# Ch 8 - Models for light .pdf - scunderwood - home

Ch 8 - Models for light .pdf - scunderwood - home

Ch 8 - Models for light .pdf - scunderwood - home

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Light sources, in conjunction with other optical elements such aslenses or mirrors, can produce rays of <strong>light</strong> that diverge, converge or travelparallel to each other (Figure 8.2b). In each case the rays are an indicationof the direction of travel of the <strong>light</strong>; essentially <strong>light</strong> is being modelledas a stream of particles. The incandescent (filament) and fluorescent <strong>light</strong>globes in your <strong>home</strong> emit <strong>light</strong> in all directions. A point source of <strong>light</strong> is anidealised <strong>light</strong> source that emits <strong>light</strong> equally in all directions from a singlepoint (Figure 8.2c). No single point source of <strong>light</strong> exists in reality, but asmall filament lamp can be considered a good approximation.Lasers and special arrangements of <strong>light</strong> sources with mirrors or lensescan produce parallel rays of <strong>light</strong> in a beam. Very distant point sources of<strong>light</strong> can also be considered to be sources of parallel <strong>light</strong> rays. For example,on the Earth we treat the <strong>light</strong> rays that reach us from the Sun as thoughthey were parallel to each other. This is because at such a large distancefrom the source, the angle between adjacent rays would be so tiny as to beconsidered negligible (Figure 8.2d).Although a particle description <strong>for</strong> <strong>light</strong> and the accompanying raymodel are convenient <strong>for</strong> representing the behaviour of <strong>light</strong> in all of thesecases, it has long been understood that <strong>light</strong> is not made up of ordinaryparticles. Light involves the transfer of energy from a source, but there areno tangible particles carrying this energy. With developing technology overthe last two centuries physicists have been able to make more and moresophisticated observations of <strong>light</strong>. Later in the chapter we will see that amore refined electromagnetic wave model of <strong>light</strong> incorporates the wavelikeproperties of <strong>light</strong>. The ray approach is still useful. If <strong>light</strong> is consideredto be a wave emanating from its source, then rays may simply be used torepresent the direction of travel of the wavefronts (see Figure 8.3). The pointsource of <strong>light</strong> discussed above may be considered to be a point source ofspherical wavefronts, like the ripples that travel out from a stone droppedinto a pond (Figure 8.4).Modelling reflectionThe reflection of waves was discussed in <strong>Ch</strong>apter 7. Light has beenobserved to obey the same laws of reflection that apply to waves and soevidence is provided <strong>for</strong> the argument that <strong>light</strong> is a wave. Using a wavemodel, the reflection of <strong>light</strong> would be represented as a series of wavefrontsstriking a surface and reflecting as shown in Figure 8.5. However it is farmore common to model the reflection of <strong>light</strong> using ray diagrams and theconventions associated with them.rays traveloutwardfromtorchFigure 8.3 Rays can be used to represent thedirection of travel of <strong>light</strong> waves leaving thetorch.Figure 8.4 A point source of <strong>light</strong> may beconsidered to be a point source of sphericalwaves. Both the particle and the wave modelare consistent with the observation thatthe intensity reduces with the square of thedistance from the source.(c)wavefrontstraveloutward fromtorchsphericalwavefrontstravel outwardspoint source(a)(b)Figure 8.5 When studying reflection, ray diagrams are the most convenient way ofrepresenting the path of <strong>light</strong>.<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>251

In Year 12 you will study the photonmodel of <strong>light</strong> developed last century.The photon model draws on aspects ofthe wave and particle models. It refersto the <strong>light</strong> travelling in the <strong>for</strong>m of tinypackets of energy called photons. Theenergy of the photon is determined bythe frequency of the <strong>light</strong>.normalPhysics fileFigure 8.6 When <strong>light</strong> reflects from a planemirror, the angle of incidence equals the angleof reflection: i = r.incident rayglass layermetal layerpaint layerFigure 8.7 Most of the incident <strong>light</strong> on amirror is reflected from the silvered surface atthe back of the mirror. The glass on the frontand the paint on the back serve to protect thereflective surface from damage.(a)(b)incident rayangle of incidence iangle of reflection rreflected rayplane (flat) mirror~4%~96%Consider the plane mirror drawn in Figure 8.6. We define a normal tothe surface of the mirror as the line perpendicular (at 90°) to the mirror’ssurface at the point where an incoming or incident ray strikes the surfaceof the mirror. The angle between an incident ray and the normal is theangle of incidence, denoted i. The ray strikes the mirror and reflects withan angle of reflection, r, which is the angle between the reflected ray andthe normal.Experiment shows that whenever reflection occurs, the angle of incidencealways equals the angle of reflection. In addition, the <strong>light</strong> reflects in sucha way that the incident ray, the normal and the reflected ray all lie in thesame plane. The law of reflection can then be re-stated using a ray model<strong>for</strong> <strong>light</strong>.iThe LAW OF R…FL…CTION states that the angle of incidence, i, is equal to theangle of reflection, r (i = r). The incident ray, the normal and the reflected raywill all lie in the same plane.A normal household mirror is constructed with three separate layers: alayer of transparent glass, a thin coating of aluminium or silver depositedonto the glass to reflect the <strong>light</strong> and a backing layer of protective paint(Figure 8.7). When a beam of <strong>light</strong> strikes the surface of the mirror a tinyamount of the <strong>light</strong> energy (about 4%) is reflected from the front surface ofthe glass, but most of the <strong>light</strong> continues to travel through the glass and isreflected from the metal surface at the back. These reflected rays producethe image that is seen in the mirror.Regular and diffuse reflectionTo some extent at least, <strong>light</strong> will reflect from all surfaces, but only somesurfaces will produce a clearly defined image. If parallel rays of <strong>light</strong> areincident on a plane mirror or a flat polished metal surface, they will remainparallel to each other on reflection (Figure 8.8a). This is regular reflection(sometimes called specular reflection) and, as a result, a clear image can beproduced. Common examples of regular reflection include the reflection of<strong>light</strong> from plane mirrors, glossy painted surfaces and still water such as ina lake.When <strong>light</strong> is reflected from a roughened or uneven surface, it is scatteredin all directions as shown in Figure 8.8b. This is diffuse reflection. Parallelrays of incident <strong>light</strong> will be reflected in what seem to be unpredictabledirections. Each ray obeys the law of reflection, but the surface is irregular sothat normals drawn at adjacent points have completely different directions.Thus, <strong>light</strong> is reflected in many different directions. Most materials producediffuse reflection. For example, when looking at this page, you can see theprinting because the <strong>light</strong>ing in the room is reflected in all directions due todiffuse reflection. If the page behaved as a regular reflector, you would alsosee the (reflected) images of other objects in the room.Diffuse and regular reflection are the two extreme cases of how <strong>light</strong> canbe reflected. In reality most surfaces display an intermediate behaviour. ForFigure 8.8 (a) Regular reflection from a smooth surface occurs when parallel raysof incident <strong>light</strong> are reflected parallel to each other. (b) Diffuse reflection occurs at anirregular surface. Here, the incoming parallel rays are reflected at all angles.252Wave-like properties of <strong>light</strong>

Eclipses (continued)Sun(c)SunFigure 8.10 (a) During a total solar eclipse the Sun disappears behind the disc of the Moon. Depending on the relative positions of the Earth,Moon and Sun, this can last <strong>for</strong> as long as 7 minutes. (b) and (c) When the Moon is at its furthest from the Earth, its disc is no longer largeenough to cover the Sun, and an annular eclipse occurs, in which a thin ring (or annulus) of the Sun’s disc remains visible and the Moon blocksout only the central region.Physics fileWhen <strong>light</strong> travels past our eyes itcannot be seen. Light is invisible unlesssome of it is reflected into our eyes bytiny particles in the air. The particlesmight be dust, fog or smoke. An effectivedemonstration of this is to mark thepath of a laser beam in a darkened roomwith chalk dust.Reflection, absorption and transmissionAfter a beam of <strong>light</strong> strikes an object, there are three processes that canoccur: some of the <strong>light</strong> may be reflected from the surface, some may betransmitted through the material, and some may be absorbed into thesurface. This behaviour of <strong>light</strong> recommends a wave model, as viewing<strong>light</strong> as a particle would make it difficult to explain the <strong>light</strong> energyundergoing three different processes.Most materials are opaque to visible <strong>light</strong>; that is, they do not allow any<strong>light</strong> to pass through them. For example, brick, plaster and cardboard areimpervious to <strong>light</strong>. Opaque materials will reflect some <strong>light</strong> and absorbthe rest.Other materials are transparent. A transparent material will allow asignificant amount of <strong>light</strong> to pass through it. It may absorb some, andsome may even be reflected from the surface of the material. Clear glass,perspex, water and plastic food wrap are common examples of transparentmaterials.Some materials classified as transparent allow some of the incoming<strong>light</strong> to pass through but distort the path of this <strong>light</strong> so that no clear imagecan be seen through the material. Although the rays of <strong>light</strong> have passedthrough the material, the relationship between them has been altered. Suchmaterials are called translucent, and examples include frosted glass, tissuepaper and fine porcelain. Translucent materials are particularly useful ifan area needs to be illuminated but privacy is required. Frosted or mottledglass is often used <strong>for</strong> bathroom windows. In other situations, a translucentmaterial is used to deliberately scatter <strong>light</strong>. For example, the cover arounda fluorescent lamp or the ‘pearl’ finish of an incandescent globe can softenhousehold <strong>light</strong>ing by diffusing it, thereby producing less harsh shadows.Light is a <strong>for</strong>m of energy, and when <strong>light</strong> is absorbed by a material, theenergy it carries is converted directly into heat, warming the material up.254Wave-like properties of <strong>light</strong>

Some of the <strong>light</strong> energy will also be reflected, bouncing directly from thesurface. Experience tells us that a shiny, smooth surface tends to reflect agreater proportion of an incoming <strong>light</strong> beam than a roughened surface.Figure 8.11 illustrates the three possibilities <strong>for</strong> the behaviour of <strong>light</strong>falling, or incident, on a transparent material. Many transparent materialswill only absorb tiny amounts of the <strong>light</strong> energy falling on them. For thisreason, we will choose to ignore absorption in our discussions. However,it is important to note that no material is able to allow 100% of the incident<strong>light</strong> to pass through. There are no perfectly transparent materials; somereflection and absorption of the incident <strong>light</strong> will always occur.Figure 8.12 shows the effect of the transmission and reflection of <strong>light</strong>occurring simultaneously. If you look into a shop window you can oftensee an image of your own face and the streetscape behind you as well as theitems on display in the window. The image of your face and the streetscapeare the result of reflection: the window is acting as a mirror. However, youalso see the items inside the shop as a consequence of the transmissionthrough the glass of the <strong>light</strong> reflected from objects inside the shop.incident<strong>light</strong>reflected <strong>light</strong>boundary betweentwo surfacesabsorbed <strong>light</strong>transmitted <strong>light</strong>Figure 8.11 Light incident on the surfaceof a transparent material is partly reflected,partly transmitted and partly absorbed bythe material. The relative amounts of the <strong>light</strong>experiencing these processes will depend onthe nature of the material in question.Physics fileEuclid, a philosopher and mathematician(330–260 BCE) described the law ofreflection in his book Catoptrics.However, Euclid also upheld Plato andPtolemy in their misguided belief inextramission. Euclid claimed that visionwas possible because rays from our eyesspread out in all directions and fell onthe objects that we see. He proposed thatmore rays fell on closer objects and sothey were seen more clearly. Very smallor distant objects were supposed to bedifficult to see because they would liebetween adjacent rays.Echoes of this idea continued well intothe 14th century when vision was stilldescribed in terms of ‘extramitted’visual rays emanating from the eye.Roger Bacon in the late 16th century,however, proposed that <strong>light</strong> actuallytravelled from the object to theobserver’s eyes.Figure 8.12 Multiple images are <strong>for</strong>med by the window, which is simultaneouslyreflecting and transmitting <strong>light</strong> from outside and inside the shop respectively.<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>255

