Lecture 30: Chapter 36 Monday 27 March 2006

Diffraction II Review: Double Slit Path Differences • Today For point P at angle θ triangle – Single-slit diffraction review shows – Multiple slit diffraction review ∆L = d sinθ – Xray diffraction For constructive interference – Diffraction intensities we need ∆L = mλ where m=0,1,2,… is any integer.

So the bright fringes are at angles given by d sinθ = mλ

Single slit: Pattern on screen Double-slit interference fringes

So the bright fringes are at angles given by d sinθ = mλ

And the dark fringes d sinθ = (m + 1 )λ are at angles given by 2 Bright and dark fringes appear behind a single very thin slit. As the slit is made narrower the pattern of fringes becomes wider.

Dark Fringes Single Slit dark in Diffraction fringes Destructive interference: (a / 2m)sinθ = (λ / 2) asinθ = 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!

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Summary of single-slit diffraction Example: Problem 36-6 The distance between the first and fifth minima of a single- • Given light of wavelength λ passing through a slit of slit diffraction pattern is 0.35 mm with the screen 40 cm width a. away from the slit, with light of wavelength 550 nm. • There are dark fringes (diffraction minima) at Find the slit width. sinθ ≈ tanθ = y / D angles θ given by a sin θ = mλ where m is an integer. asinθ = mλ sinθ = mλ / a • Note this exactly the condition for constructive 5λ 1λ 4λ interference between the rays from the top and sinθ5 − sinθ1 = − = bottom of the slit. a a a 4λD • Also note the pattern gets wider as the slit gets y5 − y1 = Dsinθ5 − Dsinθ1 = narrower. a • The bright fringes are roughly half-way between the 4λD 4 × 550 ×10−9 × 0.4 a = = dark fringes. (Not exactly but close enough.) −4 y5 − y1 3.5 ×10 = 2.5 ×10−3 m = 2.5 mm

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

Multiple-Slit Diffraction Double-Slit Now we can finally put together our Diffraction interference and diffraction results to see what really happens with two or more slits. a = slit width d = slit separation RESULT: We get the two-slit (or multiple-slit) θ = angle on screen pattern as in chapter 35, but modified by the single-slit intensity as an envelope. Bright fringes due to 2-slit interference: θ = mλ / d Instead of all peaks being of the same height, they get weaker at larger angles. Zero due to diffraction: θ = λ / a, 2λ / a, "

2 Lecture 30: Chapter 36 Monday 27 March 2006

Double-slit diffraction Two-slit and one-slit patterns Actual photograph: 2 slits of zero width (a)= two slits 1 slit of width a = 5λ (b)=one list covered

2 slits of width a = 5λ

(Figure 36-15 from text page 1003.)

Example:Sample Problem 36-5 Scaling of diffraction patterns Two slits: d=19.44 nm, a=4.05 µm, λ=405 nm. Notice a common feature of interference and diffraction patterns: The large-scale features of the pattern are determined by (a)How many bright the small-scale regularities of the object, fringes within the and vice-versa. central peak? (b)How many in the Holograms and X-ray diffraction patterns are examples. first side peak?

Example:Sample Problem 36-5 Q.36-2 When red laser light is diffracted by two slits of equal width, there are Two slits: d=19.44 nm, many closely spaced bands of light inside wider a=4.05 µm, λ=405 nm. bands. What features of the slits determine the separation x between the closely-spaced bands? Solution: One-slit: θ = mλ / a x = .10, .20, .30, .04, " (1) Width of the individual slits. Two-slit: (2) Distance between the two slits. θ = mλ / d = .02083m (3) Ratio of distance to width. = .0208, .0416, .0625, .0833, .1042, .1250, " (4) None of the above: it’s more complicated.

3 Lecture 30: Chapter 36 Monday 27 March 2006

Q.36-2 When red laser light is diffracted by two slits of equal width, there are X-Rays many closely spaced bands of light inside wider bands. What features of the slits determine the •X-rays are just light waves with shorter separation x between the closely-spaced bands? wavelengths and higher photon energies. •Since X-ray wavelengths are comparable to atomic sizes, they are perfect for studying atoms x and the arrangement of atoms in crystals.

(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.

X-Rays and Crystals X-ray crystallogrphy A crystal surface acts like a diffraction grating In this way, using xrays of known for X rays. wavelength we can measure the distances between atoms in a crystal and determine the crystal Bragg condition structure. for a bright spot: 2d sinθ = mλ

Single-slit Intensity Phasors for Single Slit •We know where to find the dark fringes in the Break up the slit into many tiny zones, giving many single-slit pattern. But can we calculate the rays of light, which come together on the screen. actual intensity at a general point? •Yes, using the phasor method. 2 •Book gives result on page 998: I  sinα  θ =   I  α  m ∆φ = Phase difference between adjacent rays Here I is the intensity at angle θ on the screen. θ φ = Phase difference between top and bottom rays Im is the intensity at the central maximum. Amplitude at center = Sum of all phasors The angle α= φ /2, and φ is the phase difference Em = between the rays from top and bottom of slit. Eθ = Amplitude at angle θ, get from diagram

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First Maximum and Minimum Intensity for Single Slit Remember of course Em = Rφ Eθ = 2Rsin(φ / 2) the relation between  2π  phase difference and φ =  (asinθ ) E 2sin(φ / 2) λ θ = path difference   Em φ φ = 0 I 4sin2 (φ / 2) θ = Eθ = Em 2 Im φ

Eθ = 0 Which gives the textbook result: 2 φ = m(2π ) I  sinα  θ =   asinθ = mλ Im  α 

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