`
`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`ZL
`
`27.2
`
`Young's Dowhble-Shit Experiment
`
`785
`
`* Conditions for interference
`
`n the preceding chapter, we used the concept of light rays to examine
`what happens when light passes through a lens or reflects from a mirror,
`This chapter is concerned with the subject ofwave optics, which addresses
`the related optical phenomena of interference and diffraction, These phe-
`nomena cannot be adequately explained with ray (geometric) optics, but we
`shall discuss how the wave nature of light leads to satisfying descriptions of
`such events.
`
`27.1 + CONDITIONS FOR INTERFERENCE
`
`CH
`
`272
`
`273
`
`274
`27527
`276
`
`217
`278
`
`"THR OUTLINE
`(Conditions for Interference
`Young's Double-Slit
`Experiment
`Change of Phase Duc to
`Reflection
`Interference in Thin Films
`
`Diffract
`Resolution of Single-Slit and
`‘Circular Apertures
` ction Crating
`The Di
`Diffraction of X-Rays by
`(Crystals (Optional)
`
`Interference effects in visible electromagnetic waves are not easy to observe be-
`cause of their short wavelengths (from about 4 * 10-? m to 7 * 10°7 m). In order
`to observe sustained interference in light waves, three conditions must be met:
`
`
`* Thesources must be coherent— thatis.they
`st maintain a constant phase
`with respect to each other.
`Wave Optics
`* The sources must be ofidentical wavelength
`* Thesuperposition principle must apply.
`
`Let us examinethe characteristics of coherent sources. Two sources (producing
`
`interference, However,
`two traveling waves) are needed to create
`in order to pro-
`duce a stable interference pattern, the individual waves must maintain a constant
`phase with one another. Whenthis simation prevails, the sources are said to be
`
`coherent. Ac an example, the sound waves emitted by two
`side-by-side loudspeakers
`
`driven bya single amplifier
`produceinterference because the two speakers
`respond to the amplifier in the same wayat the sametime.
`
`Now, if twolight sources are placed side by side, no interference effects are
`
`observed because the light waves from one source are emitted independently of
`
`the other source; hence, the emissions from the two sources do not maintain a
`constant phase relationship with each other over
`the time of observation. An or-
`dinarylight source undergoes random changes about onceevery 10~° s. Therefore,
`the conditions for constructive interference, destructive interference, or some
`
`
`
`termediate state last for times on the order of 10~* s. The result is that no interfer-
`ence effects are observed, because the eye cannot follow such short-term changes.
`Suchlight sources are said to be noncoherent.
`A common method for producing two coherent light sources is to use one
`
`single wavelength source
`to illu
`le a screen
`containing two smallslits. The light
`emerging fromthe twoslits is coherent because a single source produces the orig-
`inal light beam andthetwo slits serve only to separate the original beaminto two
`parts (whichis exactly what was done to the soundsignaljust discussed). A random
`change in the light emitted by the source will occur in the two separate beamsat
`
`the sametime, and interference effects can still be observed.
`
`In our discussion ofwave interference of mechanical waves in Chapter 14, we
`found that two waves could add together either constructively or destructively.
`In constructive interference, the amplitude of the resuliant wave is greater
`than that of either individual wave, whereas in destructive interference, the
`resultant amplitude is less than that ofeither individual wave. Electromagnetic
`waves also undergo interfer
`ence, All interference asso-
`ciated with electromagnetic
`waves arises fundamentally
`as a result of combining the
`electromagnetic fields that
`constitute
`the
`individual
`waves,
`
`Peter Arnold, Inc.)
`
`27.2 + YOUNG’S DOUBLE-SLIT EXPERIMENT
`
`Interference in light waves fromtwoslits was first demonstrated by Thomas Young
`in 1801. A schematic diagram of the apparatus used in this experiment is shownin
`
`Figure 27.1a. (Young used pinholes in his original experiments, rather than slits.)
