Diffraction - Pham Tan Thi

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1. Diffraction Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Science Ho Chi Minh University of Technology
2. Properties of Light Effects of Materials on Light • Transmission • Reflection • Refraction • Absorption • Total Internal Reflection • Interference • Diffraction • Scattering of Light • Polarization
3. Effects of Materials on Light Materials can be classified based on how it responds to light incident on them: 1. Opaque materials - absorb light; do not let light to pass through 2. Transparent materials - allow light to easily pass through them 3. Translucent materials - allow light to pass through but distort the light during the passage
4. Definition of Diffraction Diffraction is a bending of light around the edges/corners of an obstacle and subsequently spreading out in the region of geometrical shadow of an obstacle.
5. Diffraction of Light When a narrow opaque (aperture) is placed between a source of light and a screen, light bends around the corners of the aperture. This encroachment of light is called “diffraction”. For diffraction, the size of the aperture is small (comparable to the wavelength). As a result of diffraction, the edges of the shadow (or illuminated region) are not sharp, but the intensity is distributed in a certain way depending on the nature of the aperture.
6. Difference between Interference and Diffraction Interference: occurs between waves starting from two (or more) but finite numbers of coherent sources. Diffraction: occurs between secondary wavelets starting from the different points (infinite numbers) of the same waves. Both are superposition effects and often both are present simultaneously (e.g. Young’s double slit experiment). Comparison: (a) In an interference pattern, the minima are usually almost perfectly dark while in a diffraction pattern they are not so. (b) In an interference pattern, all the maxima are of same intensity but not in the diffraction pattern. (c) The interference fringes are usually equally spaced. The diffraction fringes are never equally spaced.
7. Diffraction and Hyugen’s Principle Hyugen’s principle can be used to analyze the diffraction Diffraction pattern of a razor blade
8. What is Huygens’ Principle Hyugens’ (or Huygens-Fresnel) principle states that every point on a wavefront is a source of wavelet. These wavelets spread out in the forward direction, at the same speed as the source wave. The new waveforms is in line tangential to all the wavelets.
9. Diffraction of Light No diffraction; No spreading after passing through slit Weak diffraction; Weak spreading after passing through slit Diffraction
10. • In Figure 36.3 below, the prediction of geometric optics in (a) does not occur. Instead, a diffraction pattern is produced, as in (b). • The narrower the slit, the broader the diffraction pattern.
11. Types of Diffraction Diffraction phenomena can be classified either as Fresnel diffraction or Fraunhofer diffraction The observable difference: Fresnel diffraction The viewing screen and the aperture are located close together, the image of the aperture is clearly recognizable despite slight fringing around its periphery. As the separation between the screen and the aperture increases, the image of the aperture becomes increasingly more structured; fringes become more prominent. Fraunhofer diffraction The viewing screen and the aperture separated by a large distance, the projected pattern bears little or no resemblance to the aperture. As the separation increases, the size of the pattern changes but not its shape.
12. Types of Diffraction
13. Fresnel’s Diffraction In the case of Fresnel’s diffraction, the source of light or screen or usually both are at finite distance from the diffracting aperture (obstacle) No lenses are used The incident wavefront is either spherical or cylindrical
14. Fraunhofer’s Diffraction In the case of Fraunhofer’s diffraction, the source of light or screen are effectively at infinite distance from the diffracting aperture (obstacle). This is achieved by placing the source and screen in the focal planes of two lenses (require lenses). The incident wavefront is plane.
15. Difference between Fraunhofer and Fresnel Diffraction No Fraunhofer Diffraction Fresnel Diffraction Source and screen are at infinite Source and screen are at finite 1 distances from slits distances from slits Incident wavefront on the aperture is Incident wavefront on the aperture is 2 plane either spherical or cylindrical The diffracted wavefront is either 3 The diffracted wavefront is plane spherical or cylindrical Two convex lenses are required to 4 No lenses are required study diffraction in laboratory 5 Mathematical treatment is easy Mathematical treatment is complicated It has many applications in It has less applications in designing the 6 designing the optical instruments optical instruments The maxima and minima are well The maxima and minima are not well 7 defined defined
16. Difference between Fraunhofer and Fresnel Diffraction Fraunhofer Diffraction Fresnel Diffraction intensity pattern intensity pattern The maxima and minima are well defined The maxima and minima are not well defined
17. Fraunhofer’s Diffraction at a Single Slit Let a parallel beam of monochromatic light of wavelength λ be incident normally on a narrow slit of width AB = e. Let diffracted light be focused by a convex lens L on a screen XY placed in the focal plane of the lens. The diffraction pattern obtained on the screen consists of a central bright band, having alternate dark and weak bright bands of decreasing intensity on both sides.
