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Diffraction of light

  1. 1. CHAPTER 3. DIFFRACTION OF LIGHT By Dr. Vishal Jain Assistant Professor
  2. 2. Diffraction of Light Diffraction refers to various phenomena that occur when a wave encounters an obstacle or a slit. It is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle. Diffraction pattern of red laser beam made on a plate after passing through a small circular aperture in another plate Thomas Young's sketch of two-slit diffraction, which he presented to the Royal Society in 1803.
  3. 3. Fresnel's Diffraction Fraunhofer diffraction Cylindrical wave fronts Planar wave fronts Source of screen at finite distance from the obstacle Observation distance is infinite. In practice, often at focal point of lens. Move in a way that directly corresponds with any shift in the object. Fixed in position Fresnel diffraction patterns on flat surfaces Fraunhofer diffraction patterns on spherical surfaces. Change as we propagate them further ‘downstream’ of the source of scattering Shape and intensity of a Fraunhofer diffraction pattern stay constant. Classification of Diffraction Diffraction phenomena of light can be divided into two different classes
  4. 4. Fraunhofer Diffraction at a Single Slit Let us consider a slit be rectangular aperture whose length is large as compared to its breath. Let a parallel beam of wavelength be incident normally upon a narrow slit of AB. And each point of AB send out secondary wavelets is all direction. The rays proceeding in the same direction as the incident rays are focused at point O and which are diffracted at angle θ are focused at point B. The width of slit AB is small a. The path difference between AP and BP is calculated by draw a perpendicular BK. According to figure the path difference sinaBK  a BK AB BK sin the phase difference       sin 22 a ……………..eq 1 ……………..eq 2
  5. 5. According to Huygens wave theory each point of slit AB spread out secondary wavelets which interfere and gives diffraction phenomena. Let n be the secondary wavelets of the wave front incident on slit AB . The resultant amplitude due to all equal parts of slit AB at the point P can be determine by the method of vector addition of amplitude. This method is known as polygon method For this construct a polygon of vector that magnitude Ao represent the amplitude of each wavelets and direction of vector represented the phase of each wavelets nɸ=δ δ/2 δ/2 N A B C δ 2δ nδ r nɸ=δ A B R R/2 R/2
  6. 6. r nA radius arc o   onA r  Now a perpendicular CN is draw from the center C of an arc on the line A, which will divide the amplitude R in two parts from triangle ACN and BCN By assuming polygon as a arc of a circle of radius r we can calculate the angle , 2/ 2 sin AC R   BC R 2/ 2 sin   , 2 sin 2  AC R  AC=BC = r so 2 sin 2  BC R  2 sin2  rR  2 sin 2/2 sin2     oo nAnA R  By putting the values of r By assuming δ/2 =α=π/λ a sinθ  sin onAR  and the intensity I is given by 2 RI  2 2 sin   oII  22 oo AnI where 
  7. 7. Intensity distribution by single slit diffraction Central Maxima For the central point P on the screen θ = 0 and hence α = 0 Hence intensity at point P will be oo III  2 2 sin   1 sin 0     Lim Hence intensity at point P will be maximum Principal Minima For the principal minima intensity should be zero ,0 sin 2 2    oII ,0sin   n Where n = 1, 2, 3,4…… n = 0
  8. 8. Secondary Maxima To find out direction of secondary maxima we differentiate intensity equation with respect to α and equivalent to zero     0sincos 0sincos sin2 0 sin1 cossin2 0 sin 0 3 3 2 2 2 2                                  o o o I I I d d d dI  tan This is the condition for secondary maximas and can be solved by plotting a graph between y= tanα and y=α as shown The point of intersection of two curves gives the position of secondary maxima. The positions are α1 = 0, α2 = 1.43π, α3 = 2.46π, α4 = 3.47π,.. 2 )12(sin       na 22 22 2 )12( 4 , 4/)12( 2/)12(sin         n I I so n n II o o ....... 61 , 22 , 21 oo o I I I III 
  9. 9. Intensity distribution by single slit diffraction Central maxima Secondary maxima’s Principal Minima’s
  10. 10. Width of the central maximum The width of he central maximum can be derived as the separation between the first minimum on either side of the central maximum. If he first maximum is at distance x from the central maximum then x x D f xw 2 We know that a n for na        1 1 sin ,1 sin From the diagram θ1 f x D x 1tan If θ is very small sinθ1 = tanθ1 = θ1 a f x f x a     a f w 2  ……………..eq 1 …..eq 2 …..eq 3 …..eq 4
  11. 11. Diffraction Grating A diffraction grating is an arrangement equivalent to a large number of parallel slits of equal widths and separated from one another by equal opaque spaces. Construction Diffraction grating can be made by drawing a large number of equidistant and parallel lines on an optically plane glass plate with the help of a sharp diamond point. The rulings scatter the light and are effectively opaque, while the unruled parts transmit light and act as slits. The experimental arrangement of diffraction grating is shown They are two type refection and transmission gratings
  12. 12. A very large reflecting diffraction grating An incandescent light bulb viewed through a transmissive diffraction grating.
