Lecture 4. Physics 1502: Lecture 34 Today s Agenda

Transcription

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 single-slit and two-slits 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 N-slits 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 double-slit 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 soap-bubble 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 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? 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 far-field) θ Lens Incoming wave Screen 14

15 Fresnel Diffraction (or near-field) Lens P Incoming wave 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 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 Two-Slit 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 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 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 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 = nm. 2 d sin Θ = m λ m = 1, 2,.. 18