1 INTERFERENCE 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film."— Presentation transcript:

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2 1 INTERFERENCE

3 2 Topics Two source interference Double-slit interference Coherence Intensity in double slit interference Interference from thin film Michelsons Interferometer Text Book: PHYSICS VOL 2 by Halliday, Resnick and Krane (5 th Edition) BE-PHYSICS- INTERFERENCE MIT-MANIPAL

4 What is an Electromagnetic wave (EM)? The electromagnetic waves consist of the electric and magnetic field oscillations. In the electromagnetic waves, electric field is perpendicular to the magnetic field and both are perpendicular to the direction of propagation of the waves Electric field (E) Magnetic field (B)

5 Examples: Light waves Heat waves Radio and television Waves Ultraviolet waves Gamma rays, X- rays Electromagnetic waves are non-mechanical waves i.e they do not require material medium for propagation. They are transverse waves. ie. They travel in the form of crests and troughs. Properties of electromagnetic waves (EM)

6 They differ from each other in wavelength ( λ ) and frequency (f). In vacuum, all electromagnetic waves (EM) move at the same speed and differ from one another in their frequency (f). Speed=c=Frequency x wavelength i.e c= f x λ c= 3 x 10 8 m/s

7 Electromagnetic spectrum NameFrequency range (Hz)Wavelength range Gamma rays(γ-rays)5 x x nm X-rays3 x x nm-30nm Ultraviolet light1 x x nm-400nm Visible light8 x x nm-800nm Infra-red4 x x nm-30000nm Radio frequencies3 x x nm More frequency (f) more energy (E), and lesser wavelength(λ).

8 Albert Einstein proposed that light not only behaves as a wave, but as a particle too. Light is a particle in addition to a wave-Dual nature of light. Dual Nature of Light Dual nature of light - treated as 1) a wave or 2) as a particle Light as a stream of particles Dual nature of light successfully explains all the phenomena connected with light.

9 The wave nature of light dominates when light interacts with light. f When light behaves as a Wave? The wave nature of light explains the following properties of light: Refraction of light Reflection of light Interference of light Diffraction of light Polarization light

10 When light behaves as a stream of particles? The particle nature of light dominates when the light interacts with matter (like solids, liquids and gases). Particle nature - Photoelectric, Compton Effect, Black body radiation.. The light is propagated in bundles of small energy, each bundle being called a quantum. Each quantum is composed of many small particles called quanta or photon. Light as a stream of particles Quantum (bundles/packets of energy) (Photon/quantum) Photon energy E = hf h = Plancks constant = 6.626x Js f = frequency of radiation

11 Light as a wave: c= f Light as a particle: E = hf photon Energy of a photon or light wave: Where h = Plancks constant = 6.626x Js f = frequency of a light wave - c = velocity of light λ= wavelength of a light wave -distance between successive crests

12 Visible Light 400–435 nmViolet 435 nm-440nm Indigo 440–480 nmBlue 480–530 nmGreen 530–590 nmYellow 590–630 nmOrange 630–700 nmRed The color of visible light is determined by its wavelength. White light is a mixture of all colors. We can separate out individual colors with a prism.

13 Wave Function of Sinusoidal Waves y(x,t) = y m sin(kx- t) y m : amplitude kx- t : phase k: wave number : angular frequency

14 13 PRINCIPLE OF SUPERPOSITION When two waves traveling almost in the same direction superpose, the resulting displacement at a given point is the algebraic sum of the individual displacements. i.e. when waves, y 1 =A sin ωt & y 2 =A sin (ωt + ) superpose, the resultant displacement is y= y 1 +y 2 = a sin (ωt) + a sin (ωt+ )

15 14 INTERFERENCE OF LIGHT

16 TWO-SSOURCE INTERFERENCE When identical waves from two sources overlap at a point in space, the combined wave intensity at that point can be greater or less than the intensity of either of the two waves. This effect is called interference. When two waves of same frequency (or wavelength) with zero initial phase difference or constant phase difference superimpose over each other, then the resultant amplitude (or intensity) in the region of superposition is different than the amplitude (or intensity) of individual waves. OR

17 16 At certain points either two crests or two troughs interact giving rise to maximum amplitude resulting in maximum intensity. (Constructive interference). At certain points a crest and a trough interact giving rise to minimum or zero amplitude resulting in minimum or zero intensity. (Destructive interference).

18 Constructive Interference Destructive interference Two waves (of the same wavelength) are said to be in phase if the crests (and troughs) of one wave coincide with the crests (and troughs) of the other. The net intensity of the resultant wave is greater than the individual waves. (Constructive interference). If the crest of one wave coincides with the trough of the second, they are said to be completely out of phase. The net intensity of the resultant wave is less than the individual waves. (Destructive interference). TWO-SOURCE INTERFERENCE

19 INTERFERENCE PATTERN PRODUCED BY WATER WAVES IN A RIPPLE TANK 18 Maxima: where the shadows show the crests and valleys (or troughs). Minima: where the shadows are less clearly visible BE-PHYSICS- INTERFERENCE TWO-SOURCE INTERFERENCE MIT-MANIPAL

20 Phase: Phase of a vibrating particle at any instant indicates its state of vibration. Phase may be expressed in terms of angle as a fraction of 2π. PHASE AND PATH DIFFERENCE F C A G λ λ B π/2 π 3π/2 D E 2π2π O Path difference corresponds to phase difference of 2. λ/4 λ/2 3λ/4 λ t=0

21 Constructive interference path difference p= 0 or phase difference = 0 path difference p = 1 or phase difference =2 path difference p =2 or phase difference = 4 General condition: Path difference p = m or phase difference = 2m where m = 0, 1, 2, 3,………… order of interference. 2λ2λ

22 Maximal constructive interference of two waves occurs when their: path difference between the two waves is a whole number multiple of wavelength. OR Phase difference is 0, 2, 4, … (the waves are in-phase). Constructive interference

23 22 path difference p= /2 phase difference = 1 path difference p = 3 /2 or phase difference =3 DISTRUCTIVE INTERFERENCE path difference p = 5 /2 phase difference = 5 General condition: Path difference p=(m+1/2) or phase difference =(2m+1) where m = 0, 1, 2, 3, ………

24 23 Complete destructive interference of two waves occur when the path difference between the two waves is an odd number multiple of half wavelength. Or the phase difference is, 3, 5, … (the waves are 180 o out of phase).

