Fresnel Fringes Thermal electron source. Few fringes. http://em-outreach.ucsd.edu/web-course/Sec-I.D/Sec-I.D.html High brightness electron beam from a multi-walled carbon nanotube: Small source many fringes: N. Jonge, et al. Nature 420, 393 (2002) J. Elec. Microscopy, 22 (1973) 141 Maxima when constructive interference occurs between straight beam and scattering beam at edge of sample when defocused. sample α D = Defocus screen x I Fresnel Fringes (W&C 27.7, Fig. 5.13, 9.20; C. Hall, “Intro. to E.M.”) d source • Fringes at sample edges or boundaries seen only in out of focus images • Fresnel zone theory gives the position of fringe maxima: x m = 2( L + r )λ ( n − 5 / 8) n = 1, 2, 3 ... • Would continue indefinitely laterally if electron source was a point d = 0; • If d finite: The distance x at which the fringes disappear will be such d that the path difference ∆ is on the order of a wavelength. Hence: θ x ∆ λ < = L+r d d ( L + r )λ x< d ( L + r )λ d< x r θ sample xm I ∆ L screen So, the smaller the source the larger x and hence the number of fringes n. x Phase Contrast W&C Chap. 27 Electron optics and sample introduce phase differences between transmitted and scattered beams which then are recombined by objective lens, producing lateral amplitude fluctuations. Lattice Fringes (section 27.2) • Fringes that occur with a period comparable to the atomic lattice < 1 nm. • Obtained by allowing more than one beam to interfere by using a larger objective aperture (or no aperture at all) • Assuming a thin sample where we can ignore dynamical scattering effects: ψ = ϕ o ( z )e 2πi ( k •r ) + ϕ g ( z )e 2πi ( k I DI •r ) where k D = k I + g + s = k I + g ' assuming ϕ o ( z ) ≈ A a constant and ϕ g ( z ) = Beiδ where B = π π sin πts and δ = − πts 2 ξ g πs ψ = e 2πi ( k •r ) [ A + Be 2πi ( g '• r +δ ) ] I I = A2 + B 2 + AB[e 2πi ( g '• r +δ ) + e − 2πi ( g '• r +δ ) ] I = A2 + B 2 + 2 AB cos(2πg '•r + δ ) I = A2 + B 2 − 2 AB sin(2πg ' x − πts ) I = A2 + B 2 − 2 AB sin(2πg ' x − πts ) s = 0 only a sinusoidal function parallel to g and x of spacing 1/g’ = dhkl s ≠ 0 then a small shift depending on thickness and |s| No direct relationship to the position of actual atoms in the sample. Need computer simulation to interpret a lattice image and then must have a very thin sample (no absorption, no multiple scattering) – Weak phase object. Good example of problem see Si <110> lattice image: Fig. 27.3 W&C. SF InGaAs InGaAs(14% In)/GaAs(001) <110> TEM cross-section (JEOL 4000, 0.18 point to point resolution) SF has a small shift along the line intercepting the two regions. <111> <112> GaAs Interface somewhere here Lab #1 polycrystalline Aluminum, 200 keV • Example from a Tecnai image taken by Ryan and Shawn Penson (2007 students). • Finer fringes are lattice fringes • Larger ones are moiré fringes from overlapping grains • No fringes visible does not necessarily mean amorphous since grains probably misaligned ZnSe nanowire: zincblende – wurtzite stacking faults. Fringe spacings: 0.35 ± 0.02 nm Fringe spacings: 0.25 ± 0.01 nm (Al d111 = a/√3 = 0.234 nm) Angles between the two rows 70.5o also correct for {111} planes. Moiré Fringes Two overlapping fringe images, translational, rotational or a combination: Translational: ∆gtm = g2 – g1 I = A2 + B 2 − 2 AB sin[2π ( g 2 − g1 ) x − πts ] g1 g2 Rotational: ∆grm = 2gsin(β/2) g1 β I = A2 + B 2 − 2 AB sin [ 4π(g sin (β / 2 ))x − πts] ∆gr g2 GaSb on GaAs Fine fringes perpendicular to g = 004 translational Moiré spacing 2.1 nm Fe/GaAs (001) translational Moiré 10 nm Moiré Fringe and Dislocation Spacing Prediction GaSb/GaAs GaAs: ao = 5.65 A GaSb: ao = 6.10 A Mismatch = f = ∆ a/a = 0.08 = 8% If the interfacial dislocations are pure edge type: a b =  [110] 2  a  a b =   12 + 12 =   2  2 Spacing of dislocations to relieve all mismatch strain along <110>: b a = f 2f 5.6 = = 50Α 0.08 2 D= Black lines are dislocations with average spacing 5 