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Ph 101-1

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Ph 101-1

  1. 1. S N Dash Dept. of Physics, NIT Rourkela Phone: 0661-2462733, Email: dsuryanarayan@gmail.com dashsurya@nitrkl.ac.in WAVE OPTICS/MODERN PHYSICS Course: PHYSICS-I (PH101) 4 Credits (3-1-0)
  2. 2. Topics Superposition of waves, Interference, Diffraction, Electromagnetic waves, Polarization, Fiber Optics (mid term) Special Relativity, Particle properties of Waves, Wave properties of Particles, Quantum Mechanics. Mark Distributions: Mid Term: 30% End Term: 50% TA: 20% GRADE Distributions:  90  Ex  80  A  70  B  60  C  50  D  35  P  35  F TA: ATTENDANCE (5) ASSIGNMENTS (10) OVER ALL BEHAVIOUR (5)
  3. 3.  Education is what remains after one has forgotten what one has learned in school Albert Einstein
  4. 4. Assignments Assignment-1 Assignment-2 Assignment-3 Assignment-4 Assignment-5 51 2 3 4 Assignments of PH-101 Faculty: Dr. S.N. Dash Name: XYZ Roll no.: 0000 Date: 00.00.2012 Assignments of PH 101 Faculty: Dr. S.N. Dash Name: XYZ Roll no.: 0000 Date: 00.00.2012 Assignments of PH 101 Faculty: Dr. S.N. Dash Name: XYZ Roll no.: 0000 Date: 00.00.2012
  5. 5. Assigned Part (i) Interference: Condition for interference, Division of wave front (two beam interference): Young's double slit experiment, fringe pattern on transverse and longitudinal planes, intensity distribution, Fresnel's biprism, interference with white light, displacement of fringes; Division of amplitude (two beam interference): cosine law, Newton's rings experiment, Michelson Interferometer, fringes of equal inclination and equal thickness. (ii) Diffraction: Fraunhofer, Fresnel's diffraction, single slit (infinite beam interference), two and N slits( Grating) Fraunhofer diffraction pattern and intensity distribution.
  6. 6. Books [1] Fundamentals of Optics, F. A. Jenkins, H. E. White, Tata McGraw-Hill (2011). [2] Optics, A. Ghatak, Tata McGraw-Hill (2011). [3] A Textbook of Optics, N. Subrahmanyam, B. Lal, M. N. Avadhanulu, S. Chand and company (2012). [4] Geometrical and Physical Optics, P. K. Chakrabarti, New Central Book Agency (2009). WEB REFERENCES 1.www. google.co.in 2.http://en.wikipedia.org 3.www.youtube.com
  7. 7. Why Optics ? • Windows • Camera • Display System • Microscope and Telescopes • Solar Cell • Light Emitting Diodes (LED) • X-ray, LASER • Material Characterization (UV-Visible spectroscopy, FTIR, PL etc) • Remote Sensing • Signal processing (optical fiber) • Bio-medical applications
  8. 8. Optics History The word optics comes from the ancient Greek word ὀπτική, meaning appearance or look. Optics began with the development of lenses by the ancient Egyptians and Mesopotamians as early as 700 BC. The ancient Romans and Greeks filled glass spheres with water to make lenses. Involves the behavior and properties of light, including its interactions with matter and the construction of instruments that use or detect it. Geometrical optics (ray optics): Light as a collection of rays that travel in straight lines and bend when they pass through or reflect from surfaces. Physical optics (wave optics): wave effects such as interference and diffraction that cannot be accounted for in geometrical optics.
  9. 9. Continued…. 1284, Salvino D'Armate (Itally) invented the first wearable eyeglasses. This was the start of the optical industry of grinding and polishing lenses for these "spectacles“ in Venice and Florence. Early 17th century Johannes Kepler expanded on geometric optics through reflection by flat and curved mirrors and the optical explanations of astronomical phenomena such as lunar and solar eclipses. He was also able to correctly deduce the role of the retina as the actual organ that recorded images. Until middle of 17th century: light consisted of a stream of corpuscles (Newton) Middle of 17th century: Wave motion of light 1670: Reflection and refraction of light by wave theory (Huygen) 1663-1665: Diffraction (Grimaldi, Hooke) 1827: Interference (Young, Fresnel) 1873: Electromagnetic nature of light (Maxwell) 1901: Quantization of light (Planck)
  10. 10. Light Wave
  11. 11. Long wavelength (Low frequency) Short wavelength (High frequency) Light Wave E = h × f Where, E = light energy h = Planck’s constant = 6.626 x 10-34 J.s f = frequency of light C = λ × f Where, C = velocity of light (3 ×108 m/s) λ = wavelength of light f = frequency of light
  12. 12. Lights of different color RED ORANGE YELLOW GREEN BLUE INDIGO VIOLET Long wavelength (Low frequency) Short wavelength (High frequency) Violet light has higher energy
  13. 13. (a) Constructive interference: If a crest of one wave meets a crest of another wave, the resultant intensity increases. (b) Destructive interference: If a crest of one wave meets a trough of another wave, the resultant intensity decreases. Types of Interference Interference of light Interference is a phenomenon in which two coherent light waves superpose to form a resultant wave of greater or lower amplitude.
