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Oscillation and waves lecture notes

Oscillation and waves lecture notes

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- 1. Topic 4.1 Waves, Interference and Optics 1 UEEP1033 Oscillations and Waves Topic 7: Interference and Diffraction
- 2. Topic 4.1 Waves, Interference and Optics 2 UEEP1033 Oscillations and Waves • When a wavefront encounters an aperture in an opaque barrier, the barrier suppresses all propagation of the wave except through the aperture • Following Huygen’s principle, the points on the wavefront across the aperture act as sources of secondary wavelets • When the width of the aperture is comparable with the wavelength, the aperture acts like a point source and the outgoing wavefronts are semicircular Huygen’s Principle
- 3. Topic 4.1 Waves, Interference and Optics 3 UEEP1033 Oscillations and Waves 3 • Ignores most of each secondary wavelet and only retaining the portions common to the envelope • As a result, Huygens’s principle by itself is unable to account for the details of the diffraction process • The difficulty was resolved by Fresnel with his addition of the concept of interference Huygens’s Principle
- 4. Topic 4.1 Waves, Interference and Optics 4 UEEP1033 Oscillations and Waves Augustin Jean Fresnel • 1818, Fresnel brought together the ideas of Huygens and Young and by making some arbitrary assumptions about the amplitude and phases of Huygens’ secondary sources • Fresnel able to calculate the distribution of light in diffraction patterns with excellent accuracy by allowing the various wavelet to mutually interfere Huygens-Fresnel Principle
- 5. Topic 4.1 Waves, Interference and Optics 5 UEEP1033 Oscillations and Waves Huygens-Fresnel Principle Every unobstructed point of a wavefront, at given instant, serves as a source of spherical secondary wavelets (with the same frequency as that of the primary wave) The amplitude of the optical field at any point beyond is the superposition of all these wavelets (considering their amplitudes and relative phases)
- 6. Topic 4.1 Waves, Interference and Optics 6 UEEP1033 Oscillations and Waves Christian Huygens Huygens’s Principle Each point on the wavefront of a disturbance were considered to be a new source of a “secondary” spherical disturbance, then the wavefront at a later instant could be found by constructing the “envelope” of the secondary wavelets”
- 7. Topic 4.1 Waves, Interference and Optics 7 UEEP1033 Oscillations and Waves Huygens’s Principle Every point on a propagation wavefront serves as the source of spherical secondary wavelets, such that the wavefront at some later time is the envelope of these wavelets Plane wave Spherical wave
- 8. Topic 4.1 Waves, Interference and Optics 8 UEEP1033 Oscillations and Waves
- 9. Topic 4.1 Waves, Interference and Optics 9 UEEP1033 Oscillations and Waves Huygens’s Principle Plane wave Spherical wave Every point on a propagation wavefront serves as the source of spherical secondary wavelets the wavefront at some later time is the envelope of these wavelets
- 10. Topic 4.1 Waves, Interference and Optics 10 UEEP1033 Oscillations and Waves ri Law of Reflection Law of Refraction (Snell’s law) ttii nn sinsin Interface Incident medium ni Refracting medium ni Surface normal
- 11. Topic 4.1 Waves, Interference and Optics 11 UEEP1033 Oscillations and Waves Law of Reflection When a ray of light is reflected at an interface dividing two uniform media, the reflected ray remains within the plane of incidence, and the angle of reflection equals the angle of incidence. The plane of incidence includes the incident ray and the normal to the point of incidence Law of Refraction (Snell’s law) When a ray of light is refracted at an interface dividing two uniform media, the transmitted ray remains within the plane of incidence and the sine of the angle of refraction is directly proportional to the sine of the angle of incidence
