2The Lunatic Fringe: The waves arriving at the screen from the two slits will interfere constructivelyor destructively depending on the differentlength they have travelled.Maximum fringe at q=0, (central maxima)other maxima atSimilarly for dark fringes: d sin q= (m+1/2) l
3The ReturnOf the Phasor!Phase Difference2=360°=180°
4ExampleRed laser light (l=633nm) goes through two slits 1cm apart, and produces a diffraction pattern on a screen 55cm away. How far apart are the fringes near the center?If the fringes are near the center, we can usesin q ~ q, and thenml=dsinq~dq => q=m/d is the angle for eachmaximum (in radians)Dq= l/d =is the “angular separation”.The distance between the fringes is thenDx=LDq=Ll/d=55cmx633nm/1cm=35 mmFor the spacing to be 1mm, we need d~ Ll/1mm=0.35mm
5ExampleIn a double slit experiment, we can measure the wavelength of the light if we know the distances between the slits and the angular separation of the fringes.If the separation between the slits is 0.5mm and the first order maximum of the interference pattern is at an angle of 0.059o from the center of the pattern, what is the wavelength and color of the light used?qd sinq=ml =>l=0.5mm sin(0.059o)= m=515nm ~ green
6ExampleA double slit experiment has a screen 120cm away from the slits, which are 0.25cm apart. The slits are illuminated with coherent 600nm light. At what distance above the central maximum is the average intensity on the screen 75% of the maximum?I/I0=4cos2f/2 ; I/Imax=cos2f/2 =0.75 => f=2cos–1 (0.75)1/2=60o=p/3 radf=(2pd/l)sinq => q= sin-1(l/2pd)f=0.0022o=40 mrad (small!)y=Lq=48mm
7Radio WavesTwo radio towers, separated by 300 m as shown the figure, broadcast identical signals at the same wavelength.A radio in the car, which traveling due north, receives the signals. If the car is positioned at the second maximum, what is the wavelength of the signals? Where will the driver find a minimum in reception?dsinq=ml =>l= dsinq/2sin q= 400/( )1/2=0.37l=55.7m“Dark” fringes: dsinq=(m+1/2)lsinq=(m+1/2)l/d=2.5x55.7m/300m=0.464 =>q= 28otanq=y/1000m => y=1000m x tan(28o)=525m
8Interference: Example A red light beam with wavelength l=0.625mm travels through glass (n=1.46) a distance of 1mm. A second beam, parallel to the first one and originally in phase with it, travels the same distance through sapphire (n=1.77).How many wavelengths are there of each beam inside the material?In glass, lg=0.625mm/1.46= mm and Ng=D/ lg=In sapphire, ls=0.625mm/1.77= mm (UV!) and Ns=D/ ls=What is the phase difference in the beams when they come out?The difference in wavelengths is Ns-Ng= Each wavelength is 360o, so DN= means Df=DNx360o=0.41x360o=148oHow thick should the glass be so that the beams are exactly out of phase at the exit (destructive interference!)DN=D/ ls- D/ lg= (D/ l)(n2-n1)=0.31 (D/ l)=m+1/2A thickness D=(m+0.5) 2.02 mm would make the waves OUT of phase.For example, mm makes them in phase, and mm makes them OUT of phase.
9Thin Film Interference: The patterns of colors that one seesin oil slicks on water or in bubblesis produced by interference of thelight bouncing off the two sides ofthe film.To understand this weneed to discuss thephase changes thatoccur when a wavemoves from one mediumto the another where thespeed is different. Thiscan be understood witha mechanical analogy.
10Reflection, Refraction and Changes of Phase: Consider an UP pulse moving in a rope, that reaches a juncturewith another rope of different density. A reflected pulse is generated.The reflected pulse is also UP if the speed of propagation in the rope of the right is faster than on the left. (Low impedance.)The reflected pulse is DOWN if the speed of propagation in the right is slower than on the left. (High impedance.)The extreme case of ZERO speed on the right corresponds to a ropeanchored to a wall. (Highest impedance.)If we have a wave instead of a pulse “DOWN” means 180 degrees OUT of phase, and UP means 360° or IN PHASE.
