2. What are the types of diffraction and give the differences between them? (June 2005, June 2011)

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1 UNIT-1 b DIFFRACTION Diffrction:A) Distinction between Fresnel nd Frunhofer diffrction, B) diffrction due to single slit, N-slits,C) Diffrction grting experiment. 1 A) Distinction between Fresnel nd Frunhofer diffrction, 1. Wht is ment by diffrction of light? The phenomenon of bending of light into the geometricl shdow region is clled diffrction. This is shown by ll kinds of wves, irrespective of their nture.. Wht re the types of diffrction nd give the differences between them? (June 005, June 011) 3. Explin Fresnel nd frunhofer diffrction (My 017) Diffrction phenomenon cn be divided into following two generl clsses: (1) Fresnel s diffrction: In this clss of diffrction, source nd screen re plced t finite distnces from the perture of obstcle hving shrp edges. In this cse no lenses re used for mking the rys prllel or convergent. The incident wve front re either sphericl or cylindricl. () Frunhofer s diffrction: In this clss of diffrction source nd the screen or telescope (through which the imge is viewed) re plced t infinity or effectively t infinity. In this cse the wve front which is incident on the perture or obstcle is plne. OR Fresnel diffrction 1 In this clss of diffrction, source nd screen re plced t finite distnces from the perture of obstcle hving shrp edges. The incident wvefront re either sphericl or cylindricl. 3 Source used is smll in size Big or extended source is used 4 Lenses re not used Lenses re used Frunhofer diffrction In this clss of diffrction source nd the screen or telescope (through which the imge is viewed) re plced t infinity or effectively t infinity. In this cse the wve front which is incident on the perture or obstcle is plne. 4. Wht is the difference between interference nd diffrction? (June 005) 5. Differentite between interference nd diffrction. (My 007, June 011) 6. Explin wht is ment by diffrction of light. How diffrction is different from interference? (June 011) Prof. S.M.ASADULLAH NSAKCET

2 Following re the differences between interference nd diffrction phenomen: INTERFERENCE DIFFRACTION Interference is due to the Diffrction is due to the interction between interction between two seprte the secondry wvelets originting from wve fronts originting from different points of sme wve front. two coherent sources In n interference pttern ll the In diffrction pttern the intensity decreses mxim re of sme intensity on either side of the centrl mximum In interference ll fringes hve In diffrction the centrl mximum is very equl widths. In the interference pttern the regions of minimum intensity re usully lmost perfectly drk B) diffrction due to single slit, N-slits wide. The width decreses on either side In the in diffrction pttern the regions of minimum intensity re not perfectly drk. 7. ) Give the theory of Frunhofer diffrction due to single slit nd hence b) obtin the condition for primry nd secondry mxim. Using this c) obtin intensity distribution curve (June 006) 8. Obtin the condition for primry mxim in Frunhofer diffrction due to single slit nd d) derive n expression for width of centrl mximum (005, 011) 9. Briefly explin Frunhofer diffrction t single slit (My 017) 10. Sketch net digrm of Frunhofer diffrction t single slit (My 017) ) FRAUNHOFER DIFFRACTION AT SINGLE SLIT Suppose AB is nrrow slit width nd perpendiculr to the plne of the pper. Let plne wve front WW of monochromtic light of wve length propgting normlly to the slit be incident on it. Let the diffrcted light be focussed by mens of convex lens on screen plced in the focl plne of the lens. The secondry wvelets trvelling normlly to the slit, i.e., long the direction OP o re brought to focus t Po by the lens. Thus Po is bright centrl imge. This is clled zero order mximum. The secondry wvelets trvelling t n ngle θ with the norml re focussed t point P 1 on the screen. In order to find out intensity t P 1, drw perpendiculr AM to the diffrcted rys. b) Suppose the slit is divided into two hlves AO nd OB. Suppose the wves strting from the top of the two hlves hve pth difference /. These two wves interfere destructively producing minimum. For every point in the first hlf there is corresponding point in the second hlf producing wves hving pth difference /. Prof. S.M.ASADULLAH NSAKCET

