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  2. INTERFERENCE • Topics • Two source interference • Double-slit interference • Coherence • Intensity in double slit interference • Interference from thin film • Michelson’s Interferometer Text Book: PHYSICS VOL 2 by Halliday, Resnick and Krane (5th Edition) BE-PHYSICS- INTERFERENCE-2010-11

  3. What is an Electromagnetic wave (EM)? Electric field (E) 900 Magnetic field (B) 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.

  4. Properties of electromagnetic waves (EM) 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’. • Examples: • Light waves • Heat waves • Radio and television Waves • Ultraviolet waves • Gamma rays, X- rays

  5. 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 108 m/s

  6. Electromagnetic spectrum More frequency (f) more energy (E), and lesser wavelength(λ).

  7. Dual Nature of Light • 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 natureof light. Dual natureof 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.

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

  9. 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.. Light as a stream of particles 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. Photon energy E = hf h = Planck’s constant = 6.626x10-34Js f = frequency of radiation Quantum (bundles/packets of energy) (Photon/quantum)

  10. c= f  Light as a wave: E = hf photon Light as a particle: Energy of a photon or light wave: Where h = Planck’s constant = 6.626x10-34Js f = frequency of a light wave - c = velocity of light λ= wavelength of a light wave -distance between successive crests

  11. Visible Light 400–435 nm Violet 435 nm-440nm Indigo 440–480 nm Blue 480–530 nm Green 530–590 nm Yellow 590–630 nm Orange 630–700 nm Red • 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.

  12. k:wave number w:angular frequency Wave Function of Sinusoidal Waves y(x,t) = ymsin(kx-wt) ym:amplitude kx-wt :phase

  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, y1=A sin ωt & y2=A sin (ωt + ) superpose, the resultant displacement is • y= y1+y2= a sin (ωt) + a sin (ωt+)


  15. 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. OR 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.

  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).

  17. TWO-SOURCE INTERFERENCE 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).

  18. TWO-SOURCE INTERFERENCE INTERFERENCE PATTERN PRODUCED BY WATER WAVES IN A RIPPLE TANK Maxima: where the shadows show the crests and valleys (or troughs). Minima: where the shadows are less clearly visible BE-PHYSICS- INTERFERENCE-2010-11

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

  20. Constructive interference path difference p = 1 or phase difference  =2 path difference p= 0 or phase difference  = 0 General condition: Path difference p = m orphase difference  = 2m where m = 0, 1, 2, 3,………… order of interference. 2λ path difference p =2 or phase difference  = 4

  21. Constructive interference 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).

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

  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 180o out of phase).

  24. COHERENCE 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

  25. COHERENCE A SECTION OF INFINITE WAVE 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. A WAVE TRAIN OF FINITE LENGTH L 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.

  26. COHERENCE A SECTION OF INFINITE WAVE A WAVE TRAIN OF FINITE LENGTH L 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. BE-PHYSICS- INTERFERENCE-2010-11

  27. COHERENCE Laser light is highly coherent whereas a laboratory monochromatic light source (sodium vapor lamp) may be partially coherent. Common sources of visible light emit light wave trains of finite length rather than an infinite wave. BE-PHYSICS- INTERFERENCE-2010-11

  28. COHERENCE Methods of producing coherent sources: (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: Young’s double slit arrangement, b. Lloyd's single mirror method. (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 Newton’s rings, b. Michelson’s interferometer.

  29. DOUBLE SLIT INTERFERENCE If light waves did not spread out after passing through the slits, no interference would occur.. Two narrow slits (can be considered as two sources of coherent light waves). If the widths of the slits are small compared with the wavelengthdistance 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 Screen

  30. DOUBLE SLIT INTERFERENCE A monochromatic light source produces two coherent light sources by illuminating a barrier containing two small openings (slits) S1 and S2 separated by a distance ‘d’ and kept at a distance ‘D’ from the screen. Waves originating from two coherent light sources S1 and S2 because maintain a constant phase relationship. In interference phenomenon, we have assumed that slits are point sources of light. B C

  31. DOUBLE SLIT INTERFERENCE When the light from S1 and S2 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

  32. DOUBLE SLIT INTERFERENCE O At center of the screen O, we always get a bright fringe, because at this point the two waves from slit S1 and S2, interfere constructively without any path/phase difference. 32

  33. DOUBLE SLIT INTERFERENCE - Analysis of Interference pattern a-is the mid point of the slit We consider waves from each slit that combine at an arbitrary point P on the screen C. The point P is at distances of r1 and r2 from the narrow slits S1 and S2, respectively.

  34. DOUBLE SLIT INTERFERENCE • For D>>d, we can approximate rays r1 and r2 as being parallel. • The line S2b is drawn so that the lines PS2and Pb have equal lengths. • Path length (S1b) between the rays r1 and r2 reaching the point P decides the intensity at P. • (i.e. maximum/minimum).