25 24 Coherence – necessary condition for interference to occur. Two waves are called coherent when they are of : same amplitude same frequency/wavelength same phase or are at a constant phase difference COHERENCE

26 25 A SECTION OF INFINITE WAVE A WAVE TRAIN OF FINITE LENGTH L COHERENCE No two independent sources can act as coherent sources, because the emission of light by the atoms of one source is independent of that the other. If the two sources are completely independent light sources, no fringes appear on the screen (uniform illumination). This is because the two sources are completely incoherent. For interference pattern to occur, the phase difference at point on the screen must not change with time. This is possible only when the two sources are completely coherent.

27 26 A SECTION OF INFINITE WAVE A WAVE TRAIN OF FINITE LENGTH L BE-PHYSICS- INTERFERENCE COHERENCE Common sources of visible light emit light wave trains of finite length rather than an infinite wave. The degree of coherence decreases as the length of wave train decreases. MIT-MANIPAL

28 27 BE-PHYSICS- INTERFERENCE COHERENCE Common sources of visible light emit light wave trains of finite length rather than an infinite wave. MIT-MANIPAL Laser light is highly coherent whereas a laboratory monochromatic light source (sodium vapor lamp) may be partially coherent.

29 28 (2) Division of Amplitude: In this method, the amplitude of the incoming beam is divided into two or more parts by partial reflection with the help of mirrors, lenses and prisms. These divided parts travel different paths and finally brought together to produce interference. The common methods are a.Newtons rings, b. Michelsons interferometer. COHERENCE ( 1). Division of wave front: In this method, the wave front is divided into two or more parts with the help of mirrors, lenses and prisms. The common methods are: a.Youngs double slit arrangement, b. Lloyd's single mirror method. Methods of producing coherent sources:

30 29 DOUBLE SLIT INTERFERENCE Screen Two narrow slits (can be considered as two sources of coherent light waves). If the widths of the slits are small compared with the wavelength distance a (<< ) - the light waves from the two slits spread out (diffract) – overlap- produce interference fringes on a screen placed at a distance D from the slits. d If light waves did not spread out after passing through the slits, no interference would occur..

31 30 A monochromatic light source produces two coherent light sources by illuminating a barrier containing two small openings (slits) S 1 and S 2 separated by a distance d and kept at a distance D from the screen. Waves originating from two coherent light sources S 1 and S 2 because maintain a constant phase relationship. In interference phenomenon, we have assumed that slits are point sources of light. DOUBLE SLIT INTERFERENCE B C

32 When the light from S 1 and S 2 both arrive at a point on the screen such that constructive interference occurs at that location, a bright fringe appears. When the light from the two slits combine destructively at any location on the screen, a dark fringe results. B C DOUBLE SLIT INTERFERENCE

33 32 At center of the screen O, we always get a bright fringe, because at this point the two waves from slit S 1 and S 2, interfere constructively without any path/phase difference. O DOUBLE SLIT INTERFERENCE

34 - Analysis of Interference pattern We consider waves from each slit that combine at an arbitrary point P on the screen C. The point P is at distances of r 1 and r 2 from the narrow slits S 1 and S 2, respectively. DOUBLE SLIT INTERFERENCE a-is the mid point of the slit

35 34 For D>>d, we can approximate rays r 1 and r 2 as being parallel. The line S 2 b is drawn so that the lines PS 2 and Pb have equal lengths. Path length (S 1 b) between the rays r 1 and r 2 reaching the point P decides the intensity at P. (i.e. maximum/minimum). DOUBLE SLIT INTERFERENCE

36 Path difference between two waves from S 1 & S 2 (separated by a distance d) on reaching a point P on a screen at a distance D from the sources is d sin. The path difference (S 1 b=d sin ) determines whether the two waves are in phase or out of phase when they arrive at point P. If path length (S 1 b= d sin θ ) is either zero or some integer multiple of the wavelength, constructive interference results at P. (d sinθ) DOUBLE SLIT INTERFERENCE

37 Constructive Interference: Maximum at P: P O The condition for constructive interference or Maxima at point P is d sin =m ……….(maxima) Where m= 0, ±1, ±2……. m- order number. Central maximum at O has order m=0. Each maximum above has a symmetrically located maximum below O; these correspond to m= -1, -2, -3…… Central maximum m=0

38 Destructive Interference: Minimum at P: When path length (S 1 b=d sin ) is an odd multiple of λ /2, the two waves arriving at point P are (= π ) out of phase and give rise to destructive interference. The condition for dark fringes, or destructive interference, at point P is Central maximum m=0 m=0 m=1 m=2 m=3 m=2 m=1 m=0 Negative values of m locate the minima on the lower half of the screen. P O

39 Fringe width (Δy) The distance between two consecutive bright or dark fringes (for small θ )is known as fringe width Δ y. ΔyΔy ΔyΔy The spacing between the adjacent minima is same the spacing between adjacent maxima. DOUBLE SLIT INTERFERENCE

40 39 m Q O y m+1 For small value of, we can make following approximation. If d sin =m, m th order constructive interference will take place at P. DOUBLE SLIT INTERFERENCE

41 Sample 41-1:Problem 1: The double -slit arrangement is illuminated with light from a mercury vapor lamp filtered so that only the strong green line ( λ =546 nm) is visible. The slits are 0.12 mm apart, and the screen on which the interference pattern appears is 55 cm away. (a)What is the angular position of the first minimum? (b)What is the distance on the screen between the adjacent maxima? λ=546 nm D=0.55 m d=0.12 mm a)θ =? for first minimum. b)Fringe width Δ y=? DOUBLE SLIT INTERFERENCE

42 CHECK YOURSELF Solve for First maximum : use d sinθ=mλ put m=1 DOUBLE SLIT INTERFERENCE

43 Problem 5: A double-slit arrangement produces interference fringes for sodium light ( λ =589 nm) that are apart. For what wavelength would the angular separation be 10% greater? (Assume that θ is small). d sin θ = m λ sin θ θ ( θ is very small) d θ = λ (for first maxima, m=1) Use: λ 1/ λ 2 = θ 1 / θ 2, λ 2 = λ 1 x θ 2 / θ 1, Given: θ 1 = (for 100%) θ 2 = x 1.1 = (10% more: for 110%) λ 2 = λ 1 x θ 2 / θ 1 = (589 nm)(0.253/0.23)= nm DOUBLE SLIT INTERFERENCE