3 sin θ NAWAB SHAH ALAM KHAN COLLEGE OF ENGINEERING & TECHNOLOGY or sinθ = (1) The pth difference between secondry wvelets from A nd B direction θ. The rys trvelling in this direction interfere destructively producing minimum clled first order minimum. The higher order minim cn be obtined in the direction sinθ = m where m =,3,4,.. In ddition to the centrl mximum there re secondry mxim which lie in between the secondry minim on either side. Hence for secondry mxim sinθ = (m + 1) / It should be noted tht secondry mxim do not fll exctly mid-wy between two minim, but they re displced towrds the centre of the pttern, of course, the displcement decreses s the order of mximum increses. Thus the diffrction pttern due to single slit consists of centrl bright mximum flnked by secondry mxim nd minim on both the sides. 3 c) Intensity distribution in diffrction pttern Suppose the width of the slit is divided into n equl prts nd the mplitude of the wve from ech prt is (becuse width of ech prt is sme). Suppose the phse difference between ny two consecutive wves from these prts would be δ 1 n π = δ = n 1 α (sy) sinθ Using the method of vector ddition of mplitudes, the resultnt mplitude A θ is given by A nδ sin δ sin sinα A θ for lrge vlue of n, α/n is very smll. α sin n sin α Hence A θ n α sin α Aθ Ao α The intensity in ny direction is given by sinα Iθ Io α where Io is the intensity of the principl mximum t θ = 0. Figure represents the intensity distribution. It is grph of sinα Io α (long Y-xis) s function of α or sin θ (long X-xis). It cn be seen tht most of the light is confined to the centrl mximum. The intensity of the secondry mxim flls off rpidly Io/, Io/61.. Prof. S.M.ASADULLAH NSAKCET

4 4 d) Liner width of the principl mximum Liner width of the principl mximum is distnce between the first order secondry minim on either side of the centrl mximum. If x is the distnce of the first secondry minimum from the center then the width of the centrl mximum W = x. If f is the focl length of the lens used to focus the diffrction pttern we hve sinθ = x/f But sinθ = /. Therefore x f or f x Prof. S.M.ASADULLAH NSAKCET Hence, the width of the centrl mximum is given by f W x From the bove it is cler tht s the slit nrrows the width of the centrl mximum increses s shown. 11. How do you mesure the slit width. Mesurement of slit width The slit is illuminted with lser bem nd the diffrction pttern is obtined on screen plced t distnce one meter. The distnce between the first minim on either side x is mesured. The slit width cn be mesured using f x where f is the focl length of the lens or distnce of the screen from the slit. Problem 1: A slit is illuminted with light of wve length. find the ngulr width of the centrl mximum if the width of the slit is 1) 4 ) 3 3) 4). Hint: sin θ = In the bove if the screen is t distnce m find the width of the centrl mximum. Problem : A slit of width 1.5 mm is illuminted by light of wvelength 500 nm nd diffrction pttern is observed on screen m wy. Clculte the width of the centrl mximum. [1.33 mm] Hint: D x Problem 3: A screen is plced m wy from nrrow slit. If the first minim lie t 5 mm on either side of the centrl mximum when light of wvelength 500 nm is used. Find the width of the slit.[0. mm] Hint: = D/x.(My 017) Problem 4: A lens whose focl length is 40 cm forms Frunhofer diffrction pttern of slit 0.3 mm wide. Clculte the distnces of the first drk bnd nd of the next bright bnd from the xis (wvelength of light used is 5890 Å [0.785 mm, mm] Hint: x f 1 nd x 3 f Problem 5: find the hlf ngulr width of the centrl bright mximum in the Frunhofer diffrction pttern of slit of width 1x10-5 cm when the slit is illuminted by light of wve length 6000Å.[30 o ] Hint: sinθ = n Problem 6: Find the ngulr width of the centrl mximum in the Frunhofer diffrction using slit of width 1 μm when the slit is illuminted by light of wvelength 600 nm.(june 006 ) [θ = o ][θ = o ]

5 Hint: sinθ = Problem 7: Clculte the ngulr seprtion between the first order minim on either side of centrl mximum when the slit is 6 x10-4 cm width nd light illuminting it hs wvelength 6000 Å 1. Explin with theory the Frunhofer diffrction due to N slits. (My 003, June 004, June 011) Frunhofer Diffrction due to N slits (Diffrction grting) Diffrction grting consists of very lrge number of nrrow slits side by side nd seprted by opque spces. The incident light is trnsmitted through the slits nd blocked by opque spces. Such grting is clled trnsmission grting. When light psses through the grting, ech one of the slit diffrcts the wves. All the diffrcted wves reinforce one nother producing shrper nd intense mxim on the screen. In prctice plne trnsmission grting is plne sheet of trnsprent mteril on which opque rulings re mde with dimond point. The spces between the rulings re equl nd trnsprent nd constitute the prllel slits. The rulings re opque nd re of equl width. The combined width of ruling nd slit is clled grting element. Theory of plne trnsmission grting Let ABC.H represent the section of grting norml to the plne of the pper. Let the width of ech slit be nd tht of opque ruling be b. Now ( + b) which is the combined width of ruling nd slit is clled grting element. It is lso the distnce between two successive slits. Any two points on successive slits seprted by distnce ( + b) re clled corresponding points. Let plne wve front be incident normlly on the grting. The points in the slits ct s secondry sources of light giving rise to secondry wves. These wves spred in ll directions on the other side of the grting. These wves re brought to focus on screen with the help of lens. The secondry wves trvelling in the sme direction s tht of the incident wve re focused t Po. Since ll these secondry wves hve trvelled equl distnce to rech Po, they reinforce constructively nd hence the point Po is the position of centrl bright mximum. Now let us consider secondry wves trvelling t n ngle θ with the direction of incidence nd reching P 1. The intensity t P 1 depends on the pth difference between the secondry wves originting from the corresponding points of two djcent slits. Since the distnce between corresponding points is ( + b) the pth difference is ( + b) sinθ. The intensity t P 1 will be mximum if ( + b) sinθ = n Prof. S.M.ASADULLAH NSAKCET 5