- 12. Topic 4.1 Waves, Interference and Optics 12 UEEP1033 Oscillations and Waves Huygens’ construction to prove the law of reflection Narrow, parallel ray of light Plane of interface XY Angle of incidence Angle of reflection
- 13. Topic 4.1 Waves, Interference and Optics 13 UEEP1033 Oscillations and Waves Huygens’ construction to prove the law of reflection • Since points along the plane wavefront do not arrive at the interface simultaneously, allowance is made for these differences in constructing the wavelets that determine the reflected wavefront • If the interface XY were not present, the Huygens construction would produce the wavefront GI at the instant ray CF reached the interface at I • The intrusion of the reflecting surface, means that during the same time interval required for ray CF to progress from F to I, ray BE has progressed from E to J and then a distance equivalent to JH after reflection
- 14. Topic 4.1 Waves, Interference and Optics 14 UEEP1033 Oscillations and Waves Huygens’ construction to prove the law of reflection • Wavelet of radius JN = JH centered at J is drawn above the reflecting surface • Wavelet of radius DG is drawn centered at D to represent the propagation after reflection of the lower part of the light • The new wavefront, which must now be tangent to these wavelets at points M and N, and include the point I, is shown as KI in the figure • A representative reflected ray is DL, shown perpendicular to the reflected wavefront • The normal PD drawn for this ray is used to define angles of incidence and reflection for the light
- 15. Topic 4.1 Waves, Interference and Optics 15 UEEP1033 Oscillations and Waves The Law of Refraction Use Huygen’s principle to derive the law of refraction The refraction of a plane wave at an air-glass interface Figures show three successive stages of the refraction of several wavefronts at a plane interface between air (medium 1) and glass (medium 2) 1 = wavelength in medium 1 v1 = speed of light in medium 1 v2 = speed of light in medium 2 < v1 1 = angle of incidence
- 16. Topic 4.1 Waves, Interference and Optics 16 UEEP1033 Oscillations and Waves As the wave moves into the glass, a Huygens wavelet at point e will expand to pass through point c, at a distance of 1 from point e. The time interval required for this expansion is that distance divided by the speed of the wavelet = 1/v1 In the same time interval, a Huygens wavelet at point h will expand to pass through point g, at the reduced speed v2 and with wavelength 2, i.e. the time interval = 2/v2 2 2 1 1 vv 2 1 2 1 v v
- 17. Topic 4.1 Waves, Interference and Optics 17 UEEP1033 Oscillations and Waves According to Huygens’ principle, the refracted wavefront must be tangent to an arc of radius 2 centered on h, say at point g the refracted wavefront must also be tangent to an arc of radius 1 centered on e, say at point c 2 = angle of refraction h c e h c g hc 1 1sin hc 2 2sin 2 1 2 1 2 1 sin sin v v
- 18. Topic 4.1 Waves, Interference and Optics 18 UEEP1033 Oscillations and Waves Define: refraction index for a medium c = speed of light v = speed of light in the medium Speed of light in any medium depends on the index of refraction of the medium 1 1 v c n e.g. 2 2 v c n v c n 1 2 2 1 2 1 2 1 / / sin sin n n nc nc v v 2211 sinsin nn
- 19. Topic 4.1 Waves, Interference and Optics 19 UEEP1033 Oscillations and Waves The wavelength of light in any medium depends on the index of refraction of the medium Let a certain monochromatic light: Medium refraction index wavelength speed vacuum 1 c medium n n v 2 1 2 1 v v From slide-8: c v n The greater the index of refraction of a medium, the smaller the wavelength of light in that medium n n
- 20. Topic 4.1 Waves, Interference and Optics 20 UEEP1033 Oscillations and Waves
- 21. Topic 4.1 Waves, Interference and Optics 21 UEEP1033 Oscillations and Waves Frequency Between Media • As light travels from one medium to another, its frequency does not change. – Both the wave speed and the wavelength do change. – The wavefronts do not pile up, nor are they created or destroyed at the boundary, so ƒ must stay the same.
