2 Wave OpticsThe wave nature of light is needed to explain various phenomenaInterferenceDiffractionPolarizationThe particle nature of light was the basis for ray (geometric) optics
3 InterferenceLight waves interfere with each other much like mechanical waves doAll interference associated with light waves arises when the electromagnetic fields that constitute the individual waves combine
4 Conditions for Interference For sustained interference between two sources of light to be observed, there are two conditions which must be metThe sources must be coherentThey must maintain a constant phase with respect to each otherThe waves must have identical wavelengths
5 Producing Coherent Sources Light from a monochromatic source is allowed to pass through a narrow slitThe light from the single slit is allowed to fall on a screen containing two narrow slitsThe first slit is needed to insure the light comes from a tiny region of the source which is coherentOld method
6 Producing Coherent Sources, cont Currently, it is much more common to use a laser as a coherent sourceThe laser produces an intense, coherent, monochromatic beam over a width of several millimetersThe laser light can be used to illuminate multiple slits directly
7 Young’s Double Slit Experiment Thomas Young first demonstrated interference in light waves from two sources in 1801Light is incident on a screen with a narrow slit, SoThe light waves emerging from this slit arrive at a second screen that contains two narrow, parallel slits, S1 and S2
8 Young’s Double Slit Experiment, Diagram The narrow slits, S1 and S2 act as sources of wavesThe waves emerging from the slits originate from the same wave front and therefore are always in phase
9 Resulting Interference Pattern The light from the two slits form a visible pattern on a screenThe pattern consists of a series of bright and dark parallel bands called fringesConstructive interference occurs where a bright fringe appearsDestructive interference results in a dark fringe
10 Fringe PatternThe fringe pattern formed from a Young’s Double Slit Experiment would look like thisThe bright areas represent constructive interferenceThe dark areas represent destructive interference
11 Interference Patterns Constructive interference occurs at the center pointThe two waves travel the same distanceTherefore, they arrive in phase
12 Interference Patterns, 2 The upper wave has to travel farther than the lower waveThe upper wave travels one wavelength fartherTherefore, the waves arrive in phaseA bright fringe occurs
13 Interference Patterns, 3 The upper wave travels one-half of a wavelength farther than the lower waveThe trough of the bottom wave overlaps the crest of the upper waveThis is destructive interferenceA dark fringe occurs
14 Interference Equations The path difference, δ, is found from the tan triangleδ = r2 – r1 = d sin θThis assumes the paths are parallelNot exactly parallel, but a very good approximation since L is much greater than d
15 Interference Equations, 2 For a bright fringe, produced by constructive interference, the path difference must be either zero or some integral multiple of the wavelengthδ = d sin θbright = m λm = 0, ±1, ±2, …m is called the order numberWhen m = 0, it is the zeroth order maximumWhen m = ±1, it is called the first order maximum
16 Interference Equations, 3 The positions of the fringes can be measured vertically from the zeroth order maximumy = L tan θ L sin θAssumptionsL>>dd>>λApproximationθ is small and therefore the approximation tan θ sin θ can be used
17 Interference Equations, 4 When destructive interference occurs, a dark fringe is observedThis needs a path difference of an odd half wavelengthδ = d sin θdark = (m + ½) λm = 0, ±1, ±2, …
18 Interference Equations, final For bright fringesFor dark fringes
19 Uses for Young’s Double Slit Experiment Young’s Double Slit Experiment provides a method for measuring wavelength of the lightThis experiment gave the wave model of light a great deal of credibilityIt is inconceivable that particles of light could cancel each other
20 Lloyd’s MirrorAn arrangement for producing an interference pattern with a single light sourceWave reach point P either by a direct path or by reflectionThe reflected ray can be treated as a ray from the source S’ behind the mirror
21 Interference Pattern from the Lloyd’s Mirror An interference pattern is formedThe positions of the dark and bright fringes are reversed relative to pattern of two real sourcesThis is because there is a 180° phase change produced by the reflection
22 Example 1A pair of narrow, parallel slits separated by mm is illuminated by the green component from a mercury vapor lamp (λ = nm). The interference pattern is observed on a screen 1.20 m from the plane of the parallel slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands in the interference pattern.
