UVA PHYS 632 - Lecture 16 Diffraction Chp. 36

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Lecture 16 Diffraction Chp. 36• Topics– Young’s double slit interference experiment– Diffraction and the wave theory– Single slit diffraction– Intensity of single slit diffraction– Circular aperture and double slit diffraction– Diffraction grating– Dispersion and resolving power• Demos– laser pointer internal reflected in water flow– laser pointer interference from many reflections– Youngs double slit interference with narrow slits– Newtons rings– Laser diffracted from bead– Laser defracted from single slit– Laser diffracted from single slits of different widths– Now show diffraction and interference together using two slits whose width is a few wavelengths– Measure diameter of hair using laser .– Show diffraction using gratings with different number of lines– Set up discharge tubes and view light through a diffraction grating with telescope1i2=1f2!1p2=1!15cm!1!30cm = - 115 +130 i2= !30cm.Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It isvirtual (since i2 < 0).30The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the imagehas the same size orientation as the object.f1f1Lens 1 Lens 2f2f21040+20-1540Actual ray diagram purple.Dashed lines are virtual raysYoung’s Double SlitInterference Experimentm=0m=1m=1m=2m=2!DyConstructive interference Constructive interference Destructive interference If you now send the light from the two openings onto a screen,an interference pattern appears, due to differing path lengthsfrom each source• we have constructive interference if paths differ by any numberof full wavelengths• destructive interference if difference is half a wavelengthlonger or shorterGeometry Path length differenceConstructive interferenceDestructive interferenceConstructive interferenceDestructive interferenceHow do we locate the vertical position of the fringes on the screen?1) L >> d2) d >> _These tell us that _ is smallTherefore, ym=m!Dd d sin!= m" m = 0,1,2,3... Maximum d sin!= (m +12)" m = 0,1,2,3... Minimumtan!=ymD sin !" tan!sin!"ymD =m# d 0D"/d2D"/d3D"/d0123ym +/-mMaximaD"/2d3D"/2d5D"/2d7D"/2d0123ym +/-mMinima ym=(m + 1/2)!DdExample13E. Suppose that Young’s experiment is performed with blue-greenlight of 500 nm. The slits are 1.2mm apart, and the viewing screen is5.4 m from the slits. How far apart the bright fringes?From the table on the previous slide we see that the separation between bright fringes isS = D!dS = D!d= (5.4m)500 " 10#9m0.0012mS = 0.00225m = 2.25mmIntensity DistributionThe electric field at P is sum ofE1 and E2. However the Poyntingvector isTaking the time average, theintensity Irepresents the correlation between the two waves.For incoherent light, as there is no definite phaserelation between E1 and E2 and cross term vanishesand incoherent sum isFor coherent sources, the cross term is non-zero. In fact, forconstructive interference E1 = E2 and I = 4I1For destructive interference E1 = -E2 and and correlation term = -I1, the total intensity becomes I = I1 -2I1 + I1 = 0Suppose that the waves emerged from the slits are coherentsinusoidal plane waves. Let the electric field components of thewave from slits 1 and 2 at P be given byWe have dropped the kx term by assuming that P is at the origin (x=0) and we have acknowledged that wave E2 has traveled farther bygiving it a phase shift (_) relative to E1For constructive interference, with path difference of _ = _ wouldcorrespond to a phase shift of _=2!. This then impliesFor constructive interference, with path difference of _ = _ wouldcorrespond to a phase shift of _=2!. This then impliesSuperposition principle allowsThe intensity I is proportional tothe square of the amplitude of thetotal electric fieldI ! E2I = 4E02cos2(!2)Let I0= E02Then..I = 4 I0cos212!What about the intensity of light along the screen? I = 4I0cos212!!=2"d#sin$• Double slit experiment produces interferencepattern if light is coherentFor constructive interference:For destructive interference:Intensity:Newton’s rings are similar, but with curved glass:Why different colors? for any given difference in path length,the condition _L =n (m+1/2)_ might be satisfied for somewavelength but not for some other. A given color might ormight not be present in the visible image.Pass around Newtons RingsDiffractionIn his 1704 treatise on the theory of optical phenomena (Opticks),Sir Isaac Newton wrote that "light is never known to followcrooked passages nor to bend into the shadow".He explained this observation by describing how particles of lightalways travel in straight lines, and how objects positioned withinthe path of light particles would cast a shadow because theparticles could not spread out behind the object.True, to a point.On a much smaller scale, when light waves pass neara barrier, they tend to bend around that barrier andspread at oblique angles.This phenomenon is known as diffraction of the light,and occurs when a light wave passes very close tothe edge of an object or through a tiny opening,such as a slit or aperture.Diffraction is a wave effectInterference pattern of light and dark bandsaround the edge of the object.Diffraction is often explained in terms of theHuygens principle, which states that each pointon a wavefront can be considered as a source ofa new wave.All points on a wavefront serve aspoint sources of spherical secondarywavelets. After a time t, the newposition of the wavefront will bethat of a surface tangent to thesesecondary wavefrontsHowever, Dominique Arago, another member of the judging committee,almost immediately verified the spot experimentally. Fresnel won thecompetition, and, although it may be more appropriate to call it "theSpot of Arago," the spot goes down in history with the name "Poisson'sbright spot" like a curseIn 1818, Augustin Fresnel submitted a paper onthe theory of diffraction for a competitionsponsored by the French Academy. His theoryrepresented light as a wave, as opposed to abombardment of hard little particles, which wasthe subject of a debate that lasted sinceNewton's day. S.D. Poisson, a member of thejudging committee for the competition, was verycritical of the wave theory of light. UsingFresnel's theory, Poisson deduced the seeminglyabsurd prediction that a bright spot shouldappear behind a circular obstruction, aprediction he felt was the last nail in the coffinfor Fresnel's theory.The Fresnel bright spotBabinets Complementarity PrincipleIn the diffraction


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UVA PHYS 632 - Lecture 16 Diffraction Chp. 36

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