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. Content Multiple-slit Interference formula Diffraction Gratings Optical Spectroscopy Spectral Resolution Single-Slit Diffraction Interference + Diffraction Applications: X-ray Crystallography
3. 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
4. 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.
5. 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”
6. 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
7. 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
8. 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)
9. 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.
10. 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
11. 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.
12. 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
13. 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
14. 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
15. 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)
16. 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!
17. 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
18. 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
19. 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?