Matriculation Physics ( Physical Optics )

1PHYSICS CHAPTER 2CHAPTER 2: CHAPTER 2: Physical opticsPhysical optics(9 Hours)(9 Hours)The study of interference, diffractioninterference, diffraction and polarization ofand polarization of light light. Light is treated aswaves rather than as rays.PHYSICS CHAPTER 22At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: Explain Explain Huygens principle governing the propagation ofHuygens principle governing the propagation of wave fronts. wave fronts.Include spherical and plane wavefronts. Include spherical and plane wavefronts. Explain Explain diffraction patterns by using Huygens principle. diffraction patterns by using Huygens principle.Learning Outcome:2.1 Huygens principle (1 hour)www.kmph.matrik.edu.my/physicswww.kmph.matrik.edu.my/physicsPHYSICS CHAPTER 232.1.1 Wavefrontsis defined as a line or surface, in the path of a wave motion,a line or surface, in the path of a wave motion, on which the disturbances at every point have the sameon which the disturbances at every point have the same phase phase.Figure 2.1 shows the wavefront of the sinusoidal waves.Line joining all point of adjacent wave, e.g. A, B and C or D,E and F are in phaseWavefront always perpendicular to the direction of wave propagation.Figure2.1 Figure2.12.1 Huygens principleABCDEFwavefrontvPHYSICS CHAPTER 24Type of wavefronts Type of wavefrontsCircular wavefronts Circular wavefronts as shown in Figure 2.2are produced by a point source generates two-dimensional waves.Figure2.2 Figure2.2circular wavefrontray point sourcePHYSICS CHAPTER 25Spherical wavefronts Spherical wavefronts as shown in Figure 2.3are produced by a point source generates three-dimensional waves.Figure2.3 Figure2.3spherical wavefrontsrayspoint sourcePHYSICS CHAPTER 26Plane wavefronts Plane wavefronts as shown in Figures 2.4a and 2.4b are produced by a point source generates three-dimensional waves at large distance from the source.raysplane wavefrontFigure2.4a : (3-D) Figure2.4a : (3-D)plane wavefrontraysFigure2.4b : (2-D) Figure2.4b : (2-D)PHYSICS CHAPTER 27Ray Rayis defined as a line represents the direction of travel of aa line represents the direction of travel of a wave wave.It is at right angle to the wavefronts as shown in Figure 2.5.Beam of light Beam of lightis a collection of rays or a column of light a collection of rays or a column of light.parallel beam, e.g. a laser beam (shown in Figure 2.6a)Figure2.5 Figure2.5raywavefrontSource of light from infinityFigure2.6a Figure2.6aPHYSICS CHAPTER 28divergent beam, e.g. a lamp near you (shown in Figure 2.6b)convergent beam as shown in Figure 2.6c.Figure2.6b Figure2.6bFigure2.6c Figure2.6cPHYSICS CHAPTER 29secondary wavefrontstates that every point on a wavefront can be considered asevery point on a wavefront can be considered as a source of secondary wavelets that spread out in thea source of secondary wavelets that spread out in the forward direction at the speed of the wave. The newforward direction at the speed of the wave. The new wavefront is the envelope of all the secondary wavelets -wavefront is the envelope of all the secondary wavelets - i.e. the tangent to all of them i.e. the tangent to all of them.2.1.2 Huygens principleFigure2.7 Figure2.7waveletsPHYSICS CHAPTER 210P1P2P3P4ABABQ1Q2Q3Q4sApplication of Huygens principle Application of Huygens principlea. Construction of new wavefront for a plane waveFigure2.8 Figure2.8If the wave speed is v, hence in time t the distance travels by the wavelet is s = vt.From Huygens Principle, points P1,P2, P3 and P4 onthe wavefront ABare the sources of secondary wavelets.From the points, draw curves of radius s.Then draw a straight line AB which is tangent to the curves at points Q1,Q2,Q3 and Q4Hence, line AB is the new wavefront after t second.PHYSICS CHAPTER 211AABBsP1P2P3P4Q1Q2Q3Q4sourcerayb. Construction of new wavefront for a circular waveFigure2.9 Figure2.9Explanation as in the construction of new wavefront for a plane wavefront.Butthe wavefront ABis a curve touching points Q1,Q2,Q3 and Q4.The curve AB is the new (circular) wavefront after t second.PHYSICS CHAPTER 212c. Diffraction of wave at a single slitFigure2.10 Figure2.10Huygens principle can be used to explain the diffraction of wave.Each of the point in Figure 2.10, acts as a secondary source of wavelets (red circular arc)The tangent to the wavelets from points 2, 3 and 4 is a plane wavefront.But at the edges, points 1 and 5 are the last points that produce wavelets.Huygens principle suggest that in conforming to the curved shape of the wavelets near the edges, the new wavefront bends or diffracts around the edges - applied to all kinds of waves.Stimulation 2.1PHYSICS CHAPTER 213At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: Define Define coherence. coherence.State State the conditions to observe interference of light. the conditions to observe interference of light.State State the conditions of constructive and destructivethe conditions of constructive and destructive interference. interference. Learning Outcome:2.2 Constructive interference and destructive interference (1 hour)www.kmph.matrik.edu.my/physicswww.kmph.matrik.edu.my/physicsPHYSICS CHAPTER 2142.2.1 Interference of lightLight wave is an electromagnet waves (emw).It consists of varying electric fieldvarying electric field E E and varying magneticvarying magnetic fieldfield B B which are perpendicular to each other perpendicular to each other as shown in Figure 2.11.2.2 Constructive interference and destructive interferenceFigure2.11 Figure2.11Electric field:E = E0 sin (t-kx)Magnetic field:B = B0 sin (t-kx)PHYSICS CHAPTER 215Interference Interference is defined as the effect of interaction betweenthe effect of interaction between two or more waves which overlaps or superposed at a pointtwo or more waves which overlaps or superposed at a point and at a particular time from the sources and at a particular time from the sources.For light For light the Interference is occurred when two light waves meet at a point, a bright or a dark region bright or a dark region will be produced produced in accordance to the Principle of superposition. Principle of superposition Principle of superposition states the resultant displacementthe resultant displacement at any point is the vector sum of the displacements due toat any point is the vector sum of the displacements due to the two light waves the two light waves.Constructive interference Constructive interference is defined as a reinforcement ofa reinforcement of amplitudes of light waves that will produce a bright fringeamplitudes of light waves that will produce a bright fringe (maximum) (maximum).Destructive interference Destructive interference is defined as a total cancellation ofa total cancellation of amplitudes of light waves that will produce a dark fringeamplitudes of light waves that will produce a dark fringe (minimum) (minimum).PHYSICS CHAPTER 216Permanent interference between two sources of light only take place if they are coherent coherent sources. It meansthe sources must have the same wavelength or frequencysame wavelength or frequency (monochromatic).the sources must have a constant phase difference constant phase difference between them.The light waves that are interfering must have the same orsame or approximately of amplitude approximately of amplitude to obtain total cancellation total cancellation at minimum or to obtain a good contrast good contrast at maximum.The distance between the coherent sourcesdistance between the coherent sources should be as smallsmall as possible of the light wavelength ( of the light wavelength ( ) ).2.2.2 Conditions for permanent interferencePHYSICS CHAPTER 217x2x1is defined as the difference in distance from each sourcethe difference in distance from each sourceto a particular pointto a particular point.2.2.3 Path difference, LPath difference, L = |S2P S1P|= |x2 x1|Figure2.12 Figure2.12S1S2screenPL PHYSICS CHAPTER 218Interference of two coherent sources in phase Interference of two coherent sources in phasePath difference for constructive interference Path difference for constructive interferenceS1 and S2 are two coherent sources in phaseS1S2x1x2P (maximum)Figure2.13 Figure2.13+ PHYSICS CHAPTER 219A bright fringe is observed at P thusAt P,thenthereforeNote :m 2 where ,... 2 , 1 , 0 t t m) sin(1 0 P 1kx t E E ) sin(2 0 P 2kx t E E ) ( ) ( 1 2kx t kx t ) ( 2 1x x k since 2 k and L x x ) (2 1L 2L m 22,..... 2 , 1 , 0 t t m m L wherewavelength : Central bright fringem = 01st bright fringe (1st order bright)m = t 12ndbright fringe (2nd order bright) m = t 2When(zeroth order bright)orderPHYSICS CHAPTER 220Path difference for destructive interference Path difference for destructive interferenceS1 and S2 are two coherent sources in phaseS1S2x1x2Q (minimum)Figure2.14 Figure2.14+ PHYSICS CHAPTER 221A dark fringe is observed at Q thusAt P,thenthereforeNote :( ) m 1 2 + where ,... 2 , 1 , 0 t t m) sin(1 0 P 1kx t E E ) sin(2 0 P 2kx t E E ) ( ) ( 1 2kx t kx t ) ( 2 1x x k L 2( ) L m +21 2,..... 2 , 1 , 0 t t m