23 Example 2If the distance between two slits is mm and the distance to a screen is 2.50 m, find the spacing between the first- and second-order bright fringes for yellow light of 600-nm wavelength.
24 Example 3White light spans the wavelength range between about 400 nm and 700 nm. If white light passes through two slits 0.30 mm apart and falls on a screen 1.5 m from the slits, find the distance between the first-order violet and the first-order red fringes.
25 Phase Changes Due To Reflection An electromagnetic wave undergoes a phase change of 180° upon reflection from a medium of higher index of refraction than the one in which it was travelingAnalogous to a reflected pulse on a string
26 Phase Changes Due To Reflection, cont There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refractionAnalogous to a pulse in a string reflecting from a free support
27 Interference in Thin Films Interference effects are commonly observed in thin filmsExamples are soap bubbles and oil on waterThe interference is due to the interaction of the waves reflected from both surfaces of the film
28 Interference in Thin Films, 2 Facts to rememberAn electromagnetic wave traveling from a medium of index of refraction n1 toward a medium of index of refraction n2 undergoes a 180° phase change on reflection when n2 > n1There is no phase change in the reflected wave if n2 < n1The wavelength of light λn in a medium with index of refraction n is λn = λ/n where λ is the wavelength of light in vacuum
29 Interference in Thin Films, 3 Ray 1 undergoes a phase change of 180° with respect to the incident rayRay 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave
30 Interference in Thin Films, 4 Ray 2 also travels an additional distance of 2t before the waves recombineFor constructive interference2nt = (m + ½ ) λ m = 0, 1, 2 …This takes into account both the difference in optical path length for the two rays and the 180° phase changeFor destruction interference2 n t = m λ m = 0, 1, 2 …
31 Interference in Thin Films, 5 Two factors influence interferencePossible phase reversals on reflectionDifferences in travel distanceThe conditions are valid if the medium above the top surface is the same as the medium below the bottom surfaceIf the thin film is between two different media, one of lower index than the film and one of higher index, the conditions for constructive and destructive interference are reversed
32 Interference in Thin Films, final Be sure to include two effects when analyzing the interference pattern from a thin filmPath lengthPhase change
33 Newton’s RingsAnother method for viewing interference is to place a planoconvex lens on top of a flat glass surfaceThe air film between the glass surfaces varies in thickness from zero at the point of contact to some thickness tA pattern of light and dark rings is observedThis rings are called Newton’s RingsThe particle model of light could not explain the origin of the ringsNewton’s Rings can be used to test optical lenses
34 Problem Solving Strategy with Thin Films, 1 Identify the thin film causing the interferenceDetermine the indices of refraction in the film and the media on either side of itDetermine the number of phase reversals: zero, one or two
35 Problem Solving with Thin Films, 2 The interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λThe conditions are reversed if one of the waves undergoes a phase change on reflection
36 Problem Solving with Thin Films, 3 Equation1 phase reversal0 or 2 phase reversals2nt = (m + ½) lconstructivedestructive2nt = m l
37 Interference in Thin Films, Example An example of different indices of refractionA coating on a solar cellThere are two phase changes
38 CD’s and DVD’s Data is stored digitally A series of ones and zeros read by laser light reflected from the diskStrong reflections correspond to constructive interferenceThese reflections are chosen to represent zerosWeak reflections correspond to destructive interferenceThese reflections are chosen to represent ones
39 CD’s and Thin Film Interference A CD has multiple tracksThe tracks consist of a sequence of pits of varying length formed in a reflecting information layerThe pits appear as bumps to the laser beamThe laser beam shines on the metallic layer through a clear plastic coating
40 Reading a CDAs the disk rotates, the laser reflects off the sequence of bumps and lower areas into a photodectorThe photodector converts the fluctuating reflected light intensity into an electrical string of zeros and onesThe pit depth is made equal to one-quarter of the wavelength of the light
41 Reading a CD, contWhen the laser beam hits a rising or falling bump edge, part of the beam reflects from the top of the bump and part from the lower adjacent areaThis ensures destructive interference and very low intensity when the reflected beams combine at the detectorThe bump edges are read as onesThe flat bump tops and intervening flat plains are read as zeros
42 DVD’s DVD’s use shorter wavelength lasers The track separation, pit depth and minimum pit length are all smallerTherefore, the DVD can store about 30 times more information than a CD
43 Example 4Determine the minimum thickness of a soap film (n = 1.330) that will result in constructive interference of (a) the red Hα line (λ = nm); (b) the blue Hγ line (λ = nm).
