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Chapter 36 Diffraction In Chapter 35, we saw how light beams passing through different slits can interfere with each other and how a beam after passing through a single slit flares-diffracts- in Young's experiment. Diffraction through a single slit or past either a narrow obstacle or an edge produces rich interference patterns. The physics of diffraction plays an important role in many scientific and engineering fields. In this chapter we explain diffraction using the wave nature of light and discuss several applications of diffraction in science and technology. 1 Diffraction The bending of light around objects into what would otherwise be a “shadowed” region is known as diffraction. Diffraction occurs when light passes through very small apertures or near sharp edges. geometrical diffracted 2 Definition and Types of Diffraction • Diffraction is the bending of a wave around an object accompanied by an interference pattern • Fresnel Diffraction - curved (spherical) wave front is diffracted • Fraunhofer Diffraction - plane wave is diffracted Fresnel Diffraction from a circular obstruction Fresnel bright spot 3 Single Slit Diffraction We can get an interference pattern when there are two slits. We will also get an interference pattern with a single slit provided its size is approximately λ (neither too small nor too large). Light http://www.phys.hawaii.edu/~teb/optics/java/slitdiffr/ 4 Diffraction and the Wave Theory of Light Diffraction Pattern from a single narrow slit. Side or secondary maxima Light Central maximum Fresnel Bright Spot. These patterns cannot be explained using geometrical optics (Ch. 34)! Light Bright spot 36- 5 Diffraction by a Single Slit: Locating the Minima When the path length difference between rays r1 and r2 is λ/2, the two rays will be out of phase when they reach P1 on the screen, resulting in destructive interference at P1. The path length difference is the distance from the starting point of r2 at the center of the slit to point b. For D>>a, the path length difference between rays r1 and r2 is (a/2) sin θ. a λ sin θ = → a sin θ = λ 2 2 Fig. 36-4 (first minimum) 36- 6 Diffraction by a Single Slit: Locating the Minima, Cont'd Repeat previous analysis for pairs of rays, each separated by a vertical distance of a/2 at the slit. Setting path length difference to λ/2 for each pair of rays, we obtain the first dark fringes at: a λ sin θ = → a sin θ = λ 2 2 (first minimum) For second minimum, divide slit into 4 zones of equal widths a/4 (separation between pairs of rays). Destructive interference occurs when the path length difference for each pair is λ/2. a λ sin θ = → a sin θ = 2λ (second minimum) 4 2 Dividing the slit into increasingly larger even numbers of zones, we can find higher order minima: a sin θ = mλ , for m = 1, 2,3K (minima-dark fringes) Fig. 36-5 36- 7 Intensity in Single-Slit Diffraction 8 Intensity in Single-Slit Diffraction, Qualitatively To obtain the locations of the minima, the slit was equally divided into N zones, each with width ∆x. Each zone acts as a source of Huygens wavelets. Now these zones can be superimposed at the screen to obtain the intensity as function of θ, the angle to the central axis. To find the net electric field Eθ (intensity α Eθ2) at point P on the screen, we need the phase relationships among the wavelets arriving from different zones: phase difference = 2π path length λ difference ∆φ = 2π λ ( ∆x sin θ ) N=18 1st side max. 1st min. θ=0 θ small Fig. 36-6 36- 9 http://surendranath.tripod.com/Applets/Optics/Slits/SingleSlit/SnglSltApplet.html Intensity in Single-Slit Diffraction, Quantitatively Here we will show that the intensity at the screen due to a single slit is: I (θ ) = I m sin α 2 α 1 πa where α = φ = sin θ 2 λ (36-5) (36-6) In Eq. 36-5, minima occur when: α = mπ , for m = 1, 2,3K If we put this into Eq. 36-6 we find: sin α α → 1 as α → 0 πa mπ = sin θ , for m = 1, 2,3K λ or a sin θ = mλ , for m = 1, 2,3K Fig. 36-7 (minima-dark fringes) 36-10 Proof of Eqs. 36-5 and 36-6 If we divide slit into infinitesimally wide zones ∆x, the arc of the phasors approaches the arc of a circle. The length of the arc is Em. φ is the difference in phase between the infinitesimal vectors at the left and right ends of the arc. φ is also the angle between the 2 radii marked R. Eθ The dash line bisecting f forms two triangles, where: sin φ = 2R Em In radian measure: φ = R Em 1 Solving the previous 2 equations for Eθ one obtains: Eθ = sin 2φ 1 1 2 The intensity at the screen is therefore: I (θ ) Eθ2 sin α = 2 → I (θ ) = I m Im Em α 2 φ 2 φ is related to the path length difference across the entire slit: Fig. 36-8 φ= 2π λ ( a sin θ ) 11 12 Distant point source, e,g., star Diffraction by a Circular Aperture d lens sin θ = 1.22 λ d θ (1st min.- circ. aperture) Image is not a point, as expected from geometrical optics! Diffraction is responsible for this image pattern a Light Light sin θ = 1.22 θ λ a (1st min.