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PHYS 1030 MidTerm Average: 12.2/20 Marks by student number are on the web Friday, March 13, 2009 37 Week of March 16-20 Tutorial and Test 3: ch 22, 24, 25, 26 Week of March 23-27 Experiment 5: Spectroscopy Week of March 30 - April 3 Tutorial and Test 4: ch. 27, 28 Friday, March 13, 2009 38 Thin film interference 2 internal or 2 external reflections: (refractive indices in increasing or decreasing order) 2t = mλf ilm 1 2t = (m + )λf ilm 2 → constructive interference → destructive interference 1 internal and 1 external reflection: (refractive indices not in increasing or decreasing order) 1 2t = (m + )λf ilm 2 2t = mλf ilm → constructive interference → destructive interference λf ilm = λair nf ilm Friday, March 13, 2009 39 Young’s double slits and diffraction grating Bright fringes when: d sin ! = m", m = 0, 1, 2 . . . Dark fringes when: 1 d sin ! = (m + )", m = 0, 1, 2 . . . 2 d = distance between slits Diffraction from a single slit Dark fringes when: W sin ! = m", m = 1, 2, 3 . . . (no simple formula for bright fringes) W = width of slit Friday, March 13, 2009 m = 1 dark fringe m = 1 dark fringe m=2 m=2 0o θ 40 Prob. 27.-/20: A slit of width W = 4.3!10-5 m is located 1.32 m from a flat screen. Light shines through the slit and falls on the screen. Find the width of the central fringe of the diffraction pattern when the wavelength of the light is 635 nm. • At what angle is the first minimum found? m = 1 dark fringe m = 1 dark fringe m=2 m=2 θ! 00 Friday, March 13, 2009 41 Clicker Question A diffraction pattern is produced on a viewing screen by using a single slit with blue light. Does the pattern broaden or contract (become narrower) when (a) the blue light is replaced by red light and, (b) the slit width is increased? A) broaden, broaden B) broaden, contract C) contract, broaden D) contract, contract m = 1 dark fringe m = 1 dark fringe m=2 Minima: W sin! = m" m=2 ! Answer: B) broaden (" increases), contract (W increases) Friday, March 13, 2009 42 Diffraction by a small opaque disk Because of diffraction of light from the edge of the circular disk, the shadow cast by the disk has: • a central bright spot (constructive interference of light waves from around edge of disk) • circular bright fringes inside the shadow • bright and dark fringes outside the shadow Shadow of the disk Bright spot at centre of shadow! (Parallel beam, that is, a plane wave) Friday, March 13, 2009 43 Diffraction by a circular aperture The bright and dark fringes are circular and concentric with the aperture. The first dark fringe from the centre is at the angle shown in the diagram: ! = 1.22 " radians D D = diameter of aperture The smaller the aperture, the larger the angle of diffraction. Friday, March 13, 2009 circular hole of diameter D 44 Position of images in absence of diffraction Resolving power Light passing through any aperture, such as a camera lens, or an eye lens, is diffracted. If two objects are close together, the diffraction patterns from them may overlap. If the overlap of the patterns is large enough, it may not be possible to tell there are two objects – “resolution” is limited by diffraction. Friday, March 13, 2009 45 Resolving Power Headlights not resolved Headlights resolved Objects well-resolved Friday, March 13, 2009 46 Resolving power – Rayleigh criterion Two objects can be resolved when their angular separation is greater than: !min = 1.22 " radians (Rayleigh criterion) D Objects well-resolved Diameter, D At !min, the first minimum of one object overlaps the central maximum of the other Friday, March 13, 2009 47 Resolving Power For green light, " = 555 nm, and the diameter of the pupil of the eye, D = 2.5 mm, the angular resolution is: θmin = 1.22 λ = 2.7 × 10−4 rad D At 120 m altitude, the minimum resolvable distance on the ground is: s = Hθmin = 0.033 m Eagle/owl: D = 6.2 mm θmin = 1.1 × 10−4 rad, s = 0.013 m # better ability to find small furry animals on the ground Friday, March 13, 2009 48 Prob. 27.30: Two stars are 3.7!1011 m apart and are equally distant from the earth. A telescope has an objective lens with a diameter of 1.02 m and just detects these stars as separate objects. Assume that light of wavelength 550 nm is being observed and that diffraction effects rather than atmospheric turbulence limit the resolving power of the telescope. Find the maximum distance that these stars could be from earth. Friday, March 13, 2009 49 Diffraction grating Pairs of slits act as two Young’s slits. The bright fringes (“principal maxima”) are at the same angles as for Young’s double slits. Interference also occurs between more distant slits # sharper peaks Bright fringes (principal maxima): d sinθ = m" Friday, March 