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Waves In our everyday life there are many examples of waves for example, Sound, ocean waves, strings of musical instruments, organ pipes… These are examples of waves which need a medium to travel through and the general definition of such waves is A disturbance which propagates through a medium but the medium itself only moves locally as the wave passes. A special case of a wave is an electromagnetic wave which can propagate through vacuum, e.g. radio, light, x rays …. Waves transport energy and momentum and energy is required to generate waves The medium must have elastic properties Week 13 Physics 214 Spring 2010 1 Wave properties There are two types of waves Longitudinal – consists of the propagation of a series of compressions and rarefactions and the local movement of the medium is an oscillation back and forward along the direction of the wave Sound is one example Transverse – where the movement of the medium is at right angles to the velocity of the wave. The strings of musical instruments are an example. Week 13 Physics 214 Spring 2010 anim0019.mov 2 Wave properties Any material medium supports Longitudinal (compression) waves: solid, liquid, or gas. Fluids do NOT support transverse waves in their interior– because the restoring force needed would be shear restoring force, which clearly fluids don’t posess. You can stir water (you can’t stir set jello, and jello CAN wiggle sideways elastically. Seismic waves in the earth tell us about the intermediate liquid layer! Week 13 Physics 214 Spring 2010 anim0019.mov 3 Wave properties Water surface waves are a third type, not purely transverse nor longitudinal. The water molecules dance in place in “elliptical” orbits. Remember, ALL matter waves consist of matter dancing in place under restoring forces while the wave PATTERN carries energy. Week 13 Physics 214 Spring 2010 anim0019.mov 4 Periodic waves We will focus on regular repetitive waves λ These waves have a pattern which repeats. The length of one repeat is called the wavelength λ The number of repeats (wavelengths) which pass a point, per second, is called the frequency f T is the time for one pattern to pass. Then f = 1/T v = λ/T = fλ The wave velocity, v, is then Week 13 Physics 214 Spring 2010 5 Waves on a string If we shake the end of a rope we can send a wave along the rope. The rope must be under tension in order for the wave to propagate v = √(F/μ) F = TENSION μ = MASS/UNIT LENGTH Tighter = faster wave Lighter = faster wave Week 13 Physics 214 Spring 2010 6 Standing waves If two identical waves exist on the same string but traveling in opposite directions the result can be standing waves in which some points never have a deflection. …These are called nodes Halfway between nodes, the string oscillates with the maximum amplitude. …These are called antinodes. Standing waves provide the notes on musical instruments. When a string is secured at both ends and plucked or hit the generated waves will travel along the string and be reflected and set up standing waves. Week 13 Physics 214 Spring 2010 7 Waves on a Spring The spring in the demonstration shows many nifty simplifications as you stretch it. Mass is fixed, but if you double the length, μ is halved. And by Hooke’s law, the Tension (F) doubles because of the stretch. F = TENSION = kL; μ = MASS/ L v = √(F/μ) = √(kL/(M/L)) = √(kL2/M) = L√(k/M) If you double L, you double v. Round-trip travel time of a pulse does not change, because it has to travel twice as far, but at twice the speed! Neat. But rather a special case. Week 13 Physics 214 Spring 2010 8 Musical notes Each end of the string must be a node so the possible standing waves must be multiples of λ/2 Fundamental f = v/λ = v/2L 2nd Harmonic f = v/λ = v/L=2v/2L 3v/2L 3rd Harmonic f = v/λ = Musical sound is a mixture of harmonics modified by the body of the instrument or where the string is bowed or plucked (near an end enhances high frequency overtones, relative to low) v = √(F/μ) so a piano or a violin is tuned by changing the tension in the string. (playing the violin you change the length of the string with your finger.) Week 13 Physics 214 Spring 2010 9 “Closed” (at one end) organ pipe Node at one end and an antinode at the other Fundamental f = v/4L = 1st harmonic 2nd Harmonic f = 3v/4L = 3xFund. 