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.|+ 21m Lwhere1st dark fringe (zeroth order dark)m = 02nddark fringe (1st order dark)m = t 13rd dark fringe (2nd order dark) m = t 2WhenPHYSICS CHAPTER 222Fringe m LInterference pattern for two coherent sources in phase Interference pattern for two coherent sources in phaseFigure2.15 Figure2.15S1S2screen3240200112 210Central bright1st dark 0 211st dark1st bright2 11st bright 232nd dark3 1 232nd dark2nd bright4 22 2nd brightPHYSICS CHAPTER 223Interference of two coherent sources in antiphase Interference of two coherent sources in antiphasePath difference for constructive interference Path difference for constructive interferenceS1 and S2 are two coherent sources in antiphaseS1S2x1x2P (maximum)Figure2.16 Figure2.16+ PHYSICS CHAPTER 224A bright fringe is observed at P thusAt P,thenthereforeNote :m 2 where ,... 2 , 1 t t m) sin(1 0 P 1kx t E E ) sin(2 0 P 2 kx t E E) ( ) ( 1 2kx t kx t ) ( 2 1x x k

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.| L2

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.| L m22,..... 2 , 1 , 0 t t mwhereWhen

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.|+ 21m L1st bright fringe (zeroth order bright)m = 02ndbright fringe (1st order bright) m = t 13rd bright fringe (2nd order bright) m = t 2PHYSICS CHAPTER 225Path difference for destructive interference Path difference for destructive interferenceS1 and S2 are two coherent sources in antiphaseS1S2x1x2Q (minimum)Figure2.17 Figure2.17+ PHYSICS CHAPTER 226A dark fringe is observed at Q thusAt P,thenthereforeNote :( ) m 1 2 + where ,... 2 , 1 , 0 t t m) sin(1 0 P 1kx t E E ) sin(2 0 P 2 + kx t E E) ( ) ( 1 2kx t kx t + + ) ( 2 1x x k +

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.| L2( ) +

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.| + L m21 2,..... 2 , 1 , 0 t t m whereCentral dark fringe (zeroth order dark)m = 01st dark fringe (1st order dark) m = t 12nd dark fringe (2nd order dark) m = t 2When m L PHYSICS CHAPTER 227Fringe m LInterference pattern for two coherent sources in antiphase Interference pattern for two coherent sources in antiphaseFigure2.18 Figure2.184352200112 210Central dark1st bright2 0 211st bright1st dark3 11st dark 232nd bright4 1 232nd bright2nd dark5 22 2nd darkS1S2screenPHYSICS CHAPTER 228Two Coherent sourcesBright fringe Bright fringe Dark fringe Dark fringeIn phase In phaseAntiphase AntiphaseTable 2.1 shows the summary of chapter 2.2.3.Table2.1 Table2.1,... 2 , 1 , 0 t t m m L ,... 2 , 1 , 0 t t m

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.|+ 21m Lm 2 ,... 2 , 1 , 0 m m ) 1 2 ( + ,... 2 , 1 , 0 m,... 2 , 1 , 0 t t m

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.|+ 21m L,... 2 , 1 , 0 t t m m L m 2 ,... 2 , 1 m m ) 1 2 ( + ,... 2 , 1 , 0 mPHYSICS CHAPTER 229At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: Derive Derive with the aid of a diagram andwith the aid of a diagram and use use for bright fringes (maxima) for bright fringes (maxima)for dark fringes (minima),for dark fringes (minima), wherewhere m m = 0, 1, 2, 3, . = 0, 1, 2, 3, .Use Use expressionandexpressionand explain explain the effect of changing any of the variables. the effect of changing any of the variables.Learning Outcome:2.3 Interference of transmitted light through double-slits (2 hours)www.kmph.matrik.edu.my/physicswww.kmph.matrik.edu.my/physicsdD mym( )dD mxm21+dDy PHYSICS CHAPTER 2302.3.1 Methods of obtaining two coherent sourcesDivision of wavefront Division of wavefront2.3 Interference of transmitted lightthrough double-slits Figure2.19 Figure2.19A slit S is placed at equal distance from slits S1 and S2 as shown in figure.Light waves from S that arrived at S1 and S2 are in phase.Therefore, both slits S1 and S2 are two new coherent sources,e.g. in Youngs double slit experiment monochromatic light sourceSsingle slitS1double slitsS2PHYSICS CHAPTER 231airfilmairtDivision of amplitude Division of amplitudeFigure2.20 Figure2.20The incident wavefront is divided into two waves by partial reflection and partial transmission.Both reflected waves 1 and 2 are coherent and will result in interference when they superpose.e.g. Newtons ring, air wedge fringes and thin film interference. incident ray1 1 2 2partial reflectionpartial transmissionPHYSICS CHAPTER 232Figure 2.21 shows the schematic diagram of Youngs double-slit experiment.2.3.2 Youngs double-slit experimentFigure2.21 Figure2.21Max MaxMax MaxMax MaxMax MaxMax MaxMin MinMin MinMin MinMin MinscreenIntensityinterference patternmonochromatic light beamSsingle slitS1double slitsS2m = 2m = 1m = 0m = 1m = 2Picture 2.1PHYSICS CHAPTER 233Explanation of Youngs double-slit experiment by usingExplanation of Youngs double-slit experiment by using Huygens principle Huygens principleWavefront from light source falls on a narrow slit S and diffraction occurs.Every point on the wavefront that falls on S acts as sources ofsecondary wavelets that will produce a new wavefront that propagate to slits S1 and S2 .S1 and S2 are produced two new sources of coherent waves in phase because they originate from the same wavefront and their distance from S are equal.An interference pattern consisting of bright and dark fringes is formed on the screen as shown in Figure 2.21.The bright fringes are occurred when the light from slits S1 and S2 superposes constructively.The dark fringes are occurred when the light from slits S1 and S2 superposes destructively.PHYSICS CHAPTER 234Derivation of Youngs double-slit equations Derivation of Youngs double-slit equationsEquation for separation betweenEquation for separation between central bright fringe central bright fringe andand m mth th bright fringe bright fringeFigure2.22 Figure2.22S1S2PMmQNOdDmyy 1 + mydouble-slitscreenCentral Central brightm mth th bright(m+1) (m+1)th th brightPHYSICS CHAPTER 235SupposeP in Figure 2.22 is the mth order bright fringe, thusLet OP = ym = distance from P to O . In practice d is very small (>d,then S1N meets PQ at right angle. Hence NP = S1Pthen S2N = S2P NP = m.angle PQO = angle S2S1N =From the figure,S2S1N PQOSince is small, thus m P S P S1 2dm 1 22S SN SsinDym QOPOtantan sin DydmmPHYSICS CHAPTER 236Therefore, the separation between central bright and mth bright fringes, ym is given byNote: For bright bright fringesdD mym,... 2 , 1 , 0 order : t t m wherewavelength : screen the and slits - double betweendistance : Dslits - double betweenseparation : d0 mCentral Central bright fringe (Zeroth order Zeroth order maximum)1 1st st bright fringe (1 1st st order order maximum)2 2nd nd bright fringe (2 2nd nd order order maximum)3 3rd rd bright fringe (3 3rd rd order order maximum)1 t m2 t m3 t m(2.1) (2.1)PHYSICS CHAPTER 237Equation for separation betweenEquation for separation between central bright fringe central bright fringe andand m mth th dark fringe dark fringeFigure2.23 Figure2.23S1S2RQNOdDmxy double-slitscreen

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.|+21mCentral Central brightm mth thorderorder dark(m (m 1) 1)th th orderorder darkPHYSICS CHAPTER 238Suppose R in Figure 2.23 is the mth order dark fringe, thusLet OR = xm = distance from R to O . In practice d is very small (>d,then S1N meets RQ at right angle. Hence, NR = S1Rthenangle RQO = angle S2S1N =From the figure,S2S1N RQOSince is small, thus

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.|+ 21R S R S1 2mdm

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.|+ 21S SN Ssin1 22Dxm QOROtantan sin Dxdmm

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.|+21

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.|+ 21NR R S N S2 2mPHYSICS CHAPTER 239Therefore, the separation between central bright and mth order dark fringes, xm is given byNote: For dark dark fringesdDm xm

