AS Physics Unit 5 - Waves 1 WHAT IS WAVE MOTION? The wave motion is a means of transferring energy from one point to another without the transfer of any matter between the points. Waves may be classified as mechanical waves and electromagnetic waves. Mechanical Waves: Mechanical waves (example water waves, sound waves, waves in stretched springs) require presence of material medium or some matter for their propagation (travel). When it travels from point A to point B, it is because of the disturbance of some kind at A has caused the particles of the medium being displaced. This particle drags its neighboring particle with it, so that it too becomes displaced and has a similar effect on next particle, and so on until the disturbance reaches at point B. There are two types of mechanical waves, that is, transverse waves and longitudinal waves. o The transverse wave is the one in which the disturbance occurs perpendicular to the direction of travel of the wave. o The longitudinal wave is one in which the disturbance occurs parallel to the line of travel of the wave. Note 1: The transverse and longitudinal waves are also called periodic waves or progressive waves because they consist of cycles or patterns that are produced over and over again by the source. Note 2: A water wave is neither transverse nor longitudinal, since water particles at the surface move clockwise on nearly circular paths as the wave move from left to right. Electromagnetic waves: 1. Electromagnetic (EM) waves (example light waves, radio waves, x-rays etc) can travel through vacuum; they do not require any medium (matter) to travel through and their travel is reduced, to some extent, by the presence of any medium. The common examples of medium are glass, water, air etc. 2. EM waves are transverse in nature and comprise of oscillating electric (E) and magnetic (H) fields perpendicular to the direction of wave. Prepared by Faisal Jaffer, revised on 7-Mar-10 3. All electromagnetic waves including light waves travel with the same speed that is 3×108 m/s. They all have different frequencies and wavelengths. The band of different electromagnetic waves called electromagnetic spectrum is shown in the figure below. AS Physics Unit 5 - Waves 3 Important wave definitions: Wavelength (λ): The wavelength ‘λ’ of a wave, represented by the Greek letter λ (lambda). It is the distance between the two successive crests or troughs. Frequency (f): The frequency ‘f’ is the number of complete waves generated per second. The unit of frequency is ‘vibration per second’ or ‘hertz (Hz)’. Period (T): It is the time taken for a wave to travel one complete cycle or oscillation. Unit is seconds. Speed of wave motion (v): The speed, ‘v’ of the wave is the distance moved in the direction of travel of the wave by a crest or any point on the wave in 1 second. Displacement(s): The displacement, ‘s’ of the particles of the medium is their distance from the rest or central position in either direction. Amplitude (a): The amplitude ‘a’ is the height of a crest or the depth of trough measured from the central position. It is the maximum displacement from the rest or central position in either direction. Wavefronts: The wavefronts are represented by straight lines and can be thought as the crests or troughs of the waves. They are at the right angle to the direction of waves. The distance between the lines is called wavelength. Progressive wave: A progressive or traveling wave is the continuous disturbance of medium which carries energy from the source to another point. Phase difference: When the crests of two waves of equal wavelength are together the waves are said to be in phase that is the phase difference is zero. If a crest and a trough are together, the waves are completely out of phase that is they have a phase difference of π radians or 180o.1 Coherent Waves: Waves originate from same source, have same wavelength and have constant phase difference between them is called Coherent waves. Intensity of wave: Waves carries energy that can be used to do work. In case of sound waves the work done in forcing the eardrum to vibrate. The amount of energy transported per second is called power of the waves and is measured in joules per second (J/s) or watts (W). The intensity I of sound wave is defined as the sound power P that passes perpendicularly through a surface divided by the area A of that surface. I = P/A 1 Exercise on page no 425: Muncaster Prepared by Faisal Jaffer, revised on 7-Mar-10 The unit of wave intensity is power per unit area, or W/m2. In terms of amplitude of the wave Intensity ⋉ (amplitude) 2 Thus if the amplitude of the sound wave is doubled, it carries four times more energy per second away from the source. Graphical representation of longitudinal waves and transverse: In Longitudinal waves the motion of the individual particles of the medium is in direction which is parallel to the direction of wave or energy. The back and forth moment of the vibrating object makes the particles of the medium, for example air, move back and forth. These back and forth vibrations are passed on to adjacent neighbors