Physics in actionThe pinhole cameraThe operation of a pinhole camera provides further evidencethat <strong>light</strong> travels in straight lines. The operation of a camerais more easily explained using a ray model <strong>for</strong> <strong>light</strong>. Raysare used to represent the direction of travel of <strong>light</strong> throughthe camera. A pinhole camera consists of a sealed box witha small pinhole in the centre of one side. When open, thepinhole allows a limited amount of <strong>light</strong> into the camera and<strong>for</strong>ms an image on the opposite inside wall of the box. If theopposite wall of the box is lined with photographic film, apermanent image can be developed. The amount of <strong>light</strong> thatcan enter the camera is determined by the size of the pinhole,and because this is small, the object to be photographedmust either be well illuminated or be a luminous object itself.Typical exposure times reach several minutes, so it is onlypractical to use a pinhole camera to photograph stationaryobjects.Ray tracing can be used to determine the size and natureof the image that will be produced on the film. First, consideronly the uppermost tip of the object being photographed.An infinite number of rays can be considered to emanatefrom this point. Only a few of these rays will pass into thecamera because of its tiny aperture. These rays continue onto strike the photographic film lining the back wall of thecamera. These rays will not all fall on exactly the same pointon the photographic plate, but if the pinhole is small, theylie sufficiently close together <strong>for</strong> an image to be <strong>for</strong>med. Thegeometry of the camera dictates that rays leaving the top ofthe object strike the bottom of the film. Similarly, the raysfrom the bottom of the object strike the top of the film. Thismeans that the image on the film is inverted relative to theobject (Figure 8.13).The image in a pinhole camera is faint because only a little<strong>light</strong> has been allowed to reach the photographic film. To makea brighter image, a larger diameter pinhole might be used, butthis will not produce satisfactory results. The image will bebrighter but it will be blurred. This is because the larger holewill allow rays from one point on the object to strike differentpoints on the film. We say that the rays are not focused.On examining the geometry in the ray diagram <strong>for</strong> apinhole camera, it is clear that a relationship exists betweendistantobjectpinholescreen, film orphotographic platereal invertedimageFigure 8.13 The pinhole camera. If the object is at a distance thatis 10 times the distance from the pinhole to the film, then the sizeof the image will be one-tenth of the size of the object.the distances from the pinhole to the object and image andthe relative heights of the object and image. It can be seen byusing similar triangles that if an object is at a distance equalto 10 times the distance from the pinhole to the film, theheight of the image will be one-tenth of that of the object.You can build your own pinhole camera using anycontainer (card or metal) that can be sufficiently sealed toblock out all <strong>light</strong> except that falling on the pinhole. It mayhelp to paint the inside of the box matt black to preventscattered <strong>light</strong> from reflecting off the walls and back on tothe film. To make the pinhole, punch a nail hole in one walland cover it with aluminium foil. A pinhole in the foil of about1 mm diameter will produce good results. You will need toload the film and seal the box in darkness; it is a good idea topractise this a few times first. Alternatively, you can replacethe wall opposite the pinhole with tracing paper or anothertranslucent material to act as a viewing screen. This alsoneeds to be shielded from exterior <strong>light</strong> so that the image isnot flooded out. This can be done by surrounding this end ofthe camera with a cardboard tube. If a photograph is to betaken, mounting the camera on a stand is a good idea. Thecamera will produce best results with bright, distant objects.Outdoor scenery works well.8.1 summaryModelling simple <strong>light</strong> properties• Light travels in a straight path in a uni<strong>for</strong>m medium.• A straight ray model of <strong>light</strong> implies its particlenature, but rays can also be used to represent thedirection of travel of <strong>light</strong> waves.• When describing the reflection of <strong>light</strong>, <strong>light</strong> can bereadily modelled using rays.• The law of reflection states that the angle of incidenceis equal to the angle of reflection (i = r), and theincident ray, the normal and the reflected ray lie inthe same plane.• Smooth reflective surfaces produce regular (specular)reflection, whereas rough surfaces produce diffusereflection.• Light can be reflected, transmitted and/or absorbedat the surface of a material.• Materials can be classified as transparent, translucentor opaque to the passage of <strong>light</strong>.256Wave-like properties of <strong>light</strong>

glass, and the ray is said to have been refracted. Refraction occurs becausethe <strong>light</strong> changes speed as it enters a medium of different optical density.Later discussion will examine the speed of <strong>light</strong> in different media and howgreater changes in speed cause more significant deviation of the beam.Refraction and a ray/particle approachThe refraction of <strong>light</strong> can be represented using a ray/particle approach.The ray of <strong>light</strong> that strikes the boundary between two media is calledthe incident ray. A normal to the boundary is drawn at the point wherethe incident ray strikes. The angle between the normal and the incidentray is called the angle of incidence, i. The angle between the normal and thetransmitted or refracted ray is called the angle of refraction, r. The incidentray, the normal and the refracted ray all lie in the same plane (Figure8.17). The angle of deviation, D, is the angle through which the ray has beendeviated; hence D = (i − r).Refraction is only noticeable if the angle of incidence is other than 0°. Ifthe incident ray is perpendicular to the boundary, i.e. i = 0°, the direction oftravel of the transmitted ray will not deviate even though the speed of <strong>light</strong>has altered. An example of this can be seen in Figure 8.17 as the ray leavesthe prism and continues in a straight path.When <strong>light</strong> travelling through air enters a more optically dense mediumsuch as glass (in which it must travel more slowly), it will be refracted sothat the angle of refraction is smaller than the angle of incidence. We saythat the path of <strong>light</strong> has been deviated ‘towards the normal’. When <strong>light</strong>passes from glass to air it speeds up, as it has entered a less optically densemedium. The angle of refraction will be larger than the angle of incidence.The path of <strong>light</strong> is described as being refracted ‘away from the normal’.iThe behaviour of <strong>light</strong> undergoing refraction can be summarised by twostatements.• When a <strong>light</strong> ray passes into a medium in which it travels more slowly (amore optically dense medium), it is refracted towards the normal.• When a <strong>light</strong> ray passes into a medium in which it travels faster (a lessoptically dense medium), it is refracted away from the normal.Figure 8.16 When a <strong>light</strong> ray strikes thesurface of a glass prism the transmitted ray isrefracted because of a change in speed of the<strong>light</strong>. The bending occurs at the boundary ofthe two media.angle ofincidenceir Dangle ofdeviationangle ofrefractionFigure 8.17 The angles of incidence, refractionand deviation are defined as shown. If an incidentray is perpendicular to the boundary between twomedia, i.e. i = 0°, the direction of travel of the raydoes not deviate. An example of this can be seenas <strong>light</strong> leaves the glass prism.The path of refracted <strong>light</strong> is ‘reversible’. Figure 8.18 shows a ray of<strong>light</strong> incident on the left-hand side of a rectangular prism. It undergoesrefraction towards the normal at the air–glass boundary. It then continuesin a straight path through the glass until it strikes the glass–air boundarywhere it is refracted away from the normal. At each boundary the ray’s pathdeviates through the same-sized angle; hence, the ray that finally emergesfrom the prism is parallel to the original incident ray. If the <strong>light</strong> ray wassent in the opposite direction through the prism, i.e. if the starting andfinishing points of the <strong>light</strong> ray were swapped, the <strong>light</strong> ray would traceout this same path—but in reverse.Refraction and the wave approachAlthough the way in which <strong>light</strong> is reflected is modelled equally effectivelyusing either a particle or wave approach, this is not the case <strong>for</strong> refraction.The wave model is better able to explain the change in direction that isobserved as <strong>light</strong> enters a medium in which its speed is altered.Figure 8.18 The path of <strong>light</strong> through anyoptical element is reversible since the amountof deviation at any boundary is determined bythe change in speed of the <strong>light</strong>.<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>259

Prac 35Consider a <strong>light</strong> wave travelling at an angle towards a boundarybetween two media, as shown in Figure 8.19a. For example, the <strong>light</strong> maybe travelling from air into water. As soon as the <strong>light</strong> wave enters thewater it will slow down. At the moment shown in the diagram the sectionof wavefront AB that first enters the water will be travelling at a slowerspeed than the section of the same wavefront BC that has not yet enteredthe water. This first section of the wavefront then effectively lags behind theposition that it would have held had it been able to continue at its initialfaster speed. The overall result of this delay is that the direction of travel ofthe overall wavefront is altered. Figure 8.19b shows how, once a numberof wavefronts have passed into the second medium, the direction of travelof the overall wave has been deviated from its original course. A similararrangement can be constructed <strong>for</strong> the refraction of <strong>light</strong> as it speeds up.(a)original directionof travelSC(b)airSair (fast medium)RBwavefrontwaterRCwater (slow medium)AABnew directionof travelFigure 8.19 The change in the direction of <strong>light</strong> that is associated with a change of speedis called refraction. Refraction can be modelled by treating <strong>light</strong> as a wave.The bent strawFigure 8.20 The immersed portion of the straw isapparently shifted upwards due to refraction. This isbecause the rays appear to have come from a raisedposition in the glass of water.Objects partially immersed in water will be distorted because of refraction;that is, they will appear to have a kink in them. The photograph of the‘bent’ straw (Figure 8.15) illustrates this. Figure 8.20 shows the path ofthe rays which produce this illusion. As each ray of <strong>light</strong> emitted from thebase of the straw encounters the water–air boundary it is refracted awayfrom the normal, since the ray enters a less optically dense medium. Theobserver perceives these rays from the base of the straw to have come froma position higher up in the glass of water. Ray tracing can be carried out<strong>for</strong> each point along the straw, resulting in the image shown. The strawappears to have a bend in it because the part of the straw that is submergedappears closer to the air–water boundary.Apparent depthJust as the position of the submerged portion of the straw is apparentlyshifted, the actual depth of a body of water, or any transparent substance,cannot be judged accurately by an external observer because of refractioneffects. Young children often jump into a pool believing it to be muchshallower than it really is. Have you ever reached <strong>for</strong> an object at the bottomof a body of water and been surprised to find that you can’t reach it?Consider an object O at the bottom of a pool as shown in Figure 8.21.Rays from the object are refracted away from the normal at the water–airboundary. If the observer looks down into the water from directly above,he or she will perceive these rays to have come from a closer position as260Wave-like properties of <strong>light</strong>

shown. Since the floor of the pool will appear to be much closer than itreally is, the water seems safely shallow. The extent to which the depth ofthe water is altered is affected by the angle from which it is viewed.observerPhysics fileAlthough ray diagrams are commonlyused to represent the refraction of <strong>light</strong>,keep in mind that these describe onlythe direction of travel of the refracted<strong>light</strong> and do not tell the whole story.Although wave diagrams of refractionare more complicated, they are moreconvincing since they reveal the reasons<strong>for</strong> the change in direction of travel ofthe <strong>light</strong>.airwaterapparentdepth ofpoolOFigure 8.21 The apparent depth of a body of water appears less than its actual depthbecause of refraction. Light rays from a point at the bottom of the pool are refracted at thewater’s surface. The observer perceives these rays to have come from an elevated locationand interprets this as an indication of shallow water.Worked example 8.2APredict the approximate path of <strong>light</strong> through the following prisms. In each case identify thenormal, the angle of incidence and the angle of refraction.(a)(b)Oairperspexairglass(c)air(d)airglassglasswater<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>261

