Chapter 8: Physical Optics

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1 Chapter 8: Physical Optics Whether light is a particle or a wave had puzzled physicists for centuries. In this chapter, we only analyze light as a wave using basic optical concepts such as interference and diffraction

2 Overview Physical Optics Huygens Principle Interference Diffraction Double-slits Interference Thin Film Single-slit Diffraction Diffraction Grating Air Wedge & Newton s Ring

3 8.1 Huygens Principle State Huygens principle Sketch and explain the wavefront of light after passing through a single slit and obstacle using Huygens principle Learning Objectives

4 Wavefront Wavefront is defined as a line or surface, in the path of a wave motion, on which the disturbances at every point have the same phase. Wavefront A B C D E F Direction of movement, v A B C D E F Direction of movement, v Wave λ Line joining all point of adjacent wave, e.g. A, B and C or D,E and F are in phase Wavefront Wave front always perpendicular to the direction of wave propagation.

5 Types of Wavefront

6 Ray and Beam of Light Ray is defined as a line represents the direction of travel of a wave. Ray Wavefront λ Beam of light is a collection of rays or a column of light. Source of light from infinity

7 Huygens Principle Huygens principle states that every point on a wavefront can be considered as a source of secondary wavelets that spread out in the forward direction at the speed of the wave. The new wavefront is the envelope of all the secondary wavelets - i.e. the tangent to all of them. When applying Huygens s principle, show the centres of the wavelets the wavelets from these centres the line touching these wavelets draw an arrow to show the direction of the ray (normal to wavefront)

8 Huygens Principle At t = 0 s Ray At t = Δt s 1. Draw a wavefront of a plane wave as a straight line. 2. Choose a few points (5 points) on the wavefront as sources of wavelets. Since there is no backward-moving wave, the wavelets are hemispheres. 3. Draw 5 semicircles of equal radius, one centered on each of the five points. 4. Draw a line tangent to the five semicircles: this represents the wavefront at the later time.

9 Huygens's Principle Plane wavefront Spherical wavefront

10 Huygens's Principle Diffraction of wave at a single slit Diffraction of wave at an obstacle

11 Example 1 The figure shows a point light source P on the ground. Draw the wavefront from point P at time t = 1 s and t = 2 s.

12 Example 1 Solution Ray P t = 1 s t = 2 s

13 8.2 Constructive Interference and Destructive Interference Define coherence State the condition for interference of light State the conditions of constructive and destructive interference Learning Objectives

14 Coherence A stable interference pattern can be produced if the sources of wave are coherent. The two sources of wave are coherent if they have: the same phase difference (constant) the same wavelength (monochromatic)

15 Interference Interference of light is defined as the situation when two or more light waves meet (superposed) at a point, a bright or a dark region will be produced in accordance to the Principle of Superposition. Principle of superposition states the resultant displacement at any point is the vector sum of the displacements due to the two light waves. Conditions for fixed interference two coherent sources same or approximately same amplitude distance between the coherent sources, d λ

16 Constructive Interference Constructive interference is defined as a reinforcement of amplitudes of light waves that will produce a bright fringe (maximum). a a 2a a 2a a Two coherent sources are in phase

17 Destructive Interference Destructive interference is defined as a total cancellation of amplitudes of light waves that will produce a dark fringe (minimum). a a Two coherent sources are in anti-phase

18 Path Difference Path difference is the difference between two paths of waves from two different sources at a point (a difference in path length). Path Difference, L = S 2 P - S 1 P = x 2 x 1

19 Two Coherent Sources In Phase In phase sources Constructive interference

20 Two Coherent Sources In Phase Bright fringe (Constructive) 3.0λ Dark fringe (Destructive) 3.5λ L 0,, 2, 3,... m where m 3.0λ 0, 1, 2, 3, λ 3 5 L,,, m 2 where m 0, 1, 2, 3,...