  14. 14. Conditions for Interference [1] The two interfering waves should be coherent, i.e. the phase difference between them must remain constant with time. [2] The two waves should have same frequency. [3] If Interfering waves are polarized, they must be in same state of polarization. [4] The separation between the light sources should be as small as possible. [5] The distance of the screen from the sources should be quite large. [6] The amplitude of the interfering waves should be equal or at least very nearly equal. [7] The two sources should be narrow [8] The two sources should give monochromatic or very nearly monochromatic, or else the path difference should be very small.
  15. 15. Locus (Mathematics) In geometry, a locus is a collection of points according to certain principles. Example: A circle may be defined as the locus of points in a plane (2D) at a fixed distance from a given point (centre).
  16. 16. Wavefront In physics, a wavefront is the locus of points having the same phase (line or curve in 2D, or a surface for a wave propagating in 3D) 2-Dimensional 3-Dimensional
  17. 17. Thomas Young * In 1801, Young admitted the sunlight through a single pinhole and then directed the emerging light onto two pinholes. * The spherical waves emerging from the pinholes interfered with each other and a few colored fringes were observed on the screen. * The pinholes were latter replaced with narrow slits and the sunlight was replaced by monochromatic light ( Ex: Sodium lamp, Yellow, 5893 Å) Interference of light by division of wavefront Screen
  18. 18. Interference of Light by Division of Wavefront Light source Double slit (S1, S2) Interference pattern on dark film (screen) Intensity distribution Single slit
  19. 19. Caluculation of optical path difference between two waves As slits (S1 and S2) are equidistant from source (S), the phase of the wave at S1 will be same as the phase of the wave at S2 and therefore, S1,S2 act as coherent sources. The waves leaving from S1 and S2 interfere and produce alternative bright and dark bands on the screen. Let P is an arbitrary point on screen, which is at a distance D from the double slits. P O Screen S M x G H 2d Young’s double slit experiment 2d is the distance between S1 and S2 θ is the angle between MO and MP x is the distance between O and P S1N is the normal on to the line S2P MO bisects S1S2 and GH (i.e. S1M = S2M = GO = HO = d)
  20. 20. P O Screen S M x G H 2d The optical paths are identical with the geometrical paths, if the experiment is carried out in air. The path difference between two waves is S2P-S1P = S2N Let S1G and S2H are perpendiculars on the screen and from S2HP triangle (S2P)2 = (S2H)2 + (HP)2 = D2 + (x+d)2 Continued…..
  21. 21. Continued….. Since D>>(x+d), (x+d)2/D2 is very small After expansion, Path difference = S2P-S1P [(A+B)2 –(A-B)2 = A2 + B2 + 2AB –(A2+B2-2AB) = 4AB]
  22. 22. Continued….. The nature of the interference of the two waves at P depends simply on how many waves are contained in the length of path difference (S2N). • If the path difference (S2N) contains an integral number of wavelengths, then the two waves interfere constructively producing a maximum in the intensity of light on the screen at P. • If the path difference (S2N) contains an odd number of half- wavelengths, then the two waves interfere destructively and produce minimum intensity of light on the screen at P.
  23. 23. Continued….. For dark fringes (minima), Let xn and xn+1 denote the distances of nth and (n+1)th bright fringes. Then, Thus for bright fringes (maxima),
  24. 24. Continued….. Hence, the spacing between any consecutive bright fringe is the same. Similarly, the spacing between two dark fringes is Dλ/2d, The spacing between the fringes is independent of n The spacing between any two consecutive bright or dark fringe is called the “fringe width” and is denoted by Spacing between nth and (n+1)th bright fringe is
  25. 25. PROBLEM-1 Q: In an interference pattern, at a point we observe the 12th order maximum for λ1= 6000 Å. What order will be visible here if the source is replaced by light of wavelength λ2= 4800 Å ? A: In double slit interference, the distance x of a bright fringe from the centre (zero-order fringe) is x = (D/2d)nλ, where n = 0,1,2,…. Thus at a given point nλ is constant Or, n1λ1 = n2λ2 n2 = n1λ1/λ2 = 12 x 6000/4800 = 15
  26. 26. PROBLEM-2 Q: Two straight narrow parallel slits (2 mm apart) are illuminated with a monochromatic light of wavelength 5896 Å. Fringes are observed at a distance of 60 cm from the slits. Find the width of the fringes? 2d = 2 mm = 0.2 cm, D = 60 cm, λ = 5896 Å = 5896 x 10-8 cm Fringe-width = (60 cmx 5896 x 10-8 cm)/ 0.2 cm = 1.77 x 10-2 cm A: The interference fringe-width for a double slit is given by
  27. 27. PROBLEM-3 Q: In a two-slit interference pattern with λ = 6000 Å the zero-order and tenth-order maxima fall at 12.34 mm and 14.73 mm, respectively. If λ is changed to 5000 Å, deduce the positions of the zero-order and twentieth-order fringes, while other arrangements remaining the same. A: Fringe-width is with λ = 6000 Å, the distance between zero-order and tenth-order fringe is 14.73 mm - 12.34 mm = 2.39 mm Fringe-width = 2.39 mm/10 = 0.239 mm (Fringe-width)6000/ (Fringe-width)5000 = 6000 Å /5000 Å = 6/5 (Fringe-width)5000 = (Fringe-width)6000 x 5/6 = 0.239 mm x 5/6 = 0.199 mm Position of zero-order fringe (for λ = 5000 Å) = 12.34 mm Position of twentieth-order fringe (for λ = 5000 Å) = 12.34 mm + (0.199 mm x 20) = 16.32 mm