  13. 13. Theory for transmission grating (resultant intensity and amplitude) Let AB be the section of a grating having width of each slit as a, and b the width of each opaque space between the slits. The quantity (a + b) is called grating element, and two consecutive slits separated by the distance (a + b) are called corresponding points. The schematic ray diagram of grating has been shown in figure  sin onAR  Let a parallel beam of light of wavelength λ incident normally on the grating using the theory of single slit & Huygens principle, the amplitude of the wave diffracted at angle θ by each slit is given by ……………………eq 1
  14. 14. Diffraction by n parallel slit at an angle θ is equivalent to N parallel waves of amplitude R  sin onAR  That emitted from each slit s1, s2, s3 …..sN Where α = π/λ (a sinθ), These N parallel waves interfere and gives diffraction pattern consisting of maxima and minima on the screen. The path difference between the waves emitted from two consecutive slits given by. sin)(2 21 2 baKS KSSfrom KS    The corresponding Phase Difference       sin)( 22 ba  Thus there are N equal waves of equal amplitude and with a increasing phase difference of δ ……………………eq 2 ……………………eq 3 ……………………eq 4
  15. 15. To find the resultant amplitude of these N parallel waves we use the vector polygon method. Waves from each slit is represented by vectors where its magnitude represented by amplitude and direction represents the phase. Thus joining the N vectors tail to tip we get a polygon of N equal sides and the angle between two consecutive sides is δ The phase difference between waves from first to last slit is Nδ obtained by drawing the tangents at A and B Nδ Nδ/2 Nδ/2 A B C δ 2δ Nδ r Nδ A B RN RN N
  16. 16. Diffraction by n parallel slit at an angle θ is equivalent to N parallel waves of amplitude RN. Consider a triangle ACN and DCN C A N D δ/2 δ/2 R/2 R/2 AC R 2/ 2 sin   CD R 2/ 2 sin   Since AC=CD we can rewrite 2/sin2  R AC  ……………………eq 5 In triangle ABC         sin sinsin 2/sin 2/sinsin 2/sin2 2/sin2 2/sin2 2/sin2/ NnA R NnA R NR R NACR NACR o N o N N N N      ……………………eq 6 Here  sin onAR      sin)(2/ ba 
  17. 17. So the resultant intensity     2 2 2 2 sin sinsin NI I o  ……………………eq 7 The above equation gives the resultant intensity of N parallel waves diffracted at an angle θ. The resultant intensity is the product of two terms 2 2 sin .1  oI   2 2 sin sin ...2 N Due to diffraction from each slit Due to Interference of N slits Intensity distribution by diffraction Grating Central Maxima oo III  2 2 sin   1 sin 0     Lim Hence intensity at point P will be maximum Principal Minima ,0 sin 2 2    oII ,0sin   m Where m = 1, 2, 3,4…… 1. Due to Diffraction from Each Slit 2 2 sin .1  oI
  18. 18. Secondary Maxima  tan 2 )12(sin       ma22 )12( 4   m I I o 2. Due to Interference of N slits Principal Maxima’s Position of Principal maxima’s obtained when   2 2 sin sin ...2 N   n  0sin Where n= 0, 1, 2, 3……….. Then sinNβ is also equal to zero and becomes indeterminate so by using L’ Hosptal Rule   2 2 sin sin N N n nNNNN Lim d d N d d Lim n n      )cos( )(cos cos cos sin sin           Hence the intensity of principal maxima is given by .......2,1,0 sin 2 2 2   n where N I I o p  