18. Fraunhofer’s Diffraction at a Single Slit In terms of wave theory, a plane wavefront is incident on the slit AB. According to the Huygens’ principle, each point in AB sends out secondary wavelets in all directions. The rays proceeding in the same direction as the incident rays focused at O; while those diffracted through an angle θ are focused at P. Let us find the resultant intensity at P. Let AK be perpendicular to BP. As the optical paths from the plane AK to P are equal, the path difference between wavelets from A to B in the direction θ is BK = AB sinθ = e sinθ The corresponding phase difference = (2π/λ)e sinθ Let the width AB of the slit be divided into n equal parts. The amplitude of vibration at P due to the waves from each part will be the same (= a)
19. Fraunhofer’s Diffraction at a Single Slit The phase difference between the waves from any two consecutive parts is 1 2⇡ esin✓ = d n ✓ ◆ Hence the resultant amplitude at P is given by nd ⇡esin✓ asin 2 asin R = d = ⇡esin✓ sin 2 sin n ⇡esin✓ Let = ↵ asin↵ asin↵ ↵ R = ↵ = ↵ As is small sin n n n nasin↵ R = ↵
20. Fraunhofer’s Diffraction at a Single Slit Asin↵ Let na = A then R = ↵ The resultant intensity at P is sin↵ 2 I = R2 = A2 ↵ ✓ ◆
21. Fraunhofer’s Diffraction at a Single Slit Direction of Minima: The intensity of minimum (zero) when: sin↵ =0 ↵ or, sinα = 0 (but α ≠ 0, because for α = 0, then sinα/α = 1) α = ± mπ, where m has in integer value 1,2,3, except zero ⇡esin✓ ↵ = ⇡esin✓ = m⇡ ± esin✓ = m⇡ ± This equation gives the directions of the first, second, third, minima by putting m = 1,2,3,
22. Fraunhofer’s Diffraction at a Single Slit Direction of Maxima: To find the direction of maximum intensity, let us differentiate the intensity with respect to α and equate it to zero dI/dα = 0 2sin↵ ↵cos↵ sin↵ A2 =0 ↵ ↵2 ✓ ◆ ↵cos↵ sin↵ =0 ↵2 ↵cos↵ sin↵ =0 ↵ = tan↵ This equation is solved graphically by plotting the curves: y = α y = tan α
23. Continued Fraunhofer’s Diffraction at a Single Slit The 1st equation is a straight line through origin making an angle of The 1st equation is a straight line through origin making an angle of 45° 45 ° The 2nd equation is a discontinuous curve having a number of branches nd The 2 isThea discontinuous point of intersectioncurve of havingthe two acurvesnumber giveof thebranches value of α. These values are approximately given by α = 0, 3π/2, 5π/2, 7π/2, The point of intersection of the two curves give the value of α sin↵ 2 I = R2 = A2 ↵ ✓ ◆ These valuesSubstituteare approximately these values of αgiven, we findby α = 0, 3Ioπ =/2 A,25 π/2, 7π/2, I1 ≈ A2/22 Substituting these values of α, we can find I2 ≈ A2/61 2 I0 = A I ≈TheA2 /intensity22 of the 1st maximum 1 is about 4.96% of the central 2 I2 ≈ A /61 maximum The intensity of the 1st maximum is about 4.96% of the central maximum.
24. Fraunhofer’sContinued Diffraction at a Single Slit TheTheprincipal principal maximamaxima occursoccur atat α =α( =π (eπsin e sinθ)/θλ)/=λ 0or, θor = θ0=, i.e.,0 the principal maxima occur at the i.e. the principal maxima occurs at same direction of light. the same direction of light The diTheffractiondiffraction pattern consistspattern consistsof a brightof principala bright maximumprincipal maximumin the directionin the of incidentdirection light,of incidenthas alternativelylight, having minimaalternately and minimaweak subsidiaryand weak maximasubsidiary of rapidly decreasing intensity on maxima of rapidly decreasing either side of it. intensity on either side of it. TheTheminima minimalie atlie αat= α± =π ,±±π2, π±2, π , .