`a narrowslit, S,. The light waves
`Light is incident on a screen in which there
`
`
`emerging fromthis slit arrive at a secondscr
`i that contains two narrow, parallel
`sl
`5, and $,. These nwoslits serve as a
`f coherent light sources because
`
`
`te from the same source, S,, and therefore main-
`waves emerging from themorigi
` lain a constant phase
`relations
`The light from the twoslits produces a visible
`the glass. Studies of such pat-
`pattern on screen C; the pattern consists ofaseries of bright and dark parallel
`ternsled to the development of
`bandscalled fringes (Fig. 27.1b). When the light from S, and Sy arrives at a point
`tempered glass.
`(James | Amos,
`
`on the screen so that constructive interference occurs at
`that
`location, a bright
`fninge appears. When the light from the
`lits combines destructively at any
`location, a dark fringe results.
`
`784
`
`
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`Chapter 27
`
`Wave Optics
`
`tb) (a)
`
`(a) Scher
`Figure 27.1
`tic diagram of Young's doubbe-slit experiment. The narrow slits act as
`
`wave sources, Slits 8, and S, behave as coherent sources that produce an interference pattern
`
`we that this drawing is not to scale.) (b) Thefringe pattern formed on screen C
`on screen C.
`could look like this.
`
`Figure 27.2 is a schematic diagram of some of the ways the two waves in Young's
`experiment can combine at screen (. In Figure 27.2a, the two waves, which leave
`the twoslits in phase, strike the screen at the central point, P. Because these waves
`travel equal distances, they arrive in phase at P, and as a result constructive inter-
`ference occurs at this location and a bright fringe is observed. In Figure 27.2b, the
`
`
`
`fringe
`
`Screen
`
`fringe y Bright
`
`
`ye Dark
`
`T2
`
`fringe
`
`
`
`ib)
`
`tc)
`
`(b) Con
`(a) Constructive interference occurs at Pwhen the waves combine.
`Figure 27.2
`structive interference also occurs at @ (¢) Destructive interference occurs at A because the wave
`fromthe upperstit falls half a wavelength behind the wave fromthe lower dit. (Notethat these
`figures are not drawn to scale.)
`
`27.2
`
`Young's Double-Slit Experiment
`
`787
`
`rays is y — y = dsin O. (Note that this igure is not drawnto
`
`Viewing screen
`
`seale,)
`
`Figure 27.3 Geometric con-
`struction for describing Young's
`double-lit experiment. The
`path difference between the two
`
`two light waves again start in phase, but the upper wave has to travel one wavelength
`
`farther to reach point Q on the screen. Because the upper wave falls behind the
`lower one byexactly one wavelength, they still arrive
`in phase at Q, and so a second
`bright fringe appears at this location. Now consider point R, midway between P
`and Qin Figure 27.2c. At this location, the upper wave has fallen half a wavelength
`behind the lower wave. This means that the trough from the lower wave overlaps
`the crest from the upper wave, giving rise to destructive interference at X, For this
`reason, one observes a dark fringe at this location.
`We can obtain a quantitative description of Young's experiment with the help
`
`of Figure 27.3, Consider point P on the viewing screen;
`the screen is located a
`
`perpendicular distance of L from the screen containing
`slits 5, amd 55, which are
`
`separated by a distanceof d, and 7, and fy are the distances the waves travel from
`slit to screen. Let us assume that the source is monochromatic. Under these con-
`ditions, the waves emerging from5, and S, havethe same frequency and amplitude
`and are in phase. Thelight intensity on thescreenat /is the resultant ofthelight
`coming from both slits. Note that a wave from the lowerslit travels farther than a
`wave fromthe upper slit by an amount equal to dsin 6. This distanceis called the
`path difference, 6 (lowercase Greek delta), where
`é6=n—1r=dsne
`
`[27.1]
`
`This equation assumes that the two waves are parallel to each other, whichis ap-
`
`proximately true because £ is much greater than d. As notedearlier, the value of
` se
`or out
`
`
`this path difference determines whether or not the two waves are in ph:
`of phase whentheyarriveat P. If the pathdifference is either zero or someintegral
`multiple of the wavelength, the twowaves are in phase at Pand constructive inter-
`
`ference results. Therefore, the condition
`for bright fringes, or constructive inter-
`ference, at Pis
`
`Path difference
`
`6= dsin 6 = mA
`
`(m= 0, $1, $2
`
`[27.2]
`
`‘Conditions for constructive
`interference
`
`The number m is called the order number. The central bright fringe at @= 0
`(m = 0) is called the zeroth-order maximum. The first maximumoneither side,
`when m= +1,
`is called the first-order maximum, and so forth
`In a similar way, when the pathdiffer
`is an odd multiple of A/2, the two
`
`waves arriving at Pare 180° out of phase
`Five rise to destructive interference
`Therefore, the condition for dark fringes,
`or destructive interference, at
`? is
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`Chapter 27
`
`Wave Optics
`
`272
`
`Young's Dowhle-Slit Expertiment
`
`789
`
`Conditionsfor destructive *
`interference
`
`
`
`
`interference pattern involving
`An
`water waves is produced by two vi-
`“s Sun
`
`face. Thepatternis analogousto
`that observed in Young’
`tole
`
`slit experiment. Note the
`
`regions
`‘of com
`ctive and destructive in-
`terference.