6 Intensity distribution Suppose the wves from the grting rech point on the screen. Suppose the phse difference between two wves form the edges of slit is given by π sinθ = α (sy) nd suppose the phse difference between two wves form the djcent slits is given by π ( b) sinθ = β (sy) The intensity t ny point is due to diffrction s well s interference. It is given s Asinα I α The fctor sinnβ sinβ Asinα α while the fctor sinnβ sinβ gives the distribution of intensity due to diffrction t single slit. gives the distribution of intensity s combined effect of ll slits. Principl mxim: the most intense mxim re clled principl mxim. They re obtined for ( + b) sinθ = ± n where n = 0,1,,3.. Asinα α The intensity of principl mxim is I N Minim: in between ny two principle mxim there will be (N - 1) minim. Secondry mxim: s there re (N - 1) minim there will be (N - ) Problem 8: If grting hs 540 line/inch find the grting element [10-3 cm] Problem 9: grting produces first order spectrum for certin wvelength t 10 o. At wht ngle we second order spectrum. Problem 10: A grting hs 6000 lines/cm. find the ngulr seprtion for sodium D 1 nd D lines in the second order.(my 007)[ o ] problem 11: A grting diffrcts light of wvelength 5000Å t 30 o in the second order. Find the grting element nd the number of lines per inch. 13. Clculte the mximum number of orders possible for plne diffrction grting. (June 011) The principl mxim in grting stisfy the reltion ( + b) sinθ = n or The mximum ngle of diffrction cn be 90 o. Hence the mximum possible order is given by (n) mx o bsin90 b n b sinθ Problem 1: A grting hs 5400 lines per inch. Find the highest order tht cn be observed. Problem 13: Find the highest order tht cn be seen with grting hving lines/inch. The wvelength of light used is 600 nm (003)[] 1 Hint: n N Problem 14: How mny orders will be visible in the wvelength of incident light is 500 nm nd the number of lines on the grting is 60 in 1 inch?(my 017) 6 Prof. S.M.ASADULLAH NSAKCET

7 14. Find the missing order in grting spectrum In grting spectrum the totl intensity is controlled by diffrction t single slit nd interference due to two djcent slits. If the intensity due to ny one cuse is zero the totl intensity t tht point will be zero. Suppose t point we n th principl mximum stisfying the reltion ( + b) sin θ = n.(1) Suppose t the sme point we get m th diffrction minimum stisfying the reltion sin θ = m.() If both the conditions (1) nd () re stisfied simultneously then n th diffrction order will be missing. Dividing eqution (1) by () we get ( b)sinθ n ( b) n or.(3) sinθ m Prof. S.M.ASADULLAH NSAKCET m 15. Wht do you understnd by grting element (My 017) It is defined s the distnce between the centers of djcent slits. Or it is lso defined s the sum of the widths of n trnsprency nd opcity b C) Diffrction grting experiment. 16. Explin with theory how wvelength of spectrl line is determined using Plne diffrction grting (M 003, J 004) The wvelength of monochromtic light or wve length of spectrl lines of composite light cn be determined using diffrction grting nd spectrometer. The collimtor C of the spectrometer is djusted to produce prllel rys nd the telescope T is djusted to receive the prllel rys, by focussing distnt object. The grting G is then plced on the grting tble such tht it is norml to the xis of collimtor C. The collimtor slit is now illuminted by the monochromtic source whose wvelength is to be determined. The telescope is brought in line with the collimtor to view the un diffrcted bright imge. The telescope is slowly turned to one side viewing through it until first order diffrcted imge coincides with the verticl cross wire of the eye piece. The reding of the telescope in this position is noted. Now the telescope is turned to the other side nd the verticl cross wire is mde to coincide with the first order diffrcted 1mge. The reding of the telescope in this position is lso noted. The difference between the two redings gives us θ, where θ is the ngle of diffrction. Substituting this vlue in the eqution given below, the wvelength of the given monochromtic source of light cn be determined. ( + b) sin θ = n or where ( + b) is grying element nd n is the order of the spectrum. The bove cn lso be written s sinθ where N is the number of lines per cm. the experiment my be repeted in higher orders. n N 7