44 Example 5A coating is applied to a lens to minimize reflections. The index of refraction of the coating is 1.55, and that of the lens is If the coating is nm thick, what wavelength is minimally reflected for normal incidence in the lowest order?
45 DiffractionHuygen’s principle requires that the waves spread out after they pass through slitsThis spreading out of light from its initial line of travel is called diffractionIn general, diffraction occurs when waves pass through small openings, around obstacles or by sharp edges
46 Diffraction, 2A single slit placed between a distant light source and a screen produces a diffraction patternIt will have a broad, intense central bandThe central band will be flanked by a series of narrower, less intense secondary bandsCalled secondary maximaThe central band will also be flanked by a series of dark bandsCalled minima
47 Diffraction, 3The results of the single slit cannot be explained by geometric opticsGeometric optics would say that light rays traveling in straight lines should cast a sharp image of the slit on the screen
48 Fraunhofer Diffraction Fraunhofer Diffraction occurs when the rays leave the diffracting object in parallel directionsScreen very far from the slitConverging lens (shown)A bright fringe is seen along the axis (θ = 0) with alternating bright and dark fringes on each side
49 Single Slit Diffraction According to Huygen’s principle, each portion of the slit acts as a source of wavesThe light from one portion of the slit can interfere with light from another portionThe resultant intensity on the screen depends on the direction θ
50 Single Slit Diffraction, 2 All the waves that originate at the slit are in phaseWave 1 travels farther than wave 3 by an amount equal to the path difference (a/2) sin θIf this path difference is exactly half of a wavelength, the two waves cancel each other and destructive interference results
51 Single Slit Diffraction, 3 In general, destructive interference occurs for a single slit of width a when sin θdark = mλ / am = 1, 2, 3, …Doesn’t give any information about the variations in intensity along the screen
52 Single Slit Diffraction, 4 The general features of the intensity distribution are shownA broad central bright fringe is flanked by much weaker bright fringes alternating with dark fringesThe points of constructive interference lie approximately halfway between the dark fringes
53 Diffraction GratingThe diffracting grating consists of many equally spaced parallel slitsA typical grating contains several thousand lines per centimeterThe intensity of the pattern on the screen is the result of the combined effects of interference and diffraction
54 Diffraction Grating, cont The condition for maxima isd sin θbright = m λm = 0, 1, 2, …The integer m is the order number of the diffraction patternIf the incident radiation contains several wavelengths, each wavelength deviates through a specific angle
55 Diffraction Grating, final All the wavelengths are focused at m = 0This is called the zeroth order maximumThe first order maximum corresponds to m = 1Note the sharpness of the principle maxima and the broad range of the dark areaThis is in contrast to the broad, bright fringes characteristic of the two-slit interference pattern
56 Diffraction Grating in CD Tracking A diffraction grating can be used in a three-beam method to keep the beam on a CD on trackThe central maximum of the diffraction pattern is used to read the information on the CDThe two first-order maxima are used for steering
57 Example 6Light of wavelength 600 nm falls on a 0.40-mm-wide slit and forms a diffraction pattern on a screen 1.5 m away. (a) Find the position of the first dark band on each side of the central maximum. (b) Find the width of the central maximum.
