Chapter 33 Interference and Diffraction Conceptual Problems 3 • The spacing between Newton’s rings decreases rapidly as the diameter of the rings increases. Explain qualitatively why this occurs. Determine the Concept The thickness of the air space between the flat glass and the lens is approximately proportional to the square of d, the diameter of the ring. Consequently, the separation between adjacent rings is proportional to 1/d. 7 • A two-slit interference pattern is formed using monochromatic laser light that has a wavelength of 640 nm. At the second maximum from the central maximum, what is the path-length difference between the light coming from each of the slits? (a) 640 nm, (b) 320 nm, (c) 960 nm, (d) 1280 nm. Determine the Concept For constructive interference, the path difference is an integer multiple of λ; that is, λmr=Δ. For m = 2,()nm 6402=Δr.()dis correct. 15 • True or false: (a) When waves interfere destructively, the energy is converted into heat energy. (b) Interference patterns are observed only if the relative phases of the waves that superimpose remain constant. (c) In the Fraunhofer diffraction pattern for a single slit, the narrower the slit, the wider the central maximum of the diffraction pattern. (d) A circular aperture can produce both a Fraunhofer diffraction pattern and a Fresnel diffraction pattern. (e) The ability to resolve two point sources depends on the wavelength of the light. (a) False. When destructive interference of light waves occurs, the energy is no longer distributed evenly. For example, light from a two-slit device forms a pattern with very bright and very dark parts. There is practically no energy at the dark fringes and a great deal of energy at the bright fringe. The total energy over the entire pattern equals the energy from one slit plus the energy from the second slit. Interference re-distributes the energy. (b) True. 255 Chapter 33 256 (c) True. The width of the central maximum in the diffraction pattern is given by ammλθ1sin−= where a is the width of the slit. Hence, the narrower the slit, the wider the central maximum of the diffraction pattern. (d) True. (e) True. The critical angle for the resolution of two sources is directly proportional to the wavelength of the light emitted by the sources (Dλα22.1c=). Estimation and Approximation 19 •• (a) Estimate how close an approaching car at night on a flat, straight stretch of highway must be before its headlights can be distinguished from the single headlight of a motorcycle. (b) Estimate how far ahead of you a car is if its two red taillights merge to look as if they were one. Picture the Problem Assume a separation of 1.5 m between typical automobile headlights and tail lights, a nighttime pupil diameter of 5.0 mm, 550 nm for the wavelength of the light (as an average) emitted by the headlights, 640 nm for red taillights, and apply the Rayleigh criterion. (a) The Rayleigh criterion is given by Equation 33-25: Dcλα22.1= where D is the separation of the headlights (or tail lights). The critical angular separation is also given by: Ld=α where d is the separation of head lights (or tail lights) and L is the distance to approaching or receding automobile. Equate these expressions for αc to obtain: DLdλ22.1=⇒λ22.1DdL= Substitute numerical values and evaluate L: ()()()km 11nm 55022.1m 5.1mm 0.5≈=L (b) For red light: ()()()km .69nm 64022.1m 5.1mm 0.5≈=L Interference and Diffraction 257 21 •• Estimate the maximum distance at which a binary star system could be resolvable by the human eye. Assume the two stars are about fifty times farther apart than Earth and Sun are. Neglect atmospheric effects. (A test similar to this ″eye test″ was used in ancient Rome to test for eyesight acuity before entering the army. A person who had normal eyesight could just barely resolve two well-known stars that appear close in the sky. Anyone who could not tell there were two stars failed the test.) Picture the Problem Assume that the diameter of a pupil at night is 5.0 mm and that the wavelength of light is in the middle of the visible spectrum at about 550 nm. We can use the Rayleigh criterion for the separation of two sources and the geometry of