`(Mickard Magna, Funda:
`imental Phisteygraptis)
`
`[27.3]
`(m= 0)21, +2). 3.)
`b= dsino=(m+5)a
`It is useful to obtain expressions for the positions of the bright and dark fringes
`measured vertically from Oto P. In addition to our assumption that L > d, we shall
`assume that d > A—thatis, the distance between the twoslits is much larger than
`
`the wavelength. This situation prevails
`in practice because Lis often on the order
`of 1 m whereas d is a fraction of a millimeter and A is a fraction of a micrometer
`for visible light. Under these conditions, @ is small, and so we can use the approx-
`imation sin @= tan @. From the triangle OPQ in Figure 27.3, we see that
`sin = tan @= oI
`Using this result and making the substitution sin @ = mA/d from Equation 27.2, we
`see that the positions of the bright fringes measured from O are given by
`AL
`(27.5)
`Yorigns = > ™
`Likewise, using Equations 27.3 and 27.4, we find that the dark fringes are located
`at
`
`(27.4)
`
`1
`AL
`(27.6)
`Maas = — (m+ 9)
`As we shall demonstrate in Example 27.1, Young's double-slit experiment pro-
`vides a method for measuring the wavelength of light. In fact, Young used this
`technique
`to make thefirst measurement of the wavelength of light, In addition,
`
`the experiment gave the wave model oflight a great deal of credibility. Today we
`still use the phenomenon of interference to describe many observations of wave-
`like behavior.
`
`Thinking Physics 2
`If your stereo speakers are connected “out of phase"—thatis, with one speaker con-
`
`nected correctly and the other withits wires reversed,
`bass in the music tends to
`be weak. Whydoes this happen, and why is ita problem for the bass and not the treble
`notes?
`Reasoning This is an acoustic analog to double-slit interference, The two speakers act
`as sources of waves, just like the two slits in a Young's double-slit experiment, [f the
`speakers are connected correctly, and the same sound signal is fed to each speaker,
`both speakers move inward and outward at the sametime in response to the signal.
`
`Thus, the sound waves arein phase as they leave the speakers,
`If you are sitting ata
`
`peint in front of the speakers and midway betweenthem, you will be located at the
`
`zero-order maximum—the
`interference is constructive and the sound will be loud.
`If
`
`one speaker is wired backward, then one speaker will be moving outward while the
`
`
`other is moving inward. The sound leaves the two speakers half a wavelength out of
`
`phase. Thus, if you are sitting at the same place, you will be at an interference
`mini-
`mum. This is a particular problem for
`the bass due to the long wavelength of low
`frequency notes. This results in a very large region ofdestructive interference in front
`of the speakers, on the order ofthesize of the room. The much shorter wavelengths
`of the high-frequency notes result in closely spaced maxima and minima. The spacing
`can be on the order of the size
`of the head and smaller. Thus, if ome ear is ata
`minimum, the other might be at a maximum. What's more, small movements of the
`head will result in a shift from the position of a minimumto that of a maximum.
`
`Thinking Physics 3
` tenna rather than a cable system.
`Suppose you are watchingtelevision by means of an
`If an airplane flies near your location, you may notice wavering ghost images in the
`television picture. What might cause this?