8 8 In order to determine the wvelength of different spectrl lines of composite light source, the slit of the collimtor is illuminted by the source. In the first order diffrction itself the ngle of diffrction for different colours re mesured. Substituting those vlues in the formul, wvelength of different spectrl lines of the composite light. Thus grting is very useful in determintion of wvelength of ny spectrl line. Problem 15: A grting hs 6000 lines / cm Find the ngulr seprtion between two wvelengths 500 nm nd 510 nm in the 3rd order.(my 003/June 004). [.48 o ] Hint: sin θ = nn 17. Wht is resolving power? Wht is its formul?(my 017) The resolving power of grting is defined s the cpcity to form seprte diffrction mxim of two wvelengths 1 nd which re very close to ech other. It is given by /d where is the verge wvelength nd d is the difference in wvelengths. For grting it is given s n N where N is the totl number of lines on the grting. Prof. S.M.ASADULLAH NSAKCET d The resolving power of grting depends on the totl number of lines on the grting nd the order of the spectrum. Problem 16: Exmine if two spectrl lines of wvelengths 5890Å nd 5896Å cn be clerly resolved in the 1) first order ) second order by diffrction grting cm wide nd hving 45 lines/cm (My 017) Hint: n N [the lines will be clerly resolved in second order] d DIFFRACTION OBJECTIVE FILL IN BLANKS 1. The bending of light round the corner of obstcles is clled (diffrction). The bending of light in to the geometricl shdow region (diffrction) 3. Diffrction phenomenon indictes the (wve nture) of light. 4. Diffrction is shown by.(ll) kinds of wves. 5. Diffrction effect is clerly observed when the obstcle or opening re of the order of..(wvelength) 6. Diffrction of sound is.(more) thn tht of light. 7. Diffrction of sound is more thn tht of light becuse sound wves hve (longer) wve lengths. 8. When the source nd screen re t finite distnces from the opening or obstcle it is clled.(fresnel diffrction) 9. In Fresnel diffrction the wve fronts involved re..(sphericl or cylindricl) 10. In Fresnel diffrction the rys involved re..(diverging or converging) 11. In Fresnel diffrction the sources involved re..(point or line or smll size) 1. In Fresnel diffrction lenses re (not used) 13. Diffrction is due to superposition of wvelets from the (sme wve front) 14. When the source nd screen re t infinite distnces from the opening or obstcle it is clled.( Frunhofer diffrction) 15. In Frunhofer diffrction the wve fronts involved re..(plne) 16. In Frunhofer diffrction the rys involved re..(prllel) 17. In Frunhofer diffrction the sources involved re..(highly extended or big)

9 18. In Frunhofer diffrction lenses re (used) FRAUNHOFER DIFFRACTION AT SINGLE SLIT 19. The diffrction condition for m th order minimum is sinθ = m 0. The diffrction condition for first minimum is sinθ = 1. In single slit diffrction the first minimum is obtined t 30 o, the second minimum will be obtined t..(90 o ). In single slit diffrction the first minimum is obtined t 30 o. if the wvelength of light used is 500 nm, the slit width (1000 nm) 3. The diffrction condition for first mximum is sinθ = The diffrction condition for second mximum is sinθ =.5 5. If =, the first minimum is obtined t n ngle.(90 o ) 6. If = ( ), the diffrction pttern cnnot be observed on the screen. 7. If =, the first minimum is obtined t n ngle.(30 o ) 8. If = 4, the second order minimum is obtined t n ngle.(30 o ) 9. The ngulr width of the centrl mximum in Frunhofer diffrction is [ θ = sin -1 (/)] 30. The width of the centrl mximum. [ x ] 31. The distnces of the first drk bnd nd of the next bright bnd from the xis re f 3 f x1 nd x D 3. In the bove the distnce between first drk nd the next bright is ( f x ) 33. With decrese in slit width the width of the diffrction centrl mximum (increses.) 34. In diffrction pttern the bnds hve (different) width 35. Due to diffrction most of the light energy is concentrted in (centrl mximum) 36. With the increse in order the intensity of secondry mxim (decrese) 37. The intensities of diffrction mxim re in the rtio 1: 1/: 1/ In single slit diffrction the resultnt mplitude in ny direction is given by A nδ sin δ sin 39. In single slit diffrction the resultnt mplitude in ny direction is given by A A θ o sin α α 40. In single slit diffrction the resultnt The intensity in ny direction is given by sinα Iθ Io α 41. If slit of width slightly greter thn is used we get (diffrction) 4. If slit of width = is used diffrction pttern is (not observed). 43. If two slits ech of width = re used we get (interference only) 44. If two slits ech of width > re used we get (interference nd diffrction) 45. In grting the distnce between the centres of two successive slits is clled (grting element) 46. The totl thickness of slit nd n opcity b is clled (grting element) 9 Prof. S.M.ASADULLAH NSAKCET