Solution(a)(b)irirairglassperspex(c)(d)airiairglassglasswaterirPhysics fileWillebrord Snell (1591–1626) iscommonly accredited with the discoveryof the law of refraction in about 1621.He did not immediately publish hisfindings and meanwhile the Frenchscientist René Descartes (1596–1650)published his own derivation of the lawof refraction. This caused a disputewithin the scientific community at thetime, with some claiming that Descarteshad seen Snell’s work. It is worth notingthat in France the law of refractionknown elsewhere as Snell’s law is calledDescartes’ law.At each boundary between media the <strong>light</strong> must refract either towards or away from thenormal to the boundary. The <strong>light</strong> ray is refracted towards the normal on entering each prismand away from the normal when leaving the prism. When an incident ray meets a boundarywith an incident angle of 0°, no deviation occurs.The law of refractionrScientists spent many years trying to find the relationship between theangle of incidence and the angle of refraction produced at the boundarybetween a given pair of media. At very small angles, doubling the angle ofincidence appeared to double the angle of refraction, but this relationshipdoes not hold <strong>for</strong> larger angles of incidence. In about 1621 Willebrord Snell,a Dutch scientist, found that <strong>for</strong> a given pair of media the sine of the angleof incidence was directly proportional to the sine of the angle of refraction,i.e. sin i ∝ sin r. This relationship is now known as Snell’s law:sin isin r = constant. . . . . . (i)Each combination of a pair of materials has a different constant. Forexample, the constant <strong>for</strong> the air–water interface is different from that<strong>for</strong> the glass–water interface. The constant is called the relative refractiveindex, denoted n*. Literally, an index (or listing) of numbers was created todescribe the amount of refraction or bending occurring at the boundaryof numerous pairs of transparent media. A higher relative refractiveindex <strong>for</strong> a given pair of media indicated that more bending occurred.This was a cumbersome system which was refined, as will be explainedlater in this chapter.Figure 8.22 Willebrord Snell.262Wave-like properties of <strong>light</strong>

iR…LATIV… R…FRACTIV… IND…X: n* = sin isin rEach pair of media will have a specific relative refractive index.Worked example 8.2BA student shines a thin beam of <strong>light</strong> on to the side of a glass block. She notes that whenthe angle of incidence is 40°, the <strong>light</strong> passes into the block with an angle of refraction of25°. When the angle of incidence is 70° the angle of refraction is 39°. Determine the relativerefractive index, n*, of the air–glass boundary.SolutionThe first pair of data gives:n* = sin i sin 40°=sin r sin 25° = 1.5The second pair of data gives:n* = sin i sin 70°=sin r sin 39° = 1.5i.e. the same relative refractive index, as expected.Optical density and the speed of <strong>light</strong>Different transparent media allow <strong>light</strong> to travel at different speeds. Lighttravels fastest in a vacuum, more slowly in water and even more slowly inglass. We say that glass is more optically dense than water. Table 8.1 showsthe speed of <strong>light</strong> in various media. The amount of refraction occurringat any boundary depends upon the extent to which the speed of <strong>light</strong> hasbeen altered, i.e. the ratio of the two speeds of <strong>light</strong> in the two differentmedia.Figure 8.23 represents <strong>light</strong> travelling in air and meeting three substancesof different optical density. Light bends most when its speed is mostsignificantly altered. The medium carrying the incident ray is identifiedas medium 1 and the medium carrying the refracted ray is medium 2. Theangle of refraction depends on the speed of <strong>light</strong> in the two media and theangle of incidence.sin isin r = v 1. . . . . . (ii)v 2where v 1is the speed of <strong>light</strong> in medium 1 and v 2is the speed of <strong>light</strong> inmedium 2.The index of refractionSince it is only the ratio of the speeds of <strong>light</strong> in the two different mediathat determines the degree of refraction, each medium can be allocated anabsolute refractive index, n. This is obtained by comparing the speed of<strong>light</strong> in the medium in question with the speed of <strong>light</strong> in a vacuum:speed of <strong>light</strong> in a vacuumn =speed of <strong>light</strong> in the mediumc = speed of <strong>light</strong> in a vacuum = 3.0 × 10 8 m s −1cn =v mediumTable 8.1 The speed of <strong>light</strong> indifferent media (quoted <strong>for</strong>yellow <strong>light</strong> λ = 589 nm)Prac 36Medium Speed of <strong>light</strong> (m s −1 )Vacuum 3.00 × 10 8Air 3.00 × 10 8Water 2.25 × 10 8Glass 2.00 × 10 8Diamond 1.24 × 10 8airwaterleast refractionairglassmost refractionFigure 8.23 In each case <strong>light</strong> is entering amedium of greater optical density. The bendingof the path of the <strong>light</strong> depends on the ratioof the speeds in the two different media. Thepath of <strong>light</strong> deviates most when the change inspeed is greatest.airdiamond<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>263

Table 8.2 Absolute index of refraction,n (quoted <strong>for</strong> yellow <strong>light</strong> λ = 589 nm)MediumIndex (n)Vacuum 1.0000Air 1.00029Ice 1.31Water 1.33Quartz 1.46Light crown glass 1.51Heavy flint glass 1.65Diamond 2.42Physics fileThe refractive index is also dependenton the colour of the <strong>light</strong> travellingthrough the medium. Since white <strong>light</strong>is made up of <strong>light</strong> of different colours,the refractive index must be quoted <strong>for</strong>a specific wavelength of <strong>light</strong>. Typicallyyellow <strong>light</strong> of wavelength 589 nmis used as this can be considered anaverage wavelength of white <strong>light</strong>. If therefractive index <strong>for</strong> a particular sampleof crystal quartz was quoted as 1.55,red <strong>light</strong> would have a s<strong>light</strong>ly higherrefractive index of 1.54 and violet <strong>light</strong>would have a s<strong>light</strong>ly lower refractiveindex of 1.57. This observation producesdispersion, discussed later in the chapter.medium 1 = airmedium 2 = quartzi = 40°Nrboundaryangle of deviationBy definition the refractive index of a vacuum would be exactly 1. Lighttravels only marginally more slowly in air and so the refractive index of airis 1.0003, but in most cases a value of 1.00 is sufficiently accurate. Materialsin which <strong>light</strong> travels slowest will have the highest indices of refraction.For example, if a particular medium allowed <strong>light</strong> to travel at half thespeed it does in a vacuum, then the refractive index of the medium wouldbe 2. The refractive index can there<strong>for</strong>e be considered an indication of the‘bending power’ of a material. Table 8.2 lists the absolute refractive indices<strong>for</strong> various media.By definition of the absolute refractive index:v 1c × c = n 2v 2n 1Hence, Snell’s law can now be expressed as:sin isin r = n 2. . . . . . (iii)n 1Snell lived nearly 200 years be<strong>for</strong>e scientists were able to measure thespeed of <strong>light</strong> in air with any degree of accuracy, so he was not aware thatthe refractive index was linked to the speed of <strong>light</strong> in a particular medium.We can now combine the equations (i), (ii) and (iii) developed above so thatSnell’s law is fully expressed.iSN…LL’S LAW:Worked example 8.2Csin isin r = v 1v 2= n 2n 1= n*A ray of <strong>light</strong> passes from air into quartz, which has an absolute refractive index of 1.46.If the angle of incidence of the <strong>light</strong> is 40°, calculate:a the angle of refractionb the angle of deviation of the rayc the speed of <strong>light</strong> in the quartz.Assume the index of refraction of air is 1.00 and the speed of <strong>light</strong> in air is 3.0 × 10 8 m s −1 .SolutionDraw a diagram to model the situation. As the <strong>light</strong> is slowed down the rays should bendtowards the normal.a List the data:i = 40°, n 1= 1.00, n 2= 1.46 and r = ?sin iThen:sin r = n 2n 1sin 40°sin r= 1.461.001.00 × sin 40°Hence sin r =1.46∴ r = 26°Note that r is smaller than i, as expected.264Wave-like properties of <strong>light</strong>

The angle of deviation is equal to the difference between i and r.D = (i − r)= 40° − 26°= 14°c n 1= 1.00, n 2= 1.46, v 1= 3.0 × 10 8 m s −1 and V 2= ?v 1= n 2v 2n 13.0 × 10 8= 1.46v 2 1.00v 2= 2.1 × 10 8 m s −1Physics in actionHuygen’s wavelets and refractionIn 1678 the Dutch mathematician <strong>Ch</strong>ristiaan Huygenspublished his ideas on the nature and propagation of <strong>light</strong>.(At about the same time, Newton developed his corpusculartheory.) Huygens’s idea was that <strong>light</strong> acted like a wave. In hismodel he suggested that each point along a wavefront of <strong>light</strong>could be considered to be a point source <strong>for</strong> small, secondarywavelets. Each wavelet was spherical, and the waveletsradiated from their point source in the general direction ofthe wave propagation (i.e. the <strong>light</strong> beam). The envelopeor common tangent of the wavelets then became the newwavefront as shown in Figure 8.24.incident<strong>light</strong> λ 1λ 1A θ 1 boundaryθ 2Bλ 2λ 2rayrefracted <strong>light</strong>sourcerayFigure 8.25 Huygens’s approach allowed the refraction of <strong>light</strong>—as quantified by Snell’s law—to be accurately modelled.Consider the wavefront that is just approaching theboundary labelled AB and the wavefront just leaving it. Usingthe right-angled triangle drawn above the boundary (withhypotenuse AB), we can state:initial wavefrontnew wavefrontFigure 8.24 The envelope of the wavelets caused the <strong>for</strong>mationof the new wavefront.Figure 8.25 shows how Huygens’s principle can be usedto explain refraction. In the initial medium the spacingbetween wavefronts is λ 1. In the second (slower) medium thewavelength, λ 2, will be reduced in correspondence with theratio of the velocity of <strong>light</strong> in each medium such that:λ 1= v 1λ 2v 2raysin θ 1= λ 1ABUsing the right-angled triangle below the boundary withhypotenuse AB, we can state:sin θ 2= λ 2ABThere<strong>for</strong>e:sin θ 1= λ 1× AB = λ 1sin θ 2AB λ 2λ 2Hence Snell’s law, which states that:sin θ 1sin θ 2= λ 1λ 2= v 1v 2can be derived from a wave model of <strong>light</strong>.<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>265