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.|+ 21,... 2 , 1 , 0 order : t t m where0 m1 1st st dark fringe (Zeroth order Zeroth order minimum)2 2nd nd dark fringe (1 1st st order order minimum)3 3rd rd dark fringe (2 2nd nd order order minimum)4 4th th dark fringe (3 3rd rd order order minimum)1 t m2 t m3 t m(2.2) (2.2)PHYSICS CHAPTER 240Equation for separation between successive (consecutive)Equation for separation between successive (consecutive) bright or dark fringes,bright or dark fringes, yy (Figure 2.22)is given by m my y y +1dD mym where and( )dDm ym11+ +dDy ( )dD mdDm y + 1wherewavelength : screen the and slits - double betweendistance : Dslits - double betweenseparation : dbright e consecutiv betweenseparation : y fringes darkor (2.3) (2.3)PHYSICS CHAPTER 241Appearance of Youngs double-slit experiment Appearance of Youngs double-slit experimentFrom the equation (2.3) (2.3), y depends on :the wavelength of light, the distance apart, d of the double slits, distance between slits and the screen, DExplanation for the above factors: Explanation for the above factors:if is short and thus y decreases for fixed D and d. Theinterference fringes are closer to each other and vice-versa. if the distance apart d of the slits diminished, y increased for fixed D and and vice-versa.if D increases y also increases for fixed and vice-versa.dDy PHYSICS CHAPTER 242if a source slit S (Figure 2.21) is widened the fringes gradually disappear. The slit S then equivalent to large number of narrow slits, each producing its own fringe system at different places. The bright and dark fringes of different systems therefore overlap, giving rise to a different illumination.if one of the slit, S1 or S2 is covered up, the fringes disappear.if the source slit S is moved nearer the double slits, y is unaffected but their intensity increases.if the experiment is carried out in a different medium, for example water, the fringe separation y decreased or increased depending onthe wavelength, of the medium.if white light is used the central bright fringe is white, and the fringes on either side are coloured. Violet is the colour nearer to the central fringe and red is farther away as shown in Figure 2.24.PHYSICS CHAPTER 243Table 2.2 shows the range of wavelength for colours of visible light.Figure2.24 Figure2.24Table2.2 Table2.2ColourRange of / nmViolet 400 450Blue 450 520Green 520 560Yellow 560 600Orange 600 625Red 625 - 700Stimulation 2.2Stimulation 2.3PHYSICS CHAPTER 244A double-slits pattern is view on a screen 1.00 m from the slits. If the third order minima are 25.0 cm apart, determine a. the ratio of wavelength and separation between the slits,b. the distance between the first order minimum and fourth order maximum on the screen.Solution : Solution :a. From the figure,Example 1 :3 m; 25 . 0 m; 00 . 13 m x DS1 S2 D D d 3 3rd rd order minimum order minimumzeroth order maximumzeroth order maximum 3 3rd rd order minimum order minimum3x3x3x 225 . 0233xxm 125 . 03 xPHYSICS CHAPTER 245Solution : Solution :a. By using the equation of separation between central bright and mth order dark fringes, thusb. The separation between central max and the 1st order min. is 3 m; 25 . 0 m; 00 . 13 m x DdDm xm

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.|+ 21dDx

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.|+ 2133( )d00 . 1213 125 . 0

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.|+ 210 57 . 3 ddDx

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.|+ 2111dDx5 . 11 PHYSICS CHAPTER 246Solution : Solution :b. and the separation between central max and the 4th order max.(m = 4) is given byTherefore the distance between the first order minimum and fourth order maximum on the screen is dD mymdDy44 1 4x y d

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.|

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.| dDdDd 5 . 1 4dDd5 . 2 ( )( ) 00 . 1 10 57 . 3 5 . 22 dm 10 93 . 82 dPHYSICS CHAPTER 247a. How would you expect the interference pattern of a double-slit experiment to change if white light is used instead ofmonochromatic light?b. Describe the changes that would be observed in a double-slit interference pattern if the entire experiment were submerged in water. (Physics, 3 (Physics, 3rd rd edition, J. S. Walker, Q4&Q6, p.963) edition, J. S. Walker, Q4&Q6, p.963)Solution : Solution :a. The locations of bright and dark fringes depends on thedepends on the wavelength of light wavelength of light.Therefore, if white light is used in adouble-slit experiment, each bright fringe will show some bright fringe will show someseparation into colours separation into colours, giving a rainbow effect rainbow effect.Example 2 :PHYSICS CHAPTER 248Solution : Solution :b. Submerging the double-slit experiment in water would reducewould reduce the wavelength of the light fromthe wavelength of the light from toto /n /n, where n = 1.33 is the refraction index of water.Therefore, the bright or dark fringe separationwould be reduced fringe separationwould be reduced, according to the equation below: It follows that the interference pattern fringes get closer to eachfringes get closer to each other other.dDy PHYSICS CHAPTER 249In a Youngs double-slit experiment, when a monochromatic light of wavelength 600 nm shines on the double slits, the fringe separation of the interference pattern produced is 7.0 mm. When another monochromatic light source is used, the fringe separation is 5.0 mm. Calculate the wavelength of the second light.Solution : Solution :1st case:By applying the fringe separation equation, thusExample 3 :m; 10 0 . 7 m; 10 6003191 y m 10 0 . 532 yS1 S2 D D d Central of interference pattern Central of interference pattern1y 1y dDy11 ( )dD9310 60010 0 . 7 (1) (1)PHYSICS CHAPTER 250Solution : Solution :2nd case:(2) (1):m; 10 0 . 7 m; 10 5003191 y m 10 0 . 532 ydDy22 dD2310 0 . 5 (2) (2)923310 600 10 0 . 710 0 . 5 m 10 42992 nm 29 4 ORS1 S2 D D d Central of interference pattern Central of interference pattern2y 2y PHYSICS CHAPTER 251Figure 2.25 shows two coherent sources (S1 and S2) of light in phase. The separation of S1 and S2 is 1.2 mmand the screen is 2.5 m from the sources.a. The frequency of the light is 5.77 1014 Hz. Calculatei.the wavelength of the light usedii. the separation between two consecutive bright fringes if the experiment is carried out in air.b. If the experiment is carried out in water of refractive index 1.33, calculate the separation of two consecutive dark fringes. (The speed of light in vacuum, c = 3.00 108 m s1)Example 4 :AB2.5 m1.2 mmS2S1Figure2.25 Figure2.25PHYSICS CHAPTER 252Solution : Solution :a. i. Given By applying the wave speed equation, thusii. By using the equation of fringe separation, thusm 5 . 2 m; 10 2 . 13 D dHz 10 77 . 514 ff c ( )14 810 77 . 5 10 00 . 3 m 10 5209 nm 20 5 ORdDy m 10 08 . 13 y( )( )3910 2 . 15 . 2 10 520 yPHYSICS CHAPTER 253Solution : Solution :b. GivenThe wavelength of light in water is given byTherefore the dark fringes separation ism 5 . 2 m; 10 2 . 13 D dw nw910 52033 . 1m 10 91 . 37w dDx m 10 15 . 84 x( )( )3710 2 . 15 . 2 10 91 . 3 x33 . 1 nPHYSICS CHAPTER 254Exercise 2.1 :1. Youngs double-slit experiment is performed with 589-nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits. (Physics for scientists and engineers,6 (Physics for scientists and engineers,6th th edition,Serway&Jewett,edition,Serway&Jewett, Q37.5, p.1198) Q37.5, p.1198)ANS. : ANS. : 1.54 mm 1.54 mm2. A Youngs interference experiment is performed with monochromatic light. The separation between the slits is 0.500 mm, and the interference pattern on a screen 3.30 m away shows the first side maximum 3.40 mm from the centre of the pattern. What is the wavelength? (Physics for scientists and engineers,6 (Physics for scientists and engineers,6th th edition,Serway&Jewett,edition,Serway&Jewett, Q37.2, p.1197) Q37.2, p.1197)ANS. : ANS. : 515 nm 515 nmPHYSICS CHAPTER 255Exercise 2.1 :3. A coherent light that contains two wavelength, 660 nm (red) and 470 nm (blue) passes through two narrow slitsseparated by 0.3 mm and the interference pattern is observed on a screen 5.00 m from the slits. Determine thedistance between the first order bright fringes for each wavelength.(University physics,11 (University physics,11th th edition, Young&Freedman, Q35.14,edition, Young&Freedman, Q35.14, p.1362) p.1362)ANS. : ANS. : 3.17 mm 3.17 mm4. A monochromatic light of wavelength 560 nm passes through a Youngs double-slit system of unknown slit separation. After that, the slits is illuminated by a monochromatic light of unknown wavelength. It was observed that the 4th order minimum of the known wavelength light overlappedwith the 5th order maximum of the unknown wavelength light on a screen. Calculate the wavelength of the unknown wavelength light.ANS. : ANS. : 504 nm 504 nmPHYSICS CHAPTER 256At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: Explain Explain with the aid of a diagram the interference of lightwith the aid of a diagram the interference of light in thin films for normal incidence. in thin films for normal incidence.For non-reflective coating: For non-reflective coating:Constructive interference : 2 Constructive interference : 2nt nt == m mDestructive interference : 2 Destructive interference : 2nt nt = ( = (m m + ) + )For reflective coating: For reflective coating:Constructive interference : 2 Constructive interference : 2nt nt = ( = (m m + ) + )Destructive interference : 2 Destructive interference : 2nt nt == m mwherewhere m m = 0, 1, 2, 3, = 0, 1, 2, 3, Learning Outcome:2.4 Interference of reflected light in thin films (1 hour)www.kmph.matrik.edu.my/physicswww.kmph.matrik.edu.my/physicsPHYSICS CHAPTER 257Interference due to reflected waves is observed in many everyday circumstances such as bright colours reflected from oil film on water and soap bubble.The reflected waves can change their phase in two ways:The phase changes in proportion to the distance of thephase changes in proportion to the distance of the waves travel waves travel.The phase changes as a result of the reflection a result of the reflection process itself.Optical path Optical path is defined as the product between a distancethe product between a distance travelled by light and the refractive index of the medium travelled by light and the refractive index of the mediumOR2.4 Interference of reflected light in thin films nl L wheremedium in the lightbytravelled distance : lpath optical : Lmedium a of indexrefractive : nPHYSICS CHAPTER 258A light wave travelling in a medium of lower refractive index ( medium of lower refractive index (n n1 1) ) when reflected from a mediums surface of higher refractivereflected from a mediums surface of higher refractive index ( index (n n2 2) ) undergoes a radian phase change radian phase change as shown in Figure 2.26a. 2.4.1 Phase changes due to reflectionFigure2.26a Figure2.26a1n2nincident wave 1n2nreflected wavetransmitted wave incident pulsereflected pulse transmitted pulseradian phase change radian phase changeFigure2.26b: string analogous Figure2.26b: string analogousNote: Note:rad 2 LPHYSICS CHAPTER 259A light wave travelling in a medium of higher refractive indexmedium of higher refractive index ( (n n2 2) ) when reflected from a mediums surface of lowerreflected from a mediums surface of lower refractive index ( refractive index (n n1 1) ) undergoes no phase change no phase change as shown in Figure 2.27a. Figure2.27a Figure2.27a1n2nincident wave 1n2nreflected wavetransmitted wave incident pulseno phase change no phase changeFigure2.27b: string analogous Figure2.27b: string analogousNote: Note:0 0 Lreflected pulse transmitted pulseStimulation 2.4PHYSICS CHAPTER 260Figure 2.28 shows the light waves reflected from the upper and lower surfaces of a thin film (refractive index, n) on a denser medium.2.4.2 Interference from thin films on a denser medium 0 . 11 n5 . 1 n5 . 32 nABECDF2 21 1of thickness : tfilm thin Figure2.28 Figure2.28rad phaserad phase change changerad phaserad phase change changePHYSICS CHAPTER 261When an incident ray falls on a thin film surface almost normal to the surface (point B)division of amplitude occurs,part of ray are reflected (ray 1 ray ABE),part of ray are refracted and reflected (ray 2 ray ABCDF),point D very close to B (BC and CD become straight line).At B,the reflected ray (ray 1) undergoes radian phase change radian phase change.because the ray 1 reflected from a surface of higher refractivesurface of higher refractive index (denser medium) index (denser medium).At C,the reflected ray (ray 2) undergoes radian phase change radian phase change.Therefore both rays 1 and 2 are two coherent sources in phase because the phase difference, is and meet at a point produces interference pattern. 0 PHYSICS CHAPTER 262The optical path difference between rays 1 and 2 is given byConstructive interference: Constructive interference:Destructive interference: Destructive interference:Example of thin film on a denser medium:Non-reflective (anti-reflective) coatingOil film on water and etcABE ABCDF LCD BC+ Lnt nt L + nt L 2 m nt 2,... 2 , 1 , 0 t t mwhere