10 47. Two points occupying similr positions in two successive slits re clled (corresponding points) In grting the number of lines per cm N = b 49. The condition of diffrction mximum in grting is ( + b) sinθ = n 50. In grting spectrum the (violet) color is devited lest. 51. In prism spectrum the (red) colour is devited lest. Asinα α 5. In grting spectrum the intensity of principl mxim is I N 53. In grting spectrum the mximum possible order is given by o bsin90 b (n) mx 54. In grting spectrum n th diffrction order will be missing if ( b) n m d 55. The resolving power of grting is given by n N ( b)sinθ sinθ n m or 10 MULTIPLE CHOICE QUESTIONS: 56. The Penetrtion of wves into the regions of the geometricl shdow Is ) interference b) diffrction c) polriztion (1) dispersion 57. light wves show ) interference b) diffrction c) polriztion d) ll 58. If wve shows diffrction it cn be ) longitudinl only b) trnsverse only c) either or b d) none 59. Sound wves cnnot show ) interference b) diffrction c) polriztion d) ll 60. In Single slit diffrction, the first diffrction minim is observed t n n ngle of 30 o, when the light of wvelength 500 nm is used. The Width of the slit is ) 5x10-5 cm b).5x10-5 cm c) 10 x 10-5 cm d) 1.5x10-5 cm 61. In Frunhofer diffrction the wve front undergoing diffrction hs to be ) sphericl b) cylindricl c) ellipticl d) plne 6. In single slit experiment if the slit width is reduced ) the fringes becomes brighter b) the fringes become nrrower c) the fringes become wider d) the colour of the fringes chnge 63. Insted of red colour source, if blue colour source is used in single slit experiment ) the diffrction pttern does not chnge b) the diffrction bnds become wider c) the diffrction pttern becomes nrrower nd crowded together d) the diffrction pttern disppers. 64. The diffrction pttern of single slit consists of ) wider drk bnd t the center with lternte bright nd drk bnds on either side. b) nrrow bright bnd t the center with lternte drk nd bright bnds of equl Prof. S.M.ASADULLAH NSAKCET

11 intensity on either side. c) wider bright bnd t the center with lternte drk nd bright bnds of equl intensity on either side. d) wider nd brighter bnd t the center with lternte drk nd bright bnds of decresing intensity on either side. 65. A bem of light of wvelength 600 nm flls on single slit 0.1 mm wide nd the resulting diffrction pttern is obtined on screen m wy. The distnce between the first drk fringes on either side of the centrl bright bnd ).4 mm b) 1. cm c) 1. mm d).4 cm 66. A single slit is illuminted with prllel bem if wvelength 500 nm. The emergent bem diverges t 30 o. The size of the perture is ) 1 μm b) 10 μm c).5 d) 5 μm 67. A prllel bem of monochormtic light flls normlly on plne diffrction grting hving 5000 lines/ cm. A second order Spectrl line is diffrcted through n ngle of 30. The wvelength of light is ) 5 x 10-7 cm b) 5 x 10-6 cm c) 5 x10-5 cm d) 5 x 10-4 cm 68. When white light is incident on diffrction grting, the light diffrcted more ) blue b) yellow c) violet (1) red 69. Monochromtic light flling normlly on grting gives rise to diffrcted second order bem t ngle 30. If the grting hs 5000 lines / cm, the wvelength of light is ) 600 nm b) 400 nm c) 500 nm d) 650 nm 70. Mximum number of orders possible with grting is ) independent of grting element b) directly proportionl to grting element. c) inversely proportionl to grting element. d) directly proportionl to wvelength. 11 Prof. S.M.ASADULLAH NSAKCET