Chapter 36 Diffraction by C.-R. Hu

Transcription

1 Chapter 36 Diffraction by C.-R. Hu 1. Meaning of the term diffraction The term diffraction means that light does not simply travel in straight lines in a uniform medium, as the ray method seems to tell you. Rather, light can actually get around obstacles, and reach the regions behind the obstacles. The underlining reason is because light is a wave, and obeys Huygen's principle. For this reason, if light shines on a circular hole of a diameter D on a wall (or a slit of width a ), it will not simply form a beam (or sheet) of light of diameter D (or thickness a). Rather, the result will depend on the ratio (D or a)/λ and how far away the exit light is examined. This is illustrated by the following figure (but note that one figure cannot say it all): Incoming plane wave D or a Intensity variation according to ray method (geometric optics). Actual intensity variation near the hole or slit if (D or a) > λ. width of central peak not simply equal to (D or a). Actual intensity variation far from the hole or slit if (D or a) > λ

2 As a matter of fact, the width of the central peak in the bottom curve can be smaller or larger than (D or a). It is essentially given by the combination (λ/d) R or (λ/a) R, if R is much bigger than (D or a), where R is the distance between the slit and where light intensity is examined. (See below.) The behavior at far distance (large R) is called Fraunhofer diffraction, whereas the behavior at short distance (small R) is called Fresnel diffraction.. Diffraction by a single slit A simple case to analyze is a plane light wave diffracted by a single slit of width a, and viewed very far away from the slit. One can imagine a screen at distance R >> a from the slit, and we wish to determine the intensity variation on this screen. Qualitatively, it is given by the third curve plotted above, but we can now analyze it quantitatively, and also understand why it is so. Since the screen is very far away, light rays reaching from all points in the slit to any single point on the screen can be regarded as essentially all parallel to each other, and making the same angle θ from the forward direction. However, they will have different phases because they have traveled different distances. The following figure explains the essential point: a B θ C 90± a sin θ θ Note: The slit is perpendicular to the paper in this figure all to a single point on a screen far away A Let θ 1 be a special value of θ so that light from the slit traveling in this particular direction will reach the first minimum on the screen. We have drawn 11 equally spaced, parallel light-rays from the slit in the above figure. (The number 11 is arbitrarily chosen, except that it must be a

3 fairly large odd number, so that the slit is divided into an even number of very narrow regions. Here the number is 10.) We see that ray 1 is longer than ray 11 by the length BC = a sin θ 1, whereas ray 6 is longer than ray 11 by half as much. Similarly, ray 5 is long than ray 10 by that much. Ray 4 is longer than ray 9 by that much. Ray 3 is longer than ray 8 by that much. Ray is longer than ray 7 by that much. Ray 1 is longer than ray 6 by that much. Thus if only a sin θ 1 = 1λ, light from regions 1,, 3, 4, 5 will cancel out light from regions 6, 7, 8, 9, 10, respectively. That is, light from the whole lower half of the slit cancel out light from the whole upper half of the slit, because differing by (1/) λ in light path-length means differing by 180± in phase. Similarly, an angle θ > θ 1 will give the direction of light toward the second minimum on the screen if a sin θ = λ. All you need is to divide the slit into four equal parts in order to see the same total cancellation. (That is, part 1 will cancel part, and part 3 will cancel part 4, but parts 1 and will no longer cancel parts 3 and 4, so dividing the whole slit into two parts will no longer work!) This argument can be generalized to the general condition: a sin θ = mλ (for the m th minimum on the screen). Here m = 1,, 3,, for the 1 st, nd, 3 rd, minima on the two sides of the forward direction. (But not 0 which actually corresponds to the tallest maximum in the forward direction.) For a given m, you have to divide the slit into m regions in order to prove the total cancellation. Notice that this same formula gives all maxima for a double-slit interference, yet it gives all minima (zero) for a single-slit diffraction (except for m = 0, which gives maximum for both cases). You can see this only if you have followed the arguments in the two cases. (In the case of two narrow slits, there are only two light rays to consider, one from each slit. In the case of single-wide-slit diffraction, there are also many other light rays that lie between the two extreme ones that emanate from the edges of the wide slit.) The light intensity actually varies with the angle θ as ( ) sin β / πa sinθ I( θ) = I 0 with β =. β / λ - 3 -