  28. 28. Let S be a narrow slit illuminated by a monochromatic (single wavelength) source S1 and S2 are two similar parallel slits (S1 and S2 are equidistant from S and very close to each other) Suppose the waves from S reach S1 and S2 in the same phase (coherent) Beyond S1 and S2, the waves proceed as if they were started from S1 and S2. Young’s double slit experiment Resultant Intensity due to superposition of two interfering waves
  29. 29. Continued…. Let us calculate the resultant intensity of light of wavelength (λ) at point P on a screen placed parallel to S1 and S2. Let A1 and A2 be the amplitude at P due to the waves from S1 and S2, respectively. The waves arrive at P, having traversed different paths S1P and S2P. Hence, they are superposed at P with a phase differences δ The displacement at P due to the simple harmonic waves from S1 and S2 can be represented by Where ω/2π is the common frequency (f) of two waves
  30. 30. Continued…. By the principle of superposition, when two or more waves simultaneously reach at point, the resultant displacement is equal to the sum of the displacements of all the waves. Hence the resultant displacement is y = y1 + y2 Let us make a change in constants A1, A2 and δ by putting sin (A+B) = sin A cos B + cos A sin B Where R and θ are new constants
  31. 31. Continued…. The resultant displacement is sin A cos B + cos A sin B = sin (A+B) Hence, the resultant displacement at P is simple harmonic and of amplitude R. Adding the above two equations,
  32. 32. Continued…. In case of interference, the resultant intensity at P is not just the sum of the intensities of the individual waves. The resultant intensity I at P, which is proportional to the square of the resultant amplitude, is given by I = R2 (by assuming proportionality constant = 1)
  33. 33. Conditions for Maxima and Minima Intensity The maximum intensity is greater than the sum of two separate intensities The resultant intensity I at P, The intensity I is a maximum, when cos δ = +1 or δ = 2nπ, n=0,1,2,... Phase difference (δ) = 2nπ Path difference (S2P-S1P) =2nπ × λ/2π = nλ The intensity I is a minimum, when cos δ = -1 or δ = (2n+1)π, n=0,1,2,... Phase difference (δ) = (2n+1)π Path difference (S2P-S1P) =(2n+1)π × λ/2π = (2n+1) λ/2 The minimum intensity is less than the sum of two separate intensities 2121 2 IIII 
  34. 34. ` PROBLEM-4 Q: Two coherent beams of wavelength 5000 Å reaching a point would individually produce intensities 1.44 and 4.00 units. If they reach there together, the intensity is 0.90 units. Calculate the lowest phase difference with which the beams reach that point. A: The resultant intensity at a point due to two coherent waves of amplitudes A1 and A2 reaching the point with a difference δ is given by Here I1 = 1.44 , I2 = 4.00 and I = 0.9 0.9 = 1.44 +4.00 +2√ (1.44 × 4.00) cos δ cos δ = - 0.9458, cos (180° - δ) = 0.9458 [As cos (180° - θ) = - cos θ] 180° - δ = cos-10.9458 δ = 161°
  35. 35. ` PROBLEM-5 Q: Find the ratio of the intensity at the centre of a bright fringe in an interference pattern to the intensity at a point one-quarter of the distance between two fringes from the centre. A: The resultant intensity at a point due to two coherent waves of amplitudes A1 and A2 reaching the point with a difference δ is given by If A be the amplitude of each of the two interfering waves, then the resultant intensity at any point on the screen is given by I = 2A2 (1 + cos δ) Where δ is the phase difference between two waves at that point At the centre of a bright fringe, δ = 0, 2π, …… So that I0 = 2A2 (1 + 1) = 4A2 The phase difference between successive fringes is 2π. Thus the phase difference at a point distance one quarter of the distance between two fringes from the centre will be π/2. I1 = 2A2 (1 + cos 90°) = 2A2 I0/I1 = 4A2 / 2A2 = 2
  36. 36. Techniques for Producing Interference * The phase relation between the waves emitted by two independent light sources rapidly changes with time and therefore they can never be coherent, thought the sources are identical in all respects. * If two sources are derived from a single source by some device, then any phase change occurring in one source is simultaneously accompanied by the same phase change in the other source. Therefore, the phase difference between the waves emerging from the two sources remains constant and the sources are coherent. The techniques used for creating coherent sources of light are (a) Wavefront splitting (b) Amplitude splitting
  37. 37. Continued…. (a) Wavefront splitting: * One of the method consists in dividing a light wavefront, emerging from a narrow slit, by passing it through two slits closely spaced side by side. * The two parts of the same wavefront travel through different paths and reunite on a screen to produce fringe pattern. This is known as interference due to the division of wavefront. * This method is useful only with narrow sources. * Examples: Young’s double slit, Fresnel’s double mirror, Fresnel’s biprism, Lloyd’s mirror etc.
