# Diffraction. Today Single-slit diffraction Diffraction by a circular aperture Use of phasors in diffraction Double-slit diffraction

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1 Diffraction Today Single-slit diffraction Diffraction by a circular aperture Use of phasors in diffraction Double-slit diffraction

2 Diffraction by a single slit

3 Single slit: Pattern on screen Bright and dark fringes appear behind a single very thin slit. As the slit is made narrower the pattern of fringes becomes wider.

4 Single Slit: finding the minima Divide the single slit into many tiny regions. If the top ray and the middle ray interfere destructively, then every pair of rays will also. So each pair cancels each other and we have total cancellation!

5 Single Slit: the first minimum So we get destructive interference if: ( a / 2) sinθ ( λ / 2) asinθ λ Note this mustn t be confused for the condition for constructive interference in the two-slit case! Also note as slit width gets smaller, angle gets larger, and vice versa.

6 Dark Fringes in Diffraction Second dark fringe? First dark fringe.

7 Dark fringes First is at ( a / 4) asinθ asinθ λ As we move up on the screen the next dark fringe occurs when the first ray interferes destructively with the one from one-fourth the way down the slit. sinθ λ / 2 2λ We can continue to one-sixth, one-eighth, etc. to get all the dark fringes at a sinθ mλ Again note as slit gets narrower, pattern gets wider.

8 Summary of single-slit diffraction Given light of wavelength λ passing through a slit of width a. There are dark fringes (diffraction minima) at angles θ given by a sin θ mλ where m is an integer. Note this exactly the condition for constructive interference between the rays from the top and bottom of the slit. Also note the pattern gets wider as the slit gets narrower. The bright fringes are roughly half-way between the dark fringes. (Not exactly but close enough.)

9 Example: Problem 37-2 Light of wavelength 441 nm is incident on a narrow slit. On a screen 2 meters away, the distance between the second diffraction minimum and the central maximum is 1.5 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit.

10 Example: (cont d) (a) Calculate the angle of diffraction θ of the second minimum. θ tanθ y D rad (b) Find the width of the slit. λ m a m µ m θ asinθ θ mλ sinθ λ m a

11 Diffraction by a circular aperture Consider a round hole of diameter d. Same idea as a long slit only the geometry is different. Angle for first minimum: Long slit: a sinθ λ Circular hole: d sin θ λ

12 The Rayleigh Criterion Using a circular instrument (telescope, human eye), when can we just resolve two distant objects? When images are separated by distance to first minimum: θ 1.22λ / R d

13 Sample Problem 36-4 Camera or eye: Smallest resolvable angle to two distant objects: θ 1.22λ / R d

14 Single slit phasor diagram Divide slit into many tiny slits. Use a tiny phasor for each. Add them together graphically. See we get destructive interference if first and last phasors interfere constructively! So the condition for the m th dark fringe is: a sinθ mλ

15 Multiple-Slit Diffraction Now we can finally put together our interference and diffraction results to see what really happens with two or more slits. RESULT: We get the two-slit (or multiple-slit) pattern as in chapter 35, but modified by the single-slit intensity as an envelope. Instead of all peaks being of the same height, they get weaker at larger angles.

16 Double-Slit Diffraction a slit width d slit separation θ angle on screen Bright fringes due to 2-slit interference: θ θ λ / a, 2λ / a, m λ / Zero due to diffraction: L d

17 Double-slit diffraction 2 slits of zero width 1 slit of width a 5λ 2 slits of width a 5λ

18 Diffraction grating: Many Slits Very sharp maximum when all rays are in phase. d sinθ mλ m order 1, 2, 3, First-order maximum: sinθ λ / d Second-order maximum: sinθ 2 λ / d

19 Diffraction grating bright lines d sinθ mλ Second-order maximum: First-order maximum: Spectroscopy: separate lines of different wavelengths.