21 Two Coherent Sources Anti Phase Anti phase sources Destructive interference

22 Two Coherent Sources Anti Phase Bright fringe (Constructive) 3.5λ Dark fringe (Destructive) 3.0λ 4.0λ 3.0λ 3 5 L,,, m 2 where m 0, 1, 2, 3,... L 0,, 2, 3,... m where m 0, 1, 2, 3,...

23 Example 2 Two point sources X and Y emit waves of wavelength 2.0 cm in phase. The point P is 6.0 cm from X and 10.0 cm from Y. Another point Q is 7.0 cm from X and 4.0 cm from Y. What is the path difference of waves from X and Y at a. The point P and b. The point Q? Hence deduce whether constructive interference or destructive interference occurs at P and Q.

24 Example 2 Solution

25 8.3 Interference of Transmitted Light Through Double-slits Derive and use: md y i. m for bright fringes (maxima) y m d 1 m D 2 d ii. for dark fringes (minima) where m = 0, ±1, ±2, ±3, Use y D d and explain the effect of changing any of the variables Learning Objectives

26 Light source S 0 Young s Double Slits Experiment S 1 S 2 The coherent waves in phase from the two slits. Screen Intensity Max Min Max Min Max Min Max Min Max Interference pattern on the screen. 1 st order dark 1 st dark Order m = 2 m = 2 m = 1 m = 1 m = 0 m = 0 m = 0 m = 1 m = 1 m = 2 m = 2 Fringe 3 rd dark 2 nd bright 2 nd dark 1 st bright 1 st dark Central bright 1 st dark 1 st bright 2 nd dark 2 nd bright 3 rd dark

27 Young s Double Slits Experiment i. Wavefront from light source falls on a narrow slit S 0 and diffraction occurs. ii. Every point on the wavefront that falls on S 0 acts as sources of secondary wavelets that will produce a new wavefront that propagate to slits S 1 and S 2. iii. S 1 and S 2 are produced two new sources of coherent waves in phase because they originate from the same wavefront and their distance from S are equal. iv. Constructive interference and destructive interference occur at different points on the screen to produce a pattern of alternating bright and dark fringes.

28 Derivation of Young s Double-slit Equations P S 1 y m d A θ θ C S 2 ΔL D

29 Derivation of Young s Double-slit Equations y m d θ θ D ΔL tan y m D sin L d since θ is very small, tan θ sin θ, hence y m D L d

30 Derivation of Young s Double-slit Equations To obtain constructive interference, L m Therefore, the separation between central bright and m th bright fringes, y m is given by y m md d m : order 0, 1, 2,...

31 Derivation of Young s Double-slit Equations To obtain destructive interference, L m 1 2 Therefore, the separation between central bright and m th bright fringes, y m is given by y m 1 m D 2 d m : order 0, 1, 2,...

32 Separation Between Two Consecutive Fringes, Δy Separation between two consecutive (successive) dark or bright fringes: y y y m1 y m m 1 D d D d md d

33 Appearance of Fringes on Young s Double-Slit Experiment From equation y D d Δy, fringes are wider Δy, fringes are narrower y depends on : i) the wavelength of light, λ y ii) the distance apart, d of the double slits y 1 d iii) distance between slits and the screen, D y D

34 Appearance of Fringes on Young s Double-Slit Experiment If white light is used the central bright fringe is white, and the fringes on either side are coloured. Blue is the colour nearer to the central fringe and red is farther away as shown in figure below. red red blue blue

35 What would happen to the interference pattern if the entire experiment were submerged in water? Submerging the double-slit experiment in water would reduce the wavelength of the light from to /n, where n = 1.33 is the refraction index of water. Therefore, the bright or dark fringe separation would be reduced, the interference pattern fringes get closer to each other. y

36 Example 3 An interference pattern is formed on a screen when light of wavelength 550 nm is incident on two parallel slits 50 μm apart. The second-order bright fringe is 4.5 cm from the center of the central maximum. How far from the slits is the screen?

37 Example 3 Solution

38 Example 4 In a Young s double experiment, the slits separation is 1.0 mm. The distance between the slits and the screen is 1.0 m. The wavelength of the sodium light used is cm. a. Calculate the separation between two consecutive dark fringes. b. If the sodium light is replaced with a blue light, what are the changes to the interference pattern on the screen?

