Physics A2 - Lecture 3: Diffraction & Spectroscopy - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 3: Diffraction & Spectroscopy - Huynh Quang Linh

1. P Lecture 3: Incident Wave (wavelength l) y Diffraction & a Spectroscopy L
2. We will discuss N-split interference (diffraction) and single-split diffraction (2 slits) 2 Multi-slit Interference, I = 4I1cos (f/2) (For point sources, I1 = constant.) and Single-slit Diffraction, I1(q) (For finite sources, I1 = I1(q).) to obtain 2 Total Interference Pattern, I = 4I1(q)cos (f/2) (Remember how f is related to q: f/2p = d/l = (dsinq)/l d q/l)
3. Content  Multiple-slit Interference formula  Diffraction Gratings  Optical Spectroscopy  Spectral Resolution  Single-Slit Diffraction  Interference + Diffraction  Applications: X-ray Crystallography
4. General properties of N-Slit Interference 9I9 1 1616I1 25I1 N=3 N=4 25 N=5 20 10 5 g(Ix) h(x) I h5(Ix) 10 0 0 0 00 0 0 10-2p 00 10 f 10 0 10 f 0 2p -2p 0 2p 10-2p 00 2p 10 f 10-l/d 0x l/d 10 q 10-l/ 0x l/ 10 q d d 10-l/d 0x l/d 10 q • The positions of the principal maxima of the intensity patterns always occur at f = 0, 2p, 4p, [f is the phase between adjacent slits] (i.e., dsinq = ml, m = 0, 1, 2, ). • The principal maxima become taller and narrower as N increases. • The intensity of a principal maximum is equal to N2 times the maximum intensity from one slit. The width of a principal maximum goes as 1/N. • The # of zeroes between adjacent principal maxima is equal to N-1. The # of secondary maxima between adjacent principal maxima is N-2.
5. exercise 1 Light interfering from 10 equally spaced slits initially illuminates a screen. Now we double the number of slits, keeping the spacing constant. What happens to the net power on the screen? a. stays the same b. doubles c. increases by 4
6. exercise 1 Light interfering from 10 equally spaced slits initially illuminates a screen. Now we double the number of slits, keeping the spacing constant. What happens to the net power on the screen? a. stays the same b. doubles c. increases by 4 If we double the number of slits, we expect the net power on the screen to double. How does it do this The location and number of the principle maxima (which have most of the power) does not change. The principle maxima become 4x brighter. But they also become only half as wide. Therefore, the net power (integrating over all the peaks) increases two-fold, as we would expect. We will soon see that we often use such an array of slits (also called a “diffraction grating”) to perform very precise metrology, e.g, spectroscopy, crystallography, etc.
7. N-Slit Interference – Summary  The Intensity for N equally spaced slits is given by: 2 * sin(Nf / 2) IN = I1 sin(f / 2) y q d  As usual, to determine the pattern at the screen (detector plane), we need to relate f to q or y = Ltanq: f d d sinq dq y = = and q 2p l l l L L f is the phase difference between adjacent slits. * Note: we can not be able to use the small angle approximations if d ~ l.
8. Example 1 In an N-slit interference pattern, at what angle qmin does the intensity first go to zero? (In terms of l, d and N.) 0 qmin ? l/d q
9. Example 1 In an N-slit interference pattern, at what angle qmin does the intensity first go to zero? (In terms of l, d and N.) 0 qmin ? l/d q 2 sin(Nf / 2) IIN = 1 has a zero when Nfmin/2 = p, or fmin = 2p/N. sin(f / 2) But f = 2p(d sinq)/l 2pd q/l = 2p/N. Therefore, qmin l /Nd. As the illuminated number of slits increases, the peak widths decrease! This is a general feature: Wider slit features narrower patterns in the “far field”
10. Optical spectroscopy – how we know about the world • Quantum mechanics definite energy levels, e.g., of electrons in atoms or molecules. • When an atom transitions between energy levels emits light of a very particular frequency. • Every substance has it’s own “signature” of what colors it can emit. • By measuring the colors, we can determine the substance, as well as things about it’s surroundings (e.g., temperature, magnetic fields), whether it’s moving (via the Doppler effect), etc. Optical spectroscopy is invaluable in materials research, engineering, chemistry, biology, medicine But how do we precisely measure wavelengths???
