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Transcript
Physics 1502: Lecture 34 Today’s Agenda • Announcements: – Midterm 2: graded soon … – Homework 09: Friday December 4 • Optics – Interference – Diffraction » Introduction to diffraction » Diffraction from narrow slits » Intensity of single-slit and two-slits diffraction patterns » The diffraction grating A wave through two slits In Phase, i.e. Maxima when DP = d sinq = nl Out of Phase, i.e. Minima when DP = d sinq = (n+1/2)l d q DP=d sinq Screen A wave through two slits In Phase, i.e. Maxima when DP = d sinq = nl + Out of Phase, i.e. Minima when DP = d sinq = (n+1/2)l + The Intensity What is the intensity at P? The only term with a t dependence is sin2( ).That term averages to ½ . If we had only had one slit, the intensity would have been, So we can rewrite the total intensity as, with The Intensity We can rewrite intensity at point P in terms of distance y Using this relation, we can rewrite expression for the intensity at point P as function of y Constructive interference occurs at where m=+/-1, +/-2 … Phasor Addition of Waves Consider a sinusoidal wave whose electric field component is E2(t) E1(t) E0 wt+f E0 wt Consider second sinusoidal wave E2(t) The projection of sum of two phasors EP is equal to EP(t) E1(t) f/2 f ER wt E0 E0 Phasor Diagrams for Two Coherent Sources ER=2E0 ER 450 E0 E0 E0 ER=0 E0 E0 E0 ER E0 E0 ER 900 E0 2700 E0 ER=2E0 E0 E0 SUMMARY 2 slits interference pattern (Young’s experiment) How would pattern be changed if we add one or more slits ? (assuming the same slit separation ) 3 slits, 4 slits, 5 slits, etc. Phasor: 1 vector represents 1 traveling wave single traveling wave 2 wave interference N-slits Interference Patterns F=0 F=90 F=180 F=270 F=360 N=2 N=3 N=4 Change of Phase Due to Reflection Lloyd’s mirror P2 S P1 L I Mirror The reflected ray (red) can be considered as an original from the image source at point I. Thus we can think of an arrangement S and I as a double-slit source separated by the distance between points S and I. An interference pattern for this experimental setting is really observed ….. but dark and bright fringes are reversed in order This mean that the sources S and I are different in phase by 1800 An electromagnetic wave undergoes a phase change by 1800 upon reflecting from the medium that has a higher index of refraction than that one in which the wave is traveling. Change of Phase Due to Reflection n1 n2 1800 phase change n1<n2 n1 no phase change n1>n2 n2 Interference in Thin Films 1800 phase change 1 Air Film Air no phase change 2 A wave traveling from air toward film undergoes 1800 phase change upon reflection. The wavelength of light ln in the medium with refraction index n is t The ray 1 is 1800 out of phase with ray 2 which is equivalent to a path difference ln/2. The ray 2 also travels extra distance 2t. Constructive interference Destructive interference Chapter 34 – Act 1 Estimate minimum thickness of a soap-bubble film (n=1.33) that results in constructive interference in the reflected light if the film is Illuminated by light with l=600nm. A) 113nm B) 250nm C) 339nm Problem Consider the double-slit arrangement shown in Figure below, where the slit separation is d and the slit to screen distance is L. A sheet of transparent plastic having an index of refraction n and thickness t is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y’. Find y’ where will the central maximum be now ? Solution Phase difference for going though plastic sheet: Corresponding path length difference: Angle of central max is approx: Thus the distance y’ is: gives Phase Change upon Reflection from a Surface/Interface Reflection from Optically Denser Medium (larger n) 180o Phase Change Reflection from Optically Lighter Medium (smaller n) No Phase Change by analogy to reflection of traveling wave in mechanics constructive: 2t = (m +1/2) ln destructive: 2t = m ln constructive: 2t = m ln destructive: 2t = (m +1/2) ln Examples : Application Reducing Reflection in Optical Instruments Experimental Observations: (pattern produced by a single slit ?) How do we understand this pattern ? First Destructive Interference: (a/2) sin Q = ± l/2 sin Q = ± l/a Second Destructive Interference: (a/4) sin Q = ± l/2 sin Q = ± 2 l/a mth Destructive Interference: sin Q = ± m l/a m=±1, ±2, … See Huygen’s Principle So we can calculate where the minima will be ! sin Q = ± m l/a m=±1, ±2, … So, when the slit becomes smaller the central maximum becomes ? Why is the central maximum so much stronger than the others ? Phasor Description of Diffraction Let’s define phase difference (b) between first and last ray (phasor) b = S (Db) = N Db central max. 1st min. 2nd max. Can we calculate the intensity anywhere on diffraction pattern ? (a/l) sin Q = 1: 1st min. Db / 2p = Dy sin (Q) / l b = N Db = N 2p Dy sin (Q) / l = 2p a sin (Q) / l Yes, using Phasors ! Let take some arbitrary point on the diffraction pattern This point can be defined by angle Q or by phase difference between first and last ray (phasor) b The resultant electric field magnitude ER is given (from the figure) by : sin (b/2) = ER / 2R The arc length Eo is given by : Eo = R b ER = 2R sin (b/2) = 2 (Eo/ b) sin (b/2) = Eo [ sin (b/2) / (b/2) ] So, the intensity anywhere on the pattern : I = Imax [ sin (b/2) / (b/2) ]2 b = 2p a sin (Q) / l Other Examples Light from a small source passes by the edge of an opaque object and continues on to a screen. A diffraction pattern consisting of bright and dark fringes appears on the screen in the region above the edge of the object. What type of an object would create a diffraction pattern shown on the left, when positioned midway between screen and light source ? • A penny, … • Note the bright spot at the center. Fraunhofer Diffraction (or far-field) q Lens Incoming wave Screen Fresnel Diffraction (or near-field) Lens Incoming wave P Screen (more complicated: not covered in this course) Resolution (single-slit aperture) Rayleigh’s criterion: • two images are just resolved WHEN: When central maximum of one image falls on the first minimum of another image sin Q = l / a Qmin ~ l / a Resolution (circular aperture) Diffraction patterns of two point sources for various angular separation of the sources Rayleigh’s criterion for circular aperture: Qmin = 1.22 ( l / a) EXAMPLE A ruby laser beam (l = 694.3 nm) is sent outwards from a 2.7m diameter telescope to the moon, 384 000 km away. What is the radius of the big red spot on the moon? a. b. c. d. e. 500 m 250 m 120 m 1.0 km 2.7 km Moon Earth Qmin = 1.22 ( l / a) R / 3.84 108 = 1.22 [ 6.943 10-7 / 2.7 ] R = 120 m ! Two-Slit Interference Pattern with a Finite Slit Size Interference (interference fringes): Iinter = Imax [cos (pd sin Q / l)]2 Diffraction (“envelope” function): Idiff = Imax [ sin (b/2) / (b/2) ]2 b = 2p a sin (Q) / l Itot = Iinter . Idiff smaller separation between slits => ? The combined effects of two-slit and single-slit interference. This is the pattern produced when 650-nm light waves pass through two 3.0- mm slits that are 18 mm apart. smaller slit size Animation => ? Example The centers of two slits of width a are a distance d apart. Is it possible that the first minimum of the interference pattern occurs at the location of the first minimum of the diffraction pattern for light of wavelength l ? d No! a a 1st minimum interference: d sin Q = l /2 1st minimum diffraction: a sin Q = l The same place (same Q) : l /2d = l /a a /d = 2 Application X-ray Diffraction by crystals Can we determine the atomic structure of the crystals, like proteins, by analyzing X-ray diffraction patters like one shown ? Yes in principle: this is like the problem of determining the slit separation (d) and slit size (a) from the observed pattern, but much much more complicated ! A Laue pattern of the enzyme Rubisco, produced with a wide-band x-ray spectrum. This enzyme is present in plants and takes part in the process of photosynthesis. Determining the atomic structure of crystals With X-ray Diffraction (basic principle) Crystals are made of regular arrays of atoms that effectively scatter X-ray Scattering (or interference) of two X-rays from the crystal planes made-up of atoms Bragg’s Law Crystalline structure of sodium chloride (NaCl). length of the cube edge is a = 0.562 nm. 2 d sin Q = m l m = 1, 2, ..