nm. Moiré fringe spacing along the <100> direction is 1.8 nm or ∆ gtm = 1/1.8 = 0.55 nm-1 Predicted Moiré spacing ∆ gtm = gGaAs – gGaSb = (4/0.565)-(4/0.61) = 0.52 nm-1 High Resolution Imaging 1. Our sample consists of a certain density of material ρ(r). The diffracted wave amplitude is mathematically the Fourier transform of ρ(r): 1 F ( ρ ( r )) = 2π +∞ ∫ ρ ( r )e − 2πi∆k • r d 3r −∞ ρ(r) This is just the structure factor again. If ρ(r) has short periodicities (eg. atomic planes) then F contains large reciprocal lattice vectors: ∆k eg. r = 0.2 nm, ∆k = 1/r = 2sinθ/λ = 5 nm-1 2. The electron microscope objective lens adds a phase distortion function to the sample scattering factor called the: contrast transfer function H(∆k ) worse for higher ∆k (larger scattering angles) Adds a phase factor to F: F ( ρ ( r ))e − iH ( ∆k ) Objective lens Diffraction pattern Image: F −1 [F ( ρ (r))e − iH ( ∆k ) ] Contrast Transfer Function H(∆k ) (W&C pg. 463) Each point on the sample will be imaged as a blurred disk with a radius δ(θ) as a function of spherical aberrations (Cs ) and defocus of the objective lens: if ∆f = 0 there still is the objective lens spherical aberation δ (θ ) = ∆fθ + Csθ 3 We average this over all scattering angles to get the average blurred disk: θ ∆fθ 2 Csθ 4 D (θ ) = ∫ δ (θ )dθ = + ( nm ) 2 0 4 Phase angle χ: 2π 2π  ∆fθ 2 Csθ 4    χ= D(θ ) = + λ λ  2 4  1 2 sin θ 2θ = ≈ ≡ u (nm -1 ) d λ λ π∆fλ πCs λ3 (radians) χ= 2 + 4 2d 1 χ = π∆fλu 2 + πCs λ3u 4 2 H ( ∆k ) = 2 A(u ) sin χ (u ) d Surfaces of constant phase Sample and lens Phase varies laterally Where A(u) is an amplitude function related to the objective aperture. Both are sample independent, depends on your microscope objective lens Cs and your choice of ∆ f. Plots of 2sin χ as a function of Cs and ∆f . 2 0 -1 -2 2 4 6 Cs = 1.2 mm ∆f = -58 nm 1 0 -1 -2 0 2 Transfer Function (2 sin 1 mm CsC=s =1.0 mm = -58 ∆f f = -58nmnm 1 Transfer Function (2 sin Transfer Function (2 sin 2 8 0 2 4 6 0 -1 -2 8 -1 -1 0 2 4 6 8 -1 Reciprocal Lattice Vector (u (nm )) Reciprocal Lattice Vector (u (nm )) mm s = 2.0 Cs =C2.0 mm f = -58 nm ∆f = -58 nm 1 Reciprocal Lattice Vector (u (nm )) 2 1 1 0 Cs = 1.2 mm ∆f = 0 -1 -2 2 4 6 8 -1 Reciprocal Lattice Vector (u (nm )) Cs = 1.2 mm ∆f = -63 nm 0 -1 -2 0 2 Transfer Function (2 sin 2 Transfer Function (2 sin Transfer Function (2 sin Tecnai 20 at 200 keV 0 2 4 6 8 -1 Reciprocal Lattice Vector (u (nm )) 1 Cs = 1.2 mm ∆f = -100 nm 0 -1 -2 0 2 4 6 8 -1 Reciprocal Lattice Vector (u (nm )) The ideal shape of sin χ as a function of u is one that is flat over the largest reciprocal frequencies Optimal condition for the best resolution? 1 2 χ = π∆fλu 2 + πCs λ3u 4 dχ = 2π∆fλu + 2πCs λ3u 3 = 0 du 0 = ∆f + Cs λ2u 2 − 2π 1 = π∆fλu 2 + πCs λ3u 4 3 2 Phase angle (combining focus and aberration constant) To find the zero slope relationship between u and ∆f, λ, and Cs (1) (2) Best curve is when χ is near -120° (-2π/3) 1/ 2 ∆f Sch 4  =  Cs λ  3  uSch = 1.5Cs rSch = −1 / 4 − 3 / 4 λ 1 1/ 4 = 0.66Cs λ3 / 4 uSch Combining (1) and (2) gives best focus value called: Scherzer defocus Cross-over u at ∆fSch when χ = 0 Scherzer resolution Tecnai 20: Cs = 1.2 mm; λ = 0.0025 nm: ∆fSch = - 63 nm; uSch = 4.0 nm-1; rSch = 0.25 nm For larger u (smaller real space distances) phase angle rapidly oscillates and will not translate sample information reliably. Minimum contrast is what we have called focus. (28.9) This occurs when: sin χ ~ 0.3 or ∆fMC = - 0.44 (Csλ)1/2 = -25 µm (Tecnai at 200 keV) On the TEM the in situ Fourier analysis of the image that you can use to adjust astigmatism is essentially this contrast transfer function. Bright rings are when 2sin χ = 2 and χ (u) = nπ/2 with n odd and dark rings when 2sin χ = 0 and χ (u) = nπ/2 with n even We can measure the microscope Cs using these fringe spacings. (Fig. 30.6) Transfer Function (2 sin 2 1 0 -1 -2 0 2 4 6 Reciprocal Lattice Vector (u (nm-1)) 8