the Earth-binary star system to derive an expression for the distance to the binary stars. If the distance between the binary stars is represented by d and the Earth-star distance by L, then their angular separation is given by: Ld=α The critical angular separation of the two sources is given by the Rayleigh criterion: Dcλα22.1= For α = αc: DLdλ22.1=⇒λ22.1DdL= Substitute numerical values and evaluate L: ()()()()y 5.9m 109.461y 1km 1059.5nm 55022.1m 105.150mm 0.5151311⋅×⋅××≈×=ccL Phase Difference and Coherence 23 •• Two coherent microwave sources both produce waves of wavelength 1.50 cm. The sources are located in the z = 0 plane, one at x = 0, y = 15.0 cm and the other at x = 3.00 cm, y = 14.0 cm. If the sources are in phase, find the difference in phase between these two waves for a receiver located at the origin. Picture the Problem The difference in phase depends on the path difference according to πλδ2rΔ=. The path difference is the difference in the distances of (0, 15.0 cm) and (3.00 cm, 14.0 cm) from the origin. Chapter 33 258 Relate a path difference Δr to a phase shift δ: πλδ2rΔ= The path difference Δr is: ()(cm682.0cm0.14cm3.00cm0.1522=+−=Δr Substitute numerical values and evaluate δ: rad .922cm50.1cm682.0≈⎟⎟⎠⎞⎜⎜⎝⎛=πδ Interference in Thin Films 25 •• The diameters of fine fibers can be accurately measured using interference patterns. Two optically flat pieces of glass that each have a length L are arranged with the fiber between them, as shown in Figure 33-40. The setup is illuminated by monochromatic light, and the resulting interference fringes are observed. Suppose that L is 20.0 cm and that yellow sodium light (590 nm) is used for illumination. If 19 bright fringes are seen along this 20.0-cm distance, what are the limits on the diameter of the fiber? Hint: The nineteenth fringe might not be right at the end, but you do not see a twentieth fringe at all. Picture the Problem The light reflected from the top surface of the bottom plate (wave 2 in the diagram) is phase shifted relative to the light reflected from the bottom surface of the top plate (wave 1 in the diagram). This phase difference is the sum of a phase shift of π (equivalent to a λ21 path difference) resulting from reflection plus a phase shift due to the additional distance traveled. The condition that one sees m bright fringes requires that the path difference between light reflected from the bottom surface of the top slide and the top surface of the bottom slide is an integer multiple of a wavelength of the light. 12fiberglass plateglass plateπdλ= 590 nmL Interference and Diffraction 259 Relate the extra distance traveled by wave 2 to the distance equivalent to the phase change due to reflection and to the condition for constructive interference: ...,3,2,221λλλλ=+d or ...,,,2252321λλλ=d and ()λ212+=md where m = 0, 1, 2, …, 0≤t≤2r and λ is the wavelength of light in air. Because the nineteenth (but not the twentieth) bright fringe can be seen, the limits on d must be: ()()222121λλ+<<−mdm where m = 19 Substitute numerical values to obtain: ()()2nm590192nm590192121+<<−d or m8.5m5.5μμ<> λ) are given approximately by dLmym4λ= where m = 1, 2, 3, 5, 6, 7, 9, 10, . . ., that is, m is not a multiple of 4. (b) For a screen distance of 2.00 m, a light wavelength of 600 nm, and a source spacing of 0.100 mm, calculate the width of the principal interference maxima (the distance between successive minima) for four sources. Compare this width with that for two sources with the same spacing. Picture the Problem We can use phasor concepts to find the phase angle δ in terms of the number of phasors N (four in this problem) forming a closed polygon of N sides at the minima and then use this information to express the path difference Δr for each of these locations. Applying a small angle approximation, we can obtain an expression for y that we can evaluate for enough of the path differences to establish the pattern given in the problem statement. (a) Express the phase angle δ in terms of the number of phasors N ⎟⎠⎞⎜⎝⎛=Nmπδ2 Interference and Diffraction 267 forming a closed polygon of N sides: where m = 1, 2, 3, 4, 5, 6, ,7, … For four equally spaced sources, the