44 43 Problem 11: Sketch the interference pattern expected from using two pinholes rather than narrow slits. In case two pinholes are used instead of slits, as in Young's original experiment, hyperbolic fringes are observed.hyperbolic If the two sources are placed on a line perpendicular to the screen, the shape of the interference fringes is circular as the individual paths travelled by light from the two sources are always equal for a given fringe. DOUBLE SLIT INTERFERENCE

45 Solution: Use d sin θ =m λ d 1 sin 15 0 = λ gives the first maximum (m=1) d 2 sin 15 0 = 2 λ put the second maximum (m=2) at the location of the first. Divide the second expression by the first and d 2 = 2d 1. This is a 100% increase in d 1. Tutorial: Problem:2 Monochromatic light illuminates two parallel slits a distance d apart. The first maximum is observed at an angular position of By what percentage should d be increased or decreased so that the second maximum will instead be observed at 15 0 ? DOUBLE SLIT INTERFERENCE

46 45 Tutorial: Problem 8 In an interference experiment in a large ripple tank, the coherent vibrating sources are placed 120 mm apart. The distance between maxima 2.0 m away is 180 mm. If the speed of ripples is 25 cm/s, calculate the frequency of the vibrating sources. Given: d = 120 x m, λ = ? y m =180 x10 -3 m D = 2 m, v= 25 x m. f=? Use: v=f x λ To find λ: Use y m = λD/d gives λ=0.0108m Use: v= f x λ, f= 23 Hz. DOUBLE SLIT INTERFERENCE

47 46 HRK:958 page: Exercise: 41-2 Problem 1: Monochromatic green light of wavelength 554 nm, illuminates two parallel narrow slits 7.7 μ m apart. Calculate the angular position of the third-order (m=3) bright fringe in radians and (b) in degrees. Given: d = 7.7 x m, m=3, λ = 554 nm = 554 x10 -9 m θ = ? in radians and in degrees. Use: for the bright fringe: d sin θ = m λ (put m=3): Answers: (a)θ =0.216 radians (b) θ =12.5 degrees CHECK YOURSELF (Solve for third order dark fringe put m=2) DOUBLE SLIT INTERFERENCE

48 Solution: Fringe width= Δ y = λ D/d = (512x m)(5.4 m)=(1.2 x m) =2.3 x m CHECK YOURSELF: How far apart are the bright fringes as seen on the screen? Fringe width is the same 2.3 x m. Problem 3 : A double-slit experiment is performed with blue-green light of wavelength 512 nm. The slits are 1.2mm apart and the screen is 5.4 m from the slits. How far apart are the bright fringes as seen on the screen? Given: λ =512 nm d = 1.2 mm D= 5.4 m Fringe width Δ y=? DOUBLE SLIT INTERFERENCE

49 Use: d = λ /sin θ = (592 x m)/sin( ) =3.39x m CHECK YOURSELF Solve this problem for dark interference fringes apart in angular separation. Problem 4: Find the slit separation of a double-slit arrangement that will produce bright interference fringes apart in angular separation. Assume a wavelength of 592 nm. Given: d=? θ = λ = 592 nm DOUBLE SLIT INTERFERENCE

50 Problem 6 A double-slit arrangement produces interference fringes for sodium light ( λ = 589 nm) that are apart. What is the angular fringe separation if the entire arrangement is immersed in water (n=1.33)? Solution: Immersing the apparatus in water will shorten the wavelengths to λ /n. Start with d sin θ 0 = λ ; and then find θ from d sin θ = λ /n. Combining the two expressions, sin θ /sin θ 0 =1/n gives θ = DOUBLE SLIT INTERFERENCE

51 Problem 7 In a double-slit experiment, the distance between slits is 5.22 mm and the slits are 1.36 m from the screen. Two interference patterns can be seen on the screen, one due to light with wavelength 480 nm and the other due to light with wavelength 612 nm. Find the separation on the screen between the third-order interference fringes of the two different patterns. Solution: The third-order fringe for a wavelength will be located at y m = m λ D/d= 3 λ D/d where y m is measured from the central maximum. Then Δ y is: y 1 - y 2 = 3( λ 1 - λ 2 )D/d = 3(612x m – 480x10 -9 m)(1.36m)/(5.22x10 -3 m) = 1.03x10 -4 m: DOUBLE SLIT INTERFERENCE

52 Problem 9: If the distance between the first and tenth minima of a double slit pattern is 18 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits. What is the wavelength of the light used? m Q O y m+1 So with small angle approximation: d sin =(m+1/2 ), sin tan = =y m /D d sin =(m+1/2 ), m th order minima will take place at P. d (y m /D)= =(m+1/2 ), m th order minima will take place at P. y m = =(m+1/2 ) D/d DOUBLE SLIT INTERFERENCE

53 We are given the distance (n the screen) between the first minima (m=0) and the tenth minima (m=9). Then y 9 -y 0 = (9+1/2) D/d- (0+1/2) D/d = 9 ( D/d) Given: y 9 -y 0 = 18 mm, d=50 cm=0.5 m, d= 0.15 mm= 0.15 x10 -3 m Solving for λ = 18 x x 0.15 x / 9 x 0.5 = 600 nm Solve this problem with first and 10th maxima. (CHECK YOURSELF: Put first maxima m=1, and tenth maximum m=10 )

54 53 Coherence Problem 14: The coherence length of a wavetrain is the distance over which the phase constant is the same. (a)If an individual atom emits coherent light for 1 x s, what is the coherence length of the wavetrain? (b) Suppose a partially reflecting mirror separates this wave train into two parts that are later reunited after one beam travels 5 m and other 10 m. do the waves produce interference fringes observable by a human eye? (a)velocity = distance/ time x = c/t = (3.0 x10 8 m/s)/1 x 10 8 s) = 3 m. (b) No.

