Physics 2102 Jonathan Dowling Lecture 39 FRI 24 APR Ch 35 Interference Christian Huygens 1629 1695 The Lunatic Fringe The waves arriving at the screen from the two slits will interfere constructively or destructively depending on the different length they have travelled L d sin Constructive Destructive L m m 0 1 2 integer 1 L m m 0 1 2 integer 2 Constructive d sin m Maximum fringe at 0 central maxima other maxima at m arcsin d Similarly for dark fringes d sin m 1 2 The Return Of the Phasor 2 2 L d sin Phase Difference 2 360 180 E E1 E2 2 E0 cos 2 E0 cos 2 I E2 2 4 cos 2 I 0 E0 2 Example Red laser light 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 use sin and then m dsin d m d is the angle for each maximum in radians q l d is the angular separation The distance between the fringes is then x L L d 55cmx633nm 1cm 35 m For the spacing to be 1mm we need d L 1mm 0 35mm Example In 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 d sin m 0 5mm sin 0 059o 5 15 10 7m 515nm green Example A 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 4cos2 2 I Imax cos2 2 0 75 2cos 1 0 75 1 2 60o 3 rad 2 d sin sin 1 2 d 0 0022 40 rad small y L 48 m Radio Waves Two 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 dsin m dsin 2 sin 400 4002 10002 1 2 0 37 55 7m Dark fringes dsin m 1 2 sin m 1 2 d 2 5x55 7m 300m 0 464 28o tan y 1000m y 1000m x tan 28o 525m Interference Example A red light beam with wavelength 0 625 m 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 0 625 m 1 46 0 428 m and Ng D g 2336 45 In sapphire s 0 625 m 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 2 A 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 sees in oil slicks on water or in bubbles is produced by interference of the light bouncing off the two sides of the film To understand this we need to discuss the phase changes that occur when a wave moves from one medium to the another where the speed is different This can be understood with a mechanical analogy Reflection Refraction and Changes of Phase Consider an UP pulse moving in a rope that reaches a juncture with 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 rope anchored 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 Films First reflected light ray comes from first interface second from second These rays interfere with each other n1 n2 n3 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 a half integer multiple of the wavelength i e 0 5 1 5 0 5 2n odd number In phase 2 L Anti phase 2 L integer n2 n2 2 Thin Films Soap Bubbles If the film is very thin then the interference is totally dominated by the 180 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 1 0 Soap n 1 Air This film is illuminated with white light therefore we see fringes of different colors corresponding to the various constructive interferences of the individual components of the white light which change as we go down The thickness increases steeply as we go down which makes the width of the fringes become narrower and narrower Reflective Coatings To 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 35 n 1 38 First ray 180deg Second ray 2L 2 n L Ceo2 4 n 4 113nm Third 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 3 L2 2n2 …
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