3rd Harmonic f = 5v/4L = 5xFund. higher overtones 7x, 9x, 11x, etc. Open pipe overtones (unlike the string) are only odd multiples of the Fundamental tone The velocity of sound in air is v = 340m/s Nth harmonic f = (2N-1)v/4L Week 13 Physics 214 Spring 2010 10 Closed organ pipe Node at one end and an antinode at the other The velocity of sound in air is v = 340m/s Nth harmonic f = (2N-1)v/4L 32 foot Organ pipe, L=10m Fundamental tone has f = v/4L = 340m/s/40m = 8.5 Hz Too low to be heard as a tone, it’s a deep throbbing in your chest. Week 13 Physics 214 Spring 2010 11 QUIZ 9 April ‘10 Closed organ pipe Node at one end and an antinode at the other Fundamental f = v/4L = 1st harmonic 2nd Harmonic f = 3v/4L Velocity of sound in air is v = 340 m/s Q: ~16 foot pipe (5m) open one end closed other end. What is the 3d harmonic, in Hz? A. 68 D. 51 B. 34 E. 85 C. 17 λ = 20 m, f = v/λ = 340/20 = 17 [1/s] 3d harmonic has f = 5*fund. = 85 Hz, (E.) . Week 13 Physics 214 Spring 2010 12 Beats If two waves have slightly different frequencies then the sum has a frequency which is f1 – f2. The human ear can detect beats and this is used to tune an instrument. For example using a tuning fork at a known frequency and adjusting a piano string until no beats heard Two notes close together in frequency may sound dissonant due to the beat tone. Week 13 Physics 214 Spring 2010 f = f1 – f 2 13 Musical Scales An Octave occurs when one frequency is twice the other frequency. There are twelve half-tones in one octave (corresponding to 12 piano keys in sequence) Two string tones an octave apart have the higher tone at the frequency of the first overtone, so the sound is similar to either of the separate tones. Week 13 Physics 214 Spring 2010 14 Musical Scales The notes of the Western musical scale are based on simple chords, and tones that are simple fractions of each other. Take two tones of frequency f and 1.5f. What is the beat frequency? It’s fbeat = 1.5f - f = 0.5 f This is one octave below f, and the chord sounds smooth and mellow. Take f and 1.25f (ratio is 5/4). fbeat = 0.25 f This is two octaves below f and is still a smooth chord Week 13 Physics 214 Spring 2010 15 Musical Scales The simple ratios can work fine for one “key signature”, for example C. But if a piano is tuned with perfect chords for this key, then chords in another key (say G) might have dissonances (buzzing beat notes, or so-called discord). A good compromise is the equal-tempered or well-tempered scale, invented around the time of J. S. Bach, where the half tones differ in frequency by a ratio = the twelfth root of 2. This gives reasonably harmonious chords, and every key has the same degree of “consonance”, or harmony. This is, however, a tradeoff since the ratios are no longer perfect 5/4 or 3/2 in any key. Week 13 Physics 214 Spring 2010 16 Musical Scales For example, the chord C-E is two full tones apart, Which is four half-tones. The frequency ratio is then 24/12 = 21/3 = 1.26 in the equal tempered scale This is quite close to 5/4, and is still a harmonious chord. If we take middle C, 261.5Hz, what frequency is the “error”? 1.26 differs by 0.01 from 5/4. So E is about 2.6 Hz higher frequency than a perfect chord. C-G is seven half tones apart. The frequency ratio is then 27/12 = 1.4983 which is very close to 3/2 – again, a very harmonious chord. Here, G is 0.0017 x 261.5 Hz flat (low) compared to a perfect chord. That’s 0.44 Hz. Week 13 Physics 214 Spring 2010 17 Ch 15 CP4 cont. 21/12 is ~1.059463: A = 440 Hz Ab = A/1.0595 = 415.2902 Hz G = Ab/1.0595 = 391.9681 Hz Gb= G/1.0595 = 369.9557 Hz F = Gb/1.0595 = 344.1796 Hz E = F/1.0595 = 329.5701 Hz Eb = E/1.0595 = 311.0620 Hz D = Eb/1.0595 = 293.5932 Hz Db = D/1.0595 = 277.1054 Hz C = Db/1.0595 = 261.5435 Hz A, Bb, B, C → up to scale A = 440 Hz Bb = A × 1.0595 = 466.18 Hz B = Bb × 1.0595 = 493.9177 Hz C = B × 1.0595 = 523.3058 Hz middle C = 261.5435 Hz high C = 523.058 Hz vs 523.087 = 2x middle C (roundoff error) high C/ middle C = 2.00; Yes, high C is twice the frequency of middle C!! Week 13 Physics 214 Spring 2010 18 Doppler effect As the source is approaching the frequency increases because the waves are crowded closer together. As an observer approaches a stationary source, she passes more wave crests per second, the frequency is increased. It’s the RELATIVE motion that matters. For relative recession, the wavelength is “stretched”, which lowers the observed frequency. http://www.physics.purdue.edu/class/applets/phe/dopplereff.htm Week 13 Physics 214 Spring 2010 19 Sonic boom Sonic boom (a shock wave) At each point on the path the sound wave expands radially and they all combine along a single wave front which is a sharp pressure wave http://www.phy.ntnu.edu.tw/java/airplane/airplane.html Week 13 Physics 214 Spring 2010 20 Sonic boom A sonic boom is analogous to the bow-wake of a boat. In each case, the emitter is travelling faster than the wave velocity. The red & thin black lines are like the crest of a bow wave, except the sonic boom is in 3-D so it is a cone instead of a pair of lines. Arrows point along the wave velocity. http://www.phy.ntnu.edu.tw/java/airplane/airplane.html Week 13 Physics 214 Spring 2010 21 Summary of Chapter 15 v = λ/T = fλ Transverse or longitudinal Standing waves Fundamental f = v/λ = v/2L 2nd Harmonic f = v/λ = v/L 3rd Harmonic f = v/λ = 3v/2L Fundamental f = v/4L 2nd Harmonic f = 3v/4L 3rd Harmonic f = 5v/4L Week 13 Physics 214 Spring 2010 22 Sound effects Doppler effect f increases as sound approaches Beats f = f1 – f2 Sonic boom Week 13 Physics 214 Spring 2010 23 Electromagnetic waves v~EXB (directions) EM waves consist of oscillating magnetic and electric fields transmitted through vacuum at a constant speed of c = 3 x 108m/s. Curl you fingers from the E direction to the B direction: your thumb points in the direction of propagation. They are produced whenever there is a changing electric or magnetic field. The acceleration of electric charge produces EM waves such as a radio broadcast antenna, AC wiring and lightning. Also electrons changing orbits in atoms. A simple EM wave has f = c/λ and λ = c/f Where c = speed of light = 3 x 108 m/s Week 13 http://www.physics.purdue.edu/class/applets/phe/emwave.htm Physics 214 Spring 2010 24 Electromagnetic waves v~EXB f = c/λ and λ = c/f (directions) Visible light with λ = 500 nm (~ green) has what frequency? f = c/λ = 3x108 m/s / 5x10-7 m = 6x1014 Hz If you can discriminate one cycle, you have “measured” one second to an accuracy of almost one part in 1015 !!! Week 13 Physics 214 Spring 2010 25 Electromagnetic spectrum Almost all the information we receive from outside the earth is in the form of EM radiation. Different parts of the spectrum correspond to different physical processes We can understand what is going on in the Universe and back in time to near the beginning of the Universe using a variety of earth- and space-based telescopes. Week 13 Physics 214 Spring 2010 26 ←(part of) Electromagnetic spectrum→ Visible light spans 1 “octave”. This figure spans factor of 1015 This is ≈ 50 “octaves”, since 210 = 1024 ≈ 103 Week 13 250 ≈ 1015 Physics 214 Spring 2010 27 A few telescopes Radio telescope at Arecibo Space telescopes don’t have to look through the atmosphere. Week 13 An array of radio telescopes – can resolve very fine angular detail due to large overall size. Physics 214 Spring 2010 28 How do we see color An image is formed at the back of the eye like a camera, on the retina. There are receptor cells called cones that respond to different wavelengths. The brain interprets the mixture of the three signals as color. If we look at an object by reflected light the color we see is the wavelengths that were reflected, not absorbed. If we are looking through a colored object then the object lets that color be transmitted and the other colors are absorbed Week 13 Physics 214 Spring 2010 S. M and L refer to short, med. and long wavelengths of light.29 How do we see color Any color we see excites the S, M, and L cones each a certain amount. The color perceived is a result of the relative proportions of the three excitations, not the overall brightness. So the color space humans see is effectively 2Dimensional. Once you know the % of L and the % of M, you know the % of S [must add to 100%] Notice how but Week 13 purple is NOT a pure wavelength,Physics 214 rather a mix of the two ends of the visible spectrum. Spring 2010 30 Two-Slit Interference When two coherent & equalamplitude monochromatic beams of light are brought together they will add just like two waves on a string. When two wave peaks coincide the light will be a maximum. When a peak coincides with a valley there will be no light. At the two slits, the waves are “in phase”. The waves