55 Thomas Young (1773–1829) YOUNGS DOUBLE SLIT INTERFERENC E Double slit experiment was first performed by Thomas Young in So double slit experiment is known as Youngs Experiment. He used sun light as source for the experiment. In youngs interference experiment, sunlight diffracted from pinhole S 0 falls on pinholes S 1 and S 2 in screen B. Light diffracted from these two pinholes overlaps on screen C, producing the interference pattern. 54

56 55 INTENSITY IN DOUBLE SLIT INTERFERENCE In this section we derive an expression for the intensity I at any point P located by the angle θ. We know, the intensity of light wave is proportional to square of its E-the electric vector. Intensity α Amplitude 2 I α E 2 Let us consider the electric components of two sinusoidal waves r 1 and r 2 from the two slits S 1 and S 2 have the same angular frequency ω (=2 π f) and a constant phase difference. E 1 = E 0 sin ω t & E 2 = E 0 sin ( ω t + ) P O E 0- amplitude of each wave.

57 Then the total magnitude of the electric field at point P on the screen is the superposition of the two waves. (Assumption: The slit separation d< { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1429885/4/slides/slide_56.jpg", "name": "Then the total magnitude of the electric field at point P on the screen is the superposition of the two waves.", "description": "(Assumption: The slit separation d<

58 57 INTENSITY IN DOUBLE SLIT INTERFERENCE Adding Wave disturbances: Phasors The combined electric field can be done algebraically, using a graphical method, which proves to be convenient in more complicated situations. Phasormagnitude E o counterclockwise The sinusoidal wave can be represented graphically by a Phasor of magnitude E o rotating about the origin counterclockwise with an angular frequency. w Phasor diagram E 1 (=E 0 sinwt) w

59 58 Phasor diagrams: (a)Phasor representation of first wave disturbance E 1 (=E 0 sinwt) is represented by the projection of the phasor on the vertical axis. The second sinusoidal wave is E 2 = E o sin ( t + ) It has the same amplitude and frequency as E 1. Its phase is with respect to E 1. (b) w

60 59 (c) The sum E of wave disturbances E 1 &E 2 is the sum of the projections of the two phasors on the vertical axis, placing the tail of one arrow at the head of the other, maintaining the proper phase difference. E is the projection on the vertical axis of a phasor of length E θ, which is the vector sum of the two phasors of magnitude E 0. ωt E1E1 E2E2 E E E0E0 E0E0

61 60 ωt E1E1 E2E2 E E E0E0 E0E0 A B C From the right angled triangle ABC

62 61 Where E- The resultant electric field E θ - Amplitude of the resultant wave β – the phase difference between amplitude of the resultant wave and first wave. β β φ EθEθ E0E0 E0E0 From the geometry of the right angled triangle The three phasors, E 0,E 0,E θ form an isosceles triangle (An any triangle, an exterior angle ( φ in this case) is equal to the sum of the opposite interior angles( β and β )). So, 2 β= φ or β =φ/2 AmplitudeWave part Phase shift

63 62 The amplitude E θ of the resultant wave disturbance is given by E θ = 2 E 0 cos β which determines the intensity of the interference fringes, depends on β, which in turn depends of the value of θ, that is, on the location of point P. θ Then the intensity of the resultant wave I θ at P is given by I θ α E θ 2 (for resultant wave) and I 0 α E 0 2 (Intensity of each single wave) I θ = 4 E 0 2 cos 2 β = 4 E 0 2 cos 2 ф /2 I θ = 4 I 0 cos 2 β Note: The intensity of the resultant wave at any point varies from zero (dark or minima) to four times (bright or maxima)the intensity I 0 (i.e. 4I 0 =I m ) of individual wave.

64 63 Intensity distribution in double slit interference I θ = 4 I 0 cos 2 ф /2 Intensity Maxima occur where cos 2 ф /2 =1, or Ф =0, 2 π, 4 π,…………..2m π ф = 2m π (maxima) Or the path difference: d sin θ = m λ m= 0, ±1, ±2……………(maxima) I θ = 4 I 0 cos 2 ф /2 Intensity Minima occur where cos 2 ф /2 =0, or Ф = π, 3 π, 5 π …………..(2m+1) π ф = (2m+1) π (minima) Or the path difference: d sin θ = (m+1/2) λ m= 0, ±1, ±2……… (minima)

65 1.The horizontal solid line is I 0 : this describes the (uniform) intensity pattern on the screen if one slit is covered up. 2.If the two sources were incoherent, the intensity would be uniform over the screen and would be 2 I 0 indicated by the horizontal dashed line. 3.For two coherent sources it would be 4I 0.

66 65 The phase difference ( φ ) between r 1 and r 2 associated with the path difference S 1 b (= d sin θ ) INTENSITY IN DOUBLE SLIT INTERFERENCE PHASE AND PATH DIFFERENCE

67 Where E θ =2 E 0 cos β: Amplitude of the resultant wave. β – the phase difference between resultant wave and first wave. β = ф/2 E = E 1 +E 2 = E o (sin ω t + sin ( ω t+ )) Using Trigonometric Identity: With = ( t + ), = t, get: It can be written as, E= E θ sin(wt+ β )

68 67 Problem: SP 41-2 Find graphically the resultant E(t) of the following wave disturbances. E 1 = E 0 sin t E 2 = E 0 sin ( t + 15 o ) E 3 = E 0 sin ( t + 30 o ) E 4 = E 0 sin ( t + 45 o ) BE-PHYSICS- INTERFERENCE MIT-MANIPAL INTENSITY IN DOUBLE SLIT INTERFERENCE E=E 1 +E 2 +E 3 +E 4 E=E θ sin(wt+ β ) The phase angle φ between successive phasors is 15 0.

69 68 In any closed n-sided polygon, the sum of the interior angles is (n-2)π. So, in the five sided polygon, abcdea: So, 2 β= (165 0 ) So, β = To find β:

70 69 ax 1 = ab cosβ x 1 x 2 = bc (cosβ-15) x 2 x 3 = cd (cosβ-15) x 3 e= de (cosβ) x1x1 b c x2x2 β -15 x3x3 β To find E θ :

71 70 E θ = ax 1 +x 1 x 2 +x 2 x 3 +x 3 e = ab cosβ+ bc (cosβ-15 0 )+ cd (cosβ-15 0 )+de (cosβ) Substitute ab=bc=cd=de= E 0 and β= E θ =3.83 E 0 The resultant E(t) is the projection of E on the vertical axis: E= 3.83 E 0 sin (wt )

72 71 Alternate method: Verification: Ey- sum of verical- components EθEθ E h - sum of horizontal components cos wt sin wt β Standard equation: E(t)= E θ sin (wt+β) We can use the phasors to find sum and free to evaluate the phasors at any time t. We choose t=0.