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.|+ 212 m nt,... 2 , 1 , 0 t t mwherePicture 2.3 Picture 2.4in vacuum lightof wavelength : Picture 2.2PHYSICS CHAPTER 2630 . 1air n33 . 1 n0 . 1air nof thickness : tfilm thin Figure 2.29 shows the light waves reflected from the upper and lower surfaces of a thin film (refractive index, n) in a less dense medium.2.4.3 Interference from thin films on a less dense mediumABECDF2 21 1Figure2.29 Figure2.29rad phaserad phase change changeno no phasephase change changePHYSICS CHAPTER 264When an incident ray falls on a thin film surface almost normal to the surface (point B)division of amplitude occurs,part of ray are reflected (ray 1 ray ABE),part of ray are refracted and reflected (ray 2 ray ABCDF),point D very close to B (BC and CD become straight line).At B,the reflected ray (ray 1) undergoes radian phaseradian phase change change.because the ray 1 reflected from a surface of highersurface of higher refractive index (denser medium) refractive index (denser medium).At C,the reflected ray (ray 2) undergoes no phase change no phase change.Therefore both rays 1 and 2 are two coherent sources antiphase because the phase difference, is and meet at a point produces interference pattern.rad 0 PHYSICS CHAPTER 265The optical path difference between rays 1 and 2 is given byConstructive interference: Constructive interference:Destructive interference: Destructive interference:Example of a thin film on a less dense medium:Soap bubblesReflective coating and etcABE ABCDF L2CD BC+ + L2+ + nt nt L22+ nt L m nt 2,... 2 , 1 , 0 t t mwhere

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.|+ 212 m nt ,... 2 , 1 , 0 t t mwherem nt +22

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.|+ +2122 m ntPicture 2.5 Picture 2.6PHYSICS CHAPTER 266A non-reflective coating of magnesium fluoride of refractive index 1.38 covers the camera lens of refractive index 1.52. The coating prevents reflection of yellow-green light of wavelength in vacuum 565 nm. Determine the minimum non zero thickness of the magnesium fluoride.(Physics,7 (Physics,7th th edition, Cutnell&Johnson, Q48, p.886) edition, Cutnell&Johnson, Q48, p.886)Solution : Solution :Example 5 :0 . 11 n38 . 1 n52 . 12 n2 21 1mintrad phaserad phase change changerad phaserad phase change change2 coherent sources in2 coherent sources in phase phasem 10 5659 PHYSICS CHAPTER 267Solution : Solution :By using the condition of destructive interference for non-reflective coating, thusFor minimum thickness of MgF2,

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.|+ 212 m nt

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.|+ 210 2minnt( ) ( )9min10 5652138 . 1 2

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.| tm 10 02 . 17min tm 10 5659 0 mORnm 102PHYSICS CHAPTER 268White light is incident on a soap film of refractive index 1.30 in air. The reflected light looks bluish because the red light of wavelength 670 nm is absent in the reflection. a. State the condition for destructive interference.b. What is the minimum thickness of the soap film?(Physics,3 (Physics,3rdrd edition, J.S.Walker, Q26, p.966) edition, J.S.Walker, Q26, p.966)Solution : Solution :Example 6 :0 . 11 n30 . 1 n0 . 11 n2 21 1mintrad phaserad phase change changeno no phasephase change change2 coherent sources2 coherent sources antiphase antiphasem 10 6709 PHYSICS CHAPTER 269Solution : Solution :a. The condition of destructive interference is given byb. For minimum thickness of soap film, m nt 2( ) 1 2min nt( ) ( )9min10 670 30 . 1 2 tm 10 58 . 27min tm 10 6709 1 mORnm 58 2PHYSICS CHAPTER 270At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: Explain Explain with the aid of a diagram the interference in airwith the aid of a diagram the interference in air wedge.wedge. Explain Explain with the aid of a diagram the formation ofwith the aid of a diagram the formation of Newtons rings.Newtons rings. Use Use 2 2t t = ( = (m m ++ ) ) for bright fringes (maxima) for bright fringes (maxima)2 2t t= = m mfor dark fringes (minima), for dark fringes (minima),wherewhere m m == 0, 1, 2, 3, 0, 1, 2, 3, Learning Outcome:2.5 Interference of reflected light in air wedge and Newtons rings (1 hour)www.kmph.matrik.edu.my/physicswww.kmph.matrik.edu.my/physicsPHYSICS CHAPTER 2712.5.1 Air wedge2.5 Interference of reflected light in air wedge and Newtons ringFigure2.30: Apparatus setup Figure2.30: Apparatus setuptravelling microscopeglass plateglass slidethin foilmonochromatic light sourceX X Y YairTLL LQ QO OB BP PtlS S x1st dark fringe0 m=0 1 1 2 2 3 3 4 4 5Figure2.32 Figure2.32Figure2.31 Figure2.31PHYSICS CHAPTER 272Ray S falls almost normal to the surface of a glass slide. At point O, Ray S ispartially reflected (ray OL)partially refracted (OB) and then reflected at B (ray PQ)The two refracted rays (OL and PQ) are coherent since both have originated from the same source O.OL and PQ produces interference pattern if it is brought together as shown in Figure 2.32.Since the incidence is nearly normal (point P very close to O), the path difference between the rays at O (ray OL and ray OBPQ) is given by,path difference, L = OB + BP = nt + nt= 2ntwhere n is refractive index of air = 1.0At X, t = 0 and thus the path difference = 0 and a bright fringe is expected, but a dark fringe is observed at X. This is due to the phase change of radian for ray PQ (reflected on a denser medium at B).PHYSICS CHAPTER 273Hence, ray PQ is in antiphase with ray OL and when brought together (by the retina or lens) to interfere, a dark fringe is obtained. Constructive interference (bright fringe): Constructive interference (bright fringe):Destructive interference (dark fringe): Destructive interference (dark fringe): 212 + m tA phase change of radian radian is equivalent to a path difference of