phase angle is: ⎟⎠⎞⎜⎝⎛=2πδm Express the path difference corresponding to this phase angle to obtain: 42λδπλmr=⎟⎠⎞⎜⎝⎛=Δ (1) Interference maxima occur for: m = 4, 8, 12, … Interference minima occur for: m = 1, 2, 3, 5, 6, 7, 9, 10, … (Note that m is not a multiple of 4.) Express the path difference Δr in terms of sinθ and the separation d of the slits: θsindr=Δ or, provided the small angle approximation is valid, Lydr=Δ⇒rdLyΔ= Substituting for Δr from equation (1) yields: ,...9,7,6,5,3,2,1 ,4min==mdLmyλ (b) For L = 2.00 m, λ = 600 nm, d = 0.100 mm, and m = 1: ()()()mm00.6mm0.1004m00.2nm60022min==yFor two slits: ()dLmyλ21min22+= For L = 2.00 m, λ = 600 nm, d = 0.100 mm, and m = 0: ()()mm0.12mm0.100m00.2nm6002min==y The width for four sources is half the width for two sources. 55 ••• Three slits, each separated from its neighbor by 60.0 μm, are illuminated at the central intensity maximum by a coherent light source of wavelength 550 nm. The slits are extremely narrow. A screen is located 2.50 m from the slits. The intensity is 50.0 mW/m2. Consider a location 1.72 cm from the central maximum. (a) Draw a phasor diagram suitable for the addition of the three harmonic waves at that location. (b) From the phasor diagram, calculate the intensity of light at that location. Chapter 33 268 Picture the Problem We can find the phase constant δ from the geometry of the diagram to the right. Using the value of δ found in this fashion we can express the intensity at the point 1.72 cm from the centerline in terms of the intensity on the centerline. On the centerline, the amplitude of the resultant wave is 3 times that of each individual wave and the intensity is 9 times that of each source acting separately. m 50.2=L cm .721yθ (a) Express δ for the adjacent slits: θλπδsin2d= For θ << 1, θθθ≈≈tansin: Ly=≈θθtansin Substitute to obtain: Ldyλπδ2= Substitute numerical values and evaluate δ : ()()()()°===270rad23m50.2nm550cm72.1m0.602πμπδ The three phasors, 270° apart, are shown in the diagram to the right. Note that they form three sides of a square. Consequently, their sum, shown as the resultant R, equals the magnitude of one of the phasors. δδAAA0000R = A (b) Express the intensity at the point 1.72 cm from the centerline: 2RI∝ Interference and Diffraction 269 Because I0 ∝ 9R2: 2209RRII= ⇒ 90II= Substitute for I0 and evaluate I: 22mW/m56.59mW/m0.50==I Diffraction and Resolution 57 • Light that has a wavelength equal to 700 nm is incident on a pinhole of diameter 0.100 mm. (a) What is the angle between the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern? (b) What is the distance between the central maximum and the first diffraction minimum on a screen 8.00 m away? Picture the Problem We can use Dλθ22.1= to find the angle between the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern and the diagram to the right to find the distance between the central maximum and the first diffraction minimum on a screen 8 m away from the pinhole. LminyθD (a) The angle between the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern is given by: Dλθ22.1= Substitute numerical values and evaluate θ : mrad54.8mm100.0nm70022.1=⎟⎟⎠⎞⎜⎜⎝⎛=θ (b) Referring to the diagram, we see that: θtanminLy= Substitute numerical values and evaluate ymin: ()()cm83.6mrad54.8tanm00.8min==y Chapter 33 270 61 •• The telescope on Mount Palomar has a diameter of 200 in. Suppose a double star were 4.00 light-years away. Under ideal conditions, what must be the minimum separation of the two stars for their images to be resolved using light that has a wavelength equal to 550 nm? Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to obtain an expression we can solve for the minimum separation Δx of the stars. Your pupilΔααxLcc Rayleigh’s criterion is satisfied provided: Dλα22.1c= Relate αc to the separation Δx of the light sources: LxΔ≈cαbecause αc << 1 Equate these expressions to obtain: DLxλ22.1=Δ⇒DLxλ22.1=Δ Substitute numerical values and evaluate Δx: ()m1000.5in1cm2.54in200y1m10461.9y4nm55022.1915×=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛×⎟⎟⎠⎞⎜⎜⎝⎛⋅××⋅=Δccx Diffraction Gratings 63 • A diffraction grating that has 2000 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. (a) At what angles in the first-order spectrum would you expect to find the two violet lines that have wavelengths of 434 nm and 410 nm? (b) What are the angles if the grating has 15 000 slits per centimeter? Interference and Diffraction 271 Picture the Problem We can solve λθmd=sinfor θ with m = 1 to express the location of the first-order maximum as a function of the wavelength of the light. (a) The interference maxima in a diffraction pattern are at angles θ given by: λθmd=sin where d is the separation of the slits and m = 0, 1, 2, … Solve for the angular location θm of the maxima : ⎟⎠⎞⎜⎝⎛=−dmmλθ1sin Relate the number of slits N per centimeter to the separation d of the slits: dN1= Substitute for d to obtain: ()λθmNm1sin−= (1) Evaluateθ for λ = 434 nm and m =1: ()()[]mrad9.86nm434cm2000sin111==−−θ Evaluateθ for λ = 410 nm and m = 1: ()()[]mrad1.82nm410cm2000sin111==−−θ (b) Use equation (1) to evaluateθ for λ = 434 nm and m = 1: ()()[]mrad 709nm434cm15000sin111==−−θ Evaluate θ for λ = 410 nm and m = 1: ()()[]mrad 662nm410cm15000sin111==−−θ 67 •• A diffraction grating that has 4800 lines per centimeter is illuminated at normal incidence with white light (wavelength range of 400 nm to 700 nm). How many orders of spectra can one observe in the transmitted light? Do any of these orders overlap? If so, describe the overlapping regions. Picture the Problem We can use the grating equation ... 3, 2, 1, sin==m,mdλθ to express the order number in terms of the slit separation d, the wavelength of the light λ, and the angle θ. Chapter 33 272 The interference maxima in the diffraction pattern are at angles θ given by: λθmd=sin⇒λθsindm= where m = 1, 2, 3, … If one is to see the complete spectrum, it must be true that: 1sin≤θ ⇒λdm≤ Evaluate mmax: 98.2nm700cm48001cm480011max1max===−−λm Because mmax = 2.98, one can see the complete spectrum only for m = 1 and 2. Express the condition for overlap: 2211λλmm≥ One can see the complete spectrum for only the first and second order spectra. That is, only for m = 1 and 2. Because 700 nm < 2 × 400 nm, there is no overlap of the second-order spectrum into the first-order spectrum; however, there is overlap of long wavelengths in the second order with short wavelengths in the third-order spectrum. 71 •• Mercury has several stable isotopes, among them 198Hg and 202Hg. The strong spectral line of mercury, at about 546.07 nm, is a composite of spectral lines from the various mercury isotopes. The wavelengths of the line for 198Hg and 202Hg are 546.07532 nm and 546.07355 nm, respectively. What must be the resolving power of a grating capable of resolving these two isotopic lines in the third-order spectrum? If the grating is illuminated over a 2.00-cm-wide region, what must be the number of lines per centimeter of the grating? Picture the Problem We can use the expression for the resolving power of a grating to find the resolving power of the grating capable of resolving these two isotopic lines in the third-order spectrum. Because the total number of the slits of the grating N is related to width w of the illuminated region and the number of lines per centimeter of the grating and the resolving power R of the grating, we can use this relationship to find the number of lines per centimeter of the grating. The resolving power of a diffraction grating is given by: mNR=Δ=λλ (1) Interference and Diffraction 273 Substitute numerical values and evaluate R: 551009.3100852.307355.54607532.54607532.546×=×=−=R Express n, be the number of lines per centimeter of the grating, in terms of the total number of slits N of the grating and the width w of the grating: wNn= From equation (1) we have: mRN= Substitute for N to obtain: mwRn= Substitute numerical values and evaluate n: ()()145cm1014.5cm00.23100852.3−×=×=n 73 ••• For a diffraction grating in which all the surfaces are normal to the incident radiation, most of the energy goes into the zeroth order, which is useless from a spectroscopic point of view, because in zeroth order all the wavelengths are at 0º. Therefore, modern reflection gratings have shaped, or blazed, grooves, as shown in Figure 33-46. This shifts the specular reflection, which contains