73 72 Sample Problem: Intensity in double-slit Interference Source A of long-range radio waves leads source B by 90 degrees. The distance r A to a detector is greater than the distance r B by 100m. What is the phase difference at the detector? Both sources have a wavelength of 400m. They are in phase and the reach the detector. Problem: E rArA rBrB 100 m= λ /4 corresponds to π/2 phase Initially, source A leads source B by 90, = 1/4 wavelength (100 m) However, source A also lags behind source B since r A is longer than r B by 100 m, which is 100m/400m = 1/4 wavelength. So, the net phase difference between A and B at the detector is zero. DETECTORDETECTOR

74 73 Find the phase difference between the waves from the two slits arriving at the m th dark fringe in a double-slit experiment. Answer: (2m + 1)π radians. Problem: E 41-16

75 74 Tutorial Problem: E Find the sum of the following quantities (a) graphically, using phasors; and (b) using trigonometry. y 1 =10 sin wt, y 2 =8.0 sin (wt+30 0 ) Similar to the equation: E 1 = E 01 sinwt & E 2 = E 02 sin (wt+ φ) Resultant electric field E= E θ sin (wt+β) Where E θ - Amplitude of the resultant wave β= phase difference between first electric field and the resultant wave.

76 75 To find E θ : Substitute E 01 =10, E 02 =8, & φ=30 0 E θ =17.39 β= The resultant electric field E= sin (wt+13.3) (a) Phasors β E 01 E 02 A B D C EθEθ y x φ

77 76 Alternate method: Trigonometry Ey- sum of verical- components EθEθ E h - sum of horizontal components cos wt sin wt β Standard equation: E(t)= E θ sin (wt+β) We can use the phasors to find sum and free to evaluate the phasors at any time t. We choose t=0.

78 77 Reflection causes a phase shift Reflection causes no phase shift REFLECTION PHASE SHIFT It has been observed that if the medium beyond the interface has a higher index of refraction, the reflected wave undergoes a phase change of (180 o ) ) or a path length λ/2. If the medium beyond the interface has a lower index of refraction, there is no phase/path length change of the reflected wave. Incident wave Reflected wave Incident wave Reflected wave

79 INTERFERENCE FROM THIN FILMS We see color when sunlight falls on a bubble, an oil slick, or a soap bubble - the interference of light waves reflected from the front and back surfaces of thin, transparent films. The film thickness is typically of the order of magnitude of the wavelength of light. A soapy water film on a vertical loop viewed by reflected light

80 79 Thin films deposited on optical components- such as camera lenses-reduce reflection and enhance the intensity of the transmitted light. Thin coatings on windows can enhance the reflectivity for infrared radiation while having less effect on the visible radiation. It is possible to reduce the heating effect of sunlight on a building.

81 80 d n1 n1 n1 n1 n2 n2 Source of light A thin film (a soap film or thin film of air between two glass plates) is viewed by light reflected from a sourer S. Waves reflected from the front surface (r 1 ) and back surface (r 2 ) interfere and enter the eye. INTERFERENCE FROM THIN FILMS

82 d n1 n1 n1 n1 n2 n2 Source of light The incident ray i from the source enters the eye as ray r 1 after reflection from the front surface of the film at a. The incident ray i also enters the film at a as refracted ray and is reflected from the back surface of the film at b. It then emerges from the front surface of the film at c and also enters the eye, as ray r 2. d- thickness of the film.

83 82 Interference in light reflected from a thin film is due to a combination of rays r 1 and r 2. As the waves originated from the same source by division of amplitude, hence they are coherent and they are close together. The region ac looks bright or dark for an observer depends on the phase difference between waves of rays r 1 and r 2. d n1 n1 n1 n1 n2 n2 Source of light

84 83 d n1 n1 n1 n1 n2 n2 Source of light As r 1 and r 2 have travelled over paths of different lengths, have traversed different media, and have suffered different kinds of reflections at a and b. The phase difference between two reflected rays r 1 and r 2 determine whether they interfere constructively or destructively.

85 84 If a wave travelling from a medium of index of refraction n 1 toward a medium of index of refraction n 2 undergoes a phase change upon reflection when n 2 >n 1 and undergoes no phase change if n 2 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1429885/4/slides/slide_84.jpg", "name": "84 If a wave travelling from a medium of index of refraction n 1 toward a medium of index of refraction n 2 undergoes a phase change 180 0 upon reflection when n 2 >n 1 and undergoes no phase change if n 2

86 85 λ The wavelength of light λ n in a medium whose index of refraction is n λ - wavelength of the light in free space.

87 To obtain Equations for thin film Interference, let us simplify by assuming near - normal incidence θ i =0. The r 2 travels a longer path (2d) than r 1 as r 2 travels twice through the film before reaching the eye. d n1 n1 n1 n1 n2 n2 Source of light

88 87 The path difference due to the travel of ray r 2 through the film is approximately 2d. Other possible contributions to the total path difference between r 1 and r 2 : the phase difference of π (or path difference of one-half wavelength) that might occur on reflection at the front/back surface of the film. d n1 n1 n1 n1 n2 n2 Source of light

89 The path difference between rays r 1 and r 2 is: Path difference = 2d + λ n /2 + λ n / 2 …………(1) Front surface Back surface Note: Depending on the relative index of refraction of the film in comparison with what is on either side of the film, we might need to include neither of the extra terms, or perhaps one of them, or perhaps both of them. ?? A general equation for Thin film Interference:

90 d Air n phase change No phase change Examples: (a) Interference in a thin soap film For the interference from a thin soap film of index of refraction n surrounded by air, we must add the extra half wavelength for the front surface reflection, but not for the back surface.