39 Example 4 Solution

40 Example 4 Solution

41 Example 5 A double-slits pattern is view on a screen 1.00 m from the slits. If the third order minima are 25.0 cm apart, determine the distance between the first order minimum and fourth order maximum on the screen. *Hint: determine the ratio of wavelength and separation between the slits first 3 rd order minimum d S 1 S 2 D y 3 y 3 zeroth order maximum y 3 3 rd order minimum

42 Example 5 Solution

43 Example 5 Solution

44 Example 5 Solution

45 Example 6 A monochromatic light of wavelength 600 nm falls on a system of double-slits of unknown slit separation. At the same time, the double-slits is illuminated by a monochromatic light of unknown wavelength. It was observed that the 4 th order maximum of the known wavelength light overlapped with the 5 th order maximum of the unknown wavelength light. Find the wavelength of the unknown wavelength light.

46 Example 6 Solution

47 Example 6 Solution

48 Example 7 Suppose you pass the light from a He-Ne laser through two slits separated by mm and find that the third bright line is formed at an angle of o relative to the incident beam. What is the wavelength of the light?

49 Example 7 Solution

50 8.4 Interference of Reflected Light in Thin Films Identify the occurrence of phase change upon reflection Explain with the aid of a diagram the interference of light in thin films at normal incidence Use the following equation: For reflected light with no phase difference: Constructive interference (or reflective coating) 2nt m Destructive interference (or anti-reflective coating) 1 2nt m 2 Learning Objectives

51 8.4 Interference of Reflected Light in Thin Films For reflected light with phase difference π rad: Constructive interference Destructive interference 1 2nt m 2 2nt m where m = 0, ±1, ±2, ±3, Learning Objectives

52 Occurrence of Phase Change Upon Reflection Less dense Denser Phase change = π rad Reflected Transmitted Reflected Transmitted Denser Less dense Phase change = 0 rad

53 Non-reflective Coating Phase difference between ray 1 and ray 2, 0 2 sources in phase Based on the diagram, ray 2 travels an extra distance of 2t before the waves combine. L 2t

54 Non-reflective Coating Since ray 1 and ray 2 are in phase, constructive interference occurs when the (path length in thin film) extra distance travelled by ray 2 is equal to a whole number of wavelength L m n and n n where λ n is the wavelength of light in medium (thin film) λ is the wavelength of light in vacuum For constructive interference: m 2t 2nt m n m 0, 1, 2,...

55 Non-reflective Coating In such case, destructive interference occurs when the (path length in thin film) extra distance traveled by ray 2 is equal to an odd number of half-wavelength, so that the ray 2 would be in phase with the ray 1: L m 1 2 n and n n For destructive interference: 1 1 2t m 2nt m 2 n 2 m 0, 1, 2,...

56 Reflective Coating Phase difference between ray 1 and ray 2, 0 2 sources anti phase Based on the diagram, ray 2 travels an extra distance of 2t before the waves combine. L 2t

57 Reflective Coating Since ray 1 and ray 2 are anti phase, constructive interference occurs when the (path length in thin film) extra distance travelled by ray 2 is equal to an odd number of half-wavelength L m 1 2 n and n n For constructive interference: 1 1 2t m 2nt m 2 n 2 m 0, 1, 2,...

58 Reflective Coating In such case, destructive interference occurs when the (path length in thin film) extra distance traveled by ray 2 is equal to a whole number of wavelength, so that the ray 2 would be in phase with the ray 1: L mn and n n For destructive interference: m 2t 2nt m n m 0, 1, 2,... Example: soap bubble

59 Example 8 Light is at normal incidence on a thin soap film of refractive index 1.30 and thickness 0.15 µm. Determine the maximum wavelength of the reflected light that undergoes a. constructive interference b. destructive interference

60 Example 8 Solution

61 Example 8 Solution

62 Example 9 White light is incident normally on a lens (n 2 = 1.52) that is coated with a film of MgF 2 (n = 1.38). For what minimum thickness of the film will yellow light (λ vacuum = 550 nm) be missing in the reflected light?