apparent position of star<strong>light</strong> from star isrefracted as it travelsthrough the atmosphereFigure 8.26 The refractive index of theatmosphere is not uni<strong>for</strong>m; hence, the pathof <strong>light</strong> from a star is refracted, apparentlyaltering its position.real positionof starlayersofincreasingopticaldensityRefraction in the atmosphereAlthough air has previously been considered to be a uni<strong>for</strong>m medium, thereare circumstances when in fact it is not uni<strong>for</strong>m. Consider the envelopeof air surrounding the Earth. The atmosphere is not uni<strong>for</strong>m: the opticaldensity of air increases closer to the Earth’s surface. Although this variationoccurs gradually, the situation can be represented by a series of horizontallayers of increasing refractive index as shown in Figure 8.26. As <strong>light</strong> froman object such as a star travels through each boundary between layers itis refracted towards the normal. The observer there<strong>for</strong>e believes the <strong>light</strong>to have come from a position higher in the sky. In reality the bendingmust occur gradually rather than at distinct intervals, and the amount ofbending has been greatly exaggerated in the diagram. A maximum amountof refraction occurs when objects are quite low in the sky. This is becausethe angle of incidence on the atmosphere is greatest and the <strong>light</strong> musttravel through a wider atmospheric band. The amount of refraction is notnoticeable to the human eye as when a maximum amount of refractionoccurs stars would only be shifted in position by less than 1°.Refraction by the atmosphere also extends the length of the day.Whenever you watch a sunset you see the Sun <strong>for</strong> a few minutes after ithas actually passed below the horizon. This is because <strong>light</strong> from the Sun isrefracted as it enters and travels through the Earth’s atmosphere, as shownin Figure 8.27. This effect is greatest at sunset and sunrise when the angle ofincidence of the Sun’s rays on the atmosphere is greatest.The atmosphere consists of moving layers of air and so the opticaldensity of the layers is continually changing. Since <strong>light</strong> rays must travelthrough this varying medium, <strong>light</strong> from the one object will follow s<strong>light</strong>lydifferent paths at different times. This is one reason <strong>for</strong> the twinkling ofstars and the apparent wriggling of distant objects on a warm day.Sun appears to be on the horizonatmosphereposition ofobserveractual path ofsun<strong>light</strong>Figure 8.27 The Sun is still visible even though it is actually below the horizon.266Wave-like properties of <strong>light</strong>

Physics in actionMirages<strong>light</strong> from skycool airhot airmirageFigure 8.28 Desert mirage. When airis heated from below, an inferior miragecan occur in which an image is displaceddownwards. This diagram is simplified.The air layers of different temperaturewill not be uni<strong>for</strong>m nor parallel to theplane of the ground. Light from the skyis gradually refracted by the air layers sothat it is travelling s<strong>light</strong>ly upwards whenit enters the eye. The diagram exaggeratesthis bending. The observer sees an imageof the sky on the distant road ahead andinterprets this as a body of water.Mirages are often believed to be the insane illusions of thirstydesert wanderers, but a mirage is the image of a real objectand can be explained in terms of the refraction of <strong>light</strong>. Amirage is a displaced and often distorted image occurringwhen layers of air of different temperature cause the path of<strong>light</strong> to bend. The severity and consistency of this temperaturegradient determines many features of the observedapparitions. An inferior mirage refers to the downwarddisplacement of an image. A superior mirage means theimage is displaced upwards. The following discussionexamines just a few of the many different types of miragesthat occur.Inferior miragesAn inferior mirage occurs when the air at ground level iswarmer than the air immediately above, i.e. the air is beingheated from below. This situation often arises in the afternoonof a hot, sunny day above a black bitumen road or above thesands of the desert. Air of higher temperature is less opticallydense and hence has a lower refractive index. Light fromthe sky is refracted as it travels through the layers of air ofdifferent optical density as shown in Figure 8.28. As a resultthe <strong>light</strong> ray is travelling upwards as it enters the observer’seyes and the image is then seen on the ground ahead. Drivingon a warm day, you often see an image of the sky on the roadahead; this is interpreted as a body of water.Floating on waterA fascinating inferior mirage often occurs above shallowbodies of water in the early morning. The water retains itsheat overnight but the surrounding land does not. Cool airfrom above the land flows over the warmer water and isheated from below. Thus air temperature decreases withheight. However, the temperature gradient is not uni<strong>for</strong>m.The temperature drops quickly near the water’s surface, but atgreater heights the decrease in temperature is more gradual.If an observer looks at a person in the distance the image ofthe person is displaced downwards but the bottom of the objectis displaced more than the top of the object since these lowerrays travel through a stronger temperature gradient. Thusthe person is irregularly enlarged. This phenomenon is calledtowering.Figure 8.29 829 The picture on the left shows a mirage of the red sky appearing on a road. In the picture on the right, the horizon shows the effectof towering.<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>267

8.2 summaryRefraction of <strong>light</strong>• Refraction is the bending of <strong>light</strong> due to a change inits speed as it enters a medium of different opticaldensity. The greater the change in the speed of <strong>light</strong>the greater the bending.• The angle between the normal and the incident ray iscalled the angle of incidence. The angle between thenormal and the transmitted or refracted ray is calledthe angle of refraction.• When a <strong>light</strong> ray passes into a medium in which ittravels more slowly (a more optically dense medium)it is refracted towards the normal.• When a <strong>light</strong> ray passes into a medium in whichit travels faster (less optically dense) it is refractedaway from the normal.• Each transparent medium is allocated an absoluterefractive index, n, determined by the speed at which<strong>light</strong> can travel in the medium compared with thespeed of <strong>light</strong> in a vacuum, c:n =cv medium• Snell’s law describes the relationship between theangle of incidence, i, and the angle of refraction, r, <strong>for</strong>a given pair of media:sin isin r = v 1= n 2= n*v 2n 18.2 questionsRefraction of <strong>light</strong>1 a The figure below represents a situation involvingthe refraction of <strong>light</strong>. Which of the lines labelledA–E is:i the boundary between two media?ii the normal?iii the incident ray?iv the refracted ray? v the reflected ray?b Explain what happens to the speed of <strong>light</strong> asit crosses the boundary between medium 1 andmedium 2. How do you know?ABCACd Which medium has a refractive index very closeto the refractive index of this sample of glass?iiglassmedium ABDiiglassmedium Bmedium 1medium 2Dglassmedium Cglassmedium D2 The following diagrams show <strong>light</strong> passing fromglass into different media labelled A, B, C and D.a Which media are more optically dense than theglass?b Which media have a refractive index less than therefractive index of glass?c Which medium has the highest refractive index?E3 Using Figure 8.25 as your reference, show how awave model can be used to explain the refractionof <strong>light</strong> as it passes through the boundary into amedium in which its speed is increased.4 Explain the following observations.a When you are standing in a shallow pool youappear shorter than usual.b On a warm day a person sees a ‘puddle’ on theroad ahead.268Wave-like properties of <strong>light</strong>

5 When <strong>light</strong> passes through a pane of glass it isrefracted. This does not cause the distortion of animage seen through the glass because:A the emerging rays are perpendicular to the incidentraysB the index of refraction of glass is too small to causedistortionC the displacement of <strong>light</strong> rays is too small to benoticed unless the glass is very thickD most of the <strong>light</strong> is reflected, not refracted.6 a The speed of <strong>light</strong> in a particular transparentplastic is 2.00 × 10 8 m s −1 . Calculate the refractiveindex of the plastic. The speed of <strong>light</strong> in a vacuumis 3.00 × 10 8 m s −1 .b What is the speed of <strong>light</strong> in water (n = 1.33)?7 A student wishes to determine the refractive indexof a particular sample of glass by experiment. Bypassing a narrow beam of <strong>light</strong> from air into the glass,she measures the angles of refraction, r, producedusing a range of incident angles, i. Her results areshown.8 a What is the relative refractive index <strong>for</strong> <strong>light</strong> passingfrom water into diamond if an incident angle of30° produces an angle of refraction of 16°?b Light travels from water (n = 1.33) into glass(n = 1.60). The incident angle is 44°. Calculate theangle of refraction.9 A ray of <strong>light</strong> is incident on the surface of waterin a fish tank. The incident ray makes an angle of32.0° with the surface of the water. The <strong>light</strong> that istransmitted makes an angle of 50.4° with the surface.Calculate:a the angle of incidenceb the angle of refraction of the transmitted <strong>light</strong>c the angle of reflectiond the angle of deviation of the transmitted ray.10 A ray of <strong>light</strong> travels from air, through a layer ofglass and then into water as shown. Calculate anglesa, b and c.air(n = 1.00)glass(n = 1.50)water(n = 1.33)i (degrees) 25 30 35 40 45 50 55 60 65r (degrees) 16 19 22 25 28 31 33 35 37a Plot a graph of sin i versus sin r.b Determine the gradient of the graph, i.e. therelative refractive index of <strong>light</strong> passing from airinto glass.c Assuming the refractive index of air is 1.00, whatis the refractive index of the glass sample used?d Calculate the velocity of <strong>light</strong> in the glass.40abcWorked Solutions<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>269

8.3Critical angle, TIR and EMRInteractivePhysics fileWhen <strong>light</strong> refracts at a surface thetransmitted ray becomes less intense asthe angle of incidence increases. Placeany small flat piece of glass on this page;a microscope slide will do. Look onto theslide from directly above and you willeasily observe the writing below.Now move your head so that youare looking through the slide from agradually decreasing angle of elevation.The page beneath the glass shouldbecome gradually darker. The amount of<strong>light</strong> transmitted through the glass andtowards your eyes is becoming less.waterairwaterairi = 45°i = 49°r = 70°r < 90°Critical angle and total internal reflectionWhen <strong>light</strong> is incident upon the boundary between two media, reflection,transmission and absorption may occur. As the angle of incidence increases,the intensity of the reflected beam increases and less <strong>light</strong> is transmitted.Consider the case of <strong>light</strong> travelling from water into air. Since thetransmitted <strong>light</strong> enters a less optically dense medium it travels faster and isrefracted away from the normal. The series of diagrams in Figure 8.30 showsthe effect that increasing the angle of incidence has on the transmitted <strong>light</strong>.As the angle of incidence is increased, the angle of refraction alsoincreases. This continues until, at a certain angle of incidence calledthe critical angle, i c, the angle of refraction will be almost 90° and thetransmitted ray travels just along the surface of the water. The angle ofrefraction can increase no further. If the angle of incidence is then increasedbeyond the critical angle no ray is transmitted and total internal reflectionwill occur. This is appropriately named, as none of the incident <strong>light</strong> energyis able to be transmitted into the next medium; it is totally reflected intothe medium carrying the incident ray. In effect, as the angle of incidenceincreases, the intensity of the reflected beam gradually becomes stronger,until at an angle of refraction of more than 90° all of the <strong>light</strong> is reflectedand no <strong>light</strong> is transmitted at all.The critical angle can be found <strong>for</strong> any boundary between two mediaby using Snell’s law. If the refractive indices of the two media are known,a presumption of an angle of refraction of 90° allows the critical (incident)angle to be calculated:sin isin r = n 2n 1If the incident angle is equal to the critical angle, i.e. i = i c, then r = 90°.The above equation becomes:sin i csin 90° = n 2n 1Now sin 90° = 1; hence the critical incident angle is given by:sin i c= n 2n 1waterairi = 50°iThe CRITICAL ANGL…, i c, is the angle of incidence that produces an angle ofrefraction of 90° as <strong>light</strong> is transmitted into a medium in which it travels at ahigher speed.sin i c= n 2n 1Figure 8.30 The critical angle <strong>for</strong> <strong>light</strong>travelling from water into air is approximately49°. If the incident angle is greater than 49°total internal reflection occurs.Worked example 8.3AAn underwater <strong>light</strong> shines upwards from the centre of a swimming pool that is1.50 m deep. Determine the radius of the circle of <strong>light</strong> that is seen from above.(n air= 1.00, n water= 1.33)270Wave-like properties of <strong>light</strong>