91 90 The total path difference: = 2d + λ n /2 =(m+1/2) λ n ……… (minima) where m=0,1,2,3….. The total path difference= 2d+ λ n /2=m λ n...(maxima) where m=1,2,3……… Where we have dropped the m=0 solution because it is not physically meaningful. Interference in a thin soap film Note: These equations apply when the index of refraction of the film is greater than the index of refraction of the material on either side.

92 91 THIN FILM INTERFERENCE -WEDGE SHAPED FILM A thin wedge of air film can be formed by two glasses slides on each other at one edge and separated by a thin spacer (a thin wire or a thin sheet) at the opposite edge. A thin film having zero thickness at one end and progressively increasing to a particular thickness at the other end is called a wedge.

93 92 The arrangement for observing interference of light in a wedge shaped film. The wedge angle is usually very small and of the order of a degree. When a parallel beam of monochromatic light falls normally on a wedge shaped film part of it is reflected from upper surface and some part from lower surface (division of amplitude). wire d

94 Ray BC reflected from the top – NO phase change. Ray DE (the back surface reflection), undergoes a π phase change and λ /2 (half wave length) at the air to glass boundary due to reflection. These two coherent waves superpose-producing constructive and destructive interferences, the positions of which depend on the thickness of the film. Back surface i r

95 Constructive Interference The total path difference: Path difference = 2d + λ n /2 = m λ n …..(maxima) where m=1,2,3…… m=0 dropped, physically not meaningful. Destructive Interference Path difference = 2d + λ n /2 = (m+1/2) λ n …(minima) where m=0,1,2,3………………………………

96 Sample problem:41-3 A soap film (n=1.33) in air is 320 nm thick. If it is illuminated with white light at normal incidence, what color will it appear to be in reflected light? No phase change d Air n= phase change No phase change Solution: The wavelengths which are maximally reflected are constructively interfered.

97 96 Constructive interference maxima occur for the following wavelengths: 1702 nm (=1), 567 nm(m=2), 340 nm (m=3) and so on. Only the maximum corresponding to m=2 lies in the visible region (between about 400 nm and 700 nm); light of wavelength 567 nm appears yellow-green.

98 Sample Problem: Page 951 Lenses are often coated with thin films of transparent substances such as MgF 2 (n=1.38) to reduce the reflection from the glass surface. How thick a coating is needed to produce a minimum reflection at the center of the visible spectrum (λ=550 nm)? Given: λ =550 nm n=1.38 Minimum reflection: (Destructive interference) Thickness of coating: d=?

99 98 Solution: Light strikes the lens at near-normal incidence ( θ ). For both the front and back surfaces of the MgF 2 film the reflection have additional path difference ( λ /2). The path difference for destructive interference is therefore Path difference: 2nd+ λ n /2+ λ n /2=(m+1/2) λ n …(minima)

100 Where m=1,2,3……………………………, Dropped m=0 solution : physically not meaningful. We seek the minimum thickness for destructive interference. For m=1, we obtain

101 100 Problem: E BE-PHYSICS- INTERFERENCE A disabled tanker leaks kerosene (n=1.20) into the Persian Gulf, creating a large slick on top of water (n = 1.33). (a)If you look straight down from aeroplane on to the region of slick where thickness is 460nm, for which wavelengths of visible light is the reflection is greatest? (b) If you are scuba diving directly under this region of slick, for which wavelengths of visible light is the transmitted intensity is strongest? MIT-MANIPAL INTERFERENCE FROM THIN FILMS

102 101 Solution: The reflected light from the film is brightest at the wavelength ( λ ) for which the reflected rays are in phase with one another.(constructive interference). d=460nm White light (400 nm- 700nm) Normal incidence Reflected light (λ=?) R.I=1.0 R.I=1.33 n=1.2 Thin film of oil water (Both Front & back surface reflections have phase change).

103 102 The total path difference for maxima: Path difference = 2d + λ n /2+ λ n /2 = m λ n, Where (m=1,2,3….) 2d=(m-1) λ /n ( λ n =λ/n) λ =2nd/(m-1) Find λ for d=460 nm, n=1.2 & m=1,2,3… λ - for m=1 (not possible) For m=2: λ =1104 nm (IR region) For m=3: λ =552 nm (Green light-visible) For m=4: λ =368 nm (UV light) So, Green light appears in the reflected light

104 103 The wavelengths which are minimally reflected are maximally transmitted, and vice versa. Maximally transmitted wavelengths is the same as finding the minimally reflected wavelengths. The total path difference for minima: Path difference = 2d + λ n /2+ λ n /2 = (m+1/2) λ n where (m=1,2,3….) substitute : ( λ n =λ/n) λ =2nd/(m-1/2) Maximum Transmitted light =? R.I=1.0 R.I=1.33 n=1.2 Thin film of oil water Minimally reflected light

105 104 Find λ for d=460 nm, n=1.2, & m=1,2,3… λ - for m=1: λ =2208 nm (not visible region) For m=2: λ =736 nm (IR region) For m=3: λ =442 nm (Blue light-visible)(maximum transmitted)

106 105 Problem: E If the wavelength of the incident light is λ =572 nm, rays A and B are out of phase by 1.50 λ. Find the thickness d of the film. The total path difference for minima: Path difference = 2d + λ n /2= (m+1/2) λ n 2d= m λ n m=0 not possible Take m=1 d= λ /2n=215nm

107 OR 106 In the given total phase of 1.50 λ n. The top surface contributes a phase difference π =0.5 λ n So, phase difference because of the thickness= 2d=λ n =2π 2d= λ n d= λ/2n=572 nm/2 x 1.33=215 nm

108 107 Problem: E A broad source of light (wavelength = 680nm) illuminates normally two glass plates 120 mm long that touch at one end and are separated by a wire 0.048mm in diameter at the other end. How many bright fringes appear over 120 mm distance? BE-PHYSICS- INTERFERENCE MIT-MANIPAL INTERFERENCE FROM THIN FILMS d=0.048 x m) λ=680 x m θ x=120 mm

109 108

110 109 Problem: E BE-PHYSICS- INTERFERENCE A thin film of acetone (n = 1.25) is coating a thick glass plate (n = 1.50). Plane light waves of variable wavelengths are incident normal to the film. When one views the reflected wave, it is noted that complete destructive interference occurs at 600nm and constructive interference at 700nm. Calculate the thickness of the acetone film? MIT-MANIPAL INTERFERENCE FROM THIN FILMS Thin film n=1.25 Glass plate 1.50 Put: λ n =λ/n

