Lecture 16 Diffraction Chp. 36PowerPoint PresentationYoung’s Double Slit Interference ExperimentSlide 4Slide 5Slide 6Slide 7Example 13E. Suppose that Young’s experiment is performed with blue-green light of 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4 m from the slits. How far apart the bright fringes?Slide 9Slide 10Slide 11What about the intensity of light along the screen?Slide 13Slide 14Slide 15Slide 16Slide 17Babinets Complementarity PrincipleSlide 19Slide 20Slide 21Slide 22Where are the diffraction Maxima?Exact solution for diffraction maximaExperiment to measure diameter of Teachers hair using first diffraction minimum.Slide 26Slide 27Example 15E. The two headlights of an approaching automobile are 1.4 m apart. Assume the pupil diameter is 5.0 mm and the wavelength of light is 550 nm. (a) At what angular distance will the eye resolve them and (b) at what distance?Slide 29Diffraction and Interference by a double slitSample problem 36-5Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Lecture 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 is virtual (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 lengths from each source• we have constructive interference if paths differ by any number of full wavelengths• destructive interference if difference is half a wavelength longer 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 >> λThese tell us that θ is smallTherefore,€ ym=mλDd€ d sinθ = mλ m = 0,1,2,3... Maximum€ d sinθ = (m +12)λ m = 0,1,2,3... Minimumtanθ =ymD sin θ ≈tanθsinθ ≈ymD =mλ d m ym +/-01230D/d2D/d3D/dMaximam ym +/-0123D/2d3D/2d5D/2d7D/2dMinima€ ym=(m +1/2)λDdExample13E. Suppose that Young’s experiment is performed with blue-green light of 500 nm. The slits are 1.2mm apart, and the viewing screen is 5.4 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 =DλdS =Dλd=(5.4m)500×10−9m0.0012mS =0.00225m=2.25mmIntensity DistributionThe electric field at P is sum of E1 and E2. However the Poynting vector is Taking the time average, the intensity I represents the correlation between the two waves. For incoherent light, as there is no definite phase relation between E1 and E2 and cross term vanishesand incoherent sum isFor coherent sources, the cross term is non-zero. In fact, for constructive 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 coherent sinusoidal plane waves. Let the electric field components of the wave from slits 1 and 2 at P be given by We 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 by giving it a phase shift (ϕ) relative to E1 For constructive interference, with path difference of δ = λ would correspond to a phase shift of ϕ=2π. This then impliesFor constructive interference, with path difference of δ = λ would correspond to a phase shift of ϕ=2π. This then implies Superposition principle allowsThe intensity I is proportional to the square of the amplitude of the total electric fieldI ∝ E2I =4E02cos2(φ2)Let I0=E02Then..I =4I0cos212φWhat about the intensity of light along the screen?€ I = 4I0cos212φφ=2πdλsinθ• Double slit experiment produces interference pattern 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 ΔL =n (m+1/2)λ might be satisfied for some wavelength but not for some other. A given color might or might 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 follow crooked passages nor to bend into the shadow".He explained this observation by describing how particles of light always travel in straight lines, and how objects positioned within the path of light particles would cast a shadow because the particles could not spread out behind the object.True, to a point.On a much smaller scale, when light waves pass near a barrier, they tend to bend around that barrier and spread at oblique angles.This phenomenon is known as diffraction of the light, and occurs when a light wave passes very close to the 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 bands around the edge of the object.Diffraction is often explained in terms of the Huygens principle, which states that each point on a wavefront can be considered as a source of a new wave.All points on a wavefront serve as point