spread out from the slits, and at places on the screen where one slit is an integer number of wavelengths farther than the other from the screen, the waves add up to give brightness. Halfway between each bright stripe is darkness from the waves cancelling. Week 13 y is the sideways distance on the screen, from a point “straight ahead” Physics 214 Spring 2010 Bright fringes are located at positions y given by y/x = n λ/d where n is an integer Dark fringes occur when yd/x = nλ/2 where n is an odd integer 31 TWO-SLIT INTERFERENCE The picture at the right is for two slits extending in and out of the picture. Picture shows only the “rays” from each wavelet that aim at a PARTICULAR POINT ON THE SCREEN. The screen is far away. The slits have equal widths, so the waves in the two rays have equal amplitudes. The two rays are pointing towards a point on the screen d (where the sideways distance is y), for which there is exactly ONE HALF WAVELENGTH path difference between the two slits. These two rays, being exactly out of phase, will cancel each other. Also for any odd-half-integer path diff. CONSTRUCTIVE interference gives BRIGHT stripes straight ahead and for path differences of λ, 2λ, 3λ, etc. There are intervening DARK fringes at odd half-integer wavelength path differences. P/d ≈ y/x (for small angles away from straight ahead) Similar triangles. Long leg ≈ hypotenuse. For small angles (in radians) P/d ≈ θ ≈ y/x Week 13 Physics 214 Spring 2010 almost x yy θ x P = Path difference For the first CANCELLATION, need a half-wavelenth path difference λ/2 = d sin(θ) where θ = deflection angle. y/x = tan(θ) sin(θ) ≈ tan(θ) ≈ for small θ θ 32 Interference Where does the energy go when light waves interfere to make a dark region? There MUST also be bright regions, and the energy is just REARRANGED in space. So the “dark bulb” (O. R. Frisch’s joke) that you turn it on and the room goes dark, is an impossibility. Cute idea, it does make you think. Week 13 Physics 214 Spring 2010 Bright fringes are located at positions given by dy/x = nλ where n is an integer Dark fringes occur when dy/x = nλ/2 where n is an odd integer 33 Thin film interference Thin film interference occurs when light is reflected from the top surface and the underneath surface of film. This gives two beams of coherent light, which interfere. Since we normally observe this with white light we see color fringes because the path difference varies with the angle of observation So different wavelengths (colors) have constructive and destructive interference at different places on the film. Week 13 Physics 214 Spring 2010 34 Thin film interference If the round-trip path in the thin film is λ / 2 (so the distance travelled one-way is λ / 4) the two reflected waves will be exactly out of step with each other, and cancel. For different colors (wavelengths) this condition will take place at different angles (if at all). Redder color with longer wavelength would need an angle farther away from the vertical, and vice versa for bluer color. Week 13 Physics 214 Spring 2010 35 Anti-reflection Coatings for lenses As light passes from one transparent medium to another a few percent of the light will be reflected. This is a particular problem in optical systems like lenses where there may be many glass elements. For example if 96% of the light is transmitted at a surface, after 18 surfaces only (0.96)18 = 48% of the light is transmitted overall, and the other 52% scatters everywhere. Thin coatings are put on lens surfaces with thickness = ¼ wavelength of green light. Then green rays entering perpendicular to the surface will have the front and rear reflections from the film λ / 2 out of phase, and cancel. At normal (perpendicular) incidence, with a “quarter-wave coating”, this is only true for a single wavelength. To reduce the reflections for a range of wavelengths requires multiple thin film layers and costs more. This is why camera lenses often have a purple cast. The antireflection coating is optimized for green, the middle of the spectrum and also the color the eye is most sensitive to. The cancellation is not perfect for red or blue, so the reflected light is purple = red+blue, the two ends of the optical spectrum. Week 13 Physics 214 Spring 2010 9 lenses = 18 surfaces! 