111 110 Constructive interference 2d + λ n /2 + λ n / 2 =m λ n Gives : 2nd=(m-1) λ 2nd=(m-1)(700nm ) ……………(1) Destructive interference 2d + λ n /2 +λ n / 2 =(m+1/2 )λ n 2nd=(m-1/2) λ 2nd=(m-1/2)(600nm ) ……………(2) Divide (1)/(2): m=4, d=840 nm

112 111 Problem 21:We wish to coat a flat slab of glass (n = 1.5) with a transparent material (n=1.25) so that light of wavelength 620nm (in vacuum) incident normally is not reflected. What should be the minimum thickness of the coating? Glass=1.5 Film= 1.25 Air=1

113 112 Both the reflected rays r 1 (front surface reflection) and r 2 (back surface reflection) have additional path difference (λ/2). m=1

114 113 Problem 22: A thin film in air is 410nm thick and is illuminated by white light normal to its surface. Its index of refraction is What wavelength in the visible spectrum will be intensified in the reflected beam? No phase change d=410 nm Air n= phase change

115 114 The result is only in the visible range when m = 3, so λ= 492 nm.

116 115 Problem: 24: In costume jewelry, rhinestones (made of glass with n =1.50) are often coated with silicon monoxide (n = 2.0) to make them more reflective. How thick should the coating be to achieve strong reflection for 560 nm light incident normally? Glass=n=1.5 Film=n= 2.0Air=n=1

117 116

118 117 Problem 26: Light of wavelength 585nm is incident normally on a thin, soapy film (n=1.33) suspended in air. If the film is mm thick, determine whether it appears bright or dark when observed from point near the light source. No phase change d= mm Air n= phase change

119 118 So the interference is NOT dark. m should be an integer. So the interference is bright.

120 119 Problem 28: White light reflected at perpendicular incidence from a soap film in air has, in the visible spectrum, an interference maximum at 600nm and a minimum at 450nm with no minimum in between. If n = 1.33 for the film, what is the film thickness? No phase change d=? Air n= phase change

121 120 With First source (λ=600nm): With second source (λ=450nm): Use equation (1) or (2) to solve for d: From equation (2) d= 2 x 450/2 x1.33=338 nm

122 Problem 31: Light of wavelength 630nm is incident normally on a thin wedge shaped film with index of refraction There are ten bright and nine dark fringes over the length of the film. By how much does the film thickness changes over the length? d1d1 d2d2 x B 10 B1B1 B2B2 B3B3 B4B4 B5B5 B6B6 B7B7 B8B8 B9B9 m 1 =1m 2 =10

123 122 Number of bright bands in x mm length. 10 bright & 9 dark bands. Film thickness over the length d 2 -d 1 =?

124 Instead of wedge shaped films, interference is possible even in curved films also. Circular interference fringes can be produced by enclosing a very thin film of air of varying thickness between a plane glass plate and a plano -convex lens of a large radius of curvature. INTERFERENCE FROM THIN FILMS

125 124 If monochromatic light is allowed to fall normally and viewed, dark and bright circular fringes known as Newtons Rings are produced. The fringes are circular because the air film has circular symmetry.

126 Newtons Rings: When the light is incident on the plano-convex lens part of the light incident on the system is reflected from glass-to-air boundary (say at point D). The reminder of the light is transmitted through the air film, and it is again reflected from the air-to- glass boundary (say from point J). The two rays are (1 and 2 ) reflected from the top and bottom of the air film interfere with each other to produce darkness and brightness.

127 126 The interference effect is due to the combination of ray 1, reflected from the flat surface, with ray 2, reflected from the curved surface of the lens. Ray 1 undergoes a phase change of upon reflection (because it is reflected from a medium of higher index of refraction), whereas ray 2 undergoes no phase change (because it is reflected from a medium of lower refractive index). Air-Glass interface Phase change (π)

128 The condition for constructive interference remains unchanged. Path difference:

129 128 Consider the section AB of the lens, wherein it forms an air film of thickness d. Let m th order bright ring of radius r forms here. Let C be the center of curvature of the plano- convex lens. R(=CB=OC) be the radius of curvature of the plano- convex lens. The radius (r) of the bright ring: O C B A D R

130 129 We can write: d= OD=OC-CD O C B A D R

131 Substituting d value in equation (1) and solving for r: …….(For bright ring) The radii/diameters of the bright rings.

132 131 1.D- is the diameter of the m th bright ring. 2.m=0 is physically not possible, 3.So, m=1,2…………………….. 4.center ring must be the dark. 5.Radii/diameters of bright rings are proportional to square root of odd numbers.

133 132 Problem 33: In Newtons ring experiment, the radius of curvature R of the lens is 5m and its diameter is 20mm. a)How many rings are produced? b)How many rings would be seen if the arrangement is immersed in water (n=1.33)? (Assume that wavelength = 589nm) Given: λ = 589 nm Radius of curvature R= 5 m diameter of the ring=d=20mm Therefore, radius r= 10 mm=0.01 m (a) m=? (b) n=1.33, m=?

134 133

135 134 Problem 34: The diameter of the tenth ring in a Newtons rings apparatus changes from 1.42 to 1.27 cm as a liquid is introduced between the lens and the plate. Find the refraction of the liquid. Given: Radius of tenth bright ring in air: r air = 1.42 cm Radius of tenth bright ring in liquid: r liquid = 1.27 cm ring Refractive index of the liquid: n liquid = ?

136 135

137 136 Problem 35: A Newtons ring apparatus is used to determine the radius of curvature of a lens. The radii of the n th and (n+20) th bright rings are found to be 0.162cm and 0.368cm, respectively, in light of wavelength 546nm. Calculate the radius of curvature of the lower surface of the lens. Given: λ = 546 nm Radius of the n th ring=r n =0.162cm Radius of the (n+20) th =r n+20 =0.368cm m=? Radius of curvature R= ?