by particle interaction; thus, other surrounding particles begin to move rightward and leftward, thus sending a wave to the right. Since air molecules (the particles of the medium) are moving in a direction which is parallel to the direction which the wave moves, the sound wave is referred to as a longitudinal wave. The result of such longitudinal vibrations is the creation of compressions (C) and rarefactions (R) within the air. A transverse wave is a wave in which the particles of the medium are displaced in a direction perpendicular to the direction of wave or energy. The snapshot of the transverse wave could be represented by diagram The crest of a wave is the point on the medium which exhibits the maximum amount of positive or upwards displacement from the rest position. Points C and J on the diagram represent the troughs of this wave. The trough of a wave is the point on the medium which exhibits the maximum amount of negative or downwards displacement from the rest position. Derivation of wave equation v = f λ: Whenever the source of wave motion does one cycle the wave moves forward by the distance (x) of one wavelength (λ). Since f is the number of cycles in each second, the wave moved forward by fλ distance in this time, and therefore the velocity (v) of the wave is given by x x = λ (when x is the distance moved equal to one wavelength) λ but x = vt (where v is the velocity of the wave and t is the time) vt = λ or v = λ/t ‘t’ here is the time for one oscillation therefore it can be written as ‘T’ the period of wave. v = λ/T But the f = 1/T (where ‘f’ is the frequency of the wave) v = f λ or = where c is the speed of light waves in vacuum which is 3×108m/s. AS Physics Unit 5 - Waves 5 POLARIZATION OF LIGHT WAVES: Light wave is an electromagnetic wave which travels through vacuum. Light waves are produced by vibrating electric charges. It is a transverse wave which has both an electric and a magnetic field component. The electric and magnetic vibrations of an electromagnetic wave occur in numerous planes. A light wave which is vibrating in more than one plane is referred to as unpolarized light. Light emitted by the sun, by a lamp in the classroom or by a candle flame is unpolarized. It is possible to transform unpolarized light into polarized light. Polarized light waves are light waves in which the vibrations occur in a single plane. The process of transforming unpolarized light into polarized light is known as polarization. http://micro.magnet.fsu.edu/optics/lightandcolor/polarization.html Demonstration of polarization using string and fence: In the example, the vibrations in the rope will pass through a narrow gap provided the longest axis of the gap is parallel to the direction of the vibrations. Thus if the rope is vibrating up and down, the motion will be transmitted through a vertical gap. However, if the rope also passes through a horizontal gap the vibrations can't pass through this. Light can be polarized by transmission, reflection, refraction, scattering. For AS course we are only going to study the polarization by reflection method. Polarization by reflection If an unpolarized beam of light is incident on a glass surface at an angle of about 57o , the light that is reflected from the surface is completely plane-polarized. This can be checked by using Polaroid film. At angles of incident other than 57o , the reflected light is partially plane-polarized. When the beam of light incident on glass at 57o and partially reflected and partially refracted light make an angle of 90o between each other, this is a particular incident angle at which light is completely plane-polarized. All other angles the light is partially plane-polarized. Let’s consider the incident angle θB, refracted through an angle θ2 and refractive index of glass is ‘n’ than we can write = sin sin ℎ ℎ ⇒ = sin sin(90 − sin(90 − ⇒ = sin cos = 90 − ) ) = cos = tan The equation = tan is called Brewster’s Law. We can find that light incident at the brewster angle has reflected and refracted rays which are perpendicular to each other. If we replace the value of n = 1.54 as refractive index of glass then the above equation gives θB = 57o. Prepared by Faisal Jaffer, revised on 7-Mar-10 DIFFRATION OF WAVES: When straight waves are incident on a small opening formed between two vertical bars placed in the path of waves, then the wavefronts emerge with a circular shape and waves spread out in all directions from the opening. The spreading of waves at the edges of obstacles is called diffraction. The extent of diffraction is depend upon the width of the opening and the wavelength of the wave. It is more noticeable if the width of the opening is almost equal to the wavelength of the wave. INTERFERENCE OF LIGHT: A set of two waves (such as light or sound) can combine with each other to produce a resultant wave. The way in which this combined wave is produced is called interference or superposition of waves. Constructive interference Destructive Interference The principle of superposition of waves: “The principle of superposition states that whenever two waves are traveling in