4 Note that β is the phase difference between the two edges of the wide slit. The derivation of this formula is given in the next section. (One can easily convince oneself that this expression vanishes whenever a sin θ = mλ, with m = 1,, 3,, but not at m = 0 where I(θ ) = I 0. Thus this expression gives the correct locations for all the intensity minima.) When θ is less than about 5 degrees, one can approximate sin θ by θ in radians. Then θm = mλ/ a ( m=± 1, ±, ± 3, ) gives the angle in radians of the light rays from the slit which will produce the m th dark fringe on the far-away screen. Now let the slit-screen distance be x. Let the screen be parallel to the opaque plate containing the slit, and a line O be on the screen parallel to the slit and at the shortest distance from the center line of the slit. This line O will mark the center of the brightest fringe of the diffraction pattern on the screen, because light rays from all parts of the slit to this line will be essentially all in phase. Away from this line there will be a sequence of alternating dark and bright fringes, but with rapidly diminishing intensities, so mainly only the center bright line is the brightest. The width of this bright line is twice the distance y 1 from O to the location of the first dark fringe on either side of O. Then, if x is very large, y xθ xλ/ a 1 1 (for x>>a) More generally if y m is the distance from O to the m th dark fringe, then y = xθ = xmλ/ a ( m=± 1, ±, ± 3, ). (for x>>a) m m The intensities of the nd, 3 rd, 4 th,... bright fringes on either side of the center bright fringe are 0.047, , ,..., times that of the center bright fringe, and are all very weak. These intensities are approximately given by the formula I I 0 m = 1 m + which may be easily obtained from the general intensity formula by setting β = ± (m + 1) π, where sin( β / ) =± 1. π - 4 -

5 3. Graphic method to determine the intensity in the single-slit diffraction pattern If the slit width a is not << λ, then it can not be approximated as a line source. We must then divide it into N thin strips, with N so large that each thin strip has a width << λ. As a good approximation, we can then treat each thin strip as a line source of no width. Let E i be the electric field at P on the screen due to the i th line source, with i running from 1 (for the strip closest to the point A in the slit), up to N (for the strip closest to the point B in the slit). Then the actual electric field at P is the sum of all these E i 's. I.e., EP = E1+ E + + E. N ( N = 10 in the above figur e). The phase difference between the two edge rays of the slit is β = πa sin θ/ λ. So the phase difference between E and E 1, between E 3 and E, between E 4 and E 3, etc., are all equal to β = β/n. Following the graphic method introduced in the previous chapter, E 1 is a the projection of a short vector of length (E 0 /N), and directional angle β, E is a the projection of a short vector of length (E 0 /N), and directional angle β, etc., until finally, E N is a the projection of a short vector of length (E 0 /N), and directional angle N β. [Here E 0 is the amplitude of the electric field at a point on the screen directly in front of the slit due to the whole slit. At this point all strips of the slit are in phase. So each strip simply contributes (E 0 /N). We have set the phase angle of E 1 to β for convenience only, since only relative phases matter here. Actually, all vectors β/ β are rotating in the counter-clockβ/ wise direction with an angular frequency ω, indicating that all elec- R sin (β/) tric fields are varying sinusoidally in time. But this variation does not R affect the intensity, which is a timeaveraged concept.] To find I P, which is ε averaged over time, one R sin (β/) 0 EP needs to draw all these short vectors connected end to end as shown, and - 5 -