`Reasoning Your television antenna will receive twosignals—thedirect signal from the
`
`
`Transmitting antenr a reflected signal from the surface of the airplane. As the
` -
`
`airplane changes position, there are some times when these nwo signais
`phase
`and other times when they are out of phase. As a result, there is a variation in the
`the combined signal received at your antenna. This variation is evidenced
`
`ng ghost images of the picture
`
`
`
`Thinking Physics 1
`Consider a double-slit experiment in which a laser beamis passed through a pair of
`
`veryclosely spaced slits, and a clear interference patternis clisplayed ona distantscreen.
`Example 27.1 Measuring the Wavelength of a Light Source
`Now, suppose you place smoke particles between the double slit and the screen. With
`(b) Caleulate
`the presenceof the smoke particles, will you see the effects ofthe interferencein the
`A viewing sereen is separated from a double-slit source by
`
`1.2 m. Thedista:
`space between theslits and the screen,or will you only see the effects on the screen?
`between the twoslits is 0.030 mm, The
`fringes.
`second-order bright fringe (m=2) is 4.5 emfrom the center
`Reasoning Youwill sec the effects in the area filled with smoke. Therewill be bright
`Solution From Equation 27.5 andtheresults to (a), we get
`line, (a) Determine the wavelength of the light
`lines directed toward the bright areas on the screen and dark lines directed toward
`the dark areas on the screen. In Figure 27.3, the geometrical constructionis important
`10°27 m,
`15 x
`Solution Wecan use Equation 27.5, with »
`ALim +
`1)
`
`
`for developing the mathematical description of interference.
`It is subject to misinver-
`
`
`
`
`
`m= 2.0=1.2m, and d= 3.0 x 10°? m Jos Yon
`
`
`tion, however, becauseitn
`
`
`
`p
`hit suggest that th
`¢ does not occur
`
`
`
`dy 1079 m) (4.5x10°* m)(3.0 % Ar
`thelight rays fromthe two slits strike the screenat the
`sameposition, The morefr
`
`
`
`ml
`?x1.2m
`at
`
`
`diagram for
`this situation is Figure 27.1, in which it
`is clear that there are paths
`of
`
`destructive and constructive interference all the way fromtheslits to the screen. These
`6.6% 10°? m=
`560 om
`paths will be made visible by the smoke.
`
`the
`
`distance
`
`adjacent bright
`
`%
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`27.2
`
`Young's Double-Stit Experiment
`
`791
`
` hs
`[= Ef = 467 cos?(b/2) sin?( ot ES )
`Because mostlight-detecting instruments measure thetime-averaged light intensity,
`and the time-averaged valueofsin*(wt + 6/2) over onecycleis 1/2, we can write
`the average intensity at Pas
`
`[27.11]
`L, = fy cos*(@b/2)
`[Note that
`intensity.
`the maximum possible time-averaged light
`is
`where J,
`fy ® (Ey + Ey)? = (24)? = 46") Substituting Equation 27.8 into Equation 27.11,
`wefind thai
`
`* Average light intensityfor the
`doubleslit interference
`pattern
`
`
`(= sin @3
`Alternatively, because sin # = / L for small values of @, we can write Equation 27.12
`q
`q
`in the form
`
`L, = fh, cos
`
`[27.12]
`
`wd
`ote;)
`Constructive interference, which produces intensity maxima, occurs when the
`quantity myd/ALis an integral multiple of 7, corresponding to y = (AL/d)m. This
`
`is consistent with Equation 27.5. Intensity distribution versus @ is plotted in Figure
`27.5. Note that the interference pattern consists of equally spacedfringes of equal
`intensity. However, the result is valid only if the sli-toscreen distance, L,
`is large
`relativeto theslit separation, and only for small yalues of #.
`
`[27.13]
`
`Intensity distribution
`Figure 27.5
`versus sin @ for
`the double-slit pat-
`tern when the screen is far
`fromthe
`twa slits (Ld).
`(Photo
`
`net, Mf. Fromcon, andfC.