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.|+ 212 m t,... 2 , 1 , 0 m where m t 2,... 2 , 1 , 0 mwhereNote: Note:(2.5) (2.5)(2.4) (2.4)PHYSICS CHAPTER 274From equation (2.4),WhenFrom equation (2.5),When410; m t ,....... , , film, air of thickness when the formed are fringes brighti.e.454341 t431; m t452; m t1 1st st bright bright fringe (Zeroth order Zeroth order maximum)2 2nd nd bright bright fringe (1 1st st order order maximum)3 3rd rd bright bright fringe (2 2nd nd order order maximum)0 0; m t ,....... , , , 0 film, air of thickness when the formed are fringes darki.e.2321 t211; m t 2; m t1 1st st dark dark fringe (Zeroth order Zeroth order minimum)2 2nd nd dark dark fringe (1 1st st order order minimum)3 3rd rd dark dark fringe (2 2nd nd order order minimum)PHYSICS CHAPTER 275Equation for separation between the 1 Equation for separation between the 1st st dark fringe and thedark fringe and the m mth th order dark fringe, order dark fringe, l lFrom Figure 2.31,Rearrange eq. (2.5):ltLT tantan tl 2 mt substitute into eq. (2.6) (2.6) (2.6)tan 2ml ,... 2 , 1 , 0 order : mwherein vacuum lightof wavelength : slide glass of ninclinatio of angle : (2.7) (2.7)PHYSICS CHAPTER 276Equation for separation between the 1 Equation for separation between the 1st st dark fringe and thedark fringe and the m mth th order bright fringe, order bright fringe, l lRearrange eq. (2.4):Equation for separation between adjacent dark fringes orEquation for separation between adjacent dark fringes or bright fringes, bright fringes, x xPut m = 1 into eq. (2.7),( )221 +mt substitute into eq. (2.6) ( )tan 221+ml,... 2 , 1 , 0 order : m where(2.8) (2.8)tan 2 x (2.9) (2.9)PHYSICS CHAPTER 2772.5.2 Newtons ringFigure2.33:Figure2.33: Apparatus setupApparatus setup travelling microscopeglass platemonochromatic light sourceplano-convex lensglass blockL LQ QS StB BO OP Pt R RdFigure2.35:Figure2.35: Newtons ring Newtons ringFigure2.34 Figure2.34X XY YC CA APHYSICS CHAPTER 278Ray S falls almost normal to the surface of a plano-convex. At point O, Ray S ispartially reflected (ray OL)partially refracted (OB) and then reflected at B (ray PQ)The two refracted rays (OL and PQ) are coherent since both have originated from the same source O.OL and PQ produces interference pattern if it is brought together as shown in Figure 2.35.The pattern is a series of circular interference fringes called Newtons ring. This because of a curved piece of glass with a spherical cross section.Since the incidence is nearly normal (point P very close to O), the path difference between the rays at O (ray OL and ray OBPQ) is given by,path difference, L = OB + BP = nt + nt= 2ntwhere n is refractive index of air = 1.0PHYSICS CHAPTER 279At X, t = 0 and thus the path difference = 0 and a bright spot is expected, but a dark spot is observed at X. This is due to the phase change of radian for ray PQ (reflected on a denser medium at B).Hence, ray PQ is in antiphase with ray OL and when brought together (by the retina or lens) to interfere, a dark spot is obtained. Constructive interference (bright ring): Constructive interference (bright ring):Destructive interference (dark ring): Destructive interference (dark ring): 212 + m t

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.|+ 212 m t,... 2 , 1 , 0 m where m t 2,... 2 , 1 , 0 mwhere(2.9) (2.9)(2.8) (2.8)PHYSICS CHAPTER 280Relationship between diameter of ring,Relationship between diameter of ring, d d and thickness ofand thickness of air gap,air gap, t tFrom Figure 2.34,t R R2dY YC CA ABy using the Phytogorean theorem, thus the distance AY is Since t is very thin thus t202 2 2YC AC AY ( )222 2t R Rd

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.|2224t Rtd+ Rtd242 (2.10) (2.10)PHYSICS CHAPTER 281Equation for diameter of dark ring Equation for diameter of dark ringRearrange eq. (2.9):When2 mt substitute into eq. (2.10) Rm d 42,... 2 , 1 , 0 order : m where(2.11) (2.11)

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.|2242 mRd0 ; 0 t mCentral dark spot Central dark spot21; 1 t m 1 1st st dark ring (1 1st st orderorder minimum) t m ; 22 2nd nd dark ring (2 2nd nd orderorder minimum)23; 3 t m 3 3rd rd dark ring (3 3rd rd orderorder minimum)(zeroth order zeroth order minimum), d = 0PHYSICS CHAPTER 282Equation for diameter of bright ring Equation for diameter of bright ringRearrange eq. (2.9):When( )221 +mtsubstitute into eq. (2.10) ( )2124 + m R d,... 2 , 1 , 0 order : m where(2.12) (2.12)( )]]]

+224212 mRd41; 0 t m1 1st st bright ringbright ring (zeroth order zeroth order maximum)43; 1 t m 2 2nd nd bright ring (1 1st st orderorder maximum)45; 2 t m3 3rd rd bright ring (2 2nd nd orderorder maximum)47; 3 t m 4 4th th bright ring (3 3rd rd orderorder maximum)PHYSICS CHAPTER 283From Figure 2.35, The rings become more closely spaced more closely spaced as one moves farther from the centre of the Newtons ring.The reason is that the convex surface of the lens movesconvex surface of the lens moves away from the lower glass block at a progressivelyaway from the lower glass block at a progressively faster faster rate therefore the thickness of air film increasesthickness of air film increases rapidly rapidly.Newtons ring can be used to test the accuracy with which atest the accuracy with which a lens has been ground lens has been ground.The rings are not circular if the surface is not sphericalThe rings are not circular if the surface is not spherical (or the glass block is not flat) (or the glass block is not flat).PHYSICS CHAPTER 284An air wedge is formed by placing a human hair between two glass slides of length 44 mm on one end, and allowing them to touch on the other end. Whenthiswedgeisilluminatedbya red light of wavelength 771 nm, it is observed to have 265 bright fringes. Determinea. the diameter of hair,b. the angle of air wedge,c. the thickness of the air film for 99th dark fringe to be observed,d. the separation between two consecutive bright fringes.Solution : Solution :Example 7 :m 10 44 m; 10 7713 9 L dL265 265th th bright fringe bright fringePHYSICS CHAPTER 285Solution : Solution :a. Assuming the diameter of the hair, d = the thickness of air film, t at265th bright fringeTherefore the diameter of the hair is given byb. The angle of air wedge is

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.|+ 212 m tm 10 02 . 14 dandm 10 44 m; 10 7713 9 L 264 m( )910 77121264 2