- 22. Topic 4.1 Waves, Interference and Optics 22 UEEP1033 Oscillations and Waves n n v f Frequency of the light in a medium with index of refraction n fv f c n nc fn / / f = frequency of the light in vacuum The frequency of the light in the medium is the same as it is in vacuum
- 23. Topic 4.1 Waves, Interference and Optics 23 UEEP1033 Oscillations and Waves The fact that the wavelength of light depends on the index of refraction is important in situations involving the interference of light waves Example: Two light rays travel through two media having different indexes of refraction • Two light rays have identical wavelength and are initially in phase in air (n 1) • One of the waves travels through medium 1 of index of refraction n1 and length L • The other travels through medium 2 of index of refraction n2 and the same length L
- 24. Topic 4.1 Waves, Interference and Optics 24 UEEP1033 Oscillations and Waves • When the waves leave the two media, they will have the same wavelength – their wavelength in air • However, because their wavelengths differed in the two media, the two waves may no longer be in phase The phase difference between two light waves can change if the waves travel through different materials having different indexes of refraction How the light waves will interfere if they reach some common point?
- 25. Topic 4.1 Waves, Interference and Optics 25 UEEP1033 Oscillations and Waves Number N1 of wavelengths in the length L of medium 1 11 / nn wavelength in medium 1: 1 1 1 LnL N n wavelength in medium 2: 22 / nn 2 2 2 LnL N n )( 1212 nn L NN Phase difference between the waves 21 nn
- 26. Topic 4.1 Waves, Interference and Optics 26 UEEP1033 Oscillations and Waves Example: phase difference = 45.6 wavelengths •i.e. taking the initially in-phase waves and shifting one of them by 45.6 wavelengths •A shift of an integers number of wavelengths (such as 45) would put the waves back in phase •Only the decimal fraction (such as 0.6) that is important •i.e. phase difference of 45.6 wavelengths 0.6 wavelengths •Phase difference = 0.5 wavelength puts two waves exactly out of phase •If the two waves had equal amplitudes and were to reach some common point, they would then undergo fully destructive interference, producing darkness at that point
- 27. Topic 4.1 Waves, Interference and Optics 27 UEEP1033 Oscillations and Waves • With the phase difference = 0 or 1wavelengths, they would undergo fully constructive interference, resulting brightness at that common point • In this example, the phase difference = 0.6 wavelengths is an intermediate situation, but closer to destructive interference, and the wave would produces a dimly illuminated common point
- 28. Topic 4.1 Waves, Interference and Optics 28 UEEP1033 Oscillations and Waves Example: = 550 nm Two light waves have equal amplitudes and are in phase before entering media 1 and 2 Medium 1 = air (n1 1) Medium 2 = transparent plastic (n2 1.60, L = 2.60 m) Phase difference of the emerging waves: o 9 6 1212 1020rad17.8 swavelength84.2 )00.160.1( 10550 1060.2 )( nn L NN
- 29. Topic 4.1 Waves, Interference and Optics 29 UEEP1033 Oscillations and Waves Effective phase difference = 0.84 wavelengths = 5.3 rad 300o • 0.84 wavelengths is between 0.5 wavelength and 1.0 wavelength, but closer to 1.0 wavelength. • Thus, the waves would produce intermediate interference that is closer to fully constructive interference, • i.e. they would produce a relatively bright spot at some common point.