11. Interference Gratings the basis for optical spectroscopy Interference gratings (usually called Diffraction gratings) allow us to resolve sharp spectral signals. 2 IN = N I1 N-slit Interference: l1 0 l1/d sin q Shift of the peak: l2 0 l2/d sin q
12. Spectroscopy Demonstration We have set up some discharge tubes with various gases. Notice that the colors of the various discharges are quite different. In fact the light emitted from the highly excited gases is composed of many “discrete wavelengths.” You can see this with the plastic “grating” we supplied. Light source Your eye Hold grating less than 1 inch from your eye. Your view through Room lights must the grating: be turned off! View spectral lines by Put light source looking at about 45o. at left side of If you don’t see grating. anything, rotate grating 90o.
13. Diffraction Gratings (1)  Diffraction gratings rely on N-slit interference.  They simply consist of a large number of evenly spaced parallel slits.  Recall that the intensity pattern produced by light of wavelength l passing through N slits with spacing d is given by: 2 25I sin(Nf / 2 ) 25 1 N=5 I N = I1  y 20 sin(f / 2 )  d q h5(Ix) where: 10 dsinq fp= 2 0 0 10-2p 00 2p 10 l 10 fx 10 L Consider very narrow slits (a << d), so I1 is roughly constant.  The position of the first principal maximum is given by sin q = l/d (can’t assume small q!) Different colors different angles.  Width of the principal maximum varies as 1/N – improves ability to resolve closely spaced lines. Slits Demo 500/550 nm
14. Diffraction Gratings (2)  How effective are diffraction gratings at resolving light of different wavelengths (i.e. separating closely-spaced ‘spectral lines’)?  Concrete example: Na lamp has a spectrum with two yellow “lines” very close together: l1 = 589.0 nm, l2 = 589.6 nm (Dl = 0.6 nm)  Are these two lines distinguishable using a particular grating? 2 IN = N I1 l1 0 l1/d sin q l2 0 l2/d sin q
15. Diffraction Gratings (3)  We assume “Rayleigh’s criterion”: the minimum wavelength separation we can resolve Dlmin  l2-l1 occurs when the maximum of l2 overlaps with the first diffraction minimum of l1. (Dqmin=l/Nd) I = N2I N 1 Dqmin l /d 0 l1/Nd 1 sin q llD sinqq= D dd Dl l Dq min = min d Nd l2/d q “Rayleigh Criterion” Dl 1 Larger N Smaller Dl N = number of illuminated min = min lines in grating. l N (Higher spectral resolution)
16. Diffraction Gratings (4)  We can squeeze more resolution out of a given grating by working in “higher order”. Remember, the principal maxima occur at sinq = ml/d, where m = 1,2,3 designates the “order”. (Dqmin≈ l/Nd still*) You can easily show: Dl 1 min = l Nm First order Second order Third order m = 1 m = 2 m = 3 0 l/d 2l/d 3l/d sin q Larger Nm Smaller Dlmin (Higher spectral resolution) * To be precise: Dqmin= l/(Nd cosq), (but Dl/l = 1/Nm is correct.)
17. Exercise 1: Diffraction Gratings  Angular splitting of the Sodium doublet: Consider the two closely spaced spectral (yellow) lines of sodium (Na), l1 = 589 nm and l2 = 589.6 nm. If light from a sodium lamp fully illuminates a diffraction grating with 4000 slits/cm, what is the angular separation of these two lines in the second- order (m=2) spectrum? Hint: First find the slit spacing d from the number of slits per centimeter.
18. Exercise 1: Solution  Angular splitting of the Sodium doublet: Consider the two closely spaced spectral (yellow) lines of sodium (Na), l1 = 589 nm and l2 = 589.6 nm, mentioned earlier. If light from a sodium lamp illuminates a diffraction grating with 4000 slits/cm, what is the angular separation of these two lines in the second-order (m=2) spectrum? Hint: First find the slit spacing d from the number of slits per centimeter. 1cm d= =2.5 10-4 cm = 2.5  m 4000 -1 l1 q1 =sin m = 28.112 d 11 ll21 Dq =sin mm - sin = 0.031  dd
19. Exercise 2 1. Assuming we fully illuminate the grating from the previous problem (d = 2.5 m), how big must it be to resolve the Na lines (589 nm, 589.6 nm)? (a) 0.13 mm (b) 1.3 mm (c) 13 mm 2. How many interference orders can be seen with this grating? (a) 2 (b) 3 (c) 4 3. Which reduces the maximum number of interference orders? (a) Increase wavelength (b) Increase slit spacing (c) Increase number of slits
20. exercise 2 1. Assuming we fully illuminate the grating from the previous problem (d = 2.5 m), how big must it be to resolve the Na lines (589 nm, 589.6 nm)? (a) 0.13 mm (b) 1.3 mm (c) 13 mm We need enough lines to narrow the diffraction peak: Dl 1 l 589nm size = N d = N = = 491 490 (2.5 m) l Nm m Dl 2(0.6nm) 1.2 mm 2. How many interference orders can be seen with this grating? (a) 2 (b) 3 (c) 4 The diffraction angle can never be more than 90˚: From sinq = ml/d, it must be that ml ≤ d, or m ≤ d/l = 2.5 m/0.589m = 4.2 m = 4 3. Which reduces the maximum number of interference orders? (a) Increase wavelength Increase l, or decrease d. (b) Increase slit spacing Changing the number of slits does (c) Increase number of slits not affect the number of orders.