6 In this figure, each short arrow has length E / N 0. The total length of the whole arc is, because it is made of N arrows connected end to end. The length of E 0 the chord is R sin ( β /), and Rβ is the total arc-leng th E 0, if β is in radians. find the length of the vector pointing from the beginning of the first short vector to the end point of the last short vector. This length squared divided by is just E P averaged over time. [The factor ½ arises from time average of cos (ωt + any constant phase).] See the graph shown above. From this graph, we can see that IP = ( ε /)[Rsin( β/)]. 0 But Rβ is just the total arc length, which is N ( E / N) = E 0 0. Thus I E P = ( ε0 /) sin( β/) = I0 β / 0 sin( β / ) β /, where I0 ( ε 0/) E0 is the intensity at θ = 0. πa sinθ Here β is λ the phase difference of the two edge rays of the slit. 4. Diffraction in the double-slit experiment In Young's double-slit interference experiment, if the width of each slit is not << λ, then the interference pattern will be superposed with a diffraction pattern due to each slit. That is, we will have IP sin( β / ) δ = I0 cos, β / β πasin θ δ πd sin θ where =, and =, with a the width of each slit (assumed λ λ equal), and d the center-to-center separation between the two slits. The first factor after I 0 is due to single-wide-slit diffraction, and the second factor after I 0 is due to double-slit interference. The intensity variation on the screen then looks like: - 6 -

7 where the heights of the interference peaks are controlled by diffraction, which is indicated by the dashed line. (Interference alone would have given all peaks the same height.) 5. Diffraction grating A differential grating is simply a large number N of equally spaced, parallel, narrow slits on an opaque plate or sheet. It is usually made with a glass plate or a plastic sheet, darkened except where the slits are located. Light shines on one side of the grating, and a diffraction pattern is supposed to form on a screen far on the other side, but usually a lens is used to refocus the pattern to a screen much closer. The diffraction of the incident light by such a grating can be analyzed in much the same way as that of a single wide-slit, except that one needs only to draw a single light ray from each narrow slit (valid if the width of each slit is much smaller than λ). One should not worry about whether there might be one last light ray that is not cancelled by destructive interference, because the contribution to intensity from one light ray is totally negligible in comparison with the total contribution from all the light rays. (There are often between 10 3 to 10 6 slits in a grating.) Thus one does not need to worry about whether there is an odd or even number of slits in a grating. In the case of a single wide slit, one can draw any number of light rays one chooses to in order to prove one's point. In a grating, the number of light rays is equal to the number of slits, and is pre-fixed. The result is: d sin θ = mλ (for the m th major maximum on the screen), where d is the separation between the neighboring slits, and here m = 0, 1,, 3, (including zero). This is because at these angles all light rays are in phase, and there are no other rays in between these rays to produce destructive interference unlike a single wide slit, where between any two rays there are many other rays, because every point in the wide slit can emit a light ray. Note that this formula for the major maxima produced by a grating is identical to the formula for the maxima produced by merely two - 7 -

8 slits. The difference between the two situations is the widths of these maxima, and whether there are other features in between, For a large-n grating, each major maximum has a very narrow width θ = λ /Ν d, which is narrower for larger N. Between every two major maxima there are actually many very weak minor maxima separated by vanishingintensity lines. The intensity formula in this case can be shown with the graphic method to be: Nπ d sinθ sin λ I( θ ) = I0 π d sinθ sin λ, where N is the number of slits in the grating, and d is the width of each slit. Graphic derivation of this result is very similar to earlier derivations of the formulae for double slits and single wide slit, only here there is only one light ray from each slit, and its contribution of electric field on the screen is represented by a single small arrow in the graph. Note that the intensities of light at the major maxima are all equal to N I 0, where I 0 is the intensity from one slit alone. You should already understand why this is true from our previous discussion. If you know a graphics software you can plot the formula for various values of d/λ and N to see its main features. The example given below is for N = 9: Here N = 9. There are then 8 zeroes between every two main peaks θ The minima are located at N d sin θ = mλ, with m not an integer multiple of N. (The main peaks are given by the same formula but with m an integer multiple of N.) - 8 -

9 To understand this result, let us begin with small N. You already know what happens to N =. It is the double-slit interference pattern (for small θ ): θ with only one minimun between every two neighboring major maxima, and no minor maxima between the major maxima. This is consistent with the general statement that there should be N-1 minima (zero) between every two neighboring major maxima. The first minimum corresponds to the two light rays from the two slits exactly differing by π in phase difference. I.e., graphically it correspond to two vectors pointing in opposite directions, which add up to zero. Now consider N = 3. The result should now appear as: with two minima (zero) and one minor maximum between every two neighboring major maxima. The first zero are now represented by three vectors forming a triangle: so their vector sum is zero. Thus the phase angle between the light rays from the neighboring slits should now be 10, or π/3. As this phase difference is achieved with a λ /3 path-length difference, this condition corresponds to d sin θ = λ/ 3, which is the same as 3d sinθ = λ. This is consistent with the - 9 -