the same region the total displacement at any point is equal to the sum of their individual displacement at that point”. When two identical waves (i.e., waves of same wavelength λ and amplitude) arrive at the same point in phase – that is, crest-to-crest and trough-to-trough then according to the principal of superposition, the waves reinforce each other and constructive interference happened. The resulting total waves at a point has an amplitude that is twice the amplitude of either of the individual waves, and in case of light waves, the brightness is greater than due to either wave alone. When two waves l1 and l2 start out in phase and are in phase when they reach at point P then l2 - l1 is equal to the λ and the interference will be constructive When l1 is the distance covered by wave 1 from the source l2 is the distance covered by wave 2 from the source λ is the wavelength of the wave In general, when the waves start out in phase, constructive interference will result whenever the distance travel by two waves are the same or differ by any number of complete wavelengths l2 – l1 = nλ where n = 0, 1, 2, 3 …. Eq. When two identical waves arrive at any point out of phase with one another, or crest-to-trough then the waves cancel each other and dark region appear on the screen. According to the principal of superposition it is called destructive interference. That is l2 - l 1 = ½ λ AS Physics Unit 5 - Waves or l2 - l1 = (n + ½) λ 7 where n = 0, 1, 2, 3… Eq. If the constructive and destructive interference is to continue occurring at a point, the sources of the wave must be coherent. Two sources are coherent if the waves they maintain a constant phase relation. Lasers are coherent sources of light, where incandescent light bulbs and fluorescent lamps are incoherent sources. The conditions for two sources of light to produce interference: 1. The sources must be coherent, i.e. there must be a constant phase difference between them. The phase difference may be zero but it does not have to be. 2. The waves that are interfering must have approximately the same amplitude otherwise the resulting interference pattern lacks contrast. 3. The two waves must have exactly same frequency and wavelength. With the waves of different wavelengths, the position where the crests coincided would always be changing. Young’s Double Slit Experiment: In 1801 the English scientist Thomas Young (1773 – 1829) performed an experiment that demonstrated the wave nature of light by showing that two overlapping light waves interfered with each other. By this experiment he was also able to determine the wavelength of light. Setup of the experiment: In an experiment the light of single wavelength (monochromatic light) passes through a single narrow slit S0 and falls on two closely spaced, narrow slits S1 and S2. These two slits act as coherent sources of light waves that interfere constructively and destructively at different points on the screen to produce patterns of alternative bright (B) and dark (D) fringes. The purpose of single slit is to ensure that only light from one direction falls on the double silt. Without it, light coming from different points on the light source would strike the source from different directions and cause the patterns on the screen to be washed out. The slits S1 and S2 act as coherent sources of light waves because the light from each originates from the same primary source – the single slit. A series of alternately bright and dark bands (interference fringes), which are equally spaced and parallel to the slits, can be observed on a screen placed anywhere in the region of overlap. Calculation of fringe separation The opposite figure represents the relative positions of the coherent sources, SI and S2, and a point P on the screen. The perpendicular distance, D, from the plane of the slits to the screen is very much greater than the slit separation, a (typically, D = 20 cm, a = 0.1 cm). The distance travel by waves along S2P is greater than the distance travel by waves along SIP. The difference of two distances is called path difference. By applying the mathematical calculation using Pythagoras theorem we find that − = Prepared by Faisal Jaffer, revised on 7-Mar-10 where x is the distance from the centre of the screen to the nth bright fringe. If a crest leaves SI at the same time as a crest leaves S2, there will be a bright fringe wherever the path difference (S2P – S1P) is equal to a whole number n of wavelength − = By combining the above two equations = = where n= 0, 1, 2, 3… for bright fringes. Similarly for dark fringes the path difference is equal to the odd number of half wavelength ( + ). =( + ) Typical values for the equation are D = 20cm = 0.2m a = 1mm = 10-3m λ = 6 x 10-7m x = 0.12 mm=1.2 x 10-4m By measuring the value of ‘D’ and ‘a’ and determining the value of ‘x’ by traveling microscope, we can find the wavelength (λ) of the light. The following points should be observed in double slit experiment. 1. The separation of the fringes can be increase by increasing ‘D’. 2. The separation of the fringes decrease by increasing ‘a’. 