63 Example 9 Solution

64 Example 10 A lens appears greenish yellow (λ vacuum = 570 nm is strongest) when white light reflects from it. What minimum thickness of coating (n = 1.25) do you think is used on such a (glass) lens?

65 Example 10 Solution

66 8.5 Interference of Reflected Light in Air Wedge and Newton s Rings Explain with the aid of a diagram the interference in air wedge Use for air wedge: i. t m for bright fringes (maxima) ii. 2t m for dark fringes (minima) where m = 0, 1, 2, 3, Use diagram to explain qualitatively the formation of Newton s rings and the centre dark spot Learning Objectives

67 Air Wedge no phase change S L Q π rad phase change Phase difference between ray LO and ray BQ, 0 X α l O B L P t air T Y 2 sources anti phase Based on the diagram, ray BQ travels an extra distance of 2t before the waves combine. 1 st dark fringe m = x The two rays (ray LO & BQ) are coherent since both have originated from the same source and produces a produces interference pattern.

68 Air Wedge Since ray LO and ray BQ are anti phase, constructive interference occurs when the (path length in thin film) extra distance travelled by ray 2 is equal to an odd number of half-wavelength 1 L m 2 where λ is the wavelength of light in vacuum For constructive interference: 1 2t m m 0, 1, 2,... 2

69 Air Wedge In such case, destructive interference occurs when the (path length in thin film) extra distance traveled by ray BQ is equal to a whole number of wavelength, so that the ray 2 would be in phase with the ray LO: L m For destructive interference: 2 t m m 0, 1, 2,...

70 Air Wedge Based on the diagram: T tan L t l tan t l Therefore, the separation between first dark and m th bright fringes, is given by 2 l m 1 2 tan Therefore, the separation between first dark and m th dark fringes, is given by l m 2 tan Equation for separation between consecutive dark fringes or bright fringes : x 2 tan

71 Newton s Rings Phase difference between ray LO and ray BQ, 0 2 sources anti phase S L Q Based on the diagram, ray BQ travels an extra distance of 2t before the waves combine. t X d O P B The two rays (ray LO & BQ) are coherent since both have originated from the same source and produces a produces interference pattern.

72 Newton s Rings The interference pattern is a series of circular interference fringes because of a curved piece of glass with a spherical cross section (plano-convex lens). Path difference = ½λ Bright ring Path difference = λ Dark ring At X, t = 0 and thus the path difference = 0. A dark spot is observed at X due to the phase change of π radian for ray BQ. (ray LO & BQ are anti phase) From Figure, the rings become more closely spaced as one moves farther from the centre of the Newton s ring because the thickness of air film increases rapidly.

73 Example 11 An air wedge is formed by placing a human hair between two glass slides of length 44 mm on one end, and allowing them to touch on the other end. When this wedge is illuminated by a red light of wavelength 771 nm, it is observed to have 265 bright fringes. Determine a. the diameter of hair b. the angle of air wedge c. the thickness of the air film for 99 th dark fringe to be observed d. the separation between two consecutive bright fringes

74 Example 11 Solution

75 Example 11 Solution

76 Example 11- Solution

77 Example 11 Solution

78 Example 12 A plate of glass 10.0 cm long is placed in contact with a second plate and held at small angle with it by a metal strip mm thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength of 635 nm. How many interference fringes are observed per centimeter in the reflected light?

79 Example 12 Solution

80 8.6 Diffraction by a Single Slit Define diffraction Explain with the aid of a diagram the diffraction of a single slit. Use: nd i. for dark fringes (minima) y n a 1 2 a n D ii. y n for bright fringes (maxima) where n = 0, ±1, ±2, ±3, Learning Objectives

81 Diffraction Diffraction of light is defined as the bending of waves as they travel around obstacles or pass through an aperture comparable to the wavelength of the waves.