  38. 38. Continued…. (b) Amplitude splitting: * The amplitude (intensity) of a light wave is divided into two parts, namely reflected and transmitted components, by partial reflection at a surface. * The two parts travel through different paths and reunite to produce interference fringes. This is known as interference due to division of amplitude. * Optical elements such as beam splitters, mirrors are used for achieving amplitude division. * Examples: Newton’s ring experiment, Michelson’s interferometer, etc.
  39. 39. Biprism * An optical device for obtaining interference fringes. * A prism whose refracting angle is very nearly 180 degrees * It is used to obtain two coherent sources for interferences. * A fusion of two triangular prisms: two prisms of very small refracting angles, placed base to base. * In practice, the biprism is made from a single plate by grinding and polishing, so that it is a single prism with one of the angles about 179 º and the two about 30each.
  40. 40. Fresnel’s Biprism * S is a narrow vertical slit illuminated by a monochromatic light. * The light from S is allowed to fall symmetrically on the biprism BP. * The light beams emerging from the upper and lower halves of the prism appears to start from two virtual images of S, namely S1 and S2. * S1 and S2 act as coherent sources. * The cones of light BS1E and AS2C, diverging from S1 and S2 are superposed and the interference fringes are formed in the overlapping region BC of the screen. BP A B C E D S S1 S2 2d
  41. 41. * Let S1 and S2 be the two virtual sources produced by the biprism. * The point O on the screen is equidistant from S1 and S2. * Therefore, the waves from S1 and S2 reach O in the same phase and reinforce each other. • Hence, the point O is the centre of a bright fringe. The illumination of any other point P can be obtained by calculating the path difference S2P-S1P. Let us join S2P and S1P, and draw perpendiculars S1G and S2H on the screen. Let S1S2 = 2d, S1G = S2H = D and OP = x Young’s double slit experiment O x G H 2d Continued….
  42. 42. Continued…. From S2PH triangle, (S2P)2 = (S2H)2 + (PH)2 = D2 + (x+d)2 O x G H 2d
  43. 43. Continued…. The resultant intensity at a point is a maximum or a minimum according as the path difference between the waves is an integral multiple of wavelength or an odd multiple of half-wavelength, respectively. For P to be centre of a dark fringe, we must have Thus, for P to be the centre of a bright fringe, we must have
  44. 44. ` Continued…. Let xn and xn+1 denote the distances of the nth and (n+1)th bright fringes. Then the distance between (n+1)th and nth bright fringes is The wavelength of unknown light (λ) can be calculated by measuring the values of D, 2d, and . * This is independent of n * The distance between any two consecutive bright fringe is the same i.e. Dλ/2d. * Similarly, the distance between any two consecutive dark fringes is the same i.e. Dλ/2d. * The distance Dλ/2d is called the “ Fringe-width” and is denoted by
  45. 45. INTERFERENCE with WHITE LIGHT  The wavelength : 4 to 7 x 10-5 cm.  The central fringe produced at O will be white because all wavelengths will constructively interfere.  Now slightly below (or above) the point O, the fringes will be colored. For a nearly monochromatic source, (sodium lamp) a large number of interference fringes are obtained and it is extremely difficult to determine the position of the central fringe
  46. 46. ` Displacement of Fringes by Thin Lamina When a thin transparent plate (glass or mica) is introduced in the path of one of the two interfering beams, the entire fringe-pattern is displaced to a point towards the beam in which the plate is introduced. Let S1 and S2 be two coherent monochromatic sources giving light of wavelength λ. The thin plate of thickness “t” is introduced in the path of light from S1. For a particular λ, the refractive index of the plate is µ. Now, light from S1 travel partly in air and partly in the plate. For the light path from S1P, the distance travelled in air is S1P - t Young’s double slit experiment O xn 2d t
  47. 47. ` Continued…. If vo and v1 are the velocities of light in air and in the plate respectively, then the time taken by light to travel from S1 to P is Hence, the effective path in air from S1 to P is [S1P+(µ-1)t], i.e. the air path S1P is increased by an amount (µ-1)t due to the introduction of the plate.