11Thin Films First reflected light ray comes from first n1 interface, second from second. These rays interfere with each other.n1n2n3How they interfere will depend on the relative indices of refraction.In the example above the first ray suffers a 180 degree phase change(1/2 a wavelength) upon reflection. The second ray does not changephase in reflection, but has to travel a longer distance to come back up.The distance is twice the thickness of the layer of oil.For constructive interference the distance 2L must therefore be ahalf-integer multiple of the wavelength, i.e. 0.5 l, 1.5 l,…,(0.5+2n)l.
12Thin Films: Soap Bubbles If the film is very thin, then the interference is totally dominated by the180° phase shift in the reflection.At the top the film is thinnest (due to gravity it lumps at the bottom), so one sees thefilm dark at the top.180°Air: n=10°Soap: n>1AirThis film is illuminated with white light, therefore we see fringes ofdifferent colors corresponding to the various constructive interferencesof the individual components of the white light, which change as wego down. The thickness increases steeply as we go down, which makesthe width of the fringes become narrower and narrower.
13Reflective CoatingsTo make mirrors that reflects light of only a given wavelength, a coating of a specific thickness is used so that there is constructive interference of the given wavelength. Materials of different index of refraction are used, most commonly MgFe2 (n=1.38) and CeO2 (n=2.35), and are called “dielectric films”. What thickness is necessary for reflecting IR light with l=1064nm?First ray: Df=180deg=pSecond ray: Df=2L(2p/(l/n))=p=> L= lCeo2/4=(l/n)/4=113nmThird ray? If wafer has the same thickness (and is of the same material), Df=4L(2p/(l/n)=2p: destructive!n=2.35n=1.38Choose MgFe2 wafer so thatDf=(2n1L1+2n2L2) (2p/l)= p+ 2n2L2 (2p/l)=3p=> L2= l/2n2 = 386 nmWe can add more layers to keep reflecting the light, until no light is transmitted: all the light is either absorbed or reflected.
14Anti-Reflective Coatings Semiconductors such a silicon are usedto build solar cells. They are coatedwith a transparent thin film, whoseindex of refraction is 1.45, in order tominimize reflected light. If the index ofrefraction of silicon is 3.5, what is the minimum width of the coatingthat will produce the least reflection at a wavelength of 552nm?Both rays undergo 180 phase changes atreflection, therefore for destructiveinterference (no reflection), the distancetravelled (twice the thickness) should beequal to half a wavelength in the coatingn=1.45
15Anti-Reflective Coatings Radar waves have a wavelength of 3cm.Suppose the plane is made of metal(speed of propagation=0, n is infinite andreflection on the polymer-metal surfacetherefore has a 180 degree phase change).The polymer has n=1.5. Same calculation as in previous examplegives,Stealth FighterOn the other hand, if one coated a plane with the same polymer(for instance to prevent rust) and for safety reasons wanted to maximizeradar visibility (reflective coating!), one would have
16Michelson Interferometers: As we saw in the previous example, interference is a spectacular wayof measuring small distances (like the thickness of a soap bubble), sincewe are able to resolve distances of the order of the wavelength of thelight (for instance, for yellow light, we are talking about 0.5 of amillionth of a meter, 500nm). This has therefore technologicalapplications.In the Michelson interferometer, light from asource (at the left, in the picture) hits a semi-plated mirror. Half of it goes through to the right and half goes upwards. The two halves are bounced back towards the half plated mirror, interfere, and the interference can be seen by the observer at the bottom. The observer will see light if the two distances travelled d1 and d2 are equal, and will see darkness if they differ by half a wavelength.
17Michelson-Morley Experiment Michelson won the Nobel prize in 1907, "for his optical precision instruments and the spectroscopic and metrological investigations carried out with their aid""The interpretation of these results is that there is no displacement of the interference bands. ... The result of the hypothesis of a stationary ether is thus shown to be incorrect." (A. A. Michelson, Am. J. Sci, 122, 120 (1881))
18The largest Michelson interferometer in the world is in Livingston, LA, in LSU owned land (it is operated by a project funded by the National Science Foundation run by Caltech and MIT, and LSU collaborates in the project).Mirrors are suspended with wires and will movedetecting ripples inthe gravitational field due to astronomical events.
19Gravitational Waves Interferometry: an International Dream GEO600 (British-German)Hannover, GermanyLIGO (USA)Hanford, WA and Livingston, LATAMA (Japan) MitakaAIGO (Australia),Wallingup Plain, 85km north of PerthVIRGO (French-Italian)Cascina, Italy