Lecture 4. Physics 1502: Lecture 35 Today s Agenda. Homework 09: Wednesday December 9

Transcription

1 Physics 1502: Lecture 35 Today s Agenda Announcements: Midterm 2: graded soon» solutions Homework 09: Wednesday December 9 Optics Diffraction» Introduction to diffraction» Diffraction from narrow slits» Intensity of single-slit and two-slits diffraction patterns» The diffraction grating Diffraction 1

2 Fraunhofer Diffraction (or far-field) θ Lens Incoming wave Screen Fresnel Diffraction (or near-field) Lens P Incoming wave Screen (more complicated: not covered in this course) 2

3 Experimental Observations: (pattern produced by a single slit?) 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 3

4 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? 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? 4

5 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 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. 5

6 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 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) 6

7 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! 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 7

8 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 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. 8

9 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,.. 9