3. Increasing the width of any of the three slits increases the intensity of the fringes but more blurred. 4. Moving single slit S closer to S1 and S2 increases the intensity of the fringes but does not affect their separation. Dark and bright fringes To better understand the dark and light fringes consider the figure which shows three places on the screen where the interference happened. 1. Figure (a) shows how the bright fringe arises directly opposite the midpoint between the two slits. Here the light from the two slits travel exactly same distance and l1 is equal to l2 , therefore each containing the same number of wavelengths and constructive interference happened. 2. In figure (b) the distance l2 is larger than l1 by exactly one wavelength and therefore a bright fringe appear. Similarly whenever the difference between l1 and l2 is equal to complete number of wavelengths that is λ, 2λ, 3λ … the constructive interference will happen. AS Physics Unit 5 - Waves 9 3. Figure (c) shows how the first dark fringe arises. Here the distance l2 is larger than l1 by exactly one-half a wavelength, so the waves interfere destructively, giving rise to the dark fringe. Wherever the difference between l1 and l2 equals to one half of wavelength the destructive interference will happened. THE DIFFRACTION GRATING A diffraction grating is an arrangement of a large number of parallel lines of equal width ruled on glass. In diffracting grating there are clear spaces between the rulings of about 600 lines per millimeter. Diffraction gratings are used to produce optical spectra. The figure represents a section of a diffraction grating which is being illuminated normally by light of wavelength λ. Each of the gap acts like a very narrow slit and diffracts the incident light to an appreciable extent in all the forward directions. Consider that light which is diffracted at some angle to the normal. The slits are equally spaced and therefore if is such that light from all the gaps are in phase. Thus, if the light from one slit interferes constructively with the light from the other slit whose values of would be given by the equation = Where n = 0, 1, 2, 3….and d is the distance between the two lines in meters. The effect of the grating, therefore, is to produce a series of bright images on the screen, known as principal maxima, the angular positions , of which are given by the above equation. Each value of applies to either side of the normal. For any value of d and λ the zero order principal maxima, is always at the center of the screen and given by replacing n = 0 The positions of the first order (n = 1), second order (n = 2), etc. principal maxima; however, depend on both d and λ. A typical grating for example has a grating spacing, d, of 1/600000 m (corresponding to 600 lines per millimetre, i.e. 600000 lines per metre). If such a grating is illuminated by light whose wavelength, λ is 6×10-7m, we find, by substituting these values in equation, that n = 1 then = 21.1o n = 2 then = 46.1o but when n = 3 the equation gives sin = 1.08 (which is impossible) therefore this particular grating cannot produce a third order image when illuminated by light of this particular wavelength. It follows from equation that the number of orders of principal maxima that can be produced can be increased by increasing d (i.e. reducing the number of lines per metre). Reducing the total number of lines decreases the sharpness of the principal maxima and also gives rise to faint images, subsidiary maxima, in the regions between the principal maxima. Prepared by Faisal Jaffer, revised on 7-Mar-10 STANDING OR STATIONARY WAVES: A stationary or standing wave results when: 1. two progressive waves which are traveling in opposite direction, 2. which have the same speed and frequency and approximately equal amplitudes, 3. are in same medium 4. and are overlapped or superimposed. 5. Standing waves can be produced by the reflection of incoming waves on any hard surface. In this process the phase of reflecting wave becomes reverse of incident wave. 6. Places of maximum amplitude are called antinodes (A) and places of zero amplitude are called nodes (N). Standing waves in string: Consider a string of length L that is fixed at both ends. The string has a set of natural patterns of vibration called normal modes. This can be determined very simply. First remember that the ends are fixed so they must be nodes. This means a certain number of wavelengths or half wavelengths can fit on the string determined by the length of the string. The first three possible standing waves are shown below. The wavelengths of the standing wave in string can be related by the length of the string. The frequency, then, is found from the wavelength. v is the speed of wave in the string. AS Physics Unit 5 - Waves 11 For nth harmonic frequency we the general term as: = = 2 where n= 1, 2, 3,.... The lowest frequency is called the 1st harmonic mode or the fundamental frequency. The higher frequencies are called overtones. Multiples of the 1st harmonic modes are labeled as the 2nd, 3rd and 4th harmonic modes. Standing waves in air columns Just as we have standing waves on strings we can have standing sound waves in columns of air. The organ pipe is the basic example. At the open end of a pipe we expect displacement antinodes. If the end is closed we would expect a displacement node. Using the same sort of arguments as we did for the string we can find the normal modes of the air column in the pipe. For two open ends the first four harmonics are: =2× = Which is same as it is for two fixed end string. 