`Th
`
`
`Optical
`Phenomena, Berlin, Springe
`
`
`
`
`SS
`
`a
`
`Chapter 27
`
`Wane Optics
`
`Intensity Distribution of the Double-Slit Interference Pattern
`Weshall now calculate the distribution of light intensity (the energy delivered by
`the wave per unit area per unit time) associated with the double-slit interference
`pattern. Again, suppose that the two slits represent coherent sources of sinusoidal
`waves. Hence, the two waves have the same angular frequency, #, and a constant
`phase difference, . The total electric field at the point P on the screen in Figure
`27.4 is the vector superposition of the two waves. Assuming the two waves have the
`same amplitude, &,, we can write the electric field at Pdue to each wave separately
`as
`
`£, = & sin et
`
`and
`
`EF, = & sin(wt + d)
`
`[27.7]
`
`Although the waves have equal phase attheslits, their phase difference, ¢, at P
`depends on the path difference, 6 = r, — r, = d sin @. Because a path differ-
`ence of A corresponds to a phase difference of 27 rad (constructive interference),
`whereas a path difference of A/2 corresponds to a phase difference ofm rad (de-
`structive interference), we obtain the ratio
`
`ae
`é on
`pe pa aes
`a
`A
`
`[27.3]
`
`[27.9]
`
`This equation tells us precisely how the phase difference & depends on the
`angle @,
`Using the superposition principle and Equation 27,7, we can obtain the result-
`ant electric field at the point P:
`
`Ep = & + Ey = [sin ot + sin(wt + b)]
`To simplify this expression, we use the trigonometric identity
`
`A+B
`A- 5
`)
`sin A+ sin B= 2 sin(
`) cos
`2
`:
`Taking A = wt + @ and # = wt, we can write Equation 27.9 in the form
`ob
`[27.10]
`Ep = 26 cos($) sin(ot 37 |
`Hence, the electric field at P has the same frequency w as the original two waves,
`but its amplitude is multiplied by the factor 2 cos(¢/2). To check the consistency
`of
`this result, note that
`if @ = 0, 27,4, ... , the amplitude at P is 24, cor-
`responding to the condition for constructive interference. Referring to Equation
`27.8, we find that our result
`is consistent with Equation 27.2. Likewise,
`if
`@= 7, 30,57, .
`.
`.
`, the amplitudeat P is zero, whichis consistent with Equation
`
`27.3 for destructive
`interference.
`Finally, to obtain an expression for thelight intensity at P, recall that the im-
`tensity ofa wave is proportional to the square of the resultantelectric field at
`that point (Chapter 24, Section 24.6). Using Equation 27.10, we can therefore
`express the intensity at Pas
`
`
`
`Figure 27.40 Construction for ana-
`lyzing the double-slit interference
`pattern. A bright region, or inten-
`sity maximum, is observed at Q.
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`Chapter 27
`
`Wave Optics
`
`
`
`27.3
`
`Change of Phase Due to Reflection
`
`793
`
`180° phase change
`
`—--
`—— .
`
`Free support
`
`
`
`We have seen that the interference phenomena arising from two coherent
`sources depend on the relative phase of the waves at a given point. Furthermore,
`the phase difference at a given point depends on the path difference between the
`two waves. The resultant intensity at a point is proportional to the square of
`the resultant amplitude. That is, the intensity is proportional to (£, + E,)*, It
`would be incorrect to calculate the resultant intensity by adding the intensities of the
`individual waves. This procedure would give a different quantity, namely E,? +
`E,?. Finally,
`(E, + £)° has the same average value as £\* + £,*, when the time
`Rigid support
`average is taken over all values of the phase difference between E, and £,. Hence,
`the principle ofenergy conservation is not violated.
`String analogy
`CONCEPTUAL PROBLEM 1emanateteieeeLa
`Consider a dark fringe in a two-slit interference pattern, at which almost no light energy is
`No phase change
`arriving. Waves from both slits travel to this point, but the waves cancel. Where does the
`energy gor
`
`(Bb)
`
`S—
`
`-
`
`(a) A raytraveling in medium1 reflecting fromthe surface of medium 2 under-
`Figure 27.7
`goes a 190° phase change. Theright side shows the analogy with a reflected pulse on a string
`fixed at one end. (b) A raytraveling in medium 1 reflecting from the surface of medium 2 with
`n, > #, undergoes no phase change. Theright side shows the analogy with a reflected pulse on
`a string the end ofwhichis free.