36 Anti-reflection Coatings for lenses At normal (perpendicular) incidence, with a “quarter-wave coating”, which is λ/4 thick (and hence the round trip is λ/2), exact cancellation ALSO requires equal strength of the two reflections (from front and back surfaces of the coating). It turns out that this requires the index of refraction of the coating to be the geometrical mean of air (n~1.0) and the lens glass (n~1.5) . Then the “steps-up” in index each have the ratio shown below. (We’ll define n and its properties on the next two slides.) n = Sqrt(1.5) = 1.225 is required, but materials with this low an index tend to be soft and easily scratched. Week 13 Physics 214 Spring 2010 37 REFRACTION, Speed in a Medium In transparent media, light travels slower than in a vacuum: v = c/n. n is called the “index of refraction”. Since nothing can travel faster than c, n is always greater than 1.00 (or equal to 1.00, in vacuum) For water, n=1.33; for glass, n~1.5; for diamond, ~2.4 Going from air into water or glass, light slows down and, being a wave, therefore MUST bend its direction closer to the perpendicular of the interface. This is the basis of all lenses, prisms, etc. Week 13 Physics 214 Spring 2010 38 REFRACTION, Speed in a Medium In transparent media, light travels slower than in a vacuum: v = c/n. Charged particles can travel very close to c, even in transparent media. Thus, they can travel faster than the speed of light IN THE MEDIUM. Such fast charged particles will then emit an “optical shock wave”, which is called Cherenkov radiation after the Russian physicist who conceived this idea less than 100 years ago. Cherenkov radiation wavefronts also are coneshaped, and the flow of radiation follows another cone at right angles to the wavefronts. Week 13 Physics 214 Spring 2010 39 QUIZ Two loudspeakers next to each other broadcast the same tone, same strength. One speaker is then placed 7 halfwavelengths farther away from you than the other. You then hear : A. a pulsating tone B. a rising tone C. a falling tone D. much less sound E. more intense sound The answer is D, because there is destructive interference. The three full wavelenths of shift would not make a difference, but the extra half-wavelength puts the two waves exactly out of phase and they cancel each other. Week 13 Physics 214 Spring 2010 40 Diffraction Waves travel in straight lines EXCEPT when they encounter obstacles – then the waves spread out, and can bend around corners, etc. That’s why, for example, you can hear sounds from sources you can’t see because they’re not in your line of sight. Water waves are an excellent example. Photo of wave pool at water park in Cancun, Mexico. Week 13 Physics 214 Spring 2010 41 Week 13 Physics 214 Spring 2010 42 Diffraction Waves travel in straight lines EXCEPT when they encounter obstacles – then the waves spread out, and can bend around corners, etc. If it were not for Diffraction, the two-slit interference could not take place, since it requires the coherent waves coming from the two slits to start spreading out sideways from each of the slits. Week 13 Physics 214 Spring 2010 43 INTERFERENCE Because of diffraction, interference occurs even for a single aperture. The pattern shown BELOW is from a square aperture and the pattern can be thought of as light from different parts of the aperture interfering. As the aperture is made smaller the pattern actually expands (because a larger angle of deflection is needed to generate a full λ path difference between the edges.) This effect can limit our ability to see detail in small objects or to resolve two stars nearby to one another. Week 13 Physics 214 Spring 2010 λ/w ≈ y/x But the wavelets at the two edges are IN step !? Counter-intuitive The position of the first dark fringe is given by y = λx/w. This means the central bright fringe has a width ~ 2y which increases as w gets smaller 44 INTERFERENCE This picture is for a single slit of width w, extending in and out of the figure. Every point on the wavefront AT the aperture radiates a semi-circular “wavelet”. The picture shows only the “rays” from each wavelet that aim at a PARTICULAR POINT ON THE SCREEN. We are looking at a DARK point on the screen (where the sideways distance is y), for which there is exactly ONE WAVELENGTH difference between the two edges of the slit. To see the complete cancellation, pair up the wavelets that are HALF a wavelength out of step (bottom edge and mid-slit). Then