- 30. Topic 4.1 Waves, Interference and Optics 30 UEEP1033 Oscillations and Waves Fermat’s Principle • The ray of light traveled the path of least time from A to B • If light travels more slowly in the second medium, light bends at the interface so as to take a path that favors a shorter time in the second medium, thereby minimizing the overall transit time from A to B Construction to prove the law of refraction from Fermat’s principle
- 31. Topic 4.1 Waves, Interference and Optics 31 UEEP1033 Oscillations and Waves Fermat’s Principle • Mathematically, we are required to minimize the total time: ti v OB v AO t 22 xaAO 22 )( xcbOB ti v xcb v xa t 2222 )(
- 32. Topic 4.1 Waves, Interference and Optics 32 UEEP1033 Oscillations and Waves Fermat’s Principle 0 )( 2222 xcbv xc xav x dx dt ti • minimize the total time by setting dt / dx = 0 22 sin xa x i • From diagram: 22 )( sin xcb xc t 0 sinsin t t i i vvdx dt 0 / sin / sin t t i i ncnc ttii nn sinsin
- 33. Topic 4.1 Waves, Interference and Optics 33 UEEP1033 Oscillations and Waves Interference Young’s Double-Slit Experiment
- 34. Topic 4.1 Waves, Interference and Optics 34 UEEP1033 Oscillations and Waves Interference two waves are out of phase destructive interference two waves are in phase constructive interference amplitude of their superposition is zero amplitude of the superposition (ψ1 + ψ2) = 2A A is the amplitude of the individual waves
- 35. Topic 4.1 Waves, Interference and Optics 35 UEEP1033 Oscillations and Waves Figure (a) • Two monochromatic waves ψ1 and ψ2 at a particular point in space where the path difference from their common source is equal to an integral number of wavelengths • There is constructive interference and their superposition (ψ1 + ψ2) has an amplitude that is equal to 2A where A is the amplitude of the individual waves. Figure (b) • The two waves ψ1 and ψ2 where the path difference is equal to an odd number of half wavelengths • There is destructive interference and the amplitude of their superposition is zero Interference
- 36. Topic 4.1 Waves, Interference and Optics 36 UEEP1033 Oscillations and Waves
- 37. Topic 4.1 Waves, Interference and Optics 37 UEEP1033 Oscillations and Waves Young’s Double-Slit Experiment L >> a d d = slits separation d
- 38. Topic 4.1 Waves, Interference and Optics 38 UEEP1033 Oscillations and Waves • A monochromatic plane wave of wavelength λ is incident upon an opaque barrier containing two slits S1 and S2 • Each of these slits acts as a source of secondary wavelets according to Huygen’s Principle and the disturbance beyond the barrier is the superposition of all the wavelets spreading out from the two slits • These slits are very narrow but have a long length in the direction normal to the page, making this a two-dimensional problem • The resultant amplitude at point P is due to the superposition of secondary wavelets from the two slits Young’s Double-Slit Experiment
- 39. Topic 4.1 Waves, Interference and Optics 39 UEEP1033 Oscillations and Waves • Since these secondary wavelets are driven by the same incident wave there is a well defined phase relationship between them • This condition is called coherence and implies a systematic phase relationship between the secondary wavelets when they are superposed at some distant point P • It is this phase relationship that gives rise to the interference pattern, which is observed on a screen a distance L beyond the barrier Young’s Double-Slit Experiment
- 40. Topic 4.1 Waves, Interference and Optics 40 UEEP1033 Oscillations and Waves The secondary wavelets from S1 and S2 arriving at an arbitrary point P on the screen, at a distance x from the point O that coincides with the mid-point of the two slits Distances: S1P = l1 S2P = l2 Since L >> d it can be assumed that the secondary wavelets arriving at P have the same amplitude A The superposition of the wavelets at P gives the resultant amplitude: Young’s Double-Slit Experiment )cos()cos( 21 kltkltAR ω = angular frequency k = wave number (5) d= slits separation