  19. 19. Manima’s The intensity will be minimum when I = 0 i.e. sinNβ = 0 but sinβ ≠ 0 Nβ = ±pπ here p = 1, 2, 3………..(N-1)(N+1)…….(2N-1)(2N+1)…….. i.e. p ≠ N, 2N…… hence N p ba pN      sin)( Secondary Maxima’s To find out direction of secondary maxima we differentiate intensity equation with respect to α and equivalent to zero                        NN NNN NNNN NNNN NNNN I N I d d d dI o o tantan sin cossin 1 cos sin cos*sin2 sin cos**sin2 0 sin cos*sin2 sin cos**sin2 0 sin cos*sin2 sin cos**sin2sin 0 sin sinsin 0 3 2 2 3 2 2 3 2 22 2 2 2 2 2                          NN tantan  The solution of the above equation except p=±nπ gives the position of the secondary maxima’s
  20. 20. Intensity of Secondary Maxima’s The position of secondary maxima is given by using this equation a right triangle can be formed as shown 1 tan tan   N N  Nβ A B C Ntanβ From figure                        22 2 2 2 2 22 2 2 2 22 2 2 2 22 2 222 2 2 2 222 2 2 2 222 2 2 2 222 22 2 2 22 22 2 22 sin)1(1sin sin)1(1 sin sin)1(1 sin sin)1(1)sinsin1(sin sin )sin(cossin sin cos)tan1(sin sin sin)tan1( tan sin sin tan1 tan sin tan1 tan sin                     N N NI N N I I I N N II N N N NN N NN N NN N NN N N N N N N o o p s os 22 2 sin)1(1   N N I I p s 22 2 sin)1(1   N N I I p s As increase in number of slit the number of secondary maxima also increases
  21. 21. Intensity distribution by Diffraction Gratings
  22. 22. Formation of Spectra with Diffraction Grating With White Light With Monochromatic Light
  23. 23. Characteristics of Grating Spectra If the angle of diffraction is such that, the minima due to diffraction component in the intensity distribussion falls at the same position of principal maxima due to interference component, then the order of principal maxima then absent. If mth order minima fall on nth order principal maxima then       m n a ba nba ma     sin sin)( sin)( sin m n a ba   )( Now we consider some cases A. If b=a, i.e. width of opaque space in equal to width of slit then from equation 3. n = 2m since m=1, 2, 3 …. Then n = 2nd, 4th , 6th ….spectra will be absent B. If b=2a, i.e. width of opaque space in equal to width of slit then from equation 3. n = 3m since m=1, 2, 3 …. Then n = 3rd, 6th , 9th ….spectra will be absent 1. ABSENT SPECTRA ……………eq 1 ……………eq 2 ……………eq 3
  24. 24. 2. Maximum Number of Order Observed by Grating Principal maximum in grating spectrum is given by  nba  sin)(  sin)( ba n   Maximum possible angle of diffraction is 90 degree therefore  o ba n 90sin)( max    )( max ba n  So Q.1. A plane transmission grating has 6000 lines/cm. Calculate the higher order of spectrum which can be seen with white light of wavelength 4000 angstrom Sol. Given a+b=1/6000, Wavelength 4000X 10-8cm As we know that gratings equation written as For maximum order Maximum order will be 4th  sin)( ba n   16.4 24 100 1040006000 1)( 8max       cm cmba n  ……………..eq 1 ……………..eq 2
  25. 25. 3. Width of principal maxima The angular width of principal maxima of nth order is defined as the angular separation between the first two minima lying adjacent to principal maxima on either side θn 2δθn θn-δθn θn+δθn A O If θn is the position of nth order principal maxima θn+δθn, θn-δθn are positions of first minima adjacent to principal minima then the width of nth principal maxima = 2δθn From the Grating Equation nth order maxima And the position of minima is given by Hence equation rewritten as  nba n  sin)( ……………eq 1 )1(,,sin)(  nNpwhere N p ba n           N nN dba nn 1 )sin()( …eq 2 On dividing eq 2 by eq 1                nN d d nN dd n nn n n nnnn 1 1 )sin( sincos cos 1 1 )sin( )sincoscos(sin               nN nNd n nn 1 )sin( )sin(   If dθn is very small than cosdθn = 1, sindθn = dθn nN d nN d nN d n nnn n nn     tan , 1 cot 1 1 )sin( cos 1         nN d n n   tan2 2 