move a tiny way up and repeat until you have swept across half the slit with each wavelet of the pair. All wavelet pairs CANCEL. And this uses ALL the wavelets (don’t worry about using the middle wavelet twice, it contributes an infinitesimal amount to the overall wave sum.) Week 13 Physics 214 Spring 2010 The central bright band is 2y wide on the screen 45 Diffraction grating If we have multiple slits very close together, say several 1000 slits per centimeter the diffraction from each slit adds in a way that produces very narrow and bright fringes. Call the slit spacing d, x=dist. to screen, y = deflection For small angles the location of the bright fringes is given by y/x ~ angle ~ m(λ /d) y = mλx/d where m is an odd or even integer y Week 13 Physics 214 Spring 2010 46 Diffraction grating Diffraction gratings are excellent for resolving two nearby colors (wavelengths). They are particularly useful in astronomy, and are how we detect atomic spectral color lines and measure their redshifts. Week 13 Physics 214 Spring 2010 47 Polarization “In a normal beam of light all possible transverse polarizations are present.” A light wave is made up of zillions of photons, which randomly have all possible polarizations of their E fields. But due to the wave nature of light, this normal light beam can validly be represented as 50% vertical polarization and 50% horizontal polarization. Strangely, you can use any two orthogonal transverse directions to represent normal light. A polarizing filter lets through one component and blocks the other, thus cutting the light intensity in half. Week 13 Physics 214 Spring 2010 48 Polarization There are materials like polaroid that will only transmit light (photons) with the electric field vector along a specific direction. Light reflected by water or other shiny surfaces is also is polarized so polaroid sun glasses at the right orientation will preferentially block the reflected light. Transparent objects under stress can also cause transmitted light to be polarized. Car windshields are one example where patterns can be seen when wearing polaroid sun glasses Week 13 Physics 214 Spring 2010 At a special angle, Brewster’s angle, light reflected off a transparent interface is totally horizontally polarized, and the other polarization entirely enters the denser medium. This condition is when the reflected and transmitted rays are perpendicular to each other. At other angles, the polarization is only partial. 49 Polarization The effects of polarizing filters: Picture at top is a polarized light beam. A vertical polarizing filter allowed through only half the light: all of the vertical component and none of the horizontal. This light hits a second filter which is rotated 45o from the original filter. The E field (green) has a vector component along the new filter axis, so a reduced amplitude passes through (the red E field) Crossed polaroids (second figure) will kill ALL the light. Almost magically, an intermediate polaroid between the crossed polaroids will let some of the light through (third figure Week 13 Physics 214 DEMO Spring 2010 Filter axis Filter axis Blocks all light E after first filter E after intermed. filter E after crossed (third) filter 50 QUIZ Polarization An unpolarized light beam is passed through a linear polaroid polarizing fillter, which removes half of the light intensity. The polarized beam then passes through a polaroid filter whose axis is at 45o to the first filter. The light intensity is: A. Further reduced B. Reduced to zero D. Increased C. Not changed There is a non-zero component of E that can pass the filter, but it’s not the full E out of the 1st filter Week 13 Physics 214 Spring 2010 51 Summary of Chapter 16 f = c/λ c = 3 x 108m/s in vacuo In a medium with index n, f DOES NOT CHANGE, the wave slows down (relative to in vacuum) and λ changes (shorter). Polarization Thin film interference Week 13 Physics 214 Spring 2010 52 Summary of Chapter 16 contd. Bright fringes are located at positions given by y/x = nλ/d n is an integer Dark fringes occur when dy/x = nλ/2 where n is an odd integer 2 slits Diffraction grating dy/x = mλ where m is any integer The position of the first dark fringe is given by y = λx/w. This means the central bright fringe has a width ~ 2y and increases as w gets smaller one slit Week 13 Physics 214 Spring 2010 53 Lightning Tour of Chapt. 