10 general condition Ndsinθ = mλ for all minima, with N = 3 and m =1 (for the first minimum) here. The second minimum (zero) is obtained by having the phase angle between the light rays from the neighboring slits to be 10, or 4π/3, which is given by the condition d sinθ = λ/ 3, or Ndsinθ = mλ with N = 3 and m =. The graph is now an inverted triangle: Next, consider N = 4. That is, we now have four narrow slits in the grating. The first minimum now corresponds to four vectors forming a square: 90 so this minimum is given by the condition d sinθ = λ/ 4, or 4d sinθ = λ. It is again consistent with the general formula Ndsinθ = mλ, with now N = 4 and m = 1. The result now appears as: That is, it has three (= 4 1) minima (zero) and two minor maxima between every two neighboring major maxima, exactly as given by the general picture. In the following, I also give the results for N = 6 and N = 40, respectively, without detailed discussion: N = 6:

11 N = 40 : It needs to be emphasized that the height of the major maxima is N I 0, with I 0 the intensity due to a single slit alone. The height of the minor maxima is only in the neighborhood of 1 to N I 0. Thus when N is large, the minor minima becomes practically unobservable when we still draw the major maxima at about the same height as for low N. (We should have drawn the heights of the major maxima proportional to N, then heights of the minor peaks will not be negligibly small.) The next important thing to understand is the width of each major maximum. This is given by twice the distance from the center of the center maximum to the first minimum (or zero), or θ = ( λ/ Nd) = λ/ Nd, asuming that θ is small, so that sin θ can be approximated by θ in radians. If the incident light is white light, or any mixture of frequencies, then all major maxima except the center one (corresponding to m = 0) will be located at slightly different θ for different light frequencies, leading to separation of color. It is easy to see that for any fixed m, red color is always farthest from the center (for red corresponds to the longest wavelength in the visible range), and purple color is always closest to the center, and the larger is m, the wider is the color spread. (Color spread also occurs to some extend in the interference pattern of a double slit, or the diffraction pattern of a single wide slit, but the separation would be far less clean-cut in those cases than in the case of gratings, because each intensity maximum due to a grating is so much narrower than that due to a double slit or a single wide slit. The separation of color is the least obvious in the case of a single wide slit, because the center bright fringe does not split, whereas all other bright fringes are so much weaker than the center bright fringe that they are hard to observe.) Resolving power of a grating R = λ/ λ, where λ is the wavelength difference which can be resolved, and λ is the mean wave length. Differentiating both sides of the formula for the major maxima, d sinθ = mλ, we obtain

12 d cosθ θ = m λ. Using θ = λ / N d (from the half width of each peak), and cosθ ~1 (because all relevant angles are quite small), we obtain λ λ / m N. Hence R = mn, (which is larger for larger m and N!) Diffraction gratings (and with far less resolution, also prisms,) can be used to analyze the compositions of light emitted from a remote source such as a star, since each kind of atoms, if hot, emits a definite set of frequencies of light, and absorb light at those definite frequencies, if cold. Thus by spreading the color content in the star light we can tell whether there is hydrogen, or helium, or oxygen, etc., in a hot star, by spotting their characteristic bright lines in the color spread. Furthermore, since Doppler shift means that the frequencies of emitted light will be shifted toward higher or lower frequencies if the star is moving toward or away from the earth, and the shift is bigger if the velocity of the star is bigger, so by examining the amounts of shifts of frequencies in the bright lines in the color spread of the light from a star, we can tell how fast a star is moving, toward us or away from us. From such information scientists found out that the universe is expanding because all stars are moving away from the earth, with the farthest stars moving the fastest away from the earth. The earth does not enjoy any special position in the universe; all stars are simply moving away from each other, because the universe is expanding like a balloon is expanding, and all stars are dots on the balloon! Tracing this expansion backward, scientists can then conclude that the universe was as small as a tiny dot some 13.7 billion years ago, and there was a giant explosion in the beginning. (This is the big bang theory of the universe. It really has quite a number of concrete evidences supporting it, so it is not simply a theory, if you think a theory is just a hypothesis.) Since light takes a long time to travel a very long distance, light from farthest stars, (which have the largest Doppler shifts,) have taken the longest time to reach us. So by examining light from those stars in detail we can learn information about the very early stage of the universe. Currently the best telescope (the Hubble telescope) can already let us trace backward in time to a- bout 90 to 95% of the history of the universe. So it is not a dream to learn scientifically how the universe all began! All these are made possible by the invention of diffraction gratings