20 Diffraction grating recap Position of lines is determined by separation of rulings. d sinθ mλ Sharpness of lines is determined by number of rulings. θ λ / Nd Resolving power is determined by number of rulings and order of line. RmN

21 Interference with 3 Slits Path difference between rays from adjacent slits: L d sinθ θ L / d Phase difference between rays from adjacent slits: φ L 2 2 π λ π d θ λ Get intensity from phasor diagram:

22 3 Slits Continued 1. Central maximum: 2. First minimum: φ Next maximum: φ 180 φ θ 0 0 E 3E T 2π φ 3 E T 0 I T I 9 0 T d sinθ T E0 IT 0 φ π d sinθ λ / E I 0 I 0 λ 3 2

23 3 Slits Continued 4. Next minimum: φ sin E T I T d λ θ π φ 5. Next maximum: d I I E E d T T / 9 3 sin λ θ λ θ π φ φ 360

24 Summary 9I 0 I 0 θ L / d L 0 λ 3 λ 2 2λ 3 λ θ max m λ / d

25 Five Slits θ m λ / Note we still have max But more slits makes the peaks sharper. d For many slits, we get a diffraction grating.

26 Diffraction Grating

27 Wavelength Resolution of a Grating

28 Width of sharp lines

29 Resolving Power Positions of maximums: d sinθ m λ Small angles: θ θ m λ / m d λ / d Widths of sharp maximums: Wavelengths just resolved: Resolving power definition: R λ λ So we get: θ λ / m m λ λ / d λ / Nd λ N RmN / Nd

30 Example: Yellow sodium vapor lines Problem The strong yellow lines in the sodium spectrum are at wavelengths nm and nm. How many rulings are needed in a diffraction grating to resolve these lines in second order? We need R λ λ 589 nm 0.6 nm 982 R 982 But RmN so N 491 m 2

31 Diffraction II Today Single-slit diffraction review Multiple slit diffraction review Xray diffraction Diffraction intensities

32 Review: Double Slit Path Differences For point P at angle θ triangle shows Ld sinθ For constructive interference we need L m λ where m0,1,2, is any integer. So the bright fringes are at angles given by d sinθ mλ

33 Double-slit interference fringes So the bright fringes are at angles given by d sinθ m λ And the dark fringes are at angles given by d 1 sinθ ( m+ ) 2 λ

34 Single slit: Pattern on screen Bright and dark fringes appear behind a single very thin slit. As the slit is made narrower the pattern of fringes becomes wider.

35 Dark Fringes in Diffraction

36 Single Slit dark fringes Destructive interference: ( a / 2m) sinθ ( λ / 2) a sinθ mλ Don t confuse this with the condition for constructive interference for two slits! In fact, note that there is a dark fringe when the rays from the top and bottom interfere constructively!

37 Summary of single-slit diffraction Given light of wavelength λ passing through a slit of width a. There are dark fringes (diffraction minima) at angles θ given by a sin θ mλ where m is an integer. Note this exactly the condition for constructive interference between the rays from the top and bottom of the slit. Also note the pattern gets wider as the slit gets narrower. The bright fringes are roughly half-way between the dark fringes. (Not exactly but close enough.)

38 Example: Problem 36-6 The distance between the first and fifth minima of a singleslit diffraction pattern is 0.35 mm with the screen 40 cm away from the slit, with light of wavelength 550 nm. Find the slit width. asinθ mλ sinθ mλ / a 5λ 1λ 4λ sinθ5 sinθ1 a a a 4λD y5 y1 Dsinθ5 Dsinθ1 a sin θ tanθ y / D a 4λD y y m 2.5 mm

39 Q.36-1 A slit of width 50 µm is used with monochromatic light to form a diffraction pattern. The distance between dark fringes on a distant screen is 4 mm. If the slit width is increased to 100 µm, what will be the new distance between dark fringes? Give your answer in mm. (In the range 0-9.)

