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Lecture 27

Physics 2102 Jonathan Dowling. Lecture 27. Ch. 35: Interference. Christian Huygens 1629-1695. Wave Optics: Huygen’s Principle. All points in a wavefront serve as point sources of spherical secondary waves. After a time t, the new

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Lecture 27

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  1. Physics 2102 Jonathan Dowling Lecture 27 Ch. 35: Interference Christian Huygens 1629-1695

  2. Wave Optics: Huygen’s Principle All points in a wavefront serve as point sources of spherical secondary waves. After a time t, the new wavefront will be the tangent to all the resulting spherical waves. Christian Huygens 1629-1695

  3. Reflection and Refraction Laws from Huygen’s The light travels more slowly in more dense media: v=c/n (n=index of refraction) Snell’s law! Wavelength: Frequency:

  4. Young’s Double Slit Experiment: Lasers are a recent (1960’s) invention. Before that it was challenging to demonstrate that light was indeed a wave. Young created a clever simple experiment in 1801 that clearly demonstrated that light was a wave phenomenon. He first ran a plane wave through a small slit, which produced a quite coherent spherical wave. He then interposed two slits, which produced two spherical waves that interfered with each other on a screen.

  5. 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. Maximum fringe at q=0, (central maxima) other maxima at Similarly for dark fringes: d sin q= (m+1/2) l

  6. The Return Of the Phasor! Phase Difference 2=360° =180°

  7. Example Red 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 use sin q ~ q, and then ml=dsinq~dq => q=m/d is the angle for each maximum (in radians) Dq= l/d =is the “angular separation”. The distance between the fringes is then Dx=LDq=Ll/d=55cmx633nm/1cm=35 mm For the spacing to be 1mm, we need d~ Ll/1mm=0.35mm

  8. q 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 sinq=ml => l=0.5mm sin(0.059o) = 5.15 10-7m=515nm ~ green

  9. 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=4cos2f/2 ; I/Imax=cos2f/2 =0.75 => f=2cos–1 (0.75)1/2=60o=p/3 rad f=(2pd/l)sinq => q= sin-1(l/2pd)f=0.0022o=40 mrad (small!) y=Lq=48mm

  10. 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? dsinq=ml => l= dsinq/2 sin q= 400/(4002+10002)1/2=0.37 l=55.7m “Dark” fringes: dsinq=(m+1/2)l sinq=(m+1/2)l/d=2.5x55.7m/300m =0.464 =>q= 28o tanq=y/1000m => y=1000m x tan(28o)=525m

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