- 41. Topic 4.1 Waves, Interference and Optics 41 UEEP1033 Oscillations and Waves This result can be rewritten as: Since L >> d, the lines from S1 and S2 to P can be assumed to be parallel and also to make the same angle θ with respect to the horizontal axis Young’s Double-Slit Experiment 2/)(cos[]2/)(cos2 1212 llkllktAR The line joining P to the mid-point of the slits makes an angle θ with respect to the horizontal axis 21 cos/ lLl cos/212 Lll (6) d = slits separation
- 42. Topic 4.1 Waves, Interference and Optics 42 UEEP1033 Oscillations and Waves When the two slits are separated by many wavelengths, θ is very small and cos θ 1. Hence, we can write the resultant amplitude as: Young’s Double-Slit Experiment )2/cos()cos(2 lkkLtAR = path difference of the secondary wavelets The intensity I at point P = R2 12 lll )2/(cos)(cos4 222 lkkLtAI This equation describes the instantaneous intensity at P The variation of the intensity with time is described by the cos2(ωt − kL) term (7) (8)
- 43. Topic 4.1 Waves, Interference and Optics 43 UEEP1033 Oscillations and Waves • The frequency of oscillation of visible light is of the order of 1015 Hz, which is far too high for the human eye and any laboratory apparatus to follow. • What we observe is a time average of the intensity • Since the time average of cos2(ωt − kL) over many cycles = 1/2 the time average of the intensity is given by: Young’s Double-Slit Experiment )2/(cos2 0 lkII 2 0 2AI = intensity observed at a maximum of the interference pattern described how the intensity varies with l)2/(cos2 lk (9)
- 44. Topic 4.1 Waves, Interference and Optics 44 UEEP1033 Oscillations and Waves I = maximum whenever l = n (n = 0,±1, ±2, …) I = 0 whenever l = (n + ½) Young’s Double-Slit Experiment From figure on slide-25: l d sin θ Substituting for l in Eq. (9), we obtain: (10))2/sin(cos)( 2 0 kdII When θ is small so that sinθ θ, we can write: )/(cos)( )2/(cos)( 2 0 2 0 dII kdII (11) /2where k d = slits separation
- 45. Topic 4.1 Waves, Interference and Optics 45 UEEP1033 Oscillations and Waves If there were no interference, the intensity would be uniform and equal to Io/2 as indicated by the horizontal dashed line Young’s Double-Slit Experiment Light intensity I (θ) vs angle θ d = slits separation L/d separation of the bright fringes
- 46. Topic 4.1 Waves, Interference and Optics 46 UEEP1033 Oscillations and Waves Young’s Double-Slit Experiment Intensity maxima: .....,2,1,0, n d n .....,2,1,0, n d L nLx (12) (13) (14) (15) The bright fringes occur at distances from the point O given by: Minimum intensity occur when: The distance between adjacent bright fringes is: .....,2,1,0, 2 1 n d L nx d L xx nn 1 d = slits separation
- 47. Topic 4.1 Waves, Interference and Optics 47 UEEP1033 Oscillations and Waves Point source of light is illuminating an opaque object, casting a shadow where the edge of the shadow fades gradually over a short distance and made up of bright and dark bands, the diffraction fringes. Shadow fades gradually >> Bright and Dark Bands = Diffraction Fringes Diffraction
- 48. Topic 4.1 Waves, Interference and Optics 48 UEEP1033 Oscillations and Waves Light source Aperture Observation plane Screen Arrangement used for observing diffraction of light Corpuscular Theory shadow behind the screen should be well defined, with sharp borders Observations • The transition from light to shadow was gradual rather than abrupt • Presence of bright and dark fringes extending far into the geometrical shadow of the screen
- 49. Topic 4.1 Waves, Interference and Optics 49 UEEP1033 Oscillations and Waves Francesco Grimaldi in 1665 first accurate report description of deviation of light from rectilinear propagation (diffraction) The effect is a general characteristics of wave phenomena occurring whenever a portion of a wavefront is obstructed in some way Diffraction
- 50. Topic 4.1 Waves, Interference and Optics 50 UEEP1033 Oscillations and Waves Plane wavefronts approach a barrier with an opening or an obstruction, which both the opening and the obstruction are large compared to the wavelength Opening (size = a) Obstruction (size = a) wavelength, a >>