SolutionStep 1. Determine the critical angle <strong>for</strong> the water–air boundary.n airsin i c=n watersin i c= 1.001.33i c= sin ( 1.001.33)−1i c= 48.8°Step 2. Draw a diagram to represent the situation.airwaterP1.50 m48.8°RQ48.8°(a)Physics fileSince the refractive index of any givenmedium depends on the colour of the<strong>light</strong> travelling through the medium,each colour of <strong>light</strong> will have a s<strong>light</strong>lydifferent critical angle. For example, ifthe critical angle <strong>for</strong> red <strong>light</strong> travellingfrom glass to air was 40.5°, then yellow<strong>light</strong> would have a s<strong>light</strong>ly smallercritical angle of 40.2°. Violet <strong>light</strong> wouldhave an even smaller critical angle of39.6°. Although these values are verysimilar, they are sufficiently different tocause the dispersion of white <strong>light</strong> intoits component colours—the colours of therainbow!O<strong>light</strong> sourceConsidering one-half of the cone of <strong>light</strong> produced, since the critical angle is 48.8°,the angle QOP is 48.8°.Step 3. Using trigonometry:Rtan 48.8° =1.50hence R = 1.50 tan 48.8°= 1.71 mThe radius of the circle of <strong>light</strong> is 1.71 m.Total internal reflection is used in many optical instruments, includingcameras, periscopes and binoculars. When <strong>light</strong> reflects from a mirror thereis always some loss in the intensity of the incident <strong>light</strong>, and reflection canoccur from both the front and rear surfaces of the mirror, causing problems.By contrast, almost no loss of intensity occurs with total internal reflection.Since the refractive index <strong>for</strong> glass is about 1.5, the critical angle <strong>for</strong> <strong>light</strong>travelling from glass to air is approximately 42°, and a glass prism withinternal angles of 45° can be used as a mirror in the applications discussedbelow.Figure 8.31a shows the construction of a simple periscope. Light entersthe top glass prism perpendicular to the glass surface and so no refraction(deviation) occurs at this stage. The <strong>light</strong> passes through the prism andstrikes the back surface at an angle of 45°. The angle of incidence is greaterthan the critical angle and so the <strong>light</strong> can only be totally internallyreflected. The <strong>light</strong> travels down the tube of the periscope, enters the lowerprism and is again reflected. The surfaces of the prism do not need to besilvered <strong>for</strong> reflection to occur.Binoculars use a compound prism that is constructed of four 45° prisms(Figure 8.31b). The <strong>light</strong> actually undergoes four reflections on its passagefrom the objective lens to the eyepiece. This lengthens the path that the<strong>light</strong> must travel and hence more compact binoculars can be made <strong>for</strong> agiven magnification. Today’s binoculars are as effective as the telescopesof the past, which had to be many times longer.(b)objectiveeyepieceFigure 8.31 Good quality periscopes andbinoculars use 45° glass prisms <strong>for</strong> the totalinternal reflection of <strong>light</strong> rather than mirrors.This means reduced loss in the intensity of<strong>light</strong> and eradicates the problems caused byreflection occurring at both the front and rearsurfaces of mirrors.Prac 37<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>271

Physics in actionOptical fibresAn optical fibre uses total internal reflection to carry <strong>light</strong>,with very little energy loss. Since the 1950s fibre optics hasbeen used in the flexible fibrescope, a device that allowsdoctors to see inside the human body.By 1970 the production of very pure fibres reduced energylosses, making fibre optics feasible <strong>for</strong> use in communication.Since then this technology has made it possible to transferlarge amounts of data at remarkable speeds, enabling thedream of a worldwide computer network to be realised.The transmission processEssentially optical fibres carry in<strong>for</strong>mation in a digital <strong>for</strong>mat;that is, <strong>light</strong> signals turn on and off at very fast rates. Opticalcommunication systems take in<strong>for</strong>mation from a device,convert it into a digital <strong>for</strong>m if needed, and impress this digitalsignal onto a carrier frequency that has been produced by alaser or LED. This is called modulation. The signal is thenfed into an optical fibre <strong>for</strong> transmission (coupling). A criticalfactor is the range of frequencies of <strong>light</strong> that the opticalfibre is able to carry efficiently. Each fibre has a limitingbandwidth (frequency range) and a set number of differentsignal wavelengths that are allowable in this bandwidth. Forexample, the different signal wavelengths employed may benot allowed to be any closer than about 0.3 nm in spacing.(See later discussion.)During transmission, attenuation (energy losses alongthe way) will occur. There<strong>for</strong>e repeater stations are used toreceive the weak incoming signal and boost it be<strong>for</strong>e sendingit further along the fibre. Regenerators may also be usedto remove noise and distortion in the signal at this stage.Depending on the quality of the fibre, repeater stations areinserted about every 100 km along a long-distance opticalcable.When a signal reaches its destination it must bede-modulated (removed from its carrier wave) and the signalprocessed as required by the end user. The signal must arrivewith only as much distortion as can be compensated <strong>for</strong> at thisend.(a)core8.3 micronFigure 8.32 A magnified optical fibre torch.Different types of fibreOptical fibres can be divided into two categories dependingon the manner in which in<strong>for</strong>mation is carried: single-mode(thin core) fibres and multimode (large core) fibres. The term‘mode’ is synonymous with pathway.A single-mode fibre allows only one path on which <strong>light</strong>can travel (Figure 8.33a). Note the tiny dimensions of thefibre. A micron is one-millionth of a millimetre, and a humanhair is typically about 70 microns in diameter.Single-mode fibres are used in high-speed, long-distancetelecommunications. For example, Melbourne and Sydney arelinked by optical cable containing numerous single-mode fibres.(b)core50–100 microncladding125 microncoating250, 500 or 900 microncladding125 or 140 microncoating250–900 micronFigure 8.33 Compare the dimensions and structure of the different fibres. (a) Single-mode fibres, such as those used in high-speedtelecommunications, result in much less distortion of the optical signal. (b) Multimode fibres, such as those used in local area networks, arecheaper to produce but have more distortion problems. These are adequate <strong>for</strong> use over shorter distances.272Wave-like properties of <strong>light</strong>

A multimode fibre can have two different <strong>for</strong>ms: the stepindexmultimode fibre or the graded-index multimode fibre.A step-index multimode fibre has the same structure as thesingle-mode fibre described above, but it has a much largercore made of uni<strong>for</strong>m glass (Figure 8.33b). A graded indexfibre also has a large core, but its refractive index graduallydecreases from the centre to the outer diameter of the fibre.The path of <strong>light</strong> in a step-index multimode fibreLight rays are sent down the central core fibre. If the fibreis straight, most of the rays will travel along the axis of thefibre. Some <strong>light</strong> will strike the boundary between the coreand the cladding, particularly if the fibre is bent. Any raystriking the boundary at an angle greater than the criticalangle is totally internally reflected. The size of the criticalangle is determined by the refractive indices of the core andcladding (see Figure 8.34).Worked example 8.3BA particular step-index multimode fibre has a core of refractive index1.460 and cladding of refractive index 1.440. Calculate the criticalangle of the core–cladding boundary of this optical fibre.SolutionList the data:n core= 1.460, n cladding= 1.440sin i c= sin ( 1.4401.460)−1i c= 80.51°The core and cladding are designed so that the criticalangle <strong>for</strong> an optical fibre is typically greater than 80°. Hence,only <strong>light</strong> rays undergoing glancing collisions with the core–cladding boundary are totally internally reflected. Althoughthis causes more <strong>light</strong> energy to be lost at coupling, it meansthat all of the <strong>light</strong> rays emerging from the end of the fibrehave travelled a path of approximately the same length.cladding (n = 1.440)cone ofacceptancerays strikinghere are lost<strong>light</strong>sourcerays strikinghere are lost13°13°81°9°9°81°cladding (n = 1.440)86°core (n = 1.460)86°Figure 8.34 Outer rays from the <strong>light</strong>source that enter the core at an angle ofincidence greater than 13° will continueon to strike the core–cladding boundary atan angle of incidence less than the criticalangle of 81°. These rays will there<strong>for</strong>e betransmitted into the cladding and be lost.Rays originally within the cone of acceptancestrike the cladding at an angle greater than81° and are there<strong>for</strong>e totally internallyreflected. These reflected rays carry thesignal along the fibre.Electromagnetic wavesWhat is <strong>light</strong>? In the late 1600s it was known to involve the transfer ofenergy from one place to another. In Isaac Newton’s time a corpuscular(particle) model and a wave model <strong>for</strong> <strong>light</strong> had seemed equally valid.We have discussed these two proposed models of <strong>light</strong> along with theirrespective explanations of the reflection and refraction of <strong>light</strong>. In spite ofconsiderable endeavour by scientists it was not until the early 1800s thatone model prevailed. Thomas Young discovered that sources of <strong>light</strong> wereable to interfere with each other just like sound waves and water waves do.This finding led to a universally accepted wave theory <strong>for</strong> <strong>light</strong>. Furthermore,<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>273

the speed of <strong>light</strong> could be measured <strong>for</strong> the first time, and the wave modelof refraction (discussed in section 8.2) was validated. Meanwhile anotherarea of physics had been developing. By the 1860s investigations beingcarried out on different <strong>for</strong>ms of electromagnetic radiation led to the findingthat visible <strong>light</strong> itself is just one of the many <strong>for</strong>ms of electromagneticradiation (EMR).Electricity and magnetism were once considered to be separate subjects.However, moving charges create magnetic fields. Similarly a changingmagnetic field can be used to create electricity. In 1864 James ClerkMaxwell used mathematical equations to describe how charges movingperiodically in a conductor would set up alternating electric fields andmagnetic fields in the nearby region. Maxwell knew that the magnetic andelectric fields travelled through space. He calculated their speed and foundit to be 300 000 km s −1 , exactly the same as the speed of <strong>light</strong>! Also, hedevised mathematical expressions to describe the magnetic and electricfields. The solution to these expressions was found to be the equation of awave. Maxwell had shown that <strong>light</strong> is an electromagnetic wave.Today we know that the electromagnetic spectrum includes a wide rangeof frequencies (or wavelengths). All electromagnetic waves are created byaccelerating charges which result in a rapidly changing magnetic field andelectric field travelling out from the source at the speed of <strong>light</strong>, as shownin Figure 8.35. Note that the electric field component and the magnetic fieldcomponent are at right angles to each other and to their direction of travel.Electromagnetic radiation meets the description of a transverse wave asdiscussed in <strong>Ch</strong>apter 7.xelectric fielddirection of motion of wavevelocity c = 3 × 10 8 m s –1ymagnetic fieldzwavelength λFigure 8.35 As all electromagnetic waves travel with the same velocity, the only thing thatdifferentiates one <strong>for</strong>m of EMR from another is the frequency (and, there<strong>for</strong>e, the wavelength).The many <strong>for</strong>ms of EMR are essentially the same, differing only in theirfrequency and, there<strong>for</strong>e, their wavelength. The electromagnetic spectrumis roughly divided into seven categories depending on how the radiationis produced and the frequency. The energy carried by the electromagneticradiation is proportional to the frequency. High-frequency shortwavelengthgamma rays are at the high-energy end of the spectrum. Lowfrequencylong-wavelength radio waves carry the least energy. Humanshave cells in their eyes which can respond to EMR of frequencies betweenapproximately 400 THz and 800 THz; these frequencies make up the visible<strong>light</strong> section of the electromagnetic spectrum.274Wave-like properties of <strong>light</strong>