2 Prepared by Faisal Jaffer, revised on 7-Mar-10 For air column closed at one end and open at the other the first four harmonics are: Notice that the "closed at one end" case only exhibits the odd harmonics. The frequencies of the normal modes for the two cases can be written as: = 4 (2 − 1) (2 − 1) × 4 A good reference website can be looked for further study http://openlearn.open.ac.uk/mod/resource/view.php?id=289475 = Speed of sound wave in air column: The speed with which sound travels in any medium may be determined if the frequency and the wavelength are known. The relationship between these quantities is = where λ is the wavelength, f is the frequency and v is the speed of sound waves. In this experiment the velocity of sound in air is to be found by using tuning forks of known frequency. The wavelength of the sound will be determined by finding out the position of nodes and antinodes produced in the air column above the water (resonance). The apparatus for the experiment consists of a long cylindrical plastic tube attached to a water reservoir. The length of the water column may be changed by raising or lowering the water level while the tuning fork is held over the open end of the tube. Resonance is indicated by the sudden increase in the intensity of the sound when the column is adjusted to the proper length. The resonance is a standing wave phenomenon in the air column and occurs when the column length is /4, 3/4, 5/4 where is the sound wavelength. AS Physics Unit 5 - Waves 13 The water surface constitutes a node of the standing wave since the air is not free to move longitudinally. The open end provides the conditions for an antinode, but the actual antinode has been found to occur outside the tube at a distance of about c=0.6r from the end, where r is the tube radius. This end correction ‘c’ may be added to get a more accurate value if only one resonance can be measured, but it is usually more convenient to eliminate this "end effect" by subtracting the resonance length for /4 from those for 3 /4, 5 /4, etc. 4−−−−−−−( + = 3 4−−−−−−−( Subtract equation 1 from equation 2 3 − = − 4 4 + = − 1) 2) = 2 = 2( − ) From the above equation the speed of sound waves can be found by substituting the equation in = =2 ( − ) Finding out the speed of sound waves using cathode ray oscilloscope: This requires practical demonstration in class. Prepared by Faisal Jaffer, revised on 7-Mar-10 Exercise: Q1. What is the phase difference between two waves, each of wavelength 12cm when one leads the other by a) 6cm, b) 3 cm c) 9cm d) 12cm e) 14cm f) 18 cm g) 36 cm h) 39 cm. Q2. The distance between the 1st bright fringe and 21st bright fringe in a Young’s double slit arrangement was found to be 2.7mm. The slit separation was 1.0 mm and the distance from the slit to the plane of the fringes was 25cm. What was the wavelength of the light? Q3. In a Young’s double-slit experiment a total of 23 bright fringes occupying a distance of 3.0mm were visible in the traveling microscope. The microscope was focused on a plane which was 31cm from the double slit and the wavelength of the light being used was 5.5 x 10-7 cm. What is the separation of the double slit? Q4. When a grating with 300 lines per millimeter is illuminated normally with a parallel beam of monochromatic light a second order principal maximum is observed at 18.9o to the straight through direction. Find the wavelength of the light. Q5. How many principal maxima are produced when a grating with spacing of 2.00 x 10-6 m is illuminated normally with light of wavelength 6.44 x 10-7m? Q6. Suppose that a string is 1.2 meters long and vibrates in the first, second and third harmonic standing wave patterns. Determine the wavelength of the waves for each of the three patterns. Q7. The string at the right is 1.5 meters long and is vibrating as the first harmonic. The string vibrates up and down with 33 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave. Q8. The string at the right is 6.0 meters long and is vibrating as the harmonic. The string vibrates up and down with 45 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave. Q9. The string at the right is 5.0 meters long and is vibrating as the fourth harmonic. The string vibrates up and down with 48 complete vibrational cycles in 20 seconds. Determine the frequency, period, wavelength and speed for this wave. Q10. The string at the right is 8.2 meters long and is vibrating as the fifth harmonic. The string vibrates up and down with 21 complete vibrational cycles in 5 seconds. Determine the frequency, period, wavelength and speed for this wave. third

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