`
`the wave is reflected from a boundaryleading to a mediumof lower index ofre-
`fraction. The part of
`the wave that crosses the boundary undergoes no phase
`change.
`
`
`
`
`
`The colors, produced just before the
`(left) A layer of bubbles on water produced by soapfilm
` ted from the front and back of
`bubbles burst, are duc to interference betweenlight
`
`
`fon the thickness of thefilm, rang-
`the thin film of water making the bubble, The
`ing from
`black where the film is at
`its thinnest
`
`
`he
`film gets thicker,
`(Dr
`ps
`TESS,
`Scien
`Lirer)
`(night) Thin film interference
`n
`filmof aif on water displays interfer
`ence shown by the pattern of colors when white
`
`light is incident on the film,
`The film thick
`
`ness varies, thereby producing the interesting color pattern, Tom #
`4, Phy
`
` het
`
`Mirror
`
`eo)
`
`Figure 27.6 Lloyd's mirror, An
`interference pattern is produced
`‘on a screen at Pas a result of the
`combination of the direct ray
`(blue) and the reflected ray
`(brown). The reflected ray ander-
`goes a phase change of 180°
`
`27.3 + CHANGE OF PHASE DUE TO REFLECTION
`
`Young's method of producing two coherent light sources involves illuminating a
`pair of slits with a single source. Another simple arrangement for producing an
`interference pattern with a single light source is known as Lloyd's mirror. A light
`source is placed at point Sclose to a mirror, as illustrated in Figure 27.6. Waves can
`reach the viewing point, P, either by the direct path SP or by the indirect path
`involving reflection from the mirror. The reflected ray can be treated as a ray
`onginating from a source at §’, located behind the mirror. This source 5’, which
`is the image of 5, can be considered a virtual source.
`At points far from the source, one would expect an interference pattern due
`to waves from § and 5",
`just as is observed for tworeal coherent sources. An inter
`ference patternis indeed observed. However, the positions of the dark and bright
`fringes are reversed relative to the pattern of two real coherent sources (Young's
`experiment). This is because the coherent sources at Sand S’ differ in phase by
`180°, This 180° phase change is produced on reflection.
`To illustrate this further, consider the point P’ at which the mirror meets the
`screen. This point is equidistant from Sand S’. If path difference alone were re-
`sponsible for the phase cifference, one would expectto see a bright fringe at P*
`(because the path difference is zero for this point), corresponding to the central
`fringe of the two-slit interference pattern. Instead, one observes a dark frmge at P”
`because of the 180° phase change produced by reflection, In general, an electro-
`magnetic wave undergoes a phase change of 180° on reflection from a medium
`of higher index of refraction than the one in which it was traveling,
`It is useful to draw an analogy between reflected light waves and the reflections
`of a transverse wave on a stretched stringwhen the wave meets a boundary (Ghapter
`13, Section 13,7), as in Figure 27.7. The reflected pulse on a string undergoes a
`phase changeof 180° when it is reflected from the boundaryof a denser medium,
`
`such asa heavier string,
`ino phase change whenit is reflected from the boundary
`ofa less dense medium. Ina similar way, an electromagnetic wave undergoes a 180°
`phase change when reflected from the boundaryof a mediumof higher index of
`refraction than the one in which it was traveling. There is no phase change when
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`IPR2023-00783
`
`
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`27.5
`Diffraction
`when the thickness satisfies the condition 2m
`= (m+),
`ence appears at point 0, the apex, because the upper re-
`
`corresponding to thicknesses of A/4n, $A/4n, 54/4, and so
`C ¢ Change and thelower one
`
`flected ray
`undergoes a 180°
`If white light is used, bands ofdifferent colors are ob-
`on,
`does not. According to Equation 27.17, ether dark bands
`
`
`appear when 2nt= mA,
`sot = A/2n, & = A/n, = served at different points, corresponding to the different
`
`34/2n, and so on, Ina
`r way, bright bands are observed
`wavelengths oflight
`
`797
`
`Incident
`light
`
` light. The dark blue areas correspond to pos structive interference
`
`Figure 27.10 (Example 27.4) Interference bands in reflectedlight
`can be observedby illuminating a wedge-shapedfilo with mone.