13 6. Spectrometry and spectroscopy A Spectrometer or a spectroscope is a device to measure (via recording or viewing, respectively) wavelengths accurately by using a diffraction grating (or a prism) to separate different wavelengths of light. The process is called spectrometry or spectroscopy, respectively. The collection of data recorded in such a measurement is called a spectrograph. Since d sin θ = mλ for the m th diffraction maxima, we can measure λ accurately, by measuring θ and d accurately. (If only one knows a rough range for λ, one can figure out the integer m.) A Spectrum refers to the frequency content of a light source. Each kind of atom, when hot, emits a characteristic emission spectrum. It is a line spectrum, meaning that it is made of a set of characteristic bright lines in the spectrum. Lamp light, sun light, and light from a hot solid or gas, etc. have a continuous spectrum, meaning that practically all frequencies within a wide range are emitted. A cold gas of atoms of any single kind, placed in front of such a source, produce a set of characteristic dark lines in the continuous spectrum, called absorption spectrum, because these frequencies of light in the continuous spectrum are absorbed by the cold atoms. Thus by measuring a emission or absorption spectrum, one can identify a hot or cold atom. 7. X-rays and x-ray diffraction (or crystallography) X rays are EM waves of frequencies comparable to and higher than hard ultraviolet. X rays are not visible, but can expose photographic films. They can be emitted by atoms hit by electrons accelerated to high velocity through a large voltage difference. The so-emitted x-ray is not of a single frequency, but is a mixture of a wide range of frequencies. Since x-rays have very short wavelengths, they can be used to resolve the very closely spaced atoms in a crystal. When a collimated x-ray beam hits a crystalline solid, the regularly arranged atoms in the solid act as a reflection grating. Light scattered by the individual atoms in a solid produces a diffraction pattern on a photographic film placed at a large distance away from the solid. With a single-crystal solid the film is exposed only at a distribution of isolated spots. Measuring the center positions of these spots on the film allows one to figure out the crystal structure in the solid. If a polycrystalline solid is used as the target, the film

14 will be exposed at a set of concentric rings, the radii of which also provide similar information. To understand the origin of the x-ray diffraction pattern, first notice that a crystal has atoms arranged in many sets of equally-spaced parallel planes. Two such sets of planes are shown below (with the dots representing the atoms): incom -ing x-ray If x-ray shines on the crystal along one of the Principal axes of the crystal, as φ 1 shown, it can be reflected into a particular direction A B by any one set of parallel planes. The reflected pa- C rallel rays will give rise to d a bright spot on the photo- φ graphic film if: (1) the law of reflection is obeyed, and () light reflected by all parallel planes of the set are all in phase, in order to get the strongest constructive interference. Condition (1) is needed so that the scattered light from all atoms in any single plan will be in phase, and condition () is needed so that the scattered light from all atoms in different parallel planes are also in phase. These conditions are satisfied as follows: To satisfy the first condition, we simply require the incidcent angle to be e- qual to the reflected angle. To satisfy the second condition, we require that the difference in path-length between paths 1 and to be an integer multiple