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.|+ dLd tan3410 4410 02 . 1tan 13 . 0 PHYSICS CHAPTER 286Solution : Solution :c. By applying the equation for dark fringe (air wedge), thusd. The separation between two consecutive bright fringes is m t 2m 10 78 . 35 tandm 10 44 m; 10 7713 9 L 98 m( )910 771 98 2 ttan 2 x13 . 0 tan 210 7719 xm 10 70 . 14 xPHYSICS CHAPTER 287a. Explain why the central spot in Newtons ring is dark.b. In a Newtons ring experiment, the radius of the qth bright ringis 0.32 cm and the radius of the (q+19)th dark ring is 0.67 cm.Determine the radius of curvature of the plano-convex used in the experiment if the wavelength of light used is 589 nm.Solution : Solution :a. Example 8 :A ray of light reflected from the lower surface of the convex surface has no phase change.Meanwhile, a ray of light reflected from the top surface of glass block undergoes a radian phase change. Thus the two reflected rays are two coherent sources in antiphase.At the centre of the interference pattern, the thickness of the air film is zero, hence the path difference for these two rays goes to zero.These resulting a destructive interference at the central of the Newtons ring.PHYSICS CHAPTER 288Solution : Solution :b. For qth bright ring, For (q+19)th dark ring,m 10 5899 ( )m 10 67 . 0 m; 10 32 . 0219 q2q+ r r1 q m( )2124 + m R d andq2r d ( ) ( ) [ ]212q1 q 4 2 + R r(1) (1)19 + q m Rm d 42 and( ) 19 q2+ r d( )( ) ( ) 19 q 4 2219 q+ +R r(2) (2)( )( ) ( ) 19 q 4 2219 q+ +R r( ) ( ) 5 . 0 q 4 22q R rPHYSICS CHAPTER 289Solution : Solution :b. (2) (1) :By substituting q = 6.27 into eq. (1) thusm 10 5899 ( )m 10 67 . 0 m; 10 32 . 0219 q2q+ r r( )5 . 0 q19 q 2q19 q+

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.|+rr5 . 0 q19 q 10 32 . 010 67 . 0222+

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.|27 . 6 q( ) ( ) ( ) ( )92210 589 5 . 0 27 . 6 4 10 32 . 0 2 Rm 01 . 3 RPHYSICS CHAPTER 290Exercise 2.2 :1. A thin film of gasoline floats on a puddle of water. Sunlight falls almost perpendicularly on the film and reflects into your eyes. Although the sunlight is white, since it contains all colours, the film has a yellow hue, because destructive interference has occurred eliminating the colour of blue ( =469 nm) from the reflected light. If the refractive indices for gasoline and water are 1.40 and 1.33 respectively, Calculate the minimum thickness of the film.ANS. : ANS. : 168 nm 168 nm2. White light is incident normally on a thin soap film (n =1.33) suspended in air.a. What are the two minimum thickness that will constructivelyreflect yellow light of wavelength 590 nm?b. What are the two minimum thickness that will destructively reflect yellow light of wavelength 590 nm? (Physics,3 (Physics,3rdrd edition, J.S.Walker, Q34, p.966) edition, J.S.Walker, Q34, p.966) ANS. : ANS. : 110 nm, 330 nm ; 220 nm, 440 nm ; 110 nm, 330 nm ; 220 nm, 440 nm ;PHYSICS CHAPTER 2913. Two plane glass plates which are in contact at one edge are separated by a piece of metal foil 12.5 cm from that edge. Interference fringes parallel to the line of contact are observed in reflected light of wavelength 546 nm and are found to be 1.50 mm apart. Determine the thickness of the foil.ANS. : ANS. : 2.27 2.27 10 10 5 5 m m4. Newtons rings are formed by reflection between an biconvex lens of focal length 100 cm made of glass of refractive index 1.50 and in contact with a glass block of refractive index 1.60. Calculate the diameter and thickness of air film for fifth bright ring using light of wavelength 6000 . Given 1 angstrom () = 1010 mANS. : ANS. : 3.28 mm; 1.353.28 mm; 1.35 m m5. Newtons rings are formed with light of wavelength 589 nm between the plano-convex lens of radius of curvature 100 cm and a glass block, in perfectcontact.a. Determine the radius of the 20th dark ring from the centre.b. How will this ring move and what will its radius become if the lens and the block are slowly separated to a distance apart 5.00 104 cm? ANS. : ANS. : 3.43 mm; inwards, 1.26 cm 3.43 mm; inwards, 1.26 cmPHYSICS CHAPTER 292At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: Explain Explain with the aid of a diagram the diffraction of awith the aid of a diagram the diffraction of a single slit. single slit.Derive and use Derive and use formulaformula for dark fringes (minima) for dark fringes (minima)for bright fringes (maxima), for bright fringes (maxima),wherewhere n n = 1, 2, 3, ... = 1, 2, 3, ...Explain Explain with the aid of a diagram the effect of changingwith the aid of a diagram the effect of changing wavelength on the resolution of single slit from twowavelength on the resolution of single slit from two coherent sources. coherent sources.Learning Outcome:2.6 Diffraction by a single slit (1 hour)www.kmph.matrik.edu.my/physicswww.kmph.matrik.edu.my/physicsaD nyn( )aD nyn21+PHYSICS CHAPTER 2932.6.1 Diffraction of lightis defined as the bending of light waves as they travel aroundthe bending of light waves as they travel around obstacles or pass through an aperture or slit comparable toobstacles or pass through an aperture or slit comparable to the wavelength of the light waves the wavelength of the light waves.Figures 2.36a, 2.36b and 2.36c show the bending of plane wavefront.2.6 Diffraction by a single slitFigure2.36a:Figure2.36a: obstacle obstacleFigure2.36b:Figure2.36b: slit,slit, aa >> Figure2.36c:Figure2.36c: slit,slit, aa PHYSICS CHAPTER 294Figure 2.37 shows an apparatus setup of diffraction by a single slit.2.6.2 Diffraction by a single siltFigure2.37 Figure2.37intensity intensityscreensingle slitCentral maximum1st minimum2nd minimum3rd minimum1st maximum2nd maximum1st minimum2nd minimum3rd minimum1st maximum2nd maximum2121Animation 2.1Picture 2.7Picture 2.8Sn diffractio of angle : wherePHYSICS CHAPTER 295Explanation of single slit diffraction experiment Explanation of single slit diffraction experimentWavefront from light source falls on a narrow slit S and diffraction occurs.Every point on the wavefront that falls on S acts as sources ofsecondary wavelets and superposed each another to form an interference pattern on the screen as shown in Figure 2.37.The central fringe is bright (maximum) central fringe is bright (maximum) and widen widen compare to other bright fringes.The central fringe has the highest intensity highest intensity compare to the other bright fringes.The intensity of bright fringes reduce intensity of bright fringes reduce as the distancedistance increase increase from the central bright fringe.Other rays with angle 2 and 1 will produce minimum and maximum on both sides of the central maximum.PHYSICS CHAPTER 296Derivation of single slit diffraction equations Derivation of single slit diffraction equationsEquation for separation betweenEquation for separation between central maximum (bright)central maximum (bright) andand n nth th minimum (dark) fringes minimum (dark) fringesFigure2.38 Figure2.38slit width : awherescreen and slitsingle betweendistance : D2ndstrip1st stripCentralCentral maximum maximumP PQ QDy yn n11 A A2aa2a1sin2a1sin2aC CB Bscreen screenE En nth th minimum minimumPHYSICS CHAPTER 297A single slit is split into two equal parts, AC and CB. A,C and B are new sources of secondary wavelets. (Huygens principle)When the wavelets from A, C and B superpose, interference will occur at P.As AB is very small AB is very small, thus AE is perpendicular to CP and AP = EP AE is perpendicular to CP and AP = EP, the outgoing rays are considered parallel considered parallel,and therefore the path difference at P between ray AP and CP is :Consider two narrow strips as shown in Figure 2.38, for the two strips superposed destructively two strips superposed destructively thus both strip of light must in antiphase antiphase to each another which is equivalence to a path difference of path difference of . If the 1 1st st minimum (1 minimum (1st st order minimum) order minimum) is at P, hence :1sin2CE aL 2sin21 aL 1sin aPHYSICS CHAPTER 298For the 2 2nd nd minimum minimum and 3 3rd rd minimum minimum, AB is split into 44 equal parts, 6 equal parts equal parts, 6 equal parts and so on as shown in Figures 2.39 and 2.40.2sin4a2 2 sin a2 2sin6a3 3 sin a3 224aa 336aa2nd minimum (2nd order minimum)3rd minimum (3rd order minimum)Figure2.39 Figure2.39 Figure2.40 Figure2.401st strip2nd strip3rd strip4th strip1st strip2ndstrip5thstrip3rdstrip4thstrip6thstripPHYSICS CHAPTER 299In general, for minima (dark fringes) minima (dark fringes)If the distance of single slit to the screen is D, and D>>a D>>a then:Therefore the distance of n distance of nth th minimum from central maximum is:When nn sin a3,.. 2, 1, order : t t t n whereDynn n tan sin nDyan