Recall from <strong>Ch</strong>apter 7 that <strong>for</strong> any wave the relationship between itsfrequency and its wavelength is given by v = fλ.All electromagnetic radiation travels at a speed of 3.00 × 10 8 m s −1 in avacuum and so this significant speed has been allocated the symbol c.iFor all …L…CTROMAGN…TIC RADIATION:f =cλwhere f is the frequency of the EMR (Hz)c is the speed of the EMR = 3.0 × 10 8 m s −1λ is the wavelength of the EMR (m)Figure 8.36 shows the different categories of EMR. Note the range offrequencies and wavelengths is enormous. The range of frequencies (orwavelengths) constituting visible <strong>light</strong> occurs near the middle of thespectrum. Our eyes cannot perceive any wavelengths of EMR outside ofthis range.Gamma raysFrequency10 22 HzWavelength10 –14 mX-rays10 17 Hz 10 –9 mUltraviolet10 16 Hz10 –8 mVisible spectrum800–400 THz400–800 nmInfrared10 12 Hz10 –4 mMicrowaves10 10 Hz10 –2 mTV10 8 Hz10 mRadio10 6 Hz10 2 mFigure 8.36 The electromagnetic spectrum.<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>275

Worked example 8.3CThe EMR given off by a sample of sodium as it is burned has a wavelength of 589 nm. Whatis the frequency of this radiation? How would we detect the radiation?SolutionPhysics in actionOther <strong>for</strong>ms of EMRf = c λ3.0 × 108=589 × 10 −9= 5.09 × 10 14 Hz= 510 THzThis frequency of EMR lies in the visible <strong>light</strong> section of the electromagnetic spectrum,there<strong>for</strong>e we would see it! It is actually yellow <strong>light</strong>.Radio wavesAccelerating a positive or negative charge can produce EMR.Electrons oscillating in a conducting wire, such as an antenna,produce the radio waves that bring music to your <strong>home</strong>. Thelong-wavelength low-energy electromagnetic waves blanket thesurrounding region, and aerials can receive the signal manykilometres from the source. As a result of the radio waves,electrons in the receiving aerial wire will oscillate, producing acurrent that can be amplified. Radio waves can be transmittedover very long distances, including around the Earth’s surface,by reflection from layers in the atmosphere.MicrowavesMicrowaves are EMR of wavelengths ranging from about1 mm to about 10 cm.The microwaves that cook your dinner are produced bythe spin of an electron or nucleus. Microwave links are usedto allow computer systems to communicate remotely, andradar equipment uses microwave frequencies of centimetrewavelengths.Infrared wavesInfrared or heat radiation includes the wavelengths that ourskin responds to. When you feel the warmth from the Sunor an electric bar heater you are actually detecting infraredradiation. All objects that are not at a temperature of absolutezero radiate EMR. The hotter the object the more radiationis emitted, and the further along the spectrum the radiationis. Night scopes and infrared spy satellites create an imageby sensing infrared radiation and converting it into a visiblepicture.Ultraviolet wavesUltraviolet waves have wavelengths shorter than violet <strong>light</strong>—so our eyes cannot detect them—but no greater than about10 nm. Many insects can detect the ultraviolet <strong>light</strong> that iscommonly reflected from flowers. Although ultraviolet <strong>light</strong>is less energetic than gamma rays or X-rays, it is known tocause skin cancer, particularly with increased exposure.Silicon atoms are able to absorb some frequencies in theultraviolet region of the spectrum, reducing your chances ofgetting sunburnt through glass.X-ray wavesX-rays are produced when fast-moving electrons are firedinto an atom. The name is a result of scientists not knowingwhat they were when they were first detected, hence the letter‘X’. X-rays can pass through body tissue and be detected byphotographic film, and so are used in medical diagnosis. Theyhave extensive safety testing, security and quality controlapplications in industry.Gamma-ray wavesThe highest energy, smallest wavelength radiation is thegamma ray, which is produced within the nucleus of an atom.Gamma rays are one of the three types of emissions that comefrom radioactive (unstable) atoms. Gamma rays are extremelypenetrating and require dense material to absorb them.Prac 38Coloured <strong>light</strong>, different wavelengthsOur eyes are responsive to many different colours of <strong>light</strong> from the deepestred through to the brightest violet, the visible spectrum (Figure 8.37). Eachvariation in colour or shade is caused by <strong>light</strong> of a different wavelength.Traditionally the colours quoted as making up the visible spectrum are red,orange, yellow, green, blue and violet. However, as shown in Figure 8.37,the actual allocation of separate names <strong>for</strong> the colours is difficult since theymerge into one another. The wavelengths associated with visible <strong>light</strong> arevery small: they range from approximately 390 nanometres (or 3.9 × 10 −7 m)<strong>for</strong> violet <strong>light</strong> to around 780 nanometres <strong>for</strong> red <strong>light</strong>.276Wave-like properties of <strong>light</strong>

Physics fileredwavelength 780 nmvioletwavelength 390 nmFigure 8.37 Visible <strong>light</strong> is one category of EMR. The spectrum of visible <strong>light</strong> contains amyriad of colours. Each different colour or hue is <strong>light</strong> of a different wavelength.You wouldn’t expect a person renownedas a great scientist to be superstitious.For many years the spectrum of colourwas listed as being made up of sevenseparate colours rather than the sixcolours listed today. Isaac Newton carriedout famous experiments producing thespectrum of colour and recombining itinto white <strong>light</strong>. In his writings indigo(a very dark blue) was stated as lyingbetween blue and violet. The separateidentification of indigo <strong>light</strong> is strange, asit really does not appear as prominentlyas the other six main colours. Newtonwas rather mystical in his religiousbeliefs and seven was considered to bea ‘perfect’ number somehow related tothe natural laws governing the universe,and so he deliberately identified sevencolours in the visible spectrum.The colour of an object that we see is actually a physiological responseto the particular wavelength(s) of <strong>light</strong> entering our eyes. Our coloursensingsystem, consisting of the eye, nerve conductors and the brain, candiscriminate between hundreds of thousands of different colours. However,different combinations of wavelengths of <strong>light</strong> can evoke the same responsefrom our brain. In other words, there are a number of different ways inwhich to make an object appear a particular shade of yellow, <strong>for</strong> example.We perceive <strong>light</strong> as white <strong>light</strong> if it contains roughly equal amounts ofeach of the colours of the visible spectrum. The page of this book appearswhite because it is reflecting all of the colours (wavelengths) of visible<strong>light</strong> in roughly equal proportions. Sun<strong>light</strong>, incandescent <strong>light</strong> andfluorescent <strong>light</strong> all produce the same general sensation of white <strong>light</strong>.Figure 8.38 shows their component colours. The <strong>light</strong> from incandescentand fluorescent globes does not appear to be quite as white as sun<strong>light</strong>.This is because sun<strong>light</strong> is very evenly distributed across the spectrum,but an incandescent source radiates considerably more red <strong>light</strong> than blue<strong>light</strong>, and a fluorescent source favours blue wavelengths of <strong>light</strong>.When incident <strong>light</strong> strikes the surface of an object, it may be absorbed,transmitted and/or reflected. If all of the white <strong>light</strong> falling on a surfaceis absorbed, the object will appear black as no <strong>light</strong> is reflected. The colourof an object is often determined by which colours of <strong>light</strong> it reflects andabsorbs when white <strong>light</strong> is shone upon it.Colour addition or mixing <strong>light</strong> sourcesIn 1807 Thomas Young discovered that combining red, green and blue <strong>light</strong>on a screen produced white <strong>light</strong>. In fact various combinations of thesethree colours of <strong>light</strong> could create all of the other colours of the spectrum.Red, green and blue are there<strong>for</strong>e called the primary colours of <strong>light</strong>. Noneof the primary colours can be produced by a combination of the otherprimary colours.Relative intensityRelative intensityRelative intensitySun<strong>light</strong>Incandescent <strong>light</strong>Fluorescent <strong>light</strong>Figure 8.38 The colour components ofsun<strong>light</strong>, incandescent <strong>light</strong> and fluorescent<strong>light</strong>. All are referred to as sources of white<strong>light</strong>, but their spectral compositions vary,affecting the colour of an illuminated object.Prac 39<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>277

(a)(b)(c)Blue Green Red MixturewhitecyanyellowmagentaFigure 8.40 The yellow seen on the televisionscreen is, like all other colours, composed ofonly red, green and blue dots. The tiny dotson the screen are so close together that theycannot be recognised as separate dots, butblend together to <strong>for</strong>m a continuous picture.Different areas of the television screen aremade to produce red, green and blue dots invarying proportions, thus creating the variouscolours seen on the television screen.Figure 8.39 Pairs of the primary colours of <strong>light</strong> overlap to produce the secondary coloursyellow, cyan and magenta. When all three primary colours of <strong>light</strong> overlap white <strong>light</strong> isproduced.Figure 8.39 shows the three primary colours of <strong>light</strong> overlapping toproduce other colours. The particular colours <strong>for</strong>med by the overlappingof pairs of primary colours are called cyan, magenta and yellow. Any groupof colours which combine to <strong>for</strong>m white <strong>light</strong> are called complementarycolours. All three primary colours when combined <strong>for</strong>m white <strong>light</strong>.Combining any two primary colours <strong>for</strong>ms the complementary colour ofthe remaining primary colour. So, <strong>for</strong> example, when red and green arecombined they <strong>for</strong>m the complement of blue, which is yellow. Yellow isthe complement of blue. Cyan is the complement of red, and magenta isthe complement of green.Colour televisionTelevision screens produce coloured pictures (Figure 8.40a) yet only utilisethree different colours: red, green and blue. These colours are producedwhen electron beams strike the tiny phosphor dots lining the screen.Figure 8.40b shows a greatly magnified picture of a screen. It is actuallymade up of thousands of tiny coloured dots called pixels. Different partsof the television screen appear to be different colours because of therelative abundance of the three primary colours. Figure 8.40c shows howthe different colours are created. If white is required all three colours willbe produced. Because these dots are so close, our eye interprets this as auni<strong>for</strong>m area of white <strong>light</strong>. If magenta <strong>light</strong> is required only the blue andred dots will be stimulated. In fact all colours are produced by altering theproportions of red, green and blue dots stimulated in any particular area ofthe screen.278Wave-like properties of <strong>light</strong>