21. Single-slit “Diffraction” Laser Demo simulation  So far in the N-slit problem we have assumed that each slit is a point source.  Point sources radiate equally in all directions.  Real slits have a non-zero extent – - a “slit width” a. The transmission pattern depends on the ratio of a to l.  In general, the smaller the slit width, the more the wave will diffract! Small slit: Diffraction Large slit: Diffraction profile profile Laser Light Laser Light (wavelength l) (wavelength l) I 1 I1 screen screen Let’s examine this effect quantitatively.
22. Single-Slit Diffraction P  Slit of width a. Where are the minima? Incident Wave (wavelength l) y  The first minimum is at an angle such that the light from the top and the middle of the slit destructively interfere: a l dq==sin a q min a/2 22 l d =sinq L min a  The second minimum is at an angle such that the light from the top and a point at a/4 destructively interfere: a l Location of nth-minimum: q dq==sin a/4 42min,2 nl sinq = (n = 1, 2, ) 2l min,n a d =sinq min,2 a
23. Diffraction: Exercise 3 Suppose that when we pass red light (l = 600 nm) through a slit of width a, the width of the spot (the distance between the first zeros on each side of the bright peak) is W = 1 cm on a screen that is L = 2m behind the slit. How wide is the slit? a 1 cm = W 2 m
24. Exercise 3: solution Suppose that when we pass red light (l = 600 nm) through a slit of width a, the width of the spot (the distance between the first zeros on each side of the bright peak) is W = 1 cm on a screen that is L = 2m behind the slit. How wide is the slit? a q 1 cm = W L = 2 m Solution: The angle to the first zero is: q = ± l/a W = 2 Ltan q  2 Lq = 2Ll/a (use tan q  q) Solve for a: a = 2Ll/W = (4m)(6 10-7 m) /(10-2m) = 2.4 10-4 m = 0.24 mm
25. Exercise 4 a 1 cm = W 2 m Which of the following would broaden the diffraction peak? a. reduce the laser wavelength b. reduce the slit width c. move the screen further away
26. Exercise 4: solution a 1 cm = W 2 m Which of the following would broaden the diffraction peak? a. reduce the laser wavelength b. reduce the slit width c. move the screen further away
27. Single-slit Diffraction — Summary The intensity of a single slit has the following form: 2 d q q sin(b / 2) a = a sin a Screen I1 = I0 b / 2 (far away) q 1I01 a diff( x)I0.5 P 0 00 b -4p 10 -2p 0 0 2p 10 4p L q 12.56-l/a 0x l/a 12.56 At P, the phase difference b between 1st and last source is b = angle between 1st and last phasor given by: b d a sin q aq = a = y = Lq Single Slit Diffraction Features: 2p l l l  First zero: b/2 = p q l/a Therefore, 2 (agrees with phasor analysis) sin(paq / l) I1(q ) = I0  Secondary maximum is quite small. paq / l
28. Summary: Interference + Diffraction Combine: 2 sin(Nf / 2) Multi-slit Interference, IN = I1 sin(f / 2) and (for plane-wave sources, I1 = constant) 2 Single-slit Diffraction, sin( b / 2) I1 = I0  b / 2 to obtain  Total Interference Pattern, 2 2 sin( b / 2 ) sin( Nf / 2 ) I = I0   b / 2  sin( f / 2 )  Remember: f/2p = d/l = (d sinq)/l d q/l f = angle between adjacent phasors b/2p = da/l = (a sinq)/l a q/l b = angle between 1st and last phasor
29. Exercise 5 Imax9 Light of wavelength l is incident on an N-slit system with slit width a and slit spacing d. intensityI(x) 5 1. The intensity I as a function of y at a viewing screen located a distance L from the slits is 000 shown to the right. What is N? (L >> d, y, a) -6 00 +6 18.84 Yx (cm) 18.84 (a) N = 2 (b) N = 3 (c) N =4 2. Now the slit spacing d is halved, but the slit width a is kept constant. Which of the graphs best represents the new intensity distribution? I I I max9 (a) max9 (b) max9 (c) 5 5 5 intensityI(x) intensityI(x) intensityI(x) 0 00 000 00 0 0 -6 00 +6 -6 00 +6 -6 0 +6 9.42 x 9.42 18.84 Yx (cm) 18.84 9.42 Yx (cm) 9.42 Y (cm)
30. Exercise 5: solution I Light of wavelength l is incident on an N-slit max9 system with slit width a and slit spacing d. intensity (x) 5 1. The intensity I as a function of y at a viewing I screen located a distance L from the slits is 000 shown to the right. What is N? (L >> d, y, a) -6 00 +6 18.84 Yx (cm) 18.84 (a) N = 2 (b) N = 3 (c) N =4 2. Now the slit spacing d is halved, but the slit width a is kept constant. Which of the graphs best represents the new intensity distribution? I I I max9 (a) max9 (b) max9 (c) 5 5 5 intensityI(x) intensityI(x) intensityI(x) 0 00 000 00 0 0 -6 00 +6 -6 00 +6 -6 0 +6 9.42 x 9.42 18.84 Yx (cm) 18.84 9.42 Yx (cm) 9.42 Y (cm) Decreasing d will increase The spacing between maxima This one does it all. spacing between maxima. is increased, but diffraction Increased spacing profile shouldn’t expand, as between maxima and seen here. constant diffraction.