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.|aD nyn1 t n 1 1st st minimum fringe ( (1 1st st orderorder minimum)2 t n 2 2ndnd minimum fringe (2 2nd ndorder order minimum)3 t n3 3rd rd minimum fringe (3 3rd rdorder order minimum)PHYSICS CHAPTER 21002ndstrip1st strip3rdstripEquation for separation betweenEquation for separation between central maximum (bright)central maximum (bright) andand n nth th maximum (bright) fringes maximum (bright) fringesFigure2.41 Figure2.41CentralCentral maximum maximumR RQ QDy yn n11 A A3aa3aC CB Bscreen screenE En nth th maximum maximum3a1sin3a1sin aD DPHYSICS CHAPTER 2101A single slit is split into three equal parts, AC,CD and DB. A,C,D and B are new sources of secondary wavelets. (Huygens principle)When the wavelets from A,C,D and B superpose, interference will occur at R.As AB is very small, thus AE is perpendicular to CP and AP = EP, the outgoing rays are considered parallel,and therefore the path difference at P between ray AP and CP is :Consider three narrow strips as shown in Figure 2.41, the first two strips (pair) superposed destructively at which the path difference is and leave the third strip leave the third strip. The 3 3rd rd strip produces the maximum (bright) fringe at R. strip produces the maximum (bright) fringe at R. If the 1 1st st maximum (1 maximum (1st st order maximum) order maximum) is at R, hence :1sin3CE aL 2sin31 aL23sin1 aPHYSICS CHAPTER 2102For the 2 2nd nd maximum maximum and 3 3rd rd maximum maximum, AB is split into 55 equal parts, 7 equal parts equal parts, 7 equal parts and so on as shown in Figures 2.42 and 2.43.2sin5a2 25sin a2 2sin7a3 27sin a3 2522732nd maximum (2nd order maximum)3rd maximum (3rd order minimum)Figure2.42 Figure2.42 Figure2.43 Figure2.432nd strip3rd strip4th strip5th strip1st strip5aa7aa1st strip2ndstrip5thstrip3rdstrip4thstrip6thstrip7thstripPHYSICS CHAPTER 2103( )aD nyn21+In general, for maxima (bright fringes) maxima (bright fringes)If the distance of single slit to the screen is D, and D>>a D>>a then:Therefore the distance of n distance of nth th maximumfrom central maximum is:When

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.|+ 21sin a nn 3,.. 2, 1, t t t nwhereDynn n tan sin

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.|+

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.|21nDyan1 t n 1 1st st maximum fringe ( (1 1st st orderorder maximum)2 t n 2 2nd ndmaximum fringe (2 2nd ndorder order maximum)3 t n3 3rd rd maximum fringe (3 3rd rdorder order maximum)PHYSICS CHAPTER 2104Equation forEquation for central maximum (bright) fringe central maximum (bright) fringeFigure2.44 Figure2.4411CentralCentral maximum maximumQ Q1 1st st minimum minimum1 1st st minimum minimumy1y1Figure2.45 Figure2.45A AC CB BD DE EDascreen screensingle slit single slitPHYSICS CHAPTER 2105

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.|a1wsin 2Figure 2.45 shows five sources of Huygens wavelets and the screen is to be so far from the slit ( screen is to be so far from the slit (D>>a D>>a) thus the raysrays from each source arenearly parallel from each source arenearly parallel.All the wavelets from each source travel the same distancesame distance to the point Q to the point Q (Figure 2.44) and arriving there in phase in phase.Therefore, the constructive interference is occurred at theconstructive interference is occurred at the central of the single slit diffraction pattern central of the single slit diffraction pattern.The angular width of central maximum,angular width of central maximum, w w is given by1 w2 angle ndiffractio minimum 1 :st1 andasin11PHYSICS CHAPTER 2106

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.|aDw2The width of central maximum,width of central maximum, w w is given byNote:To calculate the maximum number of orders observed maximum number of orders observed, take the diffraction angle, diffraction angle,= = 90 90 .From both equations for minima and maxima, we obtainBy using this two relations, the changes of single slit diffraction pattern can be explained.12y w and and minimum 1 of separation :st1ymaximum centrala1Dy nsin and nyPHYSICS CHAPTER 2107A sodium light of wavelength 580 nm shines through a slit and produces a diffraction pattern on a screen 0.60 m away. The width of the central maximum fringe on the screen is 5.0 cm. Determinea. the width of the slit,b. the angular width of the central maximum fringe,c. the number of minimum that can be observed on the screen.Solution : Solution :a. SinceExample 9 :m 10 0 . 5 m, 60 . 0 m; 10 5802 9 w D CentralCentral maximum maximum1 1st st minimum minimum1 1st st minimum minimumw a12y w anda1Dy

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.|aDw2( )( )a60 . 0 10 5802 10 0 . 592 m 10 39 . 15 aPHYSICS CHAPTER 2108Solution : Solution :b. The angular width of the central maximum fringe is given byc. By applying the equation for minimum fringe,

For the maximum no. of order for minimum fringe,Therefore the number of minimum that can be observed is 2323 2 = 46 fringes 2 = 46 fringes1 w2 78 . 4w anda11sinm 10 0 . 5 m, 60 . 0 m; 10 5802 9 w D

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.|a1wsin 2

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.|591w10 39 . 110 580sin 2 n a sin90 ( ) ( )9max510 580 90 sin 10 39 . 1 n97 . 23max n23 PHYSICS CHAPTER 2109a. State the similarities and differences of double-slit interference and single slit diffraction patterns.b. How many bright fringes will be produced on the screen if agreen light of wavelength 553 nm is incident on a slit of width8.00 m?Solution : Solution :a. The similarities areExample 10 :Double-slit interference pattern Single slit diffraction patternBoth patterns consist of alternating dark and brightBoth patterns consist of alternating dark and bright fringes. fringes.The central for both patterns is bright fringe. The central for both patterns is bright fringe.PHYSICS CHAPTER 2110a. The differences areb. GivenBy applying the equation for bright (maximum) fringe,

For the maximum no. of order for bright fringe,

Therefore the number of bright that can be observed is (13(13 2)+1 = 27 fringes 2)+1 = 27 fringesDouble-slit interference pattern Single slit diffraction patternThe width of each fringe isThe width of each fringe is the same. the same.The central fringe is widerThe central fringe is wider than the other fringes. than the other fringes.The intensity of each brightThe intensity of each bright fringe is constant. fringe is constant.The intensity of brightThe intensity of bright fringes reduce as a distancefringes reduce as a distance increase from the centralincrease from the central bright.bright. m 10 00 . 8 m; 10 5536 9 a