`
`chroma
`as OF de~
`
`
`
`27.5 « DIFFRACTION
`
`Supposea light beamis incident ontwoslits, as in Young's double-slit experiment.
`
`Ifthelight truly
`traveledin straight-line paths after passing throughtheslits, as in
`Figure 27.11a, the waves would not overlap and no interference pattern would be
`seen. Instead, Huygens’ principle requires that the waves spread out from theslits,
`as shown in Figure 27.11b. In other words, the light deviates froma straighttine
`path and enters the region that would otherwise be shadowed, This divergence of
`
`light from its initial line of travelis called diffraction.
`In general, diffraction occurs when waves pass through small openings, around
`obstacles, or by sharp edges. For example, when a narrowslit is placed beween a
`distantlight source (or a laser beam) anda screen, thelight producesa diffraction
`
`pattern like that in Figure 27.12.
`The pattern consists of a broad, intense central
` tense secondary
`band, the central maximum, flankedby a series of narrower,|
`
`*
`ol
`
`Lg
`
`~
`
`Noss
`
` 4
`
`\a
`
`796
`
`Chapter 27
`
`Wawe Optics
`
`Example 27.2 Interference in 2 Soap Film
`‘Calculate the minimum thickness of a soap bubble film
`(m = 1,33)
`that results in constructive interference in the
`reflected light if the filmis illuminated with light the wave-
`Jength in free space ofwhich is 600 nm.
`EXERCISE 1 What other film thicknesses produce construc-
`Solution The minimum film thickness for constructive in-
`tive interference?
`Answer
`338 nm, 564 nm, 789 nm,and
`s0.0n.
`terference in the reflected light corresponds to m= 0in
`
`Equation 27.16. This give
`f= A/2, or
`
`
`600 nm
`A
`"faa; ee
`
`Example 27.3 Nonreflecting Coatings for Solar Cells
`Semiconductors such as silicon are used to fabricate solar
`cells—devices that generate electricity when exposed to sun-
`light. Solar cells are often coated with a transparent thin film,
`such as silicon monoxide (SiO, » = 1.45), in order to mini-
`
`mize reflective bosses from the surface, A silicon solar cell
`(n= 3.5) is coated with a thin film ofsilicon monoxide for
`this purpose (Fig. 27,9), Determine the minimum film thick-
`
`180° phase
`change
`
`180” phase
`
`change
`
`Figure 27.9 (Example 27.4) Reflective losses from a silicon
`solar cell are minimized by coating it with a thin film of silicon
`monoxide,
`
`ness that produces the least reflection al a wavelength of
`550 nm, which is the center of the visible spectrum.
`Reasoning The reflected light is a minimum when rays 1
`and 2 in Figure 27.9 meet the condition of destructive inter-
`ference. Note that both rays undergo a 180" phase change on
`reflection in this case, one from the upper and one from the
`lower surface. Hence, the net change in phase is zero duc to
`reflection, and the condition for reflection minimum re-
`quires a path difference of A,,/2; hence, 2¢ = A/2n,
`Solution Because 2¢ = A/2n, the required thickness is
`
`Typically, such antireflecting coatings reduce the reflec:
`tive loss from 30% (with no coating) to 10% (with coating),
`
`thereby increasing the cell's
`ency, because morelight is
`available to create charge carriers in the cell. In reality, the
`coating is never perfectly nonreflecting because the required
`thickness is wavelength-dependent and the incident light
`covers a wide range of wavelengths.
`
`Glass lenses used in cameras and other optical instruments
`are usually coated with a transparent thin film, such as mag-
`nesium fluoride (MgF,),
`to reduce or eliminate unwanted
`reflection, and thus such coatings enhance the transmission
`oflight throughthe lenses.
`
`
`
`
`(a) If light waves did not
`Figure 27.11
`spread out after passing throughtheslits,
`iterference would occur,
`(b) The
`waves from the two shits overlap as
`read out, filling the expected shic
`“
`regions with light and producing in
`ncefringes.
`im
`af
`uN
`7
`.
`*
`Figure 27.12)
`The diffractionpat
`
`
`| 1\7a}—1-3F :
`Ly
`ternthat appears on ascreen when
`
`the
`»
`rrow verti
`Example 27.4 Interference in a Wedge-Shaped Film
`-
`i.