- 51. Topic 4.1 Waves, Interference and Optics 51 UEEP1033 Oscillations and Waves • If the size of the opening or obstruction becomes comparable to the wavelength • The waves is not allowed to propagate freely through the opening or past the obstruction • But experiences some retardation of some parts of the wavefront • The wave proceed to "bend through" or around the opening or obstruction • The wave experiences significant curvature upon emerging from the opening or the obstruction curvaturea
- 52. Topic 4.1 Waves, Interference and Optics 52 UEEP1033 Oscillations and Waves As the barrier or opening size gets smaller, the wavefront experiences more and more curvature More curvature Diffraction a
- 53. Topic 4.1 Waves, Interference and Optics 53 UEEP1033 Oscillations and Waves Fraunhofer and Fresnel Diffraction
- 54. Topic 4.1 Waves, Interference and Optics 54 UEEP1033 Oscillations and Waves Observation screen Fraunhofer and Fresnel Diffraction S Lens Plane waves Opaque shield , with a single small aperture of width a is being illuminated by plane wave of wavelength from a distant point source S Case-1 observation screen is very close to Image of aperture is projected onto the screen
- 55. Topic 4.1 Waves, Interference and Optics 55 UEEP1033 Oscillations and Waves Observation screen Fraunhofer and Fresnel Diffraction S Lens Plane waves Case-2 observation screen is moved farther away from Image of aperture become increasingly more structured as the fringes become prominent Fresnel or Near-Field Diffraction
- 56. Topic 4.1 Waves, Interference and Optics 56 UEEP1033 Oscillations and Waves Fraunhofer and Fresnel Diffraction S Lens Plane waves Case-3 observation screen is at very great distance away from Projected pattern will have spread out considerably, bearing a little or no resemblance to the actual aperture Observation screen Thereafter moving the screen away from the aperture change only the size of the pattern and not its shape Fraunhofer or Far-Field Diffraction
- 57. Topic 4.1 Waves, Interference and Optics 57 UEEP1033 Oscillations and Waves Fraunhofer and Fresnel Diffraction S Lens Plane waves Case-4 If at that point, the wavelength of the incoming radiation is reduce Observation screen the pattern would revert back to the Fresnel case If were decreased even more, so that → 0 The fringes would disappear, and the image would take on the limiting shape of the aperture
- 58. Topic 4.1 Waves, Interference and Optics 58 UEEP1033 Oscillations and Waves Fraunhofer and Fresnel Diffraction If a point source S and the observation screen are very far from S Lens Plane waves Observation screen Fraunhofer Diffraction If a point source S and the observation screen are too near Fresnel Diffraction
- 59. Topic 4.1 Waves, Interference and Optics 59 UEEP1033 Oscillations and Waves Fraunhofer and Fresnel Diffraction S Lens Plane waves Observation screen Fraunhofer Diffractiona R R R is the smaller of the two distances from S to and to 2 a R d = slit width
- 60. Topic 4.1 Waves, Interference and Optics 60 UEEP1033 Oscillations and Waves Practical realization of the Fraunhofer condition F1 F2
- 61. Topic 4.1 Waves, Interference and Optics 61 UEEP1033 Oscillations and Waves Diffraction • Any obstacle in the path of the wave affects the way it spreads out; the wave appears to ‘bend’ around the obstacle • Similarly, the wave spreads out beyond any aperture that it meets. such bending or spreading of the wave is called diffraction • The effects of diffraction are evident in the shadow of an object that is illuminated by a point source. The edges of the shadow are not sharp but are blurred due to the bending of the light at the edges of the object • The degree of spreading of a wave after passing through an aperture depends on the ratio of the wavelength λ of the wave to the size d of the aperture • The angular width of the spreading is approximately equal to λ/d; the bigger this ratio, the greater is the spreading
