Physics A2 - Lecture 4: Applications of interference and diffraction - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 4: Applications of interference and diffraction - Huynh Quang Linh

1. Lecture 4:Applications of interference and diffraction Rosalind Franklin made the first x-ray diffraction imaging of DNA; her pictures were instrumental in the discovery of the double-helix structure.
2. X-ray Diffraction for Crystallography  If we know about the grating, we can use the diffraction pattern to learn about the light source.  If instead we know about the source, we can use the diffraction pattern to learn about the “grating”.  For this to work, we need to have a source wavelength that is less than the grating spacing (otherwise, there are no orders of diffraction).  Crystals consist of regularly spaced atoms regular array of scattering centers. Typical lattice spacing is 5 angstroms = 5 x 10-10 m = 0.5 nm. use x-rays! Bragg Law for constructive interference: 2d sinq = l d = lattice spacing q = x-ray angle (with respect to plane of crystal) l = x-ray wavelength
3. X-ray Crystallography The Braggs made so many discoveries that Lawrence described the first few years as ‘like looking for gold and finding nuggets lying around everywhere’: • showed that the sodium and chloride ions were not bonded into molecules, but arranged in a lattice • could distinguish different cubic lattices • discovered the crystal structure of diamond • Lawrence Bragg was the youngest Laureate ever (25) to receive a Nobel Prize (shared with his father in 1915) • now standardly used for all kinds of materials analysis, even biological samples! • The same multi-layer interference phenomenon is now used to make highly wavelength-specific mirrors for lasers (“distributed Bragg feedback” [DBF])
4. Diffraction-limited Optics  Diffraction has important implications for optical instruments Lens-making is a craft. Even for a perfectly designed lens, however, the image of a point source will be a little “blurry” due to diffraction in passing through the circular aperture of the lens. Image Image plane 1I1 D plane 0 I q diff(x) 0.5 Point object The image of a point source through a 0 00 circular aperture is 10 00 q 10 q 12.56 x o 12.56 like a single-slit diffraction pattern. The amount of ‘smearing’ of the image is But note the difference: determined by size of the aperture D, and wavelength of incident light, l. l q Slit 0 a l q 1.22 Circular 0 D aperture
5. Transmission of light through slits and circular apertures 1 1I0 diff(x)I0.5 Observation Slit, screen: 0 00 10-l/a 00 l/a 10 q width a 12.56 x 12.56 Monochromatic light 1 source at a great 1I0 distance, or a laser. diff(x)I0.5 Pinhole, Observation 0 00 diameter D screen: 10 0 10 q 12.56-1.22l/D 0x 1.22l/D12.56 1I1 Object at any 0 distance: diff(x)I0.5 Image Plane: Lens, 0 00 diameter D 10 0 10 q 12.56-1.22l/D 0x 1.22l/D12.56 Laser with pinholes Circular-aperture diffraction pattern =“the Airy disk”. Central lobe contains 84% of power.