58 Example 7Microwaves of wavelength 5.00 cm enter a long, narrow window in a building that is otherwise essentially opaque to the incoming waves. If the window is 36.0 cm wide, what is the distance from the central maximum to the first-order minimum along a wall 6.50 m from the window?
59 Example 8Intense white light is incident on a diffraction grating that has 600 lines/mm. (a) What is the highest order in which the complete visible spectrum can be seen with this grating? (b) What is the angular separation between the violet edge (400 nm) and the red edge (700 nm) of the first-order spectrum produced by the grating?
60 Polarization of Light Waves Each atom produces a wave with its own orientation ofAll directions of the electric field vector are equally possible and lie in a plane perpendicular to the direction of propagationThis is an unpolarized wave
61 Polarization of Light, cont A wave is said to be linearly polarized if the resultant electric field vibrates in the same direction at all times at a particular pointPolarization can be obtained from an unpolarized beam byselective absorptionreflectionscattering
62 Polarization by Selective Absorption The most common technique for polarizing lightUses a material that transmits waves whose electric field vectors in the plane are parallel to a certain direction and absorbs waves whose electric field vectors are perpendicular to that direction
63 Selective Absorption, cont E. H. Land discovered a material that polarizes light through selective absorptionHe called the material PolaroidThe molecules readily absorb light whose electric field vector is parallel to their lengths and transmit light whose electric field vector is perpendicular to their lengths
64 Selective Absorption, final The intensity of the polarized beam transmitted through the second polarizing sheet (the analyzer) varies asI = Io cos2 θIo is the intensity of the polarized wave incident on the analyzerThis is known as Malus’ Law and applies to any two polarizing materials whose transmission axes are at an angle of θ to each other
65 Polarization by Reflection When an unpolarized light beam is reflected from a surface, the reflected light isCompletely polarizedPartially polarizedUnpolarizedIt depends on the angle of incidenceIf the angle is 0° or 90°, the reflected beam is unpolarizedFor angles between this, there is some degree of polarizationFor one particular angle, the beam is completely polarized
66 Polarization by Reflection, cont The angle of incidence for which the reflected beam is completely polarized is called the polarizing angle, θpBrewster’s Law relates the polarizing angle to the index of refraction for the materialθp may also be called Brewster’s Angle
67 Polarization by Scattering When light is incident on a system of particles, the electrons in the medium can absorb and reradiate part of the lightThis process is called scatteringAn example of scattering is the sunlight reaching an observer on the earth becoming polarized
68 Polarization by Scattering, cont The horizontal part of the electric field vector in the incident wave causes the charges to vibrate horizontallyThe vertical part of the vector simultaneously causes them to vibrate verticallyHorizontally and vertically polarized waves are emitted
69 Optical ActivityCertain materials display the property of optical activityA substance is optically active if it rotates the plane of polarization of transmitted lightOptical activity occurs in a material because of an asymmetry in the shape of its constituent materials
70 Liquid CrystalsA liquid crystal is a substance with properties intermediate between those of a crystalline solid and those of a liquidThe molecules of the substance are more orderly than those of a liquid but less than those in a pure crystalline solidTo create a display, the liquid crystal is placed between two glass plates and electrical contacts are made to the liquid crystalA voltage is applied across any segment in the display and that segment turns on
71 Liquid Crystals, 2Rotation of a polarized light beam by a liquid crystal when the applied voltage is zeroLight passes through the polarizer on the right and is reflected back to the observer, who sees the segment as being bright
72 Liquid Crystals, 3When a voltage is applied, the liquid crystal does not rotate the plane of polarizationThe light is absorbed by the polarizer on the right and none is reflected back to the observerThe segment is dark
73 Liquid Crystals, finalChanging the applied voltage in a precise pattern canTick off the seconds on a watchDisplay a letter on a computer display
74 Example 9The critical angle for total internal reflection for sapphire surrounded by air is 34.4°. Calculate the Brewster angle for sapphire if the light is incident from the air.
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