138 137

139 138 OPTICAL PATH Distance traveled by light in a medium in the time interval of t is d = vt Refractive index n = c/v Hence, ct = nd nd Optical path. BE-PHYSICS- INTERFERENCE MIT-MANIPAL INTERFERENCE FROM THIN FILMS d n

140 139 Purpose Interferometers are basic optical tools used to precisely measure wavelength, distance, index of refraction of optical beams. It is a device working on the principle of interference of light and is used in precise measurements of length or changes in length. MICHELSONS INTERFEROMETER

141 140 MICHELSONS INTERFEROMETER BE-PHYSICS- INTERFERENCE Light from an extended monochromatic source P falls on a half- silvered mirror M. The incident beam is divided into reflected and transmitted beams of equal intensity (Division of amplitude). These two beams travel almost in perpendicular directions and will be reflected normally from movable mirror (M 2 ) and fixed mirror (M 1 ). MIT-MANIPAL B A G

142 141 BE-PHYSICS- INTERFERENCE The two beams finally proceed towards a telescope (T) through which interference pattern of circular fringes will be seen. The interference occurs because the two light beams travel different paths between M and M 1 or M 2. Each beam travels its respective path twice. When the beams recombine, their path difference is 2 (d 2 – d 1 ). MIT-MANIPAL MICHELSONS INTERFEROMETER

143 142 BE-PHYSICS- INTERFERENCE The path difference can be changed by moving mirror M 2. As M 2 is moved, the circular fringes appear to grow or shrink depending on the direction of motion of M 2. New rings appear at the center of the interference pattern and grow outward or larger rings collapse disappear at the center as they shrink. MIT-MANIPAL MICHELSONS INTERFEROMETER

144 143 BE-PHYSICS- INTERFERENCE Each fringe corresponds to a movement of the mirror M 2 through one-half wavelength. The number of fringes is thus the same as the number of half wavelength. MIT-MANIPAL MICHELSONS INTERFEROMETER

145 144 If N fringes cross the field of view when mirror M 2 is moved by d, then d = N ( /2) d is measured by a micrometer attached to M 2. Thus microscopic length measurements can be made by this interferometer.

146 145 Michelson interferometer equation The interferometer is used to measure changes in length by counting the number of interference fringes that pass the field of view as mirror M 2 is moved. Length measurements made in this way can be accurate if large numbers of fringes are counted.

147 146 Applications Determination of wavelength The fact that whenever the movable mirror moves by a fringe originates or vanishes at the center is used to determine from the equation 2d = N, where d is the distance moved and N, the number of fringes originated or vanished.

148 147 Applications Determination of refractive index (n) When a thin film (whose refractive index n is to be determined) of thickness d is introduced on the path of one of the interfering beams, an additional path difference (nd–d)2= 2d(n-1) will be introduced. As a result there will be shift of fringes. If m fringes shift, then, 2d(n -1) = m from which n can be determined. d

149 148 Problem: SP 41-6 Yellow light (wavelength = 589nm) illuminates a Michelson interferometer. How many bright fringes will be counted as the mirror is moved through 1.0 cm? BE-PHYSICS- INTERFERENCE MIT-MANIPAL MICHELSONS INTERFEROMETER The number of fringes is the same as the number of half wavelengths in cm. N λ = 2d N= 2 d/ λ =2( x10 -2 m)/(589 x m) = 33,956 fringes

150 149 Problem: E41-39 If mirror M 2 in Michelsons interferometer is moved through 0.233mm, 792 fringes are counted with a light meter. What is the wavelength of the light used? BE-PHYSICS- INTERFERENCE MIT-MANIPAL MICHELSONS INTERFEROMETER N λ = 2d λ = 2 d/N=2(0.233 mm)/792= 588 nm = 588 nm

151 150 Problem: E41-40 An airtight chamber 5.0 cm long with glass windows is placed in one arm of a Michelsons interferometer as indicated in Fig Light of wavelength λ = 500 nm is used. The air is slowly evacuated from the chamber using a vacuum pump. While the air is being removed, 60 fringes are observed to pass through the view. From these data find the index of refraction of air at atmospheric pressure. BE-PHYSICS- INTERFERENCE MIT-MANIPAL MICHELSONS INTERFEROMETER

152 151 Answer: The change in the optical path length is 2(nd – d) 2d(n-1)=m λ n=m λ/2d+1 =

153 Problem HRK-41-38: A thin film with n=1.42 for light of wavelength 589nm is placed in one arm of a Michelson interferometer. If a shift of 7 fringes occurs, what is the film thickness? Solution: 2d(n-1) = m d = m λ / 2(n-1) d = 7 x (589 x10 -9 m)/ 2(1.41-1) = 4.9 x m

154 153BE-PHYSICS- INTERFERENCE MIT-MANIPAL QUESTIONS – INTERFERENCE What is the necessary condition on the path length difference (and phase difference) between two waves that interfere (A) constructively and (B) destructively ? [2] Obtain an expression for the fringe-width in the case of interference of light of wavelength λ, from a double-slit of slit- separation d. [5] Explain the term coherence.[2] Obtain an expression for the intensity of light in double-slit interference using phasor-diagram. [5]

155 154BE-PHYSICS- INTERFERENCE MIT-MANIPAL QUESTIONS – INTERFERENCE Draw a schematic plot of the intensity of light in a double-slit interference against phase-difference (and path-difference).[2] Explain the term reflection phase-shift.[1] Obtain the equations for thin-film interference. [2] Explain the interference-pattern in the case of wedge-shaped thin-films. [2] Obtain an expression for the radius of m TH order bright ring in the case of Newtons rings.[5] Explain Michelsons interferometer. Explain how microscopic length measurements are made in this.[4]

156 155BE-PHYSICS- INTERFERENCE MIT-MANIPAL INTERFERENCE – ANSWERS TO PROBLEMS SP41-1: 0.12degree, 2.5mm E41-1: 0.22radian, 12 degrees E41-5: 650nm E41-9: 600nm SP41-2: E(t)= 3.83 E o sin( t+22.5 o ) E41-15: 0 degree E41-19: 1.21, 2.22, 8.13m SP41-3: 567nm (Yellow-green) SP41-4: 100nm E41-23: 552nm, 442nm E41-27: 840nm E41-29: 141 E41-33: 34, 45 SP41-6: fringes E41-39: 588nm


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