Lecture 28Interference: ExamplePowerPoint PresentationSlide 4Slide 5Slide 6Reflective CoatingsSlide 8Slide 9Slide 10Michelson-Morley ExperimentSlide 12Gravitational Waves Interferometry: an International DreamLecture 28Lecture 28Physics 2102Jonathan DowlingCh. 35: InterferenceCh. 35: InterferenceInterference: ExampleInterference: ExampleA red light beam with wavelength =0.625m 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, g=0.625m/1.46= 0.428 m and Ng=D/ g=2336.45In sapphire, s=0.625m/1.77= 0.353 m (UV!) and Ns=D/ s=2832.86•What is the phase difference in the beams when they come out?The difference in wavelengths is Ns-Ng=496.41. Each wavelength is 360o, so N=496.41 means =Nx360o=0.41x360o=148o•How thick should the glass be so that the beams are exactly out of phase at the exit (destructive interference!)N=D/ s D/ g= (D/ )(n2-n1)=0.31 (D/ )=m+1/2A thickness D=(m+0.5) 2.02 m would make the waves OUT of phase.For example, 1.008 mm makes them in phase, and 1.010 mm makes them OUT of phase.Thin 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 we need to discuss the phase changes that occur when a wave moves from one mediumto the another where thespeed is different. Thiscan be understood witha mechanical analogy.Reflection, 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.Thin FilmsFirst reflected lightray comes from firstinterface, second from second. These rays interfere with each other.How 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 change phase 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 , 1.5 ,…,(0.5+2n).22number odd2 :phaseIn nLλ×=2integer2 :phase-AntinLλ×=n1n2n3If 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.This 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.Thin Films: Soap BubblesAir: n=1AirSoap: n>1180°0°Reflective CoatingsReflective 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 =1064nm?n=2.35n=1.38First ray: =180deg=Second ray: =2L(2 /(n)==> L= Ceo2 /n)/ nmThird ray? If wafer has the same thickness (and is of the same material), =4L(2/( n)=2 : destructive!Choose MgFe2 wafer so that 2n1L1+2n2L2) (2/)= 2n2L2 (2/)=3L2= n2386 nmWe can add more layers to keep reflecting the light, until no light is transmitted: all the light is either absorbed or reflected.Semiconductors such a silicon are usedto build solar cells. They are coatedwith a transparent thin film, whose index 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 destructive interference (no reflection), the distancetravelled (twice the thickness) should be equal to half a wavelength in the coating2 95.1L L nmnλ= ⇒ =n=1.45Anti-Reflective CoatingsAnti-Reflective CoatingsRadar 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 example gives,30.54 4 1.5cmL cmnλ= = ⇒×On 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 have312 2 1.5cmL cmnλ= = ⇒×Anti-Reflective CoatingsAnti-Reflective CoatingsStealth FighterMichelson 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 a millionth of a meter, 500nm). This has therefore technological applications.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
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