most of the energy, from the zeroth order to some higher order. (a) Calculate the blaze angle φm in terms of the groove separation d, the wavelength λ, and the order number m in which specular reflection is to occur for m = 1, 2, . . . . (b) Calculate the proper blaze angle for the specular reflection to occur in the second order for light of wavelength 450 nm incident on a grating with 10 000 lines per centimeter. Picture the Problem We can use the grating equation and the geometry of the grating to derive an expression for φm in terms of the order number m, the wavelength of the light λ, and the groove separation d. (a) Because θi = θr, application of the grating equation yields: ()... 2, 1, 0, where2sini==m, mdλθ (1) Because φ and θi have their left and right sides mutually perpendicular: miφθ= Substitute for θi to obtain: ()λφmd=m2sin Chapter 33 274 Solving for φm yields: ⎟⎠⎞⎜⎝⎛=−dmλφ121msin (b) For m = 2: °=⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛=−−1.32cm000,101nm4502sin11212φ General Problems 75 • Naturally occurring coronas (brightly colored rings) are sometimes seen around the moon or the Sun when viewed through a thin cloud. (Warning: When viewing a Sun corona, be sure that the entire Sun is blocked by the edge of a building, a tree, or a traffic pole to safeguard your eyes.) These coronas are due to diffraction of light by small water droplets in the cloud. A typical angular diameter for a coronal ring is about 10º. From this, estimate the size of the water droplets in the cloud. Assume that the water droplets can be modeled as opaque disks with the same radius as the droplet, and that the Fraunhofer diffraction pattern from an opaque disk is the same as the pattern from an aperture of the same diameter. (This last statement is known as Babinet’s principle.) Picture the Problem We can use Dλθ22.1sin=to relate the diameter D of the opaque-disk water droplets to the angular diameter θ of a coronal ring and to the wavelength of light. We’ll assume a wavelength of 500 nm. The angle θ subtended by the first diffraction minimum is related to the wavelength λ of light and the diameter D of the opaque-disk water droplet: Dλθ22.1sin= Because of the great distance to the cloud of water droplets, θ << 1 and: Dλθ22.1≈⇒θλ22.1=D Substitute numerical values and evaluate D: ()m5.3180rad10nm50022.1μπ=°×°=D 79 • A long, narrow horizontal slit lies 1.00 μm above a plane mirror, which is in the horizontal plane. The interference pattern produced by the slit and its image is viewed on a screen 1.00 m from the slit. The wavelength of the light is 600 nm. (a) Find the distance from the mirror to the first maximum. (b) How many dark bands per centimeter are seen on the screen? Interference and Diffraction 275 Picture the Problem We can apply the condition for constructive interference to find the angular position of the first maximum on the screen. Note that, due to reflection, the wave from the image is 180o out of phase with that from the source. (a) Because y0 << L, the distance from the mirror to the first maximum is given by: 00θLy= (1) Express the condition for constructive interference: ()λθ21sin+=md where m = 0, 1, 2, … Solving for θ yields: ()⎥⎦⎤⎢⎣⎡+=−dmλθ211sin For the first maximum, m = 0 and: ⎥⎦⎤⎢⎣⎡=−d2sin10λθ Substitute in equation (1) to obtain: ⎥⎦⎤⎢⎣⎡=−dLy2sin10λ Because the image of the slit is as far behind the mirror’s surface as the slit is in front of it, d = 2.00 μm. Substitute numerical values and evaluate y0: ()()cm1.15m00.22nm600sinm00.110=⎥⎦⎤⎢⎣⎡=−μy (b) The separation of the fringes on the screen is given by: dLyλ=Δ The number of dark bands per unit length is the reciprocal of the fringe separation: Ldynλ=Δ=1 Substitute numerical values and evaluate n: ()()1m33.3m00.1nm600m00.2−==μn 83 •• A Fabry–Perot interferometer (Figure 33-47) consists of two parallel, half-silvered mirrors that face each other and are separated by a small distance a. A half-silvered mirror is one that transmits 50% of the incident intensity and reflects 50% of the incident intensity. Show that when light is incident on the interferometer at an angle of incidence θ, the transmitted light will have maximum intensity when 