  48. 48. ` Continued…. Let O be the position of the central bright fringe in the absence of the plate, the optical paths S1O and S2O are equal. By introducing the plate , the two optical paths become unequal. Therefore, the central fringe is shifted to O/, so that at O/ the two optical paths become equal. At P, The effective path difference Young’s double slit experiment O xn 2d t Let S1S2=2d, OP=xn Then S2P-S1P = 2dxn/D
  49. 49. ` Continued…. Effective path difference at P If the point P at the centre of the nth bright fringe, the effective path difference should be equal to nλ. In the absence of the plate (t=0), the distance of the nth maximum from O is Dnλ/2d. Therefore, the displacement of the nth bright fringe is given by This is independent of n i.e. The displacement is same for all of the bright fringes. Similarly, it can be shown that the displacement of any dark fringe is also
  50. 50. ` Determination of Thickness of the Plate The introduction of the plate produces shifting in the fringe by an amount x0. This technique can be used to measure the thickness of a plate.
  51. 51. ` PROBLEM-6 Q: A Fresnel’s biprism arrangement is set with sodium light (λ = 5893 Å) and the field of view of the eyepiece 62 fringes are observed. How many fringes shall we get in the same field of view if we replace the source by mercury lamp using green filter (λ = 5461 Å) ? A: Let l be the length of the field of view and N is the number of fringes in it. Then the fringe-width = l/N = Dλ/2d Nλ = (2d/D)l =constant N1λ1 = N2 λ2 Or, N1= 62, λ1 = 5893 Å, λ2 = 5893 Å, N2 = ? N2 = N1λ1/λ2 = 67
  52. 52. PROBLEM-7 Q: Interference fringes in biprism are formed by superposition of two coherent light beams of wavelength 6 × 10-7 meter. If a thin plate of material having refractive index 1.6 is placed in the path of one of the beams, the central fringe shifts to the place previously occupied by the 10th fringe. Find the thickness of the plate. A: The shift of the central fringe is given by The fringe-width (FW) = Dλ/2d, so that D/2d = FW/λ x0 = FW(µ-1)t/ λ t = x0 λ/FW(µ-1) x0 = 10FW, λ = 6 × 10-7 m and µ = 1.6 t = 1.0 × 10-3 cm
  53. 53. ` PROBLEM-8 Q:On introducing a thin sheet of mica (thickness 12×10-5 cm) in the path of one of the interfering beams in a biprism arrangement, the central fringe is shifted through a distance equal to the spacing between successive bright fringes. Calculate the refractive index of mica. (λ = 6×10-5 cm) A: The shift x0 of the central fringe is given by x0 = FW(µ-1)t/ λ, where FW= Fringe-width (µ-1)= x0 λ/ FW t Hera, x0 = FW, λ = 6 × 10-5 cm and t = 12 × 10-5 cm µ = 1.5
  54. 54. ` PROBLEM-9 Q: A thin mica sheet (µ = 1.6) of 7 microns thickness introduced in the path of one of the interfering beams in a biprism arrangement shifts the central fringes to a position normally occupied by the 7th bright fringe from the centre. Find the wavelength of light used. A: The shift x0 of the central fringe is given by x0 = FW(µ-1)t/ λ, where FW= Fringe-width λ= FW(µ-1)t/ x0 Hera, x0 = 7 FW, µ = 1.6 and t = 7 µm = 7 × 10-4 cm λ = 6 × 10-5 cm = 6000 Å
  55. 55. Interference of Light: Division of Amplitude (a) Newton’s ring experiment (b) Michelson’s Interferometer
  56. 56. Interference of Light by Thin Film * Interference occurs between the monochromatic light waves (reflected and transmitted), when a film of oil spread over the surface of water, or a thin glass plate. * Interference occurs between the light waves reflected from the film and also between the light waves transmitted through the film. Interference involving multiple reflections Stoke’s treatment Interference of Light by Thin Film:
  57. 57. * Let a ray of monochromatic light SA incidents at an angle ψ on a parallel-sided transparent thin film of thickness t and refractive index µ (µ >1). * At A, the ray is partly reflected along AR1 and partly refracted along AB at an angle θ. S R1 R2 T1 T2 A DB C E F t µ θ ψ ϕ Interference of Light by Thin Film: Cosine Law Medium-1 Medium-2 Medium-3 N M θ
  58. 58. * At B, the ray is again partly reflected along BC and partly refracted along BT1. * Similarly, consequent reflections ( AR1, CR2, ….) and refractions (BT1, DT2, ….) light will take place. * Let us first consider the reflected rays only. At each of the points A, B, C, D, only a small part of light is reflected, the rest is refracted. Therefore, the rays AR1 and CR2 (each having one reflection) have almost equal intensities. * The rest have rapidly decreasing intensities and can be ignored. * The rays AR1 and CR2 being derived from the same incident ray are coherent and have ability to interfere. * Therefore, we have to calculate the path difference between them. Continued……
  59. 59. * Let CN and BM be perpendicular to AR1 and AC. * As the paths of the rays AR1 and CR2 beyond CN is equal, the path difference between AR1 and CR2 is p = path ABC in film-path AN in air = µ(AB+BC)-AN Continued…… Similarly, AN = AC sin ψ = (AM+MC) sin ψ = (BM tan θ + BM tan θ) sin ψ = 2t tan θ sin ψ
  60. 60. Substituting the values of AB, BC and AN, the path difference will be Continued…… At A, the ray is reflected while going from a rarer to a denser medium and suffers a phase change of π. At B, The reflection takes place when the ray is going from a denser to a rarer medium and there is no phase change. Hence, the ray AR1, having suffered a reflection at the surface of a denser medium, undergoes a phase change of π, which is equivalent to the a path difference of λ/2, because phase diff. = path diff. × 2π/ λ
  61. 61. Hence the “effective” path difference between AR1 and CR2 is Continued…… Conditions of Maxima in Reflected Light: The two rays will reinforce each other if the path difference between them is an integral multiple of λ. When this condition is satisfied, the film will appear bright in the reflected light.