- 62. Topic 4.1 Waves, Interference and Optics 62 UEEP1033 Oscillations and Waves The Mechanism of Diffraction • Diffraction arises because of the way in which waves propagate as described by the Huygens-Fresnel Principle • The propagation of a wave can be visualized by considering every point on a wavefront as a point source for a secondary radial wave • The subsequent propagation and addition of all these radial waves form the new wavefront • When waves are added together, their sum is determined by the relative phases as well as the amplitudes of the individual waves, an effect which is often known as wave interference • The summed amplitude of the waves can have any value between zero and the sum of the individual amplitudes • Hence, diffraction patterns usually have a series of maxima and minima
- 63. Topic 4.1 Waves, Interference and Optics 63 UEEP1033 Oscillations and Waves • A monochromatic plane wave is incident upon an opaque barrier containing a single slit • Replace the relatively wide slit by an increasing number of narrow subslits • Each point in the subslits acts as a point source for a secondary radial wave • When waves are added together, their sum is determined by the relative phases and the amplitudes of the individual waves, an effect which is often known as wave interference • The summed amplitude of the waves can have any value between zero and the sum of the individual amplitudes • Hence, diffraction patterns usually have a series of maxima and minima Single Slit Diffraction
- 64. Topic 4.1 Waves, Interference and Optics 64 UEEP1033 Oscillations and Waves 36.3: Diffraction from a single slit, Locating the minima: First, if we mentally divide the slit into two zones of equal widths a/2, and then consider a light ray r1 from the top point of the top zone and a light ray r2 from the top point of the bottom zone. For destructive interference at P1,
- 65. Topic 4.1 Waves, Interference and Optics 65 UEEP1033 Oscillations and Waves One can find the second dark fringes above and below the central axis as the first dark fringes were found, except that we now divide the slit into four zones of equal widths a/4, as shown in Fig. 36-6a. In general,
- 66. Topic 4.1 Waves, Interference and Optics 66 UEEP1033 Oscillations and Waves Example, Single Slit Diffraction Pattern with White Light:
- 67. Topic 4.1 Waves, Interference and Optics 67 UEEP1033 Oscillations and Waves 36.4: Intensity in Single-Slit Diffraction Pattern, Qualitatively:
- 68. Topic 4.1 Waves, Interference and Optics 68 UEEP1033 Oscillations and Waves 36.5: Intensity in Single-Slit Diffraction Pattern, Quantitatively: From the geometry, f is also the angle between the two radii marked R. The dashed line in the figure, which bisects f, forms two congruent right triangles.
- 69. Topic 4.1 Waves, Interference and Optics 69 UEEP1033 Oscillations and Waves Fig. 36-8 The relative intensity in single-slit diffraction for three values of the ratio a/. The wider the slit is, the narrower is the central diffraction maximum. The intensity pattern is: where For intensity minimum,
- 70. Topic 4.1 Waves, Interference and Optics 70 UEEP1033 Oscillations and Waves Example, Intensities of the Maximum in a Single Slit Interference Pattern:
- 71. Topic 4.1 Waves, Interference and Optics 71 UEEP1033 Oscillations and Waves 36.6: Diffraction by a Circular Aperture:
- 72. Topic 4.1 Waves, Interference and Optics 72 UEEP1033 Oscillations and Waves 36.6: Diffraction by a Circular Aperture, Resolvability: Fig. 36-11 At the top, the images of two point sources (stars) formed by a converging lens. At the bottom, representations of the image intensities. In (a) the angular separation of the sources is too small for them to be distinguished, in (b) they can be marginally distinguished, and in (c) they are clearly distinguished. Rayleigh’s criterion is satisfied in (b), with the central maximum of one diffraction pattern coinciding with the first minimum of the other. Two objects that are barely resolvable when the angular separation is given by:
- 73. Topic 4.1 Waves, Interference and Optics 73 UEEP1033 Oscillations and Waves
