Chapter 38: Diffraction (interference part 2)
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1 Chapter 38: Diffraction (interference part 2) Diffraction is an interference effect like in Ch 37, but usually refers more specifically to bending of waves around obstacles (similar to refraction). Diffraction also manifests itself when waves from a large number (or even continuous set) of sources interfere. This happens when light illuminates diffraction gratings, many slit screens, apertures, or even crystals. 1
2 Diffraction from a single slit What if slit width a >> λ? 2
3 Interference of light and bending of light around obstacles: NOT part of our everyday (pre-technological) experience Why? Interference effects require coherent light sources Diffraction most significant when obstacle sizes are comparable to wavelengths Hard to satisfy conditions for interference / diffraction in laboratories before the 19 th century! 3
4 Diffraction of light around a razor blade 4
5 Diffraction geometry d 1 d 2 s i Fresnel diffraction (near-field): s d n Fraunhofer diffraction (far-field): i d n s >> d n i >> d n Fraunhofer is simpler since rays going to same point 5 on screen parallel
6 Fraunhofer diffraction 6
7 Fresnel diffraction Poisson spot of a penny 7
8 Diffraction Gratings Spectroscope (using a diffraction grating to separate wavelength components of light) Spectrum of a a sodium-type street lamp 8
9 Multi-slit diffraction Diffraction grating: Generalization from double-slit experiment to multi-slit experiment. principal maxima: dsin θ = mλ where m= 0, ± 1, ± 2 m is the order of the principal maxima λ note θ increases (pattern spreads) as increases d 9
10 Relationship between height and width of diffraction peak If there was no interference: illumination by N slits (each has intensity I 0 ) => average over entire screen NI 0 Due to conservation of energy, average over entire screen is still even with interference NI 0 Intensity of a maximum (with maximum constructive interference from N slits) E max = NE 0 I = max => 2-slit interference: I max = 4I0 Angular separation between successive maxima: m+ 1 θm sinθm+ 1 sin Assume each maximum is θ N θ θ = m 2 wide I 0 λ d θ I max I avg λ/d 10
11 θ I avg Assume now that the intensity in a single maximum is equal to the average intensity in : I max λ/d θ NI0 λ d λ / d With θ = I = max NI dn λ I N 2 0 = 2 0 I 0 1 N λ d we get: The bigger N (the number of slits), the taller, and narrower the peaks. 11
12 For multiple slits: I ( β ) 2 sin N = I 0 2 ( β ) ( β β ) β ( sin β ) Intensity Pattern sin β 2 π d sin θ 2 β = λ note that as θ 0, β 0, sin β β β is HALF the slit-to-slit phase difference N β 2 I = I 0 = I0N β Check for two-slit system ( N = 2): I = I sin 2 2sin cos = I sin = 4 I 2 cos β 12
13 Heightening and narrowing of diffraction peaks as N increases Compare with Michelson (single interference) and Fabry-Perot (multiple interference) interferometers 13
14 Resolution of Diffraction Gratings Angular Dispersion: change in angular separation due to different wavelength dsinθ = mλ d [ dsinθ] = [ mλ] ( cos ) = m θ m = (not same θ as in previous slide!) λ d cosθ note angular that angular dispersion dispersion increases increases (improves) (improves) as order m increases Resolution: important is the width of the maxima and the separation Definition: θ θ λ as order m increases R λ λ λ : smallest observable wavelength difference For N-slit system: R = mn Resolution increases (improves) as: order m increases number of slits/lines N increases 14
15 Example 38-1: sodium doublet at nm and nm a) how many slits required to resolve doublet? b) screen is 4 m away, grating has 2000 slits/cm, what are positions of of two priniple max. of first order. 15
16 Single-slit diffraction Destructive interference: when ends of slit differ by integer number of λ mλ sin θ = where m =± 1, ± 2, ± 3 a what about m = 0? note θ increases (pattern λ spreads) as increases a 16
17 Single-slit intensity pattern Minima at sinθ=(mλ)/a Maxima approximately halfway between minima 17
18 Example 38-3: a=0.10 mm, λ=633nm, screen is 3 m away from slit What is distance between minima on either side of central maximum? 18
19 Intensity Pattern for Single-Slit Diffraction 2 sin α I = Imax 2 α πa sin θ α= λ for θ=0 recall: sin α α lim = = 1 α 0 α α for minima: πa sin θ α= nπ= λ λ sin θ= n, a where n =± 1, ± 2, ± 3 19
20 Example 38-4: intensity ratios of 1st and 2nd maxima to the intensity of central maximum for single slit? 20
21 How to get single-slit intensity pattern Nd = a sin ( Nβ) = lim I0 N sin β a π sin θ πd sin θ N α β = = = 0 as N λ λ N sin α sin α 2 sin I = lim I = I = N I sin ( ) N α ( α ) N N I 21 α 2 α
22 Diffraction and Resolution S point D Aperture of optical system I Aperture Demo 39-1 Airy disk 22
23 θ min = 1.22 λ λ D D Rayleigh Criterion: Two point sources are just resolved if the peak of the diffraction image of the first source overlies (and is no closer than ) the 23 first minimum of the second source
24 L S min = min L θ Lλ D with θ min λ D 24
25 Example 38-5: min separation between two objects so that human eye can distinguish them at a) near point (25 cm)? b) 5 m Pupil diameter=2.5 mm 25
26 Example 38-6: Hubble telescope D=2.4 m, 600 km above earth a) θ min for visible (λ=550 nm) light? b) ideal S min for two objects on earth s surface 26
27 Slit width and grating patterns I single I mult I mult I single sin sin α I = Imult Isingle = sinβ α πdsin θ πasin θ β= α= λ λ ( ) 2 2 Nβ Demo
28 Diffraction pattern for multiple slits where d=10a note missing orders 28
29 X-ray diffraction Recall that θ 0 when λ /d 0 therefore to see an effect on light due to an intermediate object where d is small, λ must be small => For diffraction on crystals use X-rays θ 3 NaCl (table salt) crystal d 0.1 nm 29
30 Bragg condition for constructive Bragg interference spot on screen Bragg's Law: 2dsin θ= nλ where n= 1, 2,3, Constructive interference Bragg peak 30
31 Bragg Planes 31
32 X-ray diffraction pattern from crystallized DNA 32
33 von Laue method for x-ray diffraction 33
34 Example 38-7: rock salt d=0.282 nm, what wavelengths will appear in first and second orders at 25 o? 34
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