40 Q.36-1 A slit of width 50 µm is used with monochromatic light to form a diffraction pattern. The distance between dark fringes on a distant screen is 4 mm. If the slit width is increased to 100 µm, what will be the new distance between dark fringes? Pattern size is inversely proportional to slit size: 2 times slit width means (1/2) times the distance between fringes. Answer: 2 mm.

41 Multiple-Slit Diffraction Now we can finally put together our interference and diffraction results to see what really happens with two or more slits. RESULT: We get the two-slit (or multiple-slit) pattern as in chapter 35, but modified by the single-slit intensity as an envelope. Instead of all peaks being of the same height, they get weaker at larger angles.

42 Double-Slit Diffraction a slit width d slit separation θ angle on screen Bright fringes due to 2-slit interference: θ θ λ / a, 2λ / a, m λ / Zero due to diffraction: L d

43 Double-slit diffraction 2 slits of zero width 1 slit of width a 5λ 2 slits of width a 5λ

44 Two-slit and one-slit patterns Actual photograph: (a) two slits (b)one list covered (Figure from text page 1003.)

45 Scaling of diffraction patterns Notice a common feature of interference and diffraction patterns: The large-scale features of the pattern are determined by the small-scale regularities of the object, and vice-versa. Holograms and X-ray diffraction patterns are examples.

46 Example:Sample Problem 36-5 Two slits: d19.44 nm, a4.05 µm, λ405 nm. (a)how many bright fringes within the central peak? (b)how many in the first side peak?

47 Example:Sample Problem 36-5 Two slits: d19.44 nm, a4.05 µm, λ405 nm. Solution: One-slit: θ mλ / a.10,.20,.30,.04, L Two-slit: θ mλ / d.02083m.0208,.0416,.0625,.0833,.1042,.1250, L

48 Q.36-2 When red laser light is diffracted by two slits of equal width, there are many closely spaced bands of light inside wider bands. What features of the slits determine the separation x between the closely-spaced bands? x (1) Width of the individual slits. (2) Distance between the two slits. (3) Ratio of distance to width. (4) None of the above: it s more complicated.

49 Q.36-2 When red laser light is diffracted by two slits of equal width, there are many closely spaced bands of light inside wider bands. What features of the slits determine the separation x between the closely-spaced bands? x (1) Width of the individual slits. (2) Distance between the two slits. (3) Ratio of distance to width. (4) None of the above: it s more complicated.

50 X-Rays X-rays are just light waves with shorter wavelengths and higher photon energies. Since X-ray wavelengths are comparable to atomic sizes, they are perfect for studying atoms and the arrangement of atoms in crystals.

51 X-Rays and Crystals A crystal surface acts like a diffraction grating for X rays. Bragg condition for a bright spot: 2d sinθ mλ

52 X-ray crystallogrphy In this way, using xrays of known wavelength we can measure the distances between atoms in a crystal and determine the crystal structure.

53 Single-slit Intensity We know where to find the dark fringes in the single-slit pattern. But can we calculate the actual intensity at a general point? Yes, using the phasor method. Book gives result on page 998: I θ I m sinα α 2 Here I θ is the intensity at angle θ on the screen. I m is the intensity at the central maximum. The angle α φ /2, and φ is the phase difference between the rays from top and bottom of slit.

54 Phasors for Single Slit Break up the slit into many tiny zones, giving many rays of light, which come together on the screen. φ Phase difference between adjacent rays φ Phase difference between top and bottom rays E Amplitude at center Sum of all phasors m E Amplitude at angle θ, get from diagram θ

55 First Maximum and Minimum Remember of course the relation between phase difference and path difference φ 2π λ ( a sin θ ) φ 0 E θ E m E θ 0 φ m(2 π ) a sin θ m λ

56 Intensity for Single Slit E E Rsin( / 2) m Rφ θ 2 φ E φ θ 2sin( / 2) φ E m I θ I m 4sin I m 2 φ ( φ / 2 2) Which gives the textbook result: 2 I sinα θ α

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