6. exercise 1: Expansion of a Laser beam In 1985, a laser beam with a wavelength of l = 500 nm was fired from the earth and reflected off the space shuttle Discovery, in orbit at a distance of L = 350 km away from the laser. d D 1. If the (circular) aperture of the laser was D = 4.7 cm, what was the beam diameter d at the space shuttle? 2. To make a smaller spot on the shuttle, what should we do to the beam diameter at the source? a. reduce it b. increase it c. cannot be made smaller
7. Exercise 1: Expansion of a Laser beam In 1985, a laser beam with a wavelength of l = 500 nm was fired from the earth and reflected off the space shuttle Discovery, in orbit at a distance of L = 350 km away from the laser. d D 1. If the (circular) aperture of the laser was D = 4.7 cm, what was the beam diameter d at the space shuttle? -9 Half-angle-width of l 500 10 -5 q 1.22 1.22 1.3 10 radians 84% of diffraction maximum: o -2 D 4.7 10 power is in central -53 d 2 qo L 2(1.3 10 )(350 10 m) 9.1 m lobe. 2. To make a smaller spot on the shuttle, what should we do to the beam diameter at the source? a. reduce it Counter-intuitive as this is, it is correct – you b. increase it reduce beam divergence by using a bigger c. cannot be made smaller beam. (Note: this will work until D ~ d)
8. exercise 2: Focusing of a laser beam There are many times you would like to focus a laser beam to as small a spot as possible. However, diffraction limits this. Dlens Dlaser d f 1. The (circular) aperture of a laser (l = 780 nm) has Dlaser = 5 mm. What is the spot-size d of the beam after passing through a (perfect) lens with focal length f=5mm, diameter Dlens = 6 mm? (Hint: light passing through lens center is undeflected.) 2. Which of the following will reduce the spot size? a. increase l b. decrease l c. increase dlens d. decrease dlens
9. Exercise 2: Focusing of a laser beam There are many times you would like to focus a laser beam to as small a spot as possible. However, diffraction limits this. Dlens Dlaser d f 1. The (circular) aperture of a laser (l = 780 nm) has Dlaser = 5 mm. What is the spot-size d of the beam after passing through a (perfect) lens with focal length f=5mm, diameter Dlens = 6 mm? (Hint: light passing through lens center is undeflected.) The angular spread of the beam is determined qlo 1.22 / D laser by the smaller of Dlaser and Dlens: Light at this angle will intercept the focal plane at d/2 ~ f qo: d 2 qo f 2.44 l f / D laser 2.44(0.78  m)(5mm) /(5mm) 1.9  m 2. Which of the following will reduce the spot size? a. increase l Since the diffraction is already limited by D, b. decrease l increasing dlens doesn’t help. c. increase d There is a huge industry devoted to developing lens cheap blue diode lasers (l ~ 400 nm) for just d. decrease dlens this purpose, i.e., to increase DVD capacity.
10. Angular Resolution  Diffraction also limits our ability to “resolve” (i.e., distinguish) two point sources. Consider two point sources (e.g., stars) with angular separation a viewed through a circular aperture or lens of diameter D. 2 pt sources a D a diffraction disks (not interference maxima) a 2ac a ac a ac/3 y I1 1 1 1 0 1I0 1I0 Rayleigh’s Criterion: define the images diff1diff2 (xI) 0.5 diff2diff1 (xI) 0.5 diff1diff2 (xI) 0.5 to be resolved if a ac , where 0 0 0 00 0 00 00 10 0 10 10 0 10 10 0 10 l 2 12.56 x 12.56 2 12.56 x 12.56 12.56 x 12.56 2I2 2I2 12.56 x 12.56 2 2 ac 1.22 0 0 2I0 D Sumdiff3 (x) diff3 (x) diff3 (x) 1 Sum1 Sum1 At ac the central max of one 0 0 00 00 000 image falls on 10 y0 10 10 y0 10 10 y0 10 12.56 x 12.56 12.56 x 12.56 12.56 x 12.56 the first 2 images ‘Diffraction limit’ 2 images minimum of the resolvable of resolution not resolvable second image
11. Example: Angular resolution Car headlights in the distance: What is the maximum distance L you can be from an oncoming car at night, and still distinguish its two headlights, which are separated by a distance d = 1.5 m? Assume that your pupils have a diameter D = 2 mm at night, and that the wavelength of light is l = 550 nm.