  35. DOUBLE SLIT INTERFERENCE Path difference between two waves from S1 & S2 (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 (S1b=d sin) determines whether the two waves are in phase or out of phase when they arrive at point P. If path length (S1b= d sinθ) is either zero or some integer multiple of the wavelength, constructive interference results at P. (d sinθ)

  36. Constructive Interference: Maximum at P: The condition for constructive interference or Maxima at point P is P O 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

  37. Destructive Interference: Minimum at P: When path length (S1b=d sin ) is an odd multiple of λ/2, the two waves arriving at point P are 1800 (=π) out of phase and give rise to destructive interference. The condition for dark fringes, or destructive interference, at point P is P O Central maximum m=0 m=0 m=3 Negative values of m locate the minima on the lower half of the screen. m=1 m=2 m=2 m=1 m=3 m=0

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

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

  40. DOUBLE SLIT INTERFERENCE 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. 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 θ=? for first minimum. Fringe width Δy=? d=0.12 mm

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

  42. DOUBLE SLIT INTERFERENCE Problem 5: A double-slit arrangement produces interference fringes for sodium light (λ=589 nm) that are 0.230 apart. For what wavelength would the angular separation be 10% greater? (Assume that θ is small). d sin θ = mλ sinθ≈ θ (θ is very small) d θ=λ (for first maxima , m=1) Use: λ1/ λ2= θ 1/ θ 2, λ2= λ1 x θ 2/ θ 1, Given: θ 1 = 0.230 (for 100%) θ 2 = 0.230 x 1.1 = 0.2530 (10% more: for 110%) λ2= λ1 x θ 2/ θ 1 = (589 nm)(0.253/0.23)= 647.9 nm

  43. Problem 11: DOUBLE SLIT INTERFERENCE 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. 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.

  44. DOUBLE SLIT INTERFERENCE Tutorial: Problem:2 Monochromatic light illuminates two parallel slits a distance ‘d’ apart. The first maximum is observed at an angular position of 150. By what percentage should ‘d’ be increased or decreased so that the second maximum will instead be observed at 150? Solution: Use d sinθ=mλ d1 sin 150=λ gives the first maximum (m=1) d2 sin 150 = 2λ put the second maximum (m=2) at the location of the first. Divide the second expression by the first and d2 = 2d1. This is a 100% increase in d1.

  45. DOUBLE SLIT INTERFERENCE 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 10 -3 m, λ= ? ym=180 x10-3 m D = 2 m, v= 25 x 10-2 m. f=? Use: v=f x λ To find λ: Use ym= λD/d gives λ=0.0108m Use: v= f x λ , f= 23 Hz.

  46. HRK:958 page: Exercise: 41-2 Problem 1: Monochromatic green light of wavelength 554 nm, illuminates two parallel narrow slits 7.7 μm apart. Calculate the angular position of the third-order (m=3) bright fringe in radians and (b) in degrees. Given: d = 7.7 x 10 -6 m, m=3, λ= 554 nm = 554 x10-9 m θ = ? in radians and in degrees. Use: for the bright fringe: d sinθ = mλ (put m=3): Answers: θ=0.216 radians (b) θ =12.5 degrees DOUBLE SLIT INTERFERENCE CHECK YOURSELF (Solve for third order dark fringe put m=2)

  47. DOUBLE SLIT INTERFERENCE 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 Δy=? Solution: Fringe width= Δy = λD/d = (512x 10-9m)(5.4 m)=(1.2 x 10-3m) =2.3 x 10-3 m CHECK YOURSELF: How far apart are the bright fringes as seen on the screen? Fringe width is the same 2.3 x 10-3 m.

  48. DOUBLE SLIT INTERFERENCE Problem 4: Find the slit separation of a double-slit arrangement that will produce bright interference fringes 1.000 apart in angular separation. Assume a wavelength of 592 nm. Given: d=? θ=1.000 λ= 592 nm Use: d =λ/sin θ = (592 x 10-9m)/sin(1.000) =3.39x 10-5m CHECK YOURSELF Solve this problem for dark interference fringes 1.000 apart in angular separation.

  49. DOUBLE SLIT INTERFERENCE Problem 6 A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.200 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 θ0 =λ; and then find θ from d sin θ= λ/n. Combining the two expressions, sin θ/sin θ0=1/n gives θ=0.150

  50. 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 ym = mλD/d= 3 λD/d where ym is measured from the central maximum. Then Δy is: y1 - y2 = 3(λ1 -λ2)D/d = 3(612x 10-9m – 480x10-9m)(1.36m)/(5.22x10-3m) = 1.03x10-4m: DOUBLE SLIT INTERFERENCE