  62. 62. Continued…… Conditions of Minima in Reflected Light: The two rays will destroy each other if the path difference between them is an odd multiple of λ/2. When this condition is satisfied, the film will appear dark in the reflected light.
  63. 63. Path difference of Transmitted Light Similarly, the path difference between the transmitted rays BT1 and DT2 can be calculated as p = µ (BC+CD) – BL = 2µtcos θ In this case, there is no phase difference due to reflection at B or C, because in either case the light is travelling from denser to rarer medium. Hence the “effective” path difference between BT1 and DT2 will be 2µtcos θ
  64. 64. Conditions for Maxima and Minima in Transmitted light Maxima: The rays BT1 and DT2 will reinforce each other if 2µtcos θ = nλ where n = 0,1,2,……. (condition of maxima) The film will then appear bright in the transmitted light Minima: Two rays will destroy each other if, Hence the “effective” path difference between BT1 and DT2 will be 2µtcos θ This is the condition of minima. The thin film will appear dark in the transmitted light
  65. 65. Maxima and Minima in Reflected and Transmitted light * Conditions for maxima and minima in the reflected light are just the reverse of those in the transmitted light. * Hence the film which appears bright in reflected light will appear dark in transmitted light and vice-versa. * The appearances in the two cases are complementary to each other.
  66. 66. Interference of Light: Division of Amplitude Fringes of equal thickness Newton’s ring experiment Fringes of equal inclination Michelson’s Interferometer
  67. 67. Fringes of Equal Thickness * The basic formula for the path difference between two interfering rays obtained due to the division of amplitude by a thin film of thickness “t” and refractive index “μ” is 2 μt cos θ, where θ is the inclination of ray inside the film. * If the thickness “t” of the film is rapidly varying, the path difference 2 μt cos θ changes mainly due to change in t. * Therefore, the occurrence of alternate maxima and minima is due to variation in the thickness of the film and each maximum and minimum is a locus of constant film thickness. * These are called “fringes of equal thickness”. * Such fringes are localized on the film itself and observed by microscope focused on the film. * Example: Newton’s ring experiment
  68. 68. Formation of Newton’s Ring When a plano-convex lens of large radius of curvature is placed with it convex surface in contact with a plane glass plate, an air film is formed between the lower surface of the lens and the upper surface of the glass plate. * The thickness of the film gradually increases from the point of contact outwards. * If monochromatic light is allowed to fall normally on this film, a system of alternate bright and dark concentric rings with their centre dark is formed in the air film. * These are called Newton’s rings S
  69. 69. Continued…. Newton’s rings are formed as a result of interference between the light waves reflected from the upper and lower surfaces of the air film. Where µ is the refractive index of the film (air), t is the thickness of the film at the point D and θ is the inclination of the ray. The factor λ/2 accounts for the phase change of π on reflection at the lower surface of the film. The effective path difference between the two rays is For air-film (µ = 1) and normal incidence of light (θ = 0º)
  70. 70. Continued…. The effective path difference between two rays for air-film (µ = 1) and normal incidence of light (θ = 0º) The condition for maximum intensity (bright-fringe) is p= nλ n = 0, 1, 2, …… The condition for minimum intensity (dark fringe) is
  71. 71. Continued…. For t = 0, at the point of contact of the lens and the plate p= λ/2 This is the condition for minimum intensity and hence the central spot is dark * It is clear that a bright or dark fringe of any particular order n will occur for a constant value of t. * Since in the air film “t” remains constant along a circle with its centre at the point of contact, the fringes are in the form of concentric circles. * Since each fringe is the locus of constant film thickness, these are called fringes of constant thickness.