2a = mλ/cos θ. Chapter 33 276 Picture the Problem The Fabry-Perot interferometer is shown in the figure. For constructive interference in the transmitted light the path difference must be an integral multiple of the wavelength of the light. This path difference can be found using the geometry of the interferometer. θθθφaABC For constructive interference we require that: ...,2,1,0==Δm,mrλ (1) The path difference between the two rays that emerge from the interferometer is given by: φθsinABcosBC AB +=+=Δar From the geometry of the interferometer: °=+902φθ⇒θφ290−°= Substituting for φ and AB and simplifying gives: ()()θθθθθθθθcos21coscos2coscos 290sincoscos +=+=−°+=Δaaaaar Using the trigonometric identity 1cos22cos2−=θθ and simplifying yields: ()θθθcos21cos21cos 2aar=−+=Δ Substitute for rΔ in equation (1) to obtain: ...,2,1,0cos2==m,maλθ Solving for 2a gives: θλcos2ma= where m = 0, 1, 2, … 85 •• A camera lens is made of glass that has an index of refraction of 1.60. This lens is coated with a magnesium fluoride film (index of refraction 1.38) to enhance its light transmission. The purpose of this film is to produce zero reflection for light of wavelength 540 nm. Treat the lens surface as a flat plane Interference and Diffraction 277 and the film as a uniformly thick flat film. (a) What minimum thickness of this film will accomplish its objective? (b) Would there be destructive interference for any other visible wavelengths? (c) By what factor would the reflection for light of 400 nm wavelength be reduced by the presence of this film? Neglect the variation in the reflected light amplitudes from the two surfaces. Picture the Problem Note that the light reflected at both the air-film and film-lens interfaces undergoes a π rad phase shift. We can use the condition for destructive interference between the light reflected from the air-film interface and the film-lens interface to find the thickness of the film. In (c) we can find the factor by which light of the given wavelengths is reduced by this film from .cos212δ∝I AirntπFilmLensπ (a) Express the condition for destructive interference between the light reflected from the air-film interface and the film-lens interface: ()()nmmtair21film212λλ+=+= (1) where m = 0, 1, 2, … Solving for t gives: ()nmt2air21λ+= Evaluate t for m = 0: ()nm8.97nm83.9738.12nm54021==⎟⎠⎞⎜⎝⎛=t (b) Solve equation (1) for λair to obtain: 21air2+=mtnλ Evaluate λair for m = 1: ()()nm180138.1nm8.97221air=+=λ No; because 180 nm is not in the visible portion of the spectrum. (c) Express the reduction factor f as a function of the phase difference δ between the two reflected waves: δ212cos=f (2) Chapter 33 278 Relate the phase difference to the path difference Δr: film2λπδrΔ= ⇒ ⎟⎟⎠⎞⎜⎜⎝⎛Δ=film2λπδr Because Δr = 2t: ⎟⎟⎠⎞⎜⎜⎝⎛=film22λπδt Substitute in equation (2) to obtain: ⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛=air2film2film2122cos2cos22cosλπλπλπntttf Evaluate f for λ = 400 nm: ()()273.0nm400nm83.9738.12cos2400=⎥⎦⎤⎢⎣⎡=πf 87 •• The Impressionist painter Georges Seurat used a technique called pointillism, in which his paintings are composed of small, closely spaced dots of pure color, each about 2.0 mm in diameter. The illusion of the colors blending together smoothly is produced in the eye of the viewer by diffraction effects. Calculate the minimum viewing distance for this effect to work properly. Use the wavelength of visible light that requires the greatest distance between dots, so that you are sure the effect will work for all visible wavelengths. Assume the pupil of the eye has a diameter of 3.0 mm. Picture the Problem We can use the geometry of the dots and the pupil of the eye and Rayleigh’s criterion to find the greatest viewing distance that ensures that the effect will work for all visible wavelengths. Dots of paintPupilθcαdL Referring to the diagram, express the angle subtended by the adjacent dots: Ld≈θ Interference and Diffraction 279 Letting the diameter of the pupil of the eye be D, apply Rayleigh’s criterion to obtain: Dλα22.1c= Set θ = αc to obtain: DLdλ22.1=⇒λ22.1DdL= Evaluate L for the shortest wavelength light in the visible portion of the spectrum: ()()()()m12nm40022.1mm0.2mm0.3==L