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.|+ 21sin n a90 ( ) ( ) ( )9max610 553 5 . 0 90 sin 10 00 . 8 + n97 . 13max n 13 Central brightCentral bright fringe fringePHYSICS CHAPTER 2111Exercise 2.3 :1. Monochromatic light of wavelength 689 nm falls on a slit. If the angle between first bright fringes on either side of the central maximum is 38, calculate the slit width. (Physics for scientist & engineers ,3 (Physics for scientist & engineers ,3rd rdedition, Giancoli, Q4, p.913) edition, Giancoli, Q4, p.913)ANS. : ANS. : 3.23.2 m m2. Light of wavelength 633 nm from a distant source is incident on a single slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. Determine the distance between the two dark fringes on either side of the central bright fringe. (University physics,11 (University physics,11th th edition, Young&Freedman, Q36.4, p.1396) edition, Young&Freedman, Q36.4, p.1396)ANS. : ANS. : 5.91 mm 5.91 mm3. A screen is placed 1.00 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.60 cm wide. What is the distance between the two second order minima?(Physics,3 (Physics,3rd rdedition, J.S.Walker, Q45, p.967) edition, J.S.Walker, Q45, p.967) ANS. : ANS. : 3.20 cm 3.20 cmPHYSICS CHAPTER 2112At the end of this chapter, students should be able to:At the end of this chapter, students should be able to: Explain Explain with the aid of a diagram the formation ofwith the aid of a diagram the formation of diffraction. diffraction.Apply Apply formula, formula,where whereDescribe Describe with the aid of diagram the formation ofwith the aid of diagram the formation of spectrum by using white light. spectrum by using white light.Learning Outcome:2.7 Diffraction grating (2 hours)www.kmph.matrik.edu.my/physicswww.kmph.matrik.edu.my/physics n dn sinNd1PHYSICS CHAPTER 2113is defined as a large number of equally spaced parallel slits a large number of equally spaced parallel slits.Diffraction grating can be made by ruling very fine parallelruling very fine parallel lines on glass or metal by a very precise machine lines on glass or metal by a very precise machine.The untouched spaces between the lines untouched spaces between the lines serve as the slits slits as shown in Figure 2.46. 2.7 Diffraction gratingFigure2.46 Figure2.46linesslitdLight passes through the slit because it istransparent.The spaces between the lines are the slits, for example : if there are four lines then we have 3 slits.PHYSICS CHAPTER 2114If there Nlines per unit length, then slit separation, dis given by:e.g. if a diffraction grating has 5000 lines per cm, thenThe light that passes through the slits are coherent coherent .The Interference pattern is narrower and sharper narrower and sharper than double-slits.There are two type of diffraction grating which aretransmission gratingtransmission grating (usual diffraction grating)reflection gratingreflection grating e.g. CD and DVDDiffraction grating is used in spectrometer to determine thedetermine the wavelength wavelength of light and to study spectra study spectra.Nd1cm 10 24 dcm 50001 1 NdPHYSICS CHAPTER 2115Figure 2.47 shows an incident lights fall on the transmission diffraction grating.2.7.1 Explanation of diffraction by using Huygens principle for diffraction gratingFigure2.47 Figure2.47C CD Dfirst order wavefrontF FE Esecond order wavefrontthird order wavefrontA AB Bzeroth order wavefrontincident lightssource of secondary waveletsgratingPHYSICS CHAPTER 2116Using Huygens principle, each maximum is located by taking the tangent of the wavelets from the slits.If the wavelets from each of the slits are drawn and a tangent AB AB is drawn, a plane wavefront parallel to the diffraction grating is obtained. This represents the zeroth-order maximum (n = 0).If the wavelets are grouped such that the first wavelet from one slit is combined with the second wavelet from the next slit, the third wavelet from the third slit and so on, the tangent CD CD will represent the first-order maximum (n =1).For the second-order maximum, the wavelets are grouped are such that the second wavelet of one slit is combined with the fourth wavelet of the next slit, the sixth wavelet from the third slit and so on. (tangent EF EF)Similarly, the third-, fourth-,. order maximum may be obtained.PHYSICS CHAPTER 2117Figure 2.48 illustrates light travels to a distant viewing screen from five slits of the grating.2.7.2 Equation of diffraction gratingFigure2.48 Figure2.48central or zeroth order maximum (n = 0)first order maximum(n = 1)first order maximum(n = 1)incoming plane wavefront of lightdiffraction gratingd sin dFigure2.49 Figure2.49PHYSICS CHAPTER 2118The maximum (bright) fringes are sometimes called the principal maxima or principal fringes since they are placed where the light intensity is a maximum.Since the screen is far so that the rays nearly parallel while the light travels toward the screen as shown in Figure 2.49.In reaching the place on the screen while the 1st order maximum is located, light from one slit travels a distance of one wavelength farther than light from adjacent slit. Therefore the path difference for maximum fringe (constructive interference) is given byWhen n dn sin3,.. 2, 1, , 0 order : t t t n whereangle ndiffractio of order:thnn0 n CentralCentral maximum fringe ( (0 0th th orderorder maximum)1 t n 1 1st st maximum fringe (1 1st st orderorder maximum)2 t n2 2nd nd maximum fringe (2 2nd nd orderorder maximum)PHYSICS CHAPTER 2119The maximum fringes produce by a grating are much narrower and sharper than those from a double-slit as the intensity graph in Figures 2.50a and 2.50b. 0 n1 2 1 2 0 n1 2 1 2 Figure2.50a Figure2.50aFigure2.50b Figure2.50bPHYSICS CHAPTER 2120Figures 2.51 shows the diffraction grating pattern.Figure2.51 Figure2.5121gratingParallel beam of monochromatic lightzero-order maximum0 nfirst-order maximumfirst-order maximum1 n1 nsecond-order maximumsecond-order maximum2 n2 nPHYSICS CHAPTER 2121If the white light is falls on the grating, a rainbow colours would be observed to either side of the central fringe on the screen which is white as shown in Figure 2.51. This because the whitewhite light contains wavelengths between violet and red light contains wavelengths between violet and red.RainbowRainbow RainbowRainbowwhite0 n1 2 1 2 White lightFigure2.51 Figure2.51PHYSICS CHAPTER 2122Note:To calculate the maximum number of orders for brightmaximum number of orders for bright fringes observed fringes observed, take the diffraction angle, diffraction angle, = = 90 90 . ThereforeFrom the equation for maxima, we obtainBy using this two relations, the changes of diffraction grating pattern can be explained.max90 sin n d dn maxwhere nmax : maximum number of orders that can be observed. nsin anddn1sin PHYSICS CHAPTER 2123A monochromatic light of unknown wavelength falls normally on a diffraction grating. The diffraction grating has 3000 lines per cm.If the angular separation between the first order maxima is 35. Calculatea. the wavelength of the light,b. the angular separation between the second-order and third-order maxima.Solution : Solution :Example 11 :1 ; 35 2 ; cm 300011 n N351 1st st order max order max.1 1st st order max order max.1PHYSICS CHAPTER 2124Solution : Solution :a. The diffraction angle for 1st order maximum is And the slit separation, d is given byTherefore the wavelength of the light isNd15 . 171 cm 10 33 . 34 d1cm 30001 d1 ; 35 2 ; cm 300011 n NOR m 10 33 . 36 35 21 n dn sin( ) 16sin 10 33 . 3( ) 5 . 17 sin 10 33 . 36m 10 00 . 16 PHYSICS CHAPTER 2125Solution : Solution :b. By using the equation of diffraction grating for maxima, 20th order maximum0 n2nd order maximum2nd order maximum2 n2 n3rd order maximum3rd order maximum3 n3 n33 2 n dn sinPHYSICS CHAPTER 2126Solution : Solution :b. For 2nd order maximum,For 3rd order maximum,Therefore the angular separation, 2 sin2 d2 n( ) ( )62610 00 . 1 2 sin 10 33 . 3 9 . 362 3 sin3 d3 n( ) ( )63610 00 . 1 3 sin 10 33 . 3 3 . 643 2 3 23 4 . 2723 9 . 36 3 . 6423 PHYSICS CHAPTER 2127The second-order maximum produced by a diffraction grating with 560 lines per centimeter is at an angle of 3.1.a. What is the wavelength of the light that illuminates the grating?b. Determine the number of maximum can be observed on a screen.c. State and giving reason, what you would expect to observe if a grating with a larger number of lines per centimeter is used.Solution : Solution :a. By applying the equation of diffraction grating for 2nd order maximum, thus Example 12 :2 ; 3.1 ; m 10 5621 3 n N 2 sin2 d N 2 sin2 andNd1( )310 56 2 1 . 3 sin m 10 83 . 47 PHYSICS CHAPTER 2128Solution : Solution :b. By applying the equation of grating for maximum,

For the maximum no. of order for maximum fringe,Therefore the number of maximum can be observed is(36(36 2)+1 = 73 fringes 2)+1 = 73 fringesc. The fringes become farther to each another fringes become farther to each another.Reason : since a larger number of lines per cm results in a largerlarger number of lines per cm results in a larger diffraction angle diffraction angle thus the distance between two consecutive maximum fringes will increase. n dn sin andNd1 nNn sin90 n( )( )7 3max10 83 . 4 10 56 90 sin n97 . 36max n 36 d1sin andNd1 N sinPHYSICS CHAPTER 2129Exercise 2.4 :1. The first-order maximum line of 589 nm light falling on a diffraction grating is observed at an angle of 15.5. Determinea. the slit separation on the grating.b. the angle of diffraction for third-order maximum line. (Physics for scientist & engineers ,3 (Physics for scientist & engineers ,3rd rdedition, Giancoli, Q32, p.914) edition, Giancoli, Q32, p.914)ANS. : ANS. : 2.202.20 m; 53.4 m; 53.4 2. A diffraction grating has 6000 lines per cm. Calculate the angularseparationbetweenwavelengths589.6 nm and 546.1 nm respectively after transmission through it at normal incidence, in the first-order spectrum (maximum line).ANS. : ANS. : 1.60 1.60 3. When blue light of wavelength 465 nm illuminates a diffraction grating, it produces a 1st order maximum but no 2nd order maximum.a. Explain the absence of 2nd order maximum.b. What is the maximum spacing between lines on this grating?(Physics,3 (Physics,3rd rdedition, J.S.Walker, Q65, p.968) edition, J.S.Walker, Q65, p.968) ANS. : ANS. : 930 nm 930 nm130PHYSICS CHAPTER 2Next ChapterCHAPTER 3 :Electrostatics