`c
`The pattern
`consi
`a
`
`Reasoning and Solution
`Athin, wedge-shaped filmof refractive index nis illu
`Theinterference patternis that of
`
`Las
`al
`ce
`band and a series of
`-
`a thin film of variable thickness surrounded by air, Hence,
`
`
`
`with monochromatic light of wavelength A,
`
`narrower side
`intense
`and
`jess
`
`the pattern is a series of alternating bright and dark parallel
`Figure 27.10. Describe the interference pal
`this case
`bands. A dark band corresponding to destructive interfer-
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`sider waves | and 3, which originateat the bottomandcenter oftheslit, respectively.
`
`To reach the same point on the viewing screen, wave | travels farther than wave 3
`by an amount equal to the path difference (a/2) sin #, where ais the width ofthe
`slit. In a similar way, the path difference between waves 3 and5 is also (a@/2) sin 4
`If the pathdifference is exactly onehalf of a wavelength (corresponding to a phase
`difference of 180°), the two waves cancel each other and destructive interference
`gpGfLens
`
`ee
`g.--
`results. This is true, in fact, for any two waves that
`originate at points separated by
`
`—e-fEl__
`half the slit width, because the phase difference between two such points is 180°,
`Therefore, waves from the upper half ofthe slit interfere destructively with waves
`Figure 27.14 (a) Fraunhoferdiffraction
`fromthe lower half of the slit when
`»
`pattern of a single slit. The pattern con-
`sists of a central bright region flanked by
`= sin0=
`muchweaker maximaalternatingwith
`|
`Stir
`a
`dark bands. (Note that this is not to
`a.
`a
`A
`2
`Figure 27.13 Diffraction partern
`scale.) (b) Photograph ofa single-slit
`Hreomingy
`of a penny, taken with the penny
`Fraunhofer diffraction pattern,
`(From
`Mais
`midway betweenscreen and
`M. Cagnet, M. Francom, ondJ.C, Thierr, Atlas of
`source,
`(Gourtery of P. M. Rinard,
`Optical Phenomenc:, Berlin, Springer-Verlag, 1962,
`from Am. J. Phys. 44:70, 1976)
`plate 18}
`
`
` Fiat
`
`(by
`
`bands (called secondary maxima) and a series of dark bands, or minima, This
`cannot be explained within the framework ofgeometric optics, which says thatlight
`rays traveling in straightlines should cast a sharp rendition ofthe slit on the screen,
`Figure 27.13 shows the diffraction pattern and shadow ofa penny. The pattern
`consists of the shadow, a bright spot at its center, and a series of bright and dark
`bandsof light near the edge of the shadow, The bright spot at the center (called
`the Arago bright spot after its discoverer, Dominique Arago) can be explained
`through the wave theory of light, which predicts constructive interference at this
`point. In contrast, from the viewpoint of geometric optics, the center of the pattern
`would be completely screened by the penny, and so one would never observe a
`central bright spot.
`Fraunhofer diffraction occurs when the rays reaching the observing sereen
`are approximately parallel. This can be achieved experimentallyeither by placing
`the observing screen far from the slit or by using a converging lens to focus the
`parallel rays on the screen,as im Figure 27.14a. A bright fringe is observed along
`the axis at @ = 0, with alternating dark and brightfringes on each side of the central
`bright fringe. Figure 27.14b is a photograph ofa single-slit Fraunhoferdiffraction
`pattern.
`
`If we divide the slit into four parts rather Uhan ovo and usesimilar reasoning,
`we find that the screen is also dark when
`
`2A
`sin 8 = —a
`
`Likewise, we candivide theslit intosix parts and showthat darkness occurs on
`the screen when
`
`SA
`sin 6 = —a
`
`Therefore, the general condition for destructive interference is
`A
`sin d= m-
`
`(m= 1, 22, 285)
`
`[27.18]
`
`* Condition for destructive
`Interference
`
`Equation 27.18 gives the values of @ for which the diffraction pattern has zera
`intensity —thatis, a dark fringeis formed. However, Equation 27.18 tells us nott
`about the variation in intensity along the scre