8.3 summaryCritical angle, TIR and EMR• As the angle of incidence of <strong>light</strong> onto a transparentsurface is increased, proportionally more <strong>light</strong> isreflected and less <strong>light</strong> is refracted.• When <strong>light</strong> enters a less optically dense medium itis refracted away from the normal. At the criticalincident angle, i c, the angle of refraction is 90°.• If the incident angle is greater than the critical angle,i c, total internal reflection occurs.• The critical angle is given by:sin i c= n 2n 1• Visible <strong>light</strong> is only a small part of the electromagneticspectrum. The many <strong>for</strong>ms of EMR are essentiallythe same, differing only in their frequency and,there<strong>for</strong>e, their wavelength.• All EMR travels at a speed of 3.0 × 10 8 m s −1 in avacuum.• White <strong>light</strong> contains red, orange, yellow, green, blueand violet <strong>light</strong> in approximately equal proportions.• The primary colours of <strong>light</strong> are red, green andblue. Combining the three primary colours of <strong>light</strong>produces white <strong>light</strong>.8.3 questionsCritical angle, TIR and EMR1 Can total internal reflection occur as <strong>light</strong> strikes theboundary from:a air (n = 1.00) to glass (n = 1.55)?b glass to air?c glass to water (n = 1.33)?d glass (n = 1.55) to glass (n = 1.58)?2 The diagram shows four rays incident on theboundary between glass and air. Ray 2 meets theboundary at the critical incident angle. For each ofthe rays 1–4 choose the option that best describeswhat happens as it strikes the boundary.12i c3 4normalA The ray is reflected only.B The ray is refracted only.C The ray is reflected and refracted.D The ray is reflected and transmitted.airglass3 Determine the critical angle <strong>for</strong> <strong>light</strong> travelling from:a diamond (n = 2.42) into airb flint glass (n = 1.60) into airc water (n = 1.33) into aird glass (n = 1.50) into water (n = 1.33).4 The critical angle <strong>for</strong> <strong>light</strong> passing from oleic acidinto air is 43.2°. Calculate the index of refraction ofoleic acid.5 The speed of <strong>light</strong> in a particular sample of clearplastic is 1.80 × 10 8 m s −1 . Determine the critical angle<strong>for</strong> <strong>light</strong> passing from this plastic into air.6 Calculate the wavelength of:a microwaves of frequency 3 × 10 10 Hzb ultraviolet radiation of frequency 10 15 Hz.7 a List three different types of electromagnetic radiationand describe a use <strong>for</strong> each.b List two properties common to all <strong>for</strong>ms of electromagneticradiation.8 The primary colours <strong>for</strong> <strong>light</strong> are:A red, green and yellowB red, blue and yellowC red, green and blueD yellow, cyan and magenta.9 Students are experimenting with the <strong>light</strong>ing <strong>for</strong> theirschool play. They want to produce some dramatic<strong>light</strong>ing effects. Determine the colour <strong>for</strong>med from amixture of:a red and blue <strong>light</strong>b red, blue and green <strong>light</strong>c blue and yellow <strong>light</strong>d green and magenta <strong>light</strong>.10 A spinning top is decorated with all the colours ofthe rainbow, yet when spun it appears almost white.Why?<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>279

8.4Dispersion and polarisation of <strong>light</strong> wavesDispersionWe have examined how the recognition of the wave nature of <strong>light</strong> allowedthe development of a full explanation of the refraction of <strong>light</strong> as it changesspeed. For example, the change in direction of travel of the <strong>light</strong> wave asit entered an optically denser medium occurred because a section of thewavefront entered a slower medium. The slowing down of this section ofthe wavefront, but not the section still travelling in the original medium,causes the overall wave to veer from its original direction of travel.Recall that white <strong>light</strong> is made up of many different frequencies (colours)of <strong>light</strong>. For some materials the speed at which <strong>light</strong> is transmitted is actuallys<strong>light</strong>ly different <strong>for</strong> different frequencies (colours) of <strong>light</strong>. This means thaton refraction different colours of <strong>light</strong> will take s<strong>light</strong>ly different paths. Thisresults in the spreading out of the white <strong>light</strong> into its component colours.This is called the dispersion of white <strong>light</strong>.Prisms split white <strong>light</strong> into its component colours. It took scientistsmany years to be able to explain this phenomenon. Prior to Isaac Newtonit was thought that glass prisms altered the incoming white <strong>light</strong> byvarying degrees to produce the spectrum of colour. Newton carried out hisinvestigations into dispersion and was the first to conclude that white <strong>light</strong>is actually made up of the colours of the spectrum and there<strong>for</strong>e recombiningthese colours would produce white <strong>light</strong>.Figure 8.41 shows the dispersion of white <strong>light</strong> as it passes through atriangular prism. The <strong>light</strong> is dispersed both on entering and leaving theprism, so that as the <strong>light</strong> emerges the range of colours spreads over quitea wide angle. There are no distinct boundaries at which one colour finishesand another begins.Figure 8.41 Dispersion of white<strong>light</strong> by a triangular glass prism.On entering and leaving the prism,the violet <strong>light</strong> is most significantlyaltered in speed and so it isrefracted through the greatestangle. Red <strong>light</strong> is slowed less andso is refracted the least.As <strong>light</strong> enters a prism, it refracts due to a change in speed. Why does<strong>light</strong> slow down when it enters a more optically dense medium? The <strong>light</strong>energy is being momentarily absorbed and then re-radiated by the atomsthat make up the medium. Different colours of <strong>light</strong> interact differentlywith these atoms. As a result they travel at different speeds within the280Wave-like properties of <strong>light</strong>

medium and so are refracted through different angles. Of the colours thatconstitute the visible spectrum, violet <strong>light</strong> is slowed down the most and sois refracted through the greatest angle. Red <strong>light</strong> is slowed least and so isrefracted the least.A similar situation occurs when <strong>light</strong> speeds up on entering a newmedium. Different colours are refracted through different angles.Effectively, a particular medium, glass <strong>for</strong> example, has a different refractiveindex <strong>for</strong> each colour of <strong>light</strong>. Light flint glass has a refractive index of 1.62<strong>for</strong> red <strong>light</strong> and 1.67 <strong>for</strong> violet <strong>light</strong>. Quartz has a refractive index of 1.45<strong>for</strong> red <strong>light</strong> and 1.47 <strong>for</strong> violet <strong>light</strong>. In a vacuum, however, all colours of<strong>light</strong> travel at the same speed of 3.00 × 10 8 m s −1 .Worked example 8.4AA narrow beam of white <strong>light</strong> enters a crystal quartz prism with an angle of incidence of 35°.In air, the white <strong>light</strong> travels at a speed of 3.00 × 10 8 m s −1 . In the prism the different coloursof <strong>light</strong> are slowed to varying degrees. The refractive index <strong>for</strong> red <strong>light</strong> in crystal quartz is1.54 and <strong>for</strong> violet <strong>light</strong> the refractive index is 1.57. Calculate:a the angle of refraction <strong>for</strong> the red <strong>light</strong>b the angle of refraction <strong>for</strong> the violet <strong>light</strong>c the angle through which the spectrum is dispersedthe speed of the red <strong>light</strong> in the crystal quartz.dSolutionsin iasin r = n 2n 1sin 35°= 1.54sin r 1.00r = 21.9°sin ibsin r = n 2n 1sin 35°= 1.57sin r 1.00r = 21.4°c Angle of dispersion = 21.9° − 21.4° = 0.5°dn 2= v 1n 1v 21.54 3.00 × 108=1.00 v 2v 2= 1.95 × 10 8 m s −1Physics fileDiamond has a relatively high refractiveindex of 2.42; hence, the critical angle<strong>for</strong> diamond is a relatively small 24°.White <strong>light</strong> is s<strong>light</strong>ly dispersed onentering a diamond. Because of theshape of a diamond, once <strong>light</strong> hasentered the diamond any ray strikingthe diamond–air boundary is likely tohave an incident angle greater than24° and it is there<strong>for</strong>e totally internallyreflected.The special shape of a diamondmeans that the dispersed beam of <strong>light</strong>is likely to undergo a number of internalreflections be<strong>for</strong>e it meets a boundaryat an incident angle of less than 24°,each reflection spreading the beam alittle wider. After a number of internalreflections the <strong>light</strong> leaves the diamond.If the diamond is appropriately shaped,single colours of <strong>light</strong> are seen to bescattered by the diamond.Figure 8.42 Flashes of coloured <strong>light</strong> can beseen emerging from a diamond as it is viewedfrom different angles.Physics in actionRainbowsWater droplets in the air disperse white <strong>light</strong> into colours toproduce a rainbow just as a glass prism does. Rainbows areseen only when you have your back to the Sun, and manywater droplets <strong>for</strong>m a cloud in front of you. White <strong>light</strong> entersthe water droplet, reflects from the back of the droplet (dueto total internal reflection) and then leaves the droplet. Onboth entering and leaving the droplet the white <strong>light</strong> is s<strong>light</strong>lydispersed since water has a s<strong>light</strong>ly different refractive index<strong>for</strong> each colour of <strong>light</strong> (Figure 8.43). To see what will happen,we will examine the path of the two extremes of the spectrum,red and violet <strong>light</strong>. The paths of all of the other colours of<strong>light</strong> will lie between these.Because of dispersion, the red and violet rays leave thedrop in different directions. There<strong>for</strong>e an observer cannot seeboth the red and violet <strong>light</strong> emitted from the one droplet. Ifthe violet <strong>light</strong> from a particular raindrop is entering your eyeContinued on next page<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>281

Rainbows (continued)then all of the other colours reflected by that droplet mustmiss your eye. The different colours observed in a rainbowmust come from different raindrops. Those raindrops sendingthe red <strong>light</strong> to your eye must be higher in the sky since thered <strong>light</strong> emerges more downward than the violet <strong>light</strong>. Infact all of the droplets which send red <strong>light</strong> to your eye lie onan arc of about 42° as measured from the original directionof travel of the Sun’s rays (Figure 8.44). The dropletsreflecting violet <strong>light</strong> lie on an arc of about 40°, causing theshape of the rainbow!white <strong>light</strong> from the SundispersionredviolettotalinternalreflectionNo two people can actually see the same rainbow becauseto view a rainbow the <strong>light</strong> reflected and dispersed by aparticular set of raindrops is directed towards your eye.Those same drops cannot send the same colours of <strong>light</strong> toyour neighbour. A portion of the particular drops producingred <strong>light</strong> <strong>for</strong> you may be <strong>for</strong>ming the green section of yourneighbour’s rainbow.direction of Sun’s raysall drops on this arcappear red toviewer’s eyedispersionvioletred4042redvioletall drops on thisarc appear violetto viewer’s eyevioletredFigure 8.43 The rainbow is commonly seen because of thedispersion and reflection of <strong>light</strong> by water droplets. Less intenserainbows sometimes accompany a brighter rainbow. These are due to<strong>light</strong> reflecting inside the droplet more than once be<strong>for</strong>e emerging.Figure 8.44 All drops lying on the outer arc reflect red <strong>light</strong> in thedirection of the observer. All droplets on the lower arc reflect violet<strong>light</strong> to the observer’s eye.(a)(b)directionof traveldirectionof travelFigure 8.45 (a) Unpolarised <strong>light</strong>, such asthat emitted by the Sun or a globe, has electricfield variations that are not in alignment withone another. (b) Polarised <strong>light</strong> waves, such aslaser <strong>light</strong> waves, have the electric fields thatvary in the same plane as one another.PolarisationFurther evidence <strong>for</strong> the wave nature of <strong>light</strong> is the finding that <strong>light</strong> canbe polarised. Consider that <strong>light</strong> is an electromagnetic wave with associatedelectric and magnetic fields that vary. Each of these fields varies at a rightangle to the direction in which the wave travels. For this discussion weneed only think about the varying electric field associated with <strong>light</strong>, as itis this that largely determines how <strong>light</strong> interacts with materials. We canthere<strong>for</strong>e represent <strong>light</strong> as shown in Figure 8.45.Think of how <strong>light</strong> is produced by a normal <strong>light</strong> globe inside a torch.Light is emitted from many different atoms in the filament and lots of <strong>light</strong>waves may be sent in a particular direction. However, the electric fields ofthese <strong>light</strong> waves will not be aligned. This is shown in Figure 8.45a. This iscalled unpolarised <strong>light</strong>. Most <strong>light</strong> sources, including our Sun, produceunpolarised <strong>light</strong>. If the <strong>light</strong> waves did have their electric fields alignedwith one another as shown in Figure 8.45b, we would call this polarised<strong>light</strong>. The fact that unpolarised <strong>light</strong> can be converted into polarised <strong>light</strong>provides strong evidence that <strong>light</strong> is actually a wave.Techniques <strong>for</strong> polarising <strong>light</strong>The most familiar way in which unpolarised (non-aligned) <strong>light</strong> can beconverted to polarised (aligned) <strong>light</strong> is by using polarising filters. These282Wave-like properties of <strong>light</strong>