  72. 72. Diameters of Bright Rings * Let LOL/ be the lens placed on the glass plate AB, the point of contact being O. * Let R be the radius of curvature of the curved surface of the lens. E O O/ P t A B ρn C R N L L/ 2R-t
  73. 73. Continued…. * Let ρn be the radius of a Newton’s ring corresponding to a point P, where the film thickness is t. * Draw a perpendicular PN. According to the property of the circle, PN2 = ON × NE Or, ρn 2= t × (2R-t) = 2Rt-t2 Since t is small compared to R, one can neglect t2 ρn 2= t × (2R-t) = 2Rt 2t = ρn 2 /R The condition for a bright ring is 2t = (2n-1)λ/2 ρn 2 /R = (2n-1)λ/2 ρn 2 = (2n-1)λR/2 E O O / P t A B ρn C R N L L/ 2R-t
  74. 74. Continued…. If Dn be the diameter of the nth bright ring, then Dn = 2ρn Dn 2= 2 (2n-1) λR Dn = (2λR)1/2 (2n-1)1/2 Dn α (2n-1)1/2 As n is an integer, (2n-1) is an odd number Thus the diameter of the bright rings are proportional to the square-root of the odd natural number. The diameters of the first few rings are in the ratio 1 : √3 : √5 : √7 ……… =1:1.732:2.236:2.446 The separation between successive rings are in the ratio = 0.732, 0.504, 0.410, ….. i.e. the separation decreases as the order increases
  75. 75. Diameters of Dark Rings The condition for a dark ring is 2t = nλ ρn 2 /R = nλ If Dn be the diameter of the nth bright ring, then ρn = Dn/2 Dn 2/4R = nλ Dn = (4nRλ)1/2 Dn = (4nRλ)1/2 (n)1/2 Dn α n1/2 Thus the diameter of the dark fringes are proportional to the square root of natural numbers. Applications You can use them to check the thickness of a surface, and to check whether a surface is uniform
  76. 76. ` PROBLEM-10 Q: Calculate the thickness of the thinnest film (μ=1.4) in which interference of violet component (λ = 4000 Å) of incident light can take place by reflection. A: The condition for constructive interference of light reflected from a film of thickness t is 2μt cosθ = (2n+1)λ/2 n = 0,1,2,…… For normal incident θ = 0 and for minimum thickness n should be 0. 2μt = λ/2 t = 714.3 Å
  77. 77. ` PROBLEM-11 Q: A thin soap film (μ=1.33) seen by sodium light (λ = 5893 Å) by normal reflection appears dark. Find the minimum thickness of the film. A: The condition for destructive interference of light reflected from the a film of thickness t is 2μt cosθ = nλ For normal incident θ = 0 and for minimum thickness n should be 1. t = λ/2μ = 2215Å
  78. 78. ` PROBLEM-12 Q: Newton’s rings are observed normally in reflected light of wavelength 5.9 × 10-5 cm. The diameter of the 10th dark ring is 0.5 cm. Find the curvature of the lens and the thickness of the film. A: The diameter of the nth dark ring is given by Dn 2 = 4nλR After putting the values of Dn = 0.5 cm, n = 10 and λ = 5.9 x 10-5 cm, R = 106 cm If t is the thickness of the film corresponding to a ring of diameter D, then we have 2t = D2/4R t = 3x10-4 cm
  79. 79. ` PROBLEM-13 Q: A convex lens of radius 350 cm placed on a flat plate and illuminated by monochromatic light gives the 6th bright ring of diameter 0.68 cm. calculate the wavelength of light used. A: The diameter of the nth bright fringe is Dn 2 = 2(2n-1)λR, where n = 1,2,3,…… R = 350 cm, n =6, Dn = 0.68 cm λ = 6.0 × 10-5 cm = 6000 Å
  80. 80. Fringes of Equal Inclination * If the thickness of the film is uniform, then the path difference 2 μt cos θ between coherent rays can change only with inclination θ. * In this case one can get wide cones of light, and each fringe corresponds to a particular value of θ. * Such fringes are called “fringes of equal inclination” and are formed at infinity. * These were first observed by Haidinger by a telescope focused at infinity and are called Haidinger fringes.
  81. 81. Interference of Light: Division of Amplitude (b) Michelson’s Interferometer Importance: Michelson, along with Edward Morley, used this interferometer in the famous Michelson-Morley experiment (1887) in a failed attempt to demonstrate the effect of the hypothetical "ether wind" on the speed of light. Their experiment left theories of light based on the existence of a luminiferous ether without experimental support, and served ultimately as an inspiration for special relativity.
  82. 82. Michelson’s Interferometer S M1 M2 P1 P2 Telescope (T) O 1 2 M2 / Movable Fixed
  83. 83. Continued…. Construction: * Its main optical parts are two plane mirrors M1 and M2 and two similar optically-plane, parallel glass plates P1 and P2. * The plane mirrors M1 and M2 are silvered on their front surfaces and are mounted vertically on two arms at right angles to each other. * Their plane can be slightly tilted about vertical and horizontal axis by adjusting screws at their back. * The mirror M1 is mounted on a carriage provided with very accurate and fine screw and can be moved in the direction of the arrows. * The plates P1 and P2 are mounted exactly parallel to each other and inclined at 45° to M1 and M2. The surface of P1 towards P2 is partially silvered.
  84. 84. Continued…. Working: * Light from a monochromatic source S falls on P1. * A ray of light incident on the partially-silvered surface of P1 is partly reflected and partly transmitted. * The reflected ray-1 and the transmitted ray-2, travel to M1 and M2, respectively. * After reflection at M1 and M2, the two rays recombine at the partially- silvered surface and enter a short-focus telescope T. * Since the rays entering the telescope are derived from the same incident ray, they are coherent and have ability to interfere. * The interference fringes can be seen in the telescope.
