2 InterferenceLight waves interfere with each other much like mechanical waves doTwo conditions which must be metThe sources must be coherent. They must maintain a constant phase with respect to each otherThe waves must have identical wavelengths
3 Producing Coherent Sources Old method: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 coherentNew method: use a laser
4 Young’s Double Slit Experiment 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
5 Interference Patterns Constructive interference occurs at the center pointThe two waves travel the same distance and they arrive in phase
6 Interference Patterns, 2 The upper wave travels one wavelength fartherTherefore, the waves arrive in phaseA bright fringe occurs
7 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
8 Interference Equations The path difference, δ, is found from the tan triangleδ = r2 – r1 = d sin θThis assumes the paths are parallel
9 Interference Equations, 2 For a bright fringe, produced by constructive interference, the path difference must beδ = m λm = 0, ±1, ±2, …δ = d sin θbright = m λm is called the order numberWhen m = 0, it is the zeroth order maximumWhen m = ±1, it is called the first order maximum
10 Interference Equations, 3 When destructive interference occurs, a dark fringe is observedThis needs a path difference of an odd half wavelength; δ = (m + ½) λδ = d sin θdark = (m + ½) λm = 0, ±1, ±2, …
11 Interference Equations, 4 The positions of the fringes can be measured vertically from the zeroth order maximumy = L tan θ ~ L sin θApproximationθ is small and therefore tanθ ~ sin θFor bright fringesFor dark fringes
12 Quick quiz 24-1In a two slit interference pattern projected on a screen the fringes are equally spaced on the screenA. everywhereB. only for large anglesC. only for small angles
13 Problem 24.10A pair of slits, separated by mm, is illuminated by light having a wavelength of λ = 643 nm. An interference pattern is observed on a screen 140 cm from the slits. Consider a point on the screen located at y = 1.80 cm from the central maximum of this pattern.What is the path difference δ for the two slits at the location y?(b) Express this path difference in terms of the wavelength.(c) Will the interference correspond to a maximum, a minimum, or an intermediate condition?
14 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
15 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
16 Interference in Thin Films Rules 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
17 Interference in Thin Films, 2 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 waveRay 2 also travels an additional distance of 2t before the waves recombine
18 Interference in Thin Films, 3 For constructive interference2 n t = (m + ½ ) λ m = 0, 1, 2 …This takes into account both the difference in optical path length (2t) for the two rays and the 180° phase change (1/2 λ)For destruction interference2 n t = m λ m = 0, 1, 2 …
19 Interference in Thin Films, 4 An example of different indices of refractionA coating on a solar cell
20 Quick quiz 24-2Supposed Young’s experiment is carried out in air, and then, in a second experiment, the apparatus is immersed in water. In what way does the distance between bright fringes change?A. they move further apartB. they move closer togetherC. no change
21 Problem 24.17A 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?
23 Conceptual questions3. Consider a dark fringe in an interference pattern, at which almost no light energy is arriving. Light from both slits is arriving at this point, but the waves are canceling. Where does the energy go?4. If Young’s double slit experiment were performed under water, how would the observed interference pattern be affected?13.Would it be possible to place a nonreflective coating on an airplane to cancel radar waves of wavelength 3 cm?
24 Reading a CDAs the disk rotates, the laser reflects off the sequence of lands and pits 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 lightland
25 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 diffraction
26 Fraunhofer Diffraction Fraunhofer Diffraction occurs when the rays leave the diffracting object in parallel directionsA bright fringe is seen along the axis (θ = 0) with alternating bright and dark fringes on each side
27 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 θWave 1 travels farther than wave 3 by an amount equal to the path difference (a/2) sin θdestructive interference occurs whensin θdark = mλ / a
28 Single Slit Diffraction, 2 A 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
29 Problem 34A screen is placed 50.0 cm from a single slit, which is illuminated with light of wavelength 680 nm. If the distance between the first and third minima in the diffraction pattern is 3.00 mm, what is the width of the slit?
30 Quick quiz 24.3In a single-slit diffraction experiment, as the width of the slit is made smaller, the width of the central maximum of the diffraction pattern becomes (a) smaller, (b) larger, (c) remains the same.
31 Diffraction Grating The condition for maxima is d 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
32 QUICK QUIZ 24.4If laser light is reflected from a phonograph record or a compact disc, a diffraction pattern appears. This occurs because both devices contain parallel tracks of information that act as a reflection diffraction grating. Which device, record or compact disc, results in diffraction maxima that are farther apart?
33 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
34 Polarization of Light Waves Each atom produces a wave with its own orientation of EThis is an unpolarized wave
35 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
36 Polarization by Selective Absorption The most common technique for polarizing lightUses a material that transmits waves whose electric field vectors in the plane parallel to a certain direction and absorbs waves whose electric field vectors are perpendicular to that directionMalus’ law: I = Io cos2 θ
37 Polarization by Reflection The angle of incidence for which the reflected beam is completely polarized is called the polarizing angle, θpθp is also called Brewster’s AngleBrewster’s Law relates the polarizing angle to the index of refraction for the material
38 Polarization by Scattering 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
39 Conceptual question14. Certain sunglasses use a polarizing material to reduce intensity of light reflected from shiny surfaces, such as water or a hood of a car. What orientation of the transmission axis should the material have to be most effective?18. Can a sound wave be polarized?17. When you receive a chest x-ray at a hospital, the ex-ray passes through a series of parallel ribs in your chest. Do the ribs act as a diffraction grating for x-rays?
40 Optical ActivityCertain materials display the property of optical activityA substance is optically active if it rotates the plane of polarization of transmitted light
41 Liquid CrystalsRotation 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
42 Liquid CrystalsWhen 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
43 MCADTwo light sources produce light with wavelength l. The sources are placed 22.5 l and 45 l away from point P. When both sources are turned on and their intensities, I, at point P are equal, the resultant intensity at point P will beA. 0B. 0.5 IC. ID. 2I
44 The process discussed in the previous question is called Diffraction b. RefractionInterference d. DispersionWhich of the following would result in greatest diffraction?Small wavelengths moving through a small openingLarge wavelengths moving through a small openingSmall wavelengths moving through a large openingLarge wavelengths moving through a large opening