12. Solution Car headlights in the distance: What is the maximum distance L you can be from an oncoming car at night, and still distinguish its two headlights, which are separated by a distance d = 1.5 m? Assume that your pupils have a diameter D = 2 mm at night, and that the wavelength of light is l = 550 nm. l Rayleigh’s ac 1.22 Criterion: D -9 550 10 -4 ac 1.22 3.36 10 2 10-3 Therefore, the maximum distance L is given by: d -4 ac L 1.5m/3.36 10 4500m L x 1mile/1600 m = 2.8 miles for ‘perfect eyes’
13. exercise 3: Resolving Stars Halley’s Comet 1. Assuming diffraction-limited optics (best possible), what is the minimum angular separation of two stars that can be resolved by a D = 5 m reflecting telescope using light of l = 500 nm? a. 0.1 rad b. 1 rad c. 10 rad 2. If the two points are not quite resolved at screen 1, will they be resolved at screen 2? screen 1 screen 2
14. exercise 3: Resolving Stars Halley’s Comet 1. Assuming diffraction-limited optics (best possible), what is the minimum angular separation of two stars that can be resolved by a D = 5 m reflecting telescope using light of l = 500 nm? a. 0.1 rad l -7 ac 1.22 1 10 0.1 rad b. 1 rad D The real limit of earth-bound telescopes is about an order of c. 10 rad magnitude larger due to atmospheric effects (\ the Hubble). 2. If the two points are not quite resolved at screen 1, will they be resolved at screen 2? screen 1 screen 2 NO! Only the angle counts.
15. Optical Interferometers  Interference arises when there are two (or more) ways for something to happen. Ex. Two slits for the light to get from the source to the screen. 2 I = 4I1 cos (/2), with  = 2p d/l, and path-length difference d.  An interferometer is a device using mirrors and “beam splitters” (half light is transmitted, half is reflected) to give two separate paths from source to detector.  Two common types: Mach-Zehnder: Michelson : beam- splitter mirror mirror beam- splitter
16. Optical Interferometers 4I1 4I1 2I 1 4I1 = + 2I1 I = ? Itop = I1 Ibot 2 = |Atop| = I1 2 = |Abot|  If the intensity contribution from each path alone is Itop = Ibot ≡ I1, the net intensity (from both paths) in one output port is again 2 I = 4I1 cos (/2), with  = 2p d/l, and path-length difference d: Can use to measure small displacements. l/2 d 2  The intensity in the other port is 4I1sin (/2). Why??
17. exercise 4 Consider the following Michelson interferometer. d Assume that for the setup shown, all the light (with l = 500 nm) comes out the bottom port. 1. How much does the top mirror need to be moved so that none of the light comes out the bottom port? (a) 125 nm (b) 250 nm (c) 500 nm 2. Where does the light go?
18. exercise 4 Consider the following Michelson interferometer. d Assume that for the setup shown, all the light (with l = 500 nm) comes out the bottom port. 1. How much does the top mirror need to be moved so that none of the light comes out the bottom port? (a) 125 nm (b) 250 nm (c) 500 nm We need to go from complete constructive to complete destructive interference Δ = 180˚ d = l/2 However when we move the mirror by d, we change d by 2d. Therefore, d = d/2 = l/4 = 500/4 = 125 nm. 2. Where does the light go? The light goes out the way it came in. Even in quantum mechanics, energy is still conserved, so the photons can’t just disappear! The Michelson interferometer is perhaps most famous for disproving the hypothesis that EM waves propagate through an “aether” – this result stimulated Einstein to develop the Special Theory of Relativity
19. Questions 1) Could you construct a diffraction grating for sound? If so, what grating spacing is suitable for a wavelength of 0.5 m? 2) Discuss this statement:”A diffraction grating can just as well be called an interference grating”
20. Exercise 5 X-rays of wavelength 0.122 nm are found to reflect in the second order from the face of a lithium fluoride crystal at a Bragg angle of 28.1o. Calculate the distance between adjacent crystal planes.
21. Simulation lab of diffraction Present and discuss
22. exercise 5 Consider the following “Sagnac” [“sahn-yack”] interferometer. Here the two possible paths are the clockwise and counter- fiber clockwise circuits around the fiber loop. loop 1. If we insert an extra piece of glass as shown, how does the relative path length change? It doesn’t! Because the interference paths completely overlap, the Sagnac is a remarkably stable interferometer, e.g., to temperature fluctuations in the fiber. 2. How could we change the relative path-length difference, and thereby change how much light exits the bottom port? Rotate the entire interferometer (in the plane of the paper). For example, if we rotate it clockwise, the light making the clockwise circuit will have farther to go (the beamsplitter is “running away”), while the counterclockwise path will be shortened. It is not difficult to show that 2 2p 4 pR  l c2