15 of λ. This path-length difference is just AB plus BC. So it is equal to d sin φ. Thus to get constructive interference between the different atomic planes we obtain the condition: d sin φ = mλ. This is called the Bragg condition, and the phenomenon is called Bragg reflection. Here m must be an integer. Note that d is fixed for a given set of planes in the crystal, but it is not known to you. φ is determined from the position of the bright dot on the photographic film and the distance of the film to the crystal. This equation can be satisfied for some integer m, if the resultant λ falls within the range of wave-lengths contained in the incoming x-ray. The whole collection of bright dots on the film lets one determine the orientations of all possible sets of planes in the crystal, which in turn let you figure out the crystal structure. This is possible even though we don't know the values of d 's of the different sets of planes in the crystal. The analysis of x-ray diffraction from a polycrystalline sample is somewhat different. We will not go into details here. 8. Limits of resolution; circular apertures Light going through a circular hole will also exhibit diffraction. At large distance behind a hole of diameter D, light intensity will vary with angle θ in a form roughly similar to the expression for a single slit, except that each minimum now forms a concentric circle, and the first minimum (a dark ring) is located at: D sin θ = 1. λ, which marks the angular radius of a bright center disk, known as the Airy disk. For D >> λ we can approximate sin θ by θ in radians. On a screen a distance x away from the hole, the bright disk will have a radius xθ 1. λ x / D. This radius can be much bigger than D, if D is very small. Ray method (geometric optics) tells us that a point in an object will correspond to a point in the image formed by lenses and mirrors. But ray method is only an approximation. We now know that wave optics (or physical optics) tells us that the so-called point in the image is actually a disc. The size of this disc is not as important as its angular size, which is intrinsic, and is independent of where the screen is placed, and is given by the above

16 θ, if we wish to find out what is the minimum separation between two points in the object which can be clearly separated (resolved) in the image. This minimum separation is called resolving power (RP). For a microscope one should place the object at the focal point of the objective lens, so it can be viewed clearly through the eye-piece (another lens near the eye in the microscope). Then its resolving power is given by RP = f θ = 1. λ f / D (resolving power of a microscope), where f is the focal length of the objective lens, and D is the diameter of the objective lens. Two points in the object being examined under the microscope can be resolved by the microscope only if their separation is larger than this RP distance. Since the smallest f is about D/, the smallest RP is about λ/. Why is the size of the eyepiece irrelevant in this discussion? Because the ratio f / D is essentially the same for the objective and the eyepiece of a microscope. (Making the diameter D of the eyepiece bigger than is given by this condition does not help is reducing the RP, since then the outer ring of the eyepiece will have no light going through, and therefore cannot be counted. One will never make the D of the eyepiece smaller than given by this condition, since it will further limit the RP. For a telescope the objects (the planets and stars) are very, very far away, very, very big, and the separation of them are also very, very large, so it makes more sense to speak about their angular size and separation, and the angular resolving power of a telescope (ARP) is simply what we have already found: ARP = θ = 1. λ / D (angular resolving power of a telescope). Two stars can be resolved by the telescope only if their angular separation is larger than this ARP value. Here D is the diameter of the objective lens or the reflecting spherical mirror of the telescope. The diameter of the eyepiece is again irrelevant for the same reason as in the case of a microscope. For a camera one speaks of (linear) resolving power RP = d o θ = 1. λ d o / D (resolving power of a camera) for whatever object distance d o. In this case D is the diameter of the circular aperture, which is smaller than that of the lens in order to control the amount of light entering the camera. Two points in the object being filmed

17 by the camera can be resolved only if their separation is larger than this RP distance. Various reasons limit the human eye ARP to about rad. Hence a strained normal eye will have a RP of (5 cm) ( rad.) º 10-4 m. The best optical microscope has a RP of º ( m) / = m, or about 00 nm. The ratio of about 500 is the best magnification one can have from a microscope without getting blurred images due to diffraction. 9. Holography The coherent light beam from a laser can be split into two beams by using a partially reflecting mirror known as a beam splitter. One of the two beams can be allowed to scatter from an object. The scattered beam is allowed to merge with the un-scattered beam. If the combined beam is used to expose a film, then the film has recorded the interference pattern generated by the two beams, which carries information about the object which did the scattering of one of the two beams. If this film is illuminated by a similar laser beam, or white light in some cases, a three-dimensional real image and a threedimensional virtual image of the object are produced, in front of the film and behind the film, respectively, which can be viewed or used to expose a film to get a picture of the object (in the former case). Unlike ordinary photographs, which is not based on interference, these holographic images are three dimensional, and can be viewed at different angles and distances to reveal different sides and perspective. For the theory behind it, see Young and Freedman, University Physics, Sec