25. The single-slit diffraction patterns corresponding to e = 0.0088, 0.0176, 0.035The, and single-slit0. 070diffractioncm, patternsrespectively corresponding to. eThe = wavelength of the light used0.0088,is 6 0.0167,.328 0.035x 10 and–5 0.070cm cm. The wavelength of the light used is 6.328 x 10-5 cm
26. Fraunhofer’s Diffraction of a Double Slit Let a parallel beam of monochromatic light of wavelength λ be incident normally on two parallel slits AB and CD, each of width e and separate by opaque space d. The distance between the corresponding points of the two slits is (e+d). Let diffracted light be focused by a convex lens L on a screen XY placed in the focal plane of the lens.
27. Fraunhofer’s Diffraction of a Double Slit The pattern obtained on the screen is the diffraction pattern due to a single slit on which a system of interference fringes is superposed. By Huygens’ principle, every point in the slits AB and CD sends out secondary wavelets in all directions. From the theory of diffraction at a single slit, the resultant amplitude due to wavelets diffracted from each slit in a direction θ is Asin↵ ↵ ⇡esin✓ Where A is a constant and ↵ = Therefore, we can consider the two slits as equivalent to two coherent sources placed at the middle points S1 and S2 of the slits, and each sending a wavelet of amplitude (Asinα)/α in the direction of θ
28. Fraunhofer’s Diffraction of a Double Slit The resultant amplitude at a point P on the screen will be the result of interference between two waves of sample amplitude (Asinα)/α, and having a phase difference δ Let us draw S1K perpendicular to S2K. The path difference between the wavelets from S1 and S2 in the direction θ is S2K = (e + d)sinθ Hence the phase difference is δ = (2π/λ) x path difference = (2π/λ) x (e + d) sinθ The resultant amplitude R at P can be determined by the vector amplitude diagram, which gives OB2 = OA2 + AB2 + 2(OA)(AB)cos(BAC)
29. Fraunhofer’s Diffraction of a Double Slit sin↵ 2 sin↵ 2 sin↵ sin↵ R2 = A2 + A2 +2A2 cos ↵ ↵ ↵ ↵ ✓ ◆ ✓ ◆ ✓ ◆✓ ◆ Asin↵ 2 R2 = (2 + 2cos) ↵ ✓ ◆ A2sin2↵ = 4cos2 ↵2 2 ✓ ◆ A2sin2↵ =4A2 cos2 ↵2 ⇡ = = (e + d)sin✓ 2 2 A2sin2↵ The resultant intensity at P is I = R2 =4A2 cos2 ↵2
30. Fraunhofer’s Diffraction of a Double Slit A2sin2↵ I = R2 =4A2 cos2 ↵2 The intensity in the resultant pattern depends of two factors: (i) (sin2α)/α2, which gives the diffraction pattern due to each individual slit (ii) cos2β, which gives interference pattern due to diffracted light waves from the two slits
31. Continued . (a) The diffractionFraunhofer’sterm, (sin Di2 αff)/ractionα2, gives of aa centralDoublemaximum Slit in the direction(i) Theθ di=ff0raction, having termalternately, (sin2α)/α2, minimawhich givesand a subsidiarycentral maximummaxima in of decreasingthe directionintensity θon = 0either, has alternativelyside. minima and subsidiary maxima of decreasing intensity on either side. The minima are obtained in the directions given by The minima are obtained in the directions given by sinαsin = 0,α = 0 , α = ±mαπ= ± mπ (π e( πsineθsin)/λ =θ )/±mλ =π± mπ e sineθsin = ±mθ =π± m (mπ = (m=1,2,3 1,2 ,except3, , except zero) zero)
32. Continued . (b) The interference term, cos2 β, gives a set of equidistant dark and bright fringes,CalculationFraunhofer’sas in ofYoung’s Opticaldouble DiPathffraction -Dislitfferenceinterference of a between Doubleexperiment TwoSlit Waves. The bright(ii) Thefringes interference(maxima) termare, cosobtained2β, which ingivesthe adirections set of equidistantgiven by dark and bright fringes, as in Young’s double slit interference 2 experiment.cos β = 1 The bright fringesβ =(maxima)± nπ are obtained in the directions given by 2 cos(βπ =/λ 1,) (e+d) sin θ = β± =n ±nππ (π/λ)(e+d)sinθ = ±nπ (e+d) sin θ = ± nλ, Where n = 0,1,2, (e+d)sinθ = ±nλ (n = 0,1,2,3 ) The various maxima corresponding to n = 0, 1,2, are zero-order, first- The various maxima corresponding to n = 0,1,2, are zero-order, order, secondfirst order,-order second order, maxima maxima
33. Fraunhofer’sDivergence Diffraction Operator of a Double Slit (iii) The intensity distribution in Continued .the resultant diffraction pattern is a plot of (c)theThe productintensity of distributionthe constantin thetermresultant 4A2, didiffractionffraction termpattern (sinis 2aαplot)/ of α2, andthe theproduct interferenceof the constant term costerm2β. 4 A2, diffraction term (sin2 α)/ α2, and the interference term cos2 β. The entire pattern may be consideredThe entireas consistingpattern may of be interferenceconsidered fringesas dueconsisting to light from bothinterference slits, thefringes intensitiesdue to of these fringeslight from are governedboth slits, bythe the diffractionintensities occurringof the atthese the individualfringes slits.being governed by the diffraction occurring at the individual slits.