13, 2009 50 Diffraction grating Much sharper fringes than Young’s double slits d sin ! = m" Principal maxima d sin ! = m" Principal maxima $ much more precise measurement of wavelength The small peaks are “subsidiary maxima” Young’s double slits with same distance between slits, d θ! Friday, March 13, 2009 51 A grating spectrometer Parallel rays fall on the grating Grating equation: d sin ! = m" for the bright fringes != Friday, March 13, 2009 d sin " → wavelength, ! m 52 Angstroms: 10 A = 1 nm http://astro.u-strasbg.fr/~koppen/discharge/helium.html Only the lines are present in the spectrum. A faint continuum was added to place the lines within the spectrum of visible light. The spectrum of light from a star allows elements in the star’s “atmosphere” to be identified. Helium was observed in the sun’s atmosphere before it had been discovered on earth. Named after the Greek word for the sun. Friday, March 13, 2009 53 Prob. 27.42: A diffraction grating has 2604 lines per centimetre and produces a principal maximum at ! = 30º. The grating is used with light that contains all wavelengths between 410 and 660 nm. What wavelengths of the incident light that could have produced this maximum? Friday, March 13, 2009 54 http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html#c2 CD as a “reflection grating” Friday, March 13, 2009 55 CDs, DVDs Information is stored along spiral tracks 1600 nm apart as binary 0’s and 1’s that are read by how much a laser beam is reflected from the surface. No pit (land) A pit t 1 Laser beam 2t = 2 Destructive interference, weak reflection ! 2 1 2 Constructive interference, strong reflection For " = 520 nm, t = "/4 = 130 nm for destructive interference. Friday, March 13, 2009 56 Keeping the laser beam on track Two tracking beams are the first order maxima produced by a diffraction grating. Land If the laser drifts off track, one of the tracking beams will start to hit pits and will be reflected less strongly. A feedback system steers the beam to centre it so that both tracking beams are once again strongly reflected. Friday, March 13, 2009 57 Summary of Chapter 27 • Principle of Linear Superposition ! interference and diffraction • Diffraction by Young’s double slits, by a single slit, and the diffraction grating • Interference in thin films, phase change on reflection (for external reflections only) • Michelson interferometer • Interference by a circular aperture ! Rayleigh criterion and resolving power Friday, March 13, 2009 58 Chapter 28: Special Relativity • Events, inertial reference frames • Postulates of special relativity • Time dilation, length contraction • Relativistic momentum • Equivalence of mass and energy • Omit 28.7, relativistic addition of velocities Friday, March 13, 2009 59 Reference frames A reference frame is a coordinate system in which positions and times may be measured: (x, y, z, t). Inertial reference frame - is not accelerated – Newton’s laws of motion apply so an object remains at rest when zero net force acts on it. Non-inertial reference frame - is accelerated. Example, car going around corner, objects move if not held in place. Special relativity applies to inertial reference frames. Friday, March 13, 2009 60 Events An event is something that happens (surprise!). Observers record the position and time (x, y, z, t) of events according to their own distance scales and clocks. L = vt x y An event occurs here y% x% y Event occurs at (x, y, z, t) in O frame x O y% Event occurs at (x%, y%, z%, t%) in O% frame x% O% Special relativity deals with events observed in inertial frames that are in relative motion Friday, March 13, 2009 61 Postulates of Special Relativity 1) Relativity Postulate: The laws of physics are the same in all inertial reference frames. $ All inertial frames are equally good. There is no way to determine whether an inertial frame is absolutely at rest. 2) Speed of Light Postulate: The speed of light in vacuum, measured in any inertial reference frame, has the same value, independent of the motion of the source of light or of the observer. # Implies that light does not need a material medium to travel through, with respect to which its speed would be fixed. Compare sound, travelling at a fixed speed relative to air. Friday, March 13, 2009 62 Prob. 28.C1: The speed of light in water is c/n, where n = 1.33 is the refractive index of water. Thus, the speed of light in water is less than c. Why doesn’t this violate the speed of light postulate? The speed of light postulate deals with the speed of light in vacuum, and says nothing about the speed in a transparent medium. The speed of light in vacuum has special significance. Its constancy leads to special consequences. Friday, March 13, 2009 63