filters have molecules that will block all electric-field-wave componentsexcept those whose plane is aligned in a particular direction. Figure 8.46demonstrates this process using a ‘slit’ to represent the filter. Keep in mindthat filters are actually solid materials, usually special plastics.A <strong>light</strong> wave that has its electric field varying in a plane aligned withthe filter will pass straight through the filter, maintaining its originalamplitude. All of the <strong>light</strong> energy passes through. Figure 8.46a shows avertical polarising filter. It allows <strong>light</strong> with vertically orientated electricfieldvariation to pass through.A <strong>light</strong> wave that has its electric-field plane completely out of alignmentwith the filter will be blocked. That is, the <strong>light</strong> will be absorbed by thefilter, as shown in Figure 8.46b.If a <strong>light</strong> wave has only a component of its electric-field planecorresponding to that of the filter, then only this component of thewave will be transmitted. The emerging wave has significantly reducedamplitude. There<strong>for</strong>e a portion of the <strong>light</strong> energy does not pass throughthe filter, as indicated in Figure 8.46c. The emerging <strong>light</strong> is described as‘vertically polarised’. Should unpolarised <strong>light</strong> be incident on the filter,only vertically polarised <strong>light</strong> would emerge. A pair of polarising filterscan there<strong>for</strong>e be placed at right angles to one another to prevent all <strong>light</strong>from passing through. One filter may block all of the horizontal electricfieldcomponents and the other filter may block the vertical components, asshown in Figure 8.47.Physics fileThe fact that <strong>light</strong> can be polarisedprovides strong evidence that <strong>light</strong> isactually a transverse wave, becauselongitudinal waves cannot be polarised.Prac 40(a) filter(b) filter(c) filterfilteramplitudeis reducedwave is blockedFigure 8.46 (a) This filter allows vertically polarised <strong>light</strong> to pass through. (b) Allhorizontally polarised <strong>light</strong> is blocked. (c) The horizontal component of the <strong>light</strong> issuppressed, resulting in vertically polarised <strong>light</strong> of reduced amplitude. The emerging <strong>light</strong>will be less bright than the incident <strong>light</strong>.Why Polaroid sunglasses workWhen outdoors on a bright, sunny day, the smooth, highly reflective,horizontal surfaces around you are a significant contributor to the amountof <strong>light</strong> entering your eyes. Bring to mind the glare that can occur from thesurface of water or snow. Fortunately <strong>light</strong> that is reflected from smooth,horizontal surfaces tends to be polarised (aligned) in a horizontal direction.An appropriately orientated polarising filter can be used. Lenses in a pair ofPolaroid sunglasses are polarising filters orientated to block the horizontalwave components, allowing only vertical components through. Hence theintensity of <strong>light</strong>—that is, the glare—is markedly reduced.Figure 8.47 Polarising materials crossingover at right angles to one another will preventany <strong>light</strong> from passing through.<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>283

8.4 summaryDispersion and polarisation of <strong>light</strong> waves• Dispersion is the spreading of white <strong>light</strong> into itscomponent colours in a spectrum.• Dispersion occurs as white <strong>light</strong> enters or leaves aprism because different colours of <strong>light</strong> are refractedby different amounts.• Violet <strong>light</strong> has the greatest refractive index and isrefracted through the greatest angle. Red <strong>light</strong> hasthe smallest refractive index and is refracted throughthe smallest angle.• Unpolarised <strong>light</strong> waves have electric-field variationsthat are not in alignment with one another. Polarised<strong>light</strong> waves have electric fields that vary in the sameplane as one another.• Polarising filters are able to convert unpolarised<strong>light</strong> into polarised <strong>light</strong>, providing strong evidencethat <strong>light</strong> is actually a wave.8.4 questionsDispersion and polarisation of <strong>light</strong> waves1 a Which colour of <strong>light</strong> travels fastest in perspex:red, green or blue?b If red, green and blue <strong>light</strong> passes from air intoa perspex block, which colour of <strong>light</strong> would beslowed down the most?c Which colour of <strong>light</strong>—red, green or blue—wouldbe refracted the most as it passes into the perspexblock?2 How does polarisation support a wave model <strong>for</strong><strong>light</strong>?3 With the use of a diagram, show how a narrow beamof white <strong>light</strong> will be dispersed on both entering andleaving a triangular glass prism.4 You have two identical pairs of sunglasses. Howcould you find out whether the sunglasses werepolarising or not polarising?5 A piece of glass and a diamond are cut to exactly thesame size and shape. They are illuminated by thesame white <strong>light</strong>. Why does the diamond appear tosparkle more colourfully than the glass?6 Explain the effect that a polarising filter has onunpolarised <strong>light</strong>.7 A particular prism of glass has a refractive index of1.55 <strong>for</strong> violet <strong>light</strong> and 1.50 <strong>for</strong> red <strong>light</strong>. A beam ofwhite <strong>light</strong> is incident on the prism at an angle of40.0° and is dispersed.a Which colour of <strong>light</strong> will have the slower speedin glass: red or violet?b Which colour is refracted more: red or violet?c What is the angle of refraction <strong>for</strong> the red <strong>light</strong> atthe air–glass boundary?d What is the angle of refraction <strong>for</strong> the violet <strong>light</strong>at the air–glass boundary?e Over what angle will the spectrum be spread?8 A particular plastic has a refractive index of 1.455 <strong>for</strong>red <strong>light</strong> and 1.650 <strong>for</strong> violet <strong>light</strong>. For <strong>light</strong> passingfrom this plastic into air, which colour of <strong>light</strong> wouldhave the greatest critical angle?9 Explain why dispersion provides evidence <strong>for</strong> awave nature of <strong>light</strong>.10 A polarising filter is positioned so that it producesvertically polarised <strong>light</strong>. If another filter is orientatedat an angle of 30° to this filter, what happens to theintensity of the <strong>light</strong>?Worked Solutions284Wave-like properties of <strong>light</strong>

<strong>Ch</strong>apter review<strong>Models</strong> <strong>for</strong> <strong>light</strong>1 As <strong>light</strong> travels from quartz to water does it:a speed up or slow down?b refract towards or away from the normal?2 Describe the practical work you would carry out if you were givena glass prism and asked to determine its refractive index andcritical angle.3 a List seven different categories of electromagnetic radiationand give an application of each.b List two features common to all types of electromagneticradiation.4 Monochromatic <strong>light</strong> with a wavelength of 590 nm is passingthrough a medium. Its frequency is measured as 3.81 × 10 14 Hz.Determine whether the medium through which the <strong>light</strong> travelsis air. Justify your decision.5 At the boundary between medium 1 and medium 2, the angle ofrefraction is smaller than the angle of incidence.a Which medium allows <strong>light</strong> to travel faster?b Which medium has the higher index of refraction?6 A ray of <strong>light</strong> exits a glass block. On striking the inside wall of theglass block, the ray makes an angle of 58.0° with the glass–airboundary. The index of refraction of the glass is 1.52.a What is the angle of incidence?b Assume n air= 1.00. What is the angle of refraction?c What is the angle of deviation of the ray?d What is the speed of <strong>light</strong> in the glass?7 List these <strong>for</strong>ms of electromagnetic radiation—X-rays,microwaves, visible <strong>light</strong>, radio waves—in order of increasing:a frequency b wavelength c energy.8 Scuba divers notice that when they are only submerged a fewmetres the ocean waters appear blue. Why?9 Which of the following boundaries between two media willresult in the greatest refraction of any <strong>light</strong> ray that crosses theboundary?A water into airB water into diamondC air into diamondD glass into air10 Explain why snowboarders and fishermen are likely to wearpolarising sunglasses.11 Explain the following observations. Use a diagram whereappropriate.a A star can still be seen even though it is actually positionedbelow the horizon.b A stone at the bottom of a shallow pond seems closer to youthan it really is.12 State three behaviours of <strong>light</strong> <strong>for</strong> which the ray model is usefulwhen explaining the behaviour.13 Explain how the spectrum of colour reflected from a black objectis different from the spectrum of <strong>light</strong> reflected by a grey object.14 In mid-afternoon a diver looks up from below the surface of thewater. His judgement of the position of the Sun will be:A higher than it really isB lower than it really isC the same as viewed from above the water.15 A narrow beam of white <strong>light</strong> enters a crown glass prism withan angle of incidence of 30°. In air, the white <strong>light</strong> travels at aspeed of 3.00 × 10 8 m s −1 . In the prism the different colours of<strong>light</strong> are slowed to varying degrees. The refractive index <strong>for</strong> red<strong>light</strong> in crown glass is 1.50 and <strong>for</strong> violet <strong>light</strong> the refractive indexis 1.53.Calculate:a the angle of refraction <strong>for</strong> the red <strong>light</strong>b the angle of refraction <strong>for</strong> the violet <strong>light</strong>c the angle through which the spectrum is dispersedd the speed of the violet <strong>light</strong> in the crown glass.16 The speed of <strong>light</strong> in air is 3.00 × 10 8 m s −1 . As <strong>light</strong> strikes anair–perspex boundary, the angle of incidence is 43.0° and theangle of refraction is 28.5°. Calculate the speed of <strong>light</strong> in theperspex.17 Which one or more of the models of <strong>light</strong> (particle model,electromagnetic wave model, ray model) are used by scientiststo explain the following behaviours of <strong>light</strong>?a reflection b dispersionc total internal reflection d polarisation18 What is the relative refractive index <strong>for</strong> <strong>light</strong> passing fromperspex into water if an incident angle of 17.0° produces an angleof refraction of 14.5°?19 a What is the critical angle <strong>for</strong> <strong>light</strong> passing from:i diamond (n = 2.42) to air?ii glass (n = 1.50) to air?iii water (n = 1.33) to air?b What effect does the relative size of the refractive index haveon the size of the critical angle?20 Why does a ray of <strong>light</strong> that passes through plate glass emergeparallel to its original direction of travel? Has the ray of <strong>light</strong> beenrefracted?Worked Solutions<strong>Ch</strong>apter Quiz<strong>Ch</strong>apter 8 <strong>Models</strong> <strong>for</strong> <strong>light</strong>285