34. The double slit intensity distribution corresponding to e = 0.0088 cm, λ = 6.328 x 10-5 cm, and d = 0.035 cm and d = 0.07 cm 2 sin 2 sin ↵ 2 I 4I0 cos 2 e = 0.0088 cm I =4Io 2 cos ↵2 d =0.035 cm = 6.328 10-5 cm e = 0.0088 cm d =0.07 cm = 6.328 10-5 cm The double-slit intensity distribution corresponding to e = 0.0088 cm, = 6.328 x 10–5 cm, and d = 0.035 cm and d = 0.07cm respectively.
35. Fraunhofer’s Diffraction at an N Slit (Grating) Diffraction grating: A diffraction grating is an arrangement equivalent to a large number of parallel slits of equal to widths and separated from one another by equal opaque spaces It is made by ruling a large number of fine, equidistant and parallel lines on an optically-plane glass plate with a diamond point The ruling scatter the light and are effectively opaque while the unruled parts transmits and act as slits USE: When there is need to separate light of different wavelengths with high resolution, then a diffraction grating is most often the tool of choice
36. Fraunhofer’s Diffraction at an N Slit (Grating) The* diTheffractiondiffraction gratinggrating is an immenselyis an immensely useful tooluseful for separationtool for ofthe the separationspectral linesof the associatedspectral lines withassociated atomic transitions.with atomic transitions. Sometimes* Sometimes this thisdiffractiondiffraction gratinggrating is alsois also calledcalled “super‘super prismprism’” . The* conditionThe condition for maximumfor maximum intensityintensity isis thethe samesame asas thatthat forfor thethe double doubleslit or slitmultiple or multipleslits, but slits,with buta large withnumber a largeof numberslits the ofintensity slits themaximum is intensityvery sharp maximumand narrow, is very providingsharp andthe narrow,high resolution providingfor thespectroscopic high resolution for spectroscopic applications. applications. The tracks of a compact disc act as a diff*ractionThe tracks grating,of a producingcompact disc a act as a diffraction separationgrating, ofproducing the colorsa separationof white light.of the colours of white light.
37. Fraunhofer’s Diffraction at an N Slit (Grating) Let AB be the section of a plane transmission grating. Let e be the width of each slit and d is the width of each opaque space between the slits. (e+d) is called the “grating element”. The points in two consecutive slits separated by the distant (e+d) are called the “corresponding points”. Let a parallel beam of monochromatic light of wavelength λ be incident normal on the grating.
38. Fraunhofer’s Diffraction at an N Slit (Grating) According to Huygens’ principle, all the points in each slit send out secondary wavelets in all directions. According to Fraunhofer’s diffraction at a single slit, the wavelets from all points in a slit diffracted in a direction θ are equivalent to a single wave of amplitude (Asinα)/α, where α = (π/λ)e sinθ If N is the total number of slits in the grating, the diffracted rays from all the slits are equivalent to N parallel rays, one each from middle points S1, S2, S3, of the slits. After calculating the path difference and corresponding phase difference, the resultant amplitude in the direction θ is sin↵ sinN R = A ↵ sin ⇡ Where = (e + d)sin✓
39. Fraunhofer’s Diffraction at an N Slit (Grating) The resultant intensity I is A2sin2↵ sin2N I = R2 = ↵2 sin2 The first factor A2sin2α/α2 gives diffraction pattern due to the single slit, while the second factor gives the interference pattern due to N slits. Conditions for maxima: (e+d)sinθ = ±nλ, where n = 0,1,2, Conditions for minima: N(e+d) = ±mλ where m takes all integer values except 0, N, 2N, nN because these values of m make sinβ = 0, which give principal maxima.