82 Diffraction by a Single Slit S 1 1 n = 2 2 nd bright n = 2 2 nd dark n = 1 1 st bright n = 1 1 st dark Central bright n = 1 1 st dark n = 1 1 st bright n = 2 2 nd dark n = 2 2 nd bright single slit intensity screen

83 Formation of Diffraction by a Single Slit According to Huygen s principle, wavefront from light source falls on a narrow slit and diffraction occurs. Every point on the wavefront that falls on the slit acts as sources of secondary wavelets and superposed each another to form an interference pattern on the screen.

84 Formation of Diffraction by a Single Slit The key idea is that every point on wave front can be paired with another point that is a 2 away. If the path different is λ 2, the wavelets will arrive at the screen anti phase and interfere destructively. a 2 sin n n 2 a sinn n For bridge fringe, the general equation is: 1 a sinn n 2

85 Formation of Diffraction by a Single Slit Distance of n th order minimum (dark fringe) from central maximum (centre bright): y n nd a Distance of n th order maximum (bright fringe) from central maximum (central bright) y n n 1 2 a D n 1, 2, 3,..

86 Central Maximum (Bright) Fringe The wavelets going straight forward all travel in the same distance to the screen. Thus they arrive in phase and interfere constructively to produce the central maximum of diffraction patter at θ = 0. θ = 0 For first minimum (n = 1), the angle of diffraction is θ 1, but the angle 2θ 1 is the angle subtended by the central maximum. The width of central maximum is 2y 1 (equation for first dark) To calculate the maximum number of orders observed, θ = 90 o

87 Example 13 A monochromatic light of wavelength m passes through a single slit of width m. a. Calculate the width of central maximum: i. in degrees; ii. in centimeters, on a screen 5 cm away from the slit b. Find the number of minimum that can be observed.

88 Example 13 Solution

89 Example 13 Solution

90 Example 14 How many bright fringes will be produced on the screen if a green light of wavelength 553 nm is incident on a slit of width 8.00 µm?

91 Example 14 Solution

92 8.7 Diffraction Grating Explain with the aid of a diagram the formation of diffraction Apply d sin n where d 1 N Learning Objectives

93 Diffraction Grating Diffraction grating is defined as a large number of equally spaced parallel slits. Pattern of diffraction grating:

94 Diffraction Grating Diffraction grating can be made by ruling very fine parallel lines on glass (transmission grating) or metal (reflection grating) by a very precise machine. The untouched spaces between the lines serve as the slits as shown in figure below. a. The spaces between the lines are the slits, for example: if there are four lines then we have 3 slits. b. If there N lines per unit length, then slit separation, d is given by: d 1 N c. The light that passes through the slits are coherent, d. Interference pattern is narrower and sharper than double slits

95 Formation of Diffraction: First order maximum (n = 1) Incoming plane wavefront of light Central or zeroth order maximum (n = 0) diffraction grating L First order maximum (n = 1) d θ θ

96 Formation of Diffraction: According to Huygen s principle, when light is incident on a diffraction grating, each slit will become a secondary source of light so that superposition of light waves from each source will produce diffraction images of regular orders on a screen Based on the diagram: L L d sin d θ θ The condition for constructive interference is the path difference between the slits must be a whole number of wavelength, λ. L n Hence d sin n n 0, 1, 2, 3,..

97 Extra Knowledge A CD surface acts like diffraction grating because it has many finely space grooves on its surface.

98

99 Example 15 A diffraction grating with 600 lines per mm is illuminated normally with a monochromatic light of wavelength 589 nm. Calculate a. the angles of the first-order and second-order maximum lines from the zero-order maximum line. b. the number of orders that can be observed.

100 Example 15 Solution

101 Example 15 Solution

102 Example 15 Solution

103 Example 16 When a blue light of wavelength 465 nm illuminates a diffraction grating, it produces a 1 st order maximum but no 2 nd order maximum. a. Explain the absence of 2 nd order maximum. b. What is the maximum spacing between lines on this grating?

104 Example 16 Solution

105 Example 17 How many bright fringes are produced when a grating with a spacing of m is illuminated normally with light of wavelength m?

106 Example 17 Solution

107

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