17 This chapter deals with refraction of light, including index of refraction, n, which I have already lectured about a little bit. n is >= 1 In the medium, vlight = c/n is < c Refraction is the basis of prisms and lenses, also of some types of lightguide optical fibers. When light passes from a less optically dense to a more dense medium (i.e. from lower n to higher n) it is bent CLOSER to the normal (perpendicular) to the interface. Week 13 Physics 214 Spring 2010 n1 < n2 θ2 θ1 54 Lightning Tour of Chapt. 17 NOTE: This chapter is not assigned material for testing, but you need to have seen just a little bit about lenses, total internal reflection, n1 and so on – to help you understand the simple principles underlying these phenomena. I also recommend θ1 reading Ch. 17. Let’s continue: The bending towards the slower medium is because in the slower medium, the wave crests must be closer together. To match up crests at the interface, the direction of flow must change Remember, f const., λ shrinks as wave slows Week 13 Physics 214 Spring 2010 < n2 θ2 55 Lightning Tour of Chapt. 17 To match up crests at the interface, the direction of flow must change Remember, f does not change. Instead, λ shrinks as wave slows down in the denser medium. λ = v/f and now v is less than c n1 < n2 θ2 θ1 f does not change because the photon energy is directly related to its frequency (basic quantum behavior) Week 13 Physics 214 Spring 2010 56 Chapt. 17 Some lenses cause bundles of light rays to become more CONVERGENT. This can focus the almost parallel rays from a very distant obejct into a point at a certain distance beyond the lens, called the focal point of the lens. If you are far-sighted, you need such a lens, of some strength, to help the lens of your eye to focus images on your retina. If you are near-sighted, you need a DIVERGING lens of some strength to move focused images further from your eye’s lens and onto your retina. Week 13 Physics 214 Spring 2010 57 Chapt. 17 CONVERGING lenses are fat in the middle and thin at the edge. So called “convex” shape, or bulging outward. DIVERGING lenses are thin at the center and thick at the edge, “dished-in” which we call concave. HOWEVER, a “hollow air lens” immersed in water reverses the effects, fat lenses are diverging and concave lenses are converging. This is because of the inversion of dense vs less dense index of refraction, compared to “normal” Week 13 Physics 214 Spring 2010 58 Chapt. 17 In media such as glass (or water) the index of refraction depends on frequency (color) so one color bends more or less than another color. This is called DISPERSION PRISM demonstration In lenses, this produces unwanted “chromatic aberration” (different colors focus at different distances from the lens, but retina or film is at a fixed distance.) This must be compensated for by complicated arrangements of several lenses using DIFFERENT glasses having different dispersions. Week 13 Physics 214 Spring 2010 n1 < n2 .. θ2 θ1 59 Ch. 17 Total Internal Reflection Light in a denser medium aimed at an interface, if it’s oblique enough, will not be able to pass into the less dense medium. This follows from Snell’s Law, which precisely describes: smaller angle in the “denser” medium, bigger angle in the less “dense” medium. n1 sin(θ1) = n2 sin(θ2) If θ1 is too large, it requires sin(θ2) > 1 which is impossible, so no wave goes into the less dense medium 2 and all energy is reflected back into medium 1. The “critical angle” has sin(θ1) =(n2/n1)*1 Because 1 = sin(90o) Week 13 Physics 214 Spring 2010 denser n1 > less dense n2 θ2=90o θ1 The diagram shows the critical angle. Bigger angles away from the normal (that is, rays skimming closer to the interface) totally 60 reflect Ch. 17 Total Internal Reflection Clad thin (1mm) plastic fibers can transmit light with reasonably small losses, using total internal reflection. We use a core of polystyrene with n = 1.59, the cladding is pmma (plexiglas) with n = 1.49. The light must aim within 20o of the fiber axis, but this traps enough light to 20 read out the light from scintillating plastic particle detectors over several meters of length. Angles bigger than 69.6o (relative to the o 70o =~ θcritical axis cladding PERP. to the interface) will totally internally reflect sin 69.6o = 1.49/1.59 = 0.937 Week 13 Physics 214 Spring 2010 61