40. Continued . Fraunhofer’s Diffraction at an N Slit (Grating) It is clear that m = 0 gives a principal maximum, m = 1,2,3, .(N-1) give It is clear that m = 0 gives a principal maximum; m = 1,2,3, (N-1) give minimaminimaand, andm m= =N Ngives givesagain againa aprincipal principalmaximum maximum ThusThus therethere areare (N-1)(N- 1minima) minima betweenbetween two consecutivetwo consecutive principalprincipal maxima. maxima.
41. Diffraction - The Central Maximum Suppose we have two sources each being allowed to emit a wave through a small opening or slit. The distance between the slits denoted by d. The distance from the slit spacing to the screen is denoted by L. If two waves go through the slit and then proceed straight ahead to the screen, they both cover the same distance and thus will have constructive interference. Their amplitude will build and leave a very bright spot on the screen. We call this the central maximum.
42. Diffraction - The Central Maximum There are several bright spots and dark areas in between. The spot in the middle if the brightest and thus the central maximum. We call these spots being fringes. We also have additional bright spots, yet the intensity is a bit less. These additional bright spots are denoted as orders. So the first bright spot on either side of the central maximum is called the first order bright fringe. The intensity of the orders as we move farther from the bright central maximum.
43. Diffraction - Bright Fringes The reason we see additional bright fringes is because the waves constructively build. There is a difference however in the intensity as shown in previous slide. The blue wave has to travel farther than the red wave to reach the screen at the position shown. For the blue wave and the red wave to build constructively, they must be in phase. How much farther did the blue wave have to travel so that they both hit the screen in phase?
44. Diffraction - Bright Fringes: Path Difference These two waves are in phase. When they hit the screen, they both hit at the same relative position at the bottom of a crest. How much farther did the red wavelength have to travel? Exactly — One Wavelength This extra difference is the path difference. The path difference and the order of a fringe help to form a pattern.
45. Diffraction - Bright Fringes: Path Difference The bright fringes you see on either side of the central maximum are multiple wavelengths from the bright central. The multiple is the order. The path different is equal to the order times the wavelength P.D. = mλ (Constructive)
46. Diffraction - Dark Fringes We see a definite decrease in the intensity between the bright fringes. In the pattern, we visible notice the dark region. These are areas where destructive interference occurs. We call these areas being dark fringes or minima.
47. Diffraction - Dark Fringes First Order First Order Dark Fringe Dark Fringe m = 1 m = 1 Zero Order Zero Order Central Dark Fringe Dark Fringe Maximum m = 0 m = 0
48. Diffraction - Dark Fringes On either side of the bright central maximum, we see areas that are dark or minimum intensity. The blue wave has to travel farther than the red wave to reach the screen. They are said to destructively build or that they are out of phase. How much farther did the blue wave have to travel so that they both hit the screen out of phase?
49. Diffraction - Dark Fringes These two waves are out of phase. When they hit the screen they both hit at the different relative position, one at the bottom of a crest and the other coming out of a trough. Thus their amplitude subtract. How much farther did the red wave has to travel? Exactly — One half of a wavelength This extra distance is called path difference. The path difference and the order of fringe help to form another pattern.
50. Diffraction - Bright Fringes: Path Difference The dark fringes you see on either side of the central maximum are multiple wavelengths from the bright central. The multiple is the order. The path different is equal to the order plus half, times the wavelength P.D. = (m+1/2)λ (Destructive)
51. Path Difference For constructive interference or maxima: P.D. = mλ For destructive interference or minima: P.D. = (m+1/2)λ
52. Problem In Fraunhofer’s diffraction due to a narrow slit, a screen is placed 2 m away from the lens to obtain the pattern. If the slit width is 0.2 mm and the first minima lie 5 mm on either side of the central maximum, find the wave length of light?
53. Problem In Fraunhofer’s diffraction due to a narrow slit, a screen is placed 2 m away from the lens to obtain the pattern. If the slit width is 0.2 mm and the first minima lie 5 mm on either side of the central maximum, find the wave length of light? A: In Fraunhofer’s diffraction pattern due to a single slit of width e, the direction of mini are given by e sinθ = ±mλ m = 1,2,3, For the 1st minimum, m = 1 ⇒ e sinθ = λ If θ is very small and measured in radian, then sinθ = 0 θ = λ/e [radian] = λ/0.02 [radian] Now for linear separation between the 1st minimum the central minimum is 0.5 cm and the distance of the screen from the slit is 2 m, then θ = 0.5/200 [radians] Equating values of θ = λ/0.02 [radians] = 0.5/200 [radians] λ = 5000 Å