- 74. Topic 4.1 Waves, Interference and Optics 74 UEEP1033 Oscillations and Waves Example, Pointillistic paintings use the diffraction of your eye:
- 75. Topic 4.1 Waves, Interference and Optics 75 UEEP1033 Oscillations and Waves Example, Rayleigh’s criterion for resolving two distant objects:
- 76. Topic 4.1 Waves, Interference and Optics 76 UEEP1033 Oscillations and Waves Fig. 36-15 (a) The intensity plot to be expected in a double-slit interference experiment with vanishingly narrow slits. (b) The intensity plot for diffraction by a typical slit of width a (not vanishingly narrow). (c) The intensity plot to be expected for two slits of width a. The curve of (b) acts as an envelope, limiting the intensity of the double-slit fringes in (a). Note that the first minima of the diffraction pattern of (b) eliminate the double-slit fringes that would occur near 12° in (c). The intensity of a double slit pattern is:
- 77. Topic 4.1 Waves, Interference and Optics 77 UEEP1033 Oscillations and Waves Example, Double slit experiment, with diffraction of each slit included:
- 78. Topic 4.1 Waves, Interference and Optics 78 UEEP1033 Oscillations and Waves Example, Double slit experiment, with diffraction of each slit included, cont. :
- 79. Topic 4.1 Waves, Interference and Optics 79 UEEP1033 Oscillations and Waves Diffraction Grating Definition A repetitive array of diffracting elements that has the effect of producing periodic alterations in the phase, amplitude, or both of an emergent wave An idealized grating consisting of only five slits Opaque surface with narrow parallel grooves e.g. made by ruling or scratching parallel notches into the surface of a flat, clean glass plate Each of the scratches serves as a source of scattered light, and together they form a regular array of parallel line sources
- 80. Topic 4.1 Waves, Interference and Optics 80 UEEP1033 Oscillations and Waves Diffraction Grating Grating Equation: d sinm = m m = specify the order of the various principal maxima The intensity plot produced by a diffraction grating consists of narrow peaks, here label with their order number m The corresponding bright fringes seen on the screen are called lines The maxima are very narrow and they separated by relatively wide dark region d = grating spacing (spacing between rulings or slits) N rulings occupy a total width w, then d = w/N
- 81. Topic 4.1 Waves, Interference and Optics 81 UEEP1033 Oscillations and Waves 36.8: Diffraction Gratings:
- 82. Topic 4.1 Waves, Interference and Optics 82 UEEP1033 Oscillations and Waves 36.8: Diffraction Gratings, Width of the Lines:
- 83. Topic 4.1 Waves, Interference and Optics 83 UEEP1033 Oscillations and Waves Diffraction Grating Application: Grating Spectroscope collimator Plane wave Diffraction grating telescope Visible emission lines of cadmium Visible emission lines from hydrogen The lines are farther apart at greater angles
- 84. Topic 4.1 Waves, Interference and Optics 84 UEEP1033 Oscillations and Waves 36.9: Gratings, Dispersion and Resolving Power: A grating spreads apart the diffraction lines associated with the various wavelengths. This spreading, called dispersion, is defined as Here is the angular separation of two lines whose wavelengths differ by . Also, To resolve lines whose wavelengths are close together, the line should also be as narrow as possible. The resolving power R, of the grating is defined as It turns out that
- 85. Topic 4.1 Waves, Interference and Optics 85 UEEP1033 Oscillations and Waves Gratings, Dispersion and Resolving Power, proofs: The expression for the locations of the lines in the diffraction pattern of a grating is: Also, If is to be the smallest angle that will permit the two lines to be resolved, it must (by Rayleigh’s criterion) be equal to the half-width of each line, which is given by :
- 86. Topic 4.1 Waves, Interference and Optics 86 UEEP1033 Oscillations and Waves 36.9: Gratings, Dispersion and Resolving Power Compared:
- 87. Topic 4.1 Waves, Interference and Optics 87 UEEP1033 Oscillations and Waves

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