Lecture 4. Physics 1502: Lecture 34 Today s Agenda


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1 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 singleslit and twoslits diffraction patterns» The diffraction grating Interference 1
2 A wave through two slits In Phase, i.e. Maxima when ΔP = d sinθ = nλ Out of Phase, i.e. Minima when ΔP = d sinθ = (n+1/2)λ d θ ΔP=d sinθ Screen A wave through two slits In Phase, i.e. Maxima when ΔP = d sinθ = nλ + Out of Phase, i.e. Minima when ΔP = d sinθ = (n+1/2)λ + 2
3 The Intensity What is the intensity at P? The only term with a t dependence is sin 2 ( ).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 We can rewrite intensity at point P in terms of distance y The Intensity 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 3
4 Phasor Addition of Waves Consider a sinusoidal wave whose electric field component is E 2 (t) E 1 (t) ωt+φ ωt Consider second sinusoidal wave The projection of sum of two phasors E P is equal to E P (t) E 2 (t) E 1 (t) E R φ/2 ωt φ Phasor Diagrams for Two Coherent Sources E R = ER 90 0 E R = E R =2 ER 4
5 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 5
6 Nslits Interference Patterns Φ=0 Φ=90 Φ=180 Φ=270 Φ=360 N=2 N=3 N=4 Change of Phase Due to Reflection S Lloyd s mirror P 2 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 doubleslit source separated by the distance between points S and I. I L P 1 Mirror 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 An electromagnetic wave undergoes a phase change by upon reflecting from the medium that has a higher index of refraction than that one in which the wave is traveling. 6
7 Change of Phase Due to Reflection n 1 n 2 n 1 n phase change no phase change n 1 <n 2 n 1 >n 2 Interference in Thin Films Air phase change 1 no phase change 2 A wave traveling from air toward film undergoes phase change upon reflection. The wavelength of light λ n in the medium with refraction index n is Film Air t The ray 1 is out of phase with ray 2 which is equivalent to a path difference λ n /2. The ray 2 also travels extra distance 2t. Constructive interference Destructive interference 7
8 Chapter 34 Act 1 Estimate minimum thickness of a soapbubble film (n=1.33) that results in constructive interference in the reflected light if the film is Illuminated by light with λ=600nm. A) 113nm B) 250nm C) 339nm Problem Consider the doubleslit 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? 8
9 Phase difference for going though plastic sheet: Solution 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) Reflection from Optically Lighter Medium (smaller n) 180 o Phase Change No Phase Change by analogy to reflection of traveling wave in mechanics 9
10 Lecture 4 constructive: 2t = (m +1/2) λn destructive: 2t = m λn Examples : constructive: 2t = m λn destructive: 2t = (m +1/2) λn Application Reducing Reflection in Optical Instruments 10
11 Diffraction Experimental Observations: (pattern produced by a single slit?) 11
12 How do we understand this pattern? First Destructive Interference: (a/2) sin Θ = ± λ/2 sin Θ = ± λ/a Second Destructive Interference: (a/4) sin Θ = ± λ/2 sin Θ = ± 2 λ/a m th Destructive Interference: sin Θ = ± m λ/a m=±1, ±2, See Huygen s Principle So we can calculate where the minima will be! sin Θ = ± m λ/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? 12
13 Phasor Description of Diffraction Let s define phase difference (β) between first and last ray (phasor) central max. 1st min. β = Σ (Δβ) = N Δβ (a/λ) sin Θ = 1: 1st min. Δβ / 2π = Δy sin (Θ) / λ 2nd max. β = N Δβ = N 2π Δy sin (Θ) / λ = 2π a sin (Θ) / λ Can we calculate the intensity anywhere on diffraction pattern? Yes, using Phasors! Let take some arbitrary point on the diffraction pattern This point can be defined by angle Θ or by phase difference between first and last ray (phasor) β The resultant electric field magnitude E R is given (from the figure) by : sin (β/2) = E R / 2R The arc length E o is given by : E o = R β E R = 2R sin (β/2) = 2 (E o / β) sin (β/2) = E o [ sin (β/2) / (β/2) ] So, the intensity anywhere on the pattern : I = I max [ sin (β/2) / (β/2) ] 2 β = 2π a sin (Θ) / λ 13
14 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 farfield) θ Lens Incoming wave Screen 14
15 Fresnel Diffraction (or nearfield) Lens P Incoming wave Screen (more complicated: not covered in this course) Resolution (singleslit 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 Θ = λ / a Θ min ~ λ / a 15
16 Resolution (circular aperture) Diffraction patterns of two point sources for various angular separation of the sources Rayleigh s criterion for circular aperture: Θ min = 1.22 ( λ / a) EXAMPLE A ruby laser beam (λ = nm) is sent outwards from a 2.7 m diameter telescope to the moon, km away. What is the radius of the big red spot on the moon? a. 500 m b. 250 m Earth c. 120 m d. 1.0 km e. 2.7 km Θ min = 1.22 ( λ / a) Moon R / = 1.22 [ / 2.7 ] R = 120 m! 16
17 TwoSlit Interference Pattern with a Finite Slit Size Interference (interference fringes): I inter = I max [cos (πd sin Θ / λ)] 2 Diffraction ( envelope function): I diff = I max [ sin (β/2) / (β/2) ] 2 β = 2π a sin (Θ) / λ I tot = I inter. I diff smaller separation between slits =>? The combined effects of twoslit and singleslit interference. This is the pattern produced when 650nm 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 λ? No! d a a 1st minimum interference: d sin Θ = λ /2 1st minimum diffraction: a sin Θ = λ The same place (same Θ) : λ /2d = λ /a a /d = 2 17
18 Application Xray Diffraction by crystals Can we determine the atomic structure of the crystals, like proteins, by analyzing Xray 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 wideband xray spectrum. This enzyme is present in plants and takes part in the process of photosynthesis. Determining the atomic structure of crystals With Xray Diffraction (basic principle) Crystals are made of regular arrays of atoms that effectively scatter Xray Scattering (or interference) of two Xrays from the crystal planes madeup of atoms Bragg s Law Crystalline structure of sodium chloride (NaCl). length of the cube edge is a = nm. 2 d sin Θ = m λ m = 1, 2,.. 18
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