  26. 26. 4. Dispersive Power of Diffraction Gratings For a definite order of spectrum, the rate of change of angle of diffraction θ with respect to the wavelength of light ray is called dispersive power of Grating. Dispersive Power = dθ/dλ We know that gratings equation Also written as  nba n  sin)(      cos)( cos)( ba n d d n d d ba    222 2222 )( sin)()(sin1)(     nba n d d baba n ba n d d       ……………..eq 1 ……………..eq 2 ……………..eq 3
  27. 27. 5. Experimental demonstration of diffraction grating to determine wavelength
  28. 28. Resolving Power Resolving Power The ability of an optical instrument to produce two distinct separate images of two objects located very close to each other is called the resolution power Limit of resolution The smallest distance between two object, when images are seen just as separate is known as limit of resolution For eye limit of resolution is 1 minutes Resolution When two objects or their images are very close to each other they appeared as a one and it not be possible for the eye to seen them separate. Thus to see two close objects just as separate is called resolution
  29. 29. Rayleigh Criterion for Resolution Lord Rayleigh (1842-1919) a British Physicist proposed a criterion which can manifest when two object are seen just separate this criterion is called Rayleigh’s Criterion for Resolution Well Resolved Just resolved Not resolved
  30. 30. Resolving power of a telescope Resolving power of telescope is defined as the reciprocal of the smallest angle sustained at the object by two distinct closely spaced object points which can be just seen as separate ones through telescope. Let a is the diameter of objective telescope as shown in fig and P1 and P2 are the positions of the central maximum of two images. According to Rayleigh criterion these two images are said to be separated if the position of central maximum of the second images coincides with the first minimum or vice versa. P1 P2 A B dθ dθ dθ a The path difference between AP2 and BP2 is zero and the path difference between AP1 and BP1 is given by If dθ is very small sin dθ = dθ C   daBC dABBC sin sin       a dRP a dad   /1 , For rectangular aperture 22.1 a RP  For circular aperture ……………..eq 1 ………eq 2
  31. 31. Resolving power of a Diffraction Grating The resolving power of a grating is the ability to separate the spectral lines which are very close to each other. When two spectral lines in spectrum produced by diffraction grating are just resolved, then in this position the ratio of the wavelength difference and the mean wave length of spectral lines are called resolution limit of diffraction Grating Q dθ Let parallel beams of light of wavelength λ and λ+dλ be incident normally on the diffraction grating. If nth principal maxima of λ and λ+dλ are formed in the direction of θn , θn+dθn respectively For the principal maximum by wavelength λ the gratings equation written as for wavelength d λ We know that for minima By eq 2 and 3   d RP  θn λ+dλ δθn A λ P  nba n  sin)( )()sin()(  dndba nn  N nN N p dba nn   )1( )sin()(   N n NN nN N nN dn        )1( )( Nnd N nndn /     nN d RP    ……………..eq 1 ..eq 2 ..eq 3 ..eq 4

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