- single slit) a θ 36-13 Resolution • The ability of an optical system to distinguish between closely spaced objects is limited due to the wave nature of light • Consider two not coherent light sources (like stars) • Because of diffraction, the images consist of bright central regions flanked by weaker bright and dark rings 14 Rayleigh’s Criterion • If the two sources are separated so that their central maxima do not overlap, their images are said to be resolved • The limiting condition for resolution is Rayleigh’s Criterion (Lord Rayleigh, 1842–1919) – When the central maximum of one image falls on the first minimum of another image, they images are said to be just resolved – The images are just resolved when their angular separation satisfies Rayleigh’s criterion 15 Resolvability Rayleigh’s Criterion: two point sources are barely resolvable if their angular separation θR results in the central maximum of the diffraction pattern of one source’s image is centered on the first minimum of the diffraction pattern of the other source’s image. Fig. 36-10 θ R = sin −1 1.22 λ d θ R small ≈ 1.22 λ d (Rayleigh's criterion) 36-16 Additive Primary Colors (adding light) All colors can be formed by adding a combination of the (additive) primary colors: RED, GREEN and BLUE When all three primary colors of light are added together, the result is WHITE light. How does a color TV work? How do you get white, orange or black on your color TV? 17 Subtractive Primary Colors (adding paints) Adding paints or dyes is very different from adding light. In this case it is easiest to think about what colors are not present. The subtractive primary colors are CYAN, MAGENTA and YELLOW These colors can be combined to form all colors. Together these three primary colors absorb all colors of light and make black. 18 Diffraction by a Double Slit Double slit experiment described in Ch. 35 where assumed that the slit width a<<λ. What if this is not the case? Two vanishingly narrow slits a<<λ Single slit a~λ Fig. 36-14 Two Single slits a~λ Diffraction envelope I (θ ) = I m ( cos β ) 2 sin α α 2 (double slit) πd β= sin θ λ πa α= sin θ λ 36-19 Envelope for other interference patterns Intensity equation for a double slit I (θ ) = I m ( cos β ) 2 sin α α 2 (double slit) πd β= sin θ λ πa α= sin θ λ 20 Diffraction Gratings Device with N slits (rulings) can be used to manipulate light, such as separate different wavelengths of light that are contained in a single beam. How does a diffraction grating affect monochromatic light? Fig. 36-17 Fig. 36-18 d sin θ = mλ for m = 0,1, 2K (maxima-lines) 36-21 Width of Lines The ability of the diffraction grating to resolve (separate) different wavelength depends on the width of the lines (maxima) Fig. 36-20 d sin θ = mλ for m = 0,1, 2K (maxima-lines) Fig. 36-19 36-22 Width of Lines, cont’d Nd sin ∆θ hw = λ , ∆θ hw = Fig. 36-21 ∆θ hw = λ Nd λ Nd cos θ sin ∆θ hw ≈ ∆θ hw (half width of central line) (half width of line at θ ) 36-23 Grating Spectrometer • • • S-source, L1-Lens, S1-slit, Ccollimator, L2-Lens, G-grating Light from source focused by L1 on vertical slit S1 placed in focal plane of L2. Emerging light is a plane wave incident on G-grating: diffracted into diffraction patter, with m=0 order diffracted at angle θ = 0 along central axis of the grating Lens L3 of telescope focuses light diffracted at angle θ onto focal plane FF’ within telescope The third order m=3 is not shown for clarity 24 Optically Variable Graphics (OVP) Gratings embedded in device send out hundreds or even thousands of diffraction orders to produce virtual images that vary with viewing angle. Complicated to design and extremely difficult to counterfeit, so makes an excellent security graphic. Fig. 36-25 36-25 Gratings: Dispersion and Resolving Power Dispersion: the angular spreading of different wavelengths by a grating ∆θ D= (dispersion defined) ∆λ m D= (dispersion of a grating) (36-30) d cos θ Resolving Power λavg R= ∆λ (resolving power defined) R = Nm (resolving power of a grating) (36-32) 36-26 Proof of Eq. 36-30 Angular position of maxima Differential of first equation (what change in angle does a change in wavelength produce?) For small angles d sin θ = mλ d ( cos θ ) dθ = md λ dθ → ∆θ and d λ → ∆λ d ( cos θ ) ∆θ = m∆λ ∆θ m = ∆λ d ( cos θ ) 36-27 Proof of Eq. 36-32 λ Rayleigh's criterion for half-width to resolve two lines ∆θ hw = Substituting for ∆θ in calculation on previous slide ∆θ hw → ∆θ → λ N Nd cos θ = m∆λ λ R= = Nm ∆λ 36-28 Dispersion and Resolving Power Compared Table 36-1 Grating N d (nm) θ D (o/µm) R A 10 000 2540 13.4o 23.2 10 000 B 20 000 2540 13.4o 23.2 20 000 C 10 000 1360 25.5o 46.3 10 000 Data are for λ = 589 nm and m = 1 Fig. 36-26 36-29 X-Ray Diffraction X-rays are electromagnetic radiation with wavelength ~1 Å = 10-10 m (visible light ~5.5x10-7 m) X-ray generation X-ray wavelengths to short to be resolved by a standard optical grating Fig. 36-27 mλ −1 (1)( 0.1 nm ) = sin = 0.0019° θ = sin d 3000 nm −1 36-30 X-Ray Diffraction, cont’d Diffraction of x-rays by crystal: spacing d of adjacent crystal planes on the order of 0.1 nm three-dimensional diffraction grating with diffraction maxima along angles where reflections from different planes interfere constructively 2d sin θ = mλ for m = 0,1, 2K (Bragg's law) Fig. 36-28 36-31 X-Ray Diffraction, cont’d interplanar spacing d is relatedto the unit cell dimensaion a0 5d = Fig. 36-29 5 4 2 0 a a0 or d = = 0.2236a0 20 Not only can crystals be used to separate different x-ray wavelengths, but x-rays in turn can be used to study crystals, for example determine the type crystal ordering and a0 36-32