31. Diffraction from Crystals (1)  Diffraction gratings are excellent tools for studying visible light because the slit spacing is on the order of the wavelength of the light (~ few tenths of microns)  Visible light is a very small part of the spectrum of electromagnetic waves. How can we study e-m waves with smaller wavelengths (e.g., x-rays with l ~ 10-10 m)?  We can’t use a standard diffraction grating to do this. Why? Calculate q for first order peak for x-rays with l = 10-10 m for a grating with d=1000 nm q = l/d = 10-4 = 0.1 mrad The first-order peak is too close to the central maximum to be measured!
32. Diffraction from Crystals (2)  Solution? We need a “diffraction” grating with a spacing d that is on the order of the wavelength of e-m waves we’d like to measure - i.e., for x-rays: l~10-10m  Where do we find such a “grating”? Nature to the rescue!!! Crystalline solids (having regularly spaced atoms with d ~ 10-10m) naturally occur. a 0 X-rays Diffraction: A Primer  Illuminate crystal with x-rays.  The X-rays are scattered by the ions.  Enhanced scattering at certain scattering a0 angles reveal constructive interference between the scattered waves from ion different regularly-spaced planes of atoms. a0  Modern Use: use x-rays to study crystal structure, e.g., DNA
33. X-ray Diffraction for Crystallography (FYI)  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
34. X-ray Crystallography, misc. 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])
35. Appendix: Single-slit Diffraction  To analyze diffraction, we treat it as da = a sinq aq interference of light from many sources (i.e. Screen (far away) the Huygens wavelets that originate from each point in the slit opening). q a  Model the single slit as M point sources with spacing between the sources of a/M. We will P let M go to infinity on the next slide. L  The phase difference b between first and last L >> a implies rays are source is given by b/2p = da/l = a sinq / l aq/l . parallel. Destructive interference occurs when the polygon is closed (b = 2p): A1 (1 slit) a sinq This means =1 b Mfa l For small q , l A q a a (1 source) f Slits Demo a Destructive 10-slits Interference
36. Single-slit Diffraction — the math  We have turned the single-slit problem into the M-slit problem that we already solved in this lecture. b/2 A1  However, as we let M , the problem becomes much R Ao simpler because the polygon becomes the arc of a circle. 2p  The radius of the circle is determined by the relation b = aq l between angle and arc length: b = Ao/R. Trigonometry: A1/2 = R sin(b/2) With R = Ao/b, A1 = (2Ao/b) sin(b/2) sin(b / 2 ) A = A 1 0 b / 2 Intensity is related to amplitude: I = A2. So, here’s the final answer: 2 sin(b / 2) I1 = I0 Graph this function b / 2 Remember: b/2p = da/l = (a sinq)/l a q/l b = angle between 1st and last phasor
37. Homework # 2  Single slit diffraction of the light with the wavelength of 600 nm. Determine the width of central light band (central maximum) obtained on the screen which is located 1 m far from the slit. The width of slit is of 0.1 mm.  N-slit interference: Light of wavelength l (=500 nm) is incident on an N-slit system with slit width a and slit spacing d. Right after slits the lens with focus distance f= 1m is located. The screen for observation of the interference is located 1m far away from the slits. If we found that the distance between 1th two maximum is of 0.202 m. Determine the slit spacing.  Why the diffraction of sound waves more evident in daily experience than that of light waves?  Why radio waves diffract around buildings, although light waves do not?