  85. 85. Continued…. Function of the plate P2: * After partial reflection and transmission at O, the ray-1 travels through the glass plate P1 twice, while ray-2 does not do so even once. * Thus, in the absence of plate P2, the paths of rays-1 and rays-2 are not equal. •To equalise these paths a glass plate P2 having same thickness as P1 is placed parallel to P1. • P2 is called the “compensating plate”.
  86. 86. Continued…. Formation of fringes: * The form of the fringes depends on the inclination of M1 and M2. Let M2 / be the image of M2 formed by the reflection at the semi-silvered surface of P1, so that OM2 / = OM2. * The interference fringes may be regarded to be formed by light reflected from the surface of M1 and M2 /. * Thus the arrangement is equivalent to an air-film enclosed between the reflecting surfaces M1 and M2 /.
  87. 87. Continued…. Circular fringes: * When M2 is exactly perpendicular to M1, the film M1M2 / is of uniform thickness and we obtain circular fringes. S1 S2 P1 P2 2d d M1 M2 / S/ P E θ P1 P2 θ 2d
  88. 88. Continued…. * M1 and M2 / are the parallel reflecting surfaces. The actual source has been replaced by its virtual image S/ formed by reflection in the partially-silvered surface. * S/ forms two virtual images S1 and S2 in the M1 and M2 /. * Light from a point such as P on the extended source appears to an eye E to come from the corresponding coherent points P1 and P2 on S1 and S2. * If d is the separation between M1 and M2 /, then 2d is the separation between S1 and S2. •The path difference between the rays entering the eye is 2d cosθ.
  89. 89. Continued…. If 2d cosθ = nλ, P appears bright If 2d cosθ = (2n+1)λ/2, P appears dark * The focus of points on the source which subtend the same angle “θ” at the axis is a circle passing through P with its centre on the axis. * Thus a series of bright and dark circular fringes is seen. * The order of the fringes decreases as “θ” increases i.e. as we move away from the centre. Since the interfering rays are parallel, these fringes are formed at infinity.
  90. 90. Continued…. Straight fringes: * When M2 is not perpendicular to M1, the air-film between M1 and M2 / is wedge shaped. * Since light is incident on the film at different angles, curved fringes with convexity towards the thin edge of the wedge obtained. * If the thickness of the film is very small, the fringes are practically straight. The fringe corresponding to t = 0 is perfectly straight. * These fringes are formed near the film and are visible upto path differences comparatively much smaller than that in case of circular fringes.
  91. 91. Determination of Wavelength of Monochromatic Light The interferometer is set for circular fringes and the position of the mirror M1 is adjusted to obtain bright spot at the centre of the field of view. Let “d” be the thickness of the film and “n” the order of the spot obtained. 2d cos θ = nλ But at the centre θ = 0, and hence cos θ = 1 2d = nλ If now the mirror M1 is moved away from M2 / by λ/2, then 2d increases by λ. Therefore, (n+1)th bright spot appears at the centre. Thus each time M1 moves through a distance λ/2, next bright spot appears at the centre.
  92. 92. Continued…… Suppose during the movement of M1 through a distance x, N new fringes appear at the centre of the field. x = N (λ/2) λ = 2x/N The value of λ can be obtained by measuring the distance x on the micrometer screw and continue the number N, the value of λ can be obtained. The determination of λ by this method is most accurate, because x can be measured to an accuracy of 10-4 mm, and the value of N can be sufficiently increased, as the circular fringes can be obtained upto large path differences.
  93. 93. White light source Formation of Fringes More Detail Diagrams of MI
  94. 94. ` PROBLEM-14 Q: When the movable mirror of Michelson’s interferometer is moved through 0.05896 mm, a shift of 200 fringes is observed. What is the wavelength of light used ? A: The distance x moved by the mirror when N fringes cross the field of view is given by x = N (λ /2) λ = 2x/N Here x = 0.05896 mm and N = 200. λ = 5896 Å
  95. 95. ` ASSIGNMENT-I Q: What is Fourier Transform Infrared (FTIR) Spectroscopy ? What are the applications of FTIR spectroscopy ? Why Michelson interferometer is used in FTIR. Explain with schematic diagrams. [Maximum pages: 4 (no back to back), Marks: 5] Instructions: (a) Only handwritten reports will be accepted (No electronic-copy or printed/photocopies are allowed) (b) Clearly mention your name/roll no. on the front page of your report (c) Write the title of reference books, authors, publishers, page no. at the end of your report (you may provide web links, e.g. www.PH101.com) (d) Bad way of presentation (poor handwriting, disordered, incomplete reports) will fetch less marks (e) Two identical reports will fetch only “0” and “0” marks (f) Submissions after 26.08.2013 will not be accepted (Last date of submission: 26.08.2013) Assignment-3 PH 101 Faculty: Dr. S.N. Dash Name: XYZ Roll no.: 0000 Date: 00.00.2012

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