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Transcript
Unit 18 Electromagnetic Waves 18.1 Electromagnetic waves 18.2 The electromagnetic spectrum 18.3 Polarization 18.4 Polarization by reflection 18.5 Polarization by scattering 18.1 Electromagnetic waves Electromagnetic waves are traveling waves of oscillating electric and magnetic fields. The electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of propagation. They are also in phase. To find the direction of propagation of an electromagnetic wave, point the fingers of your right hand in the direction of E, then curl them toward B. Your thumb will be pointing in the direction of propagation. Electromagnetic waves can be generated by an antenna, which is connected to an AC generator. 1 Direction of propagation for electromagnetic waves can be shown by using the righthand rule: Point the fingers of your right hand in the direction of E, curl your fingers toward B, and your thumb will point in the direction of propagation. In fact, another way to generate electromagnetic waves is to accelerate electric charge. As accelerated charges radiate electromagnetic waves, the intensity of radiated electromagnetic waves depends on the orientation of the acceleration relative to the viewer. For example, viewing the antenna perpendicular to its length, so that the charges accelerate at right angles to the line of sight, results in maximum intensity. Conversely, viewing the antenna straight down from above, in the same direction as the acceleration, results in zero intensity. According to Maxwell’s work in electric and magnetic fields, he proposed the light is an electromagnetic wave. Visible light is of course an electromagnetic wave. Unlike sound or a wave on a string requires a medium through which it can propagate, light can propagate through a vacuum. In fact, electromagnetic waves travel through a vacuum with the maximum speed, c, that any form of energy can have. It is also the speed of light. c = 3.00 × 10 8 m / s A beam of light can travel around the world about seven times in a single second. Light travels slower in a denser medium, e.g. glass and water. Example An electromagnetic wave propagates in the positive y direction, as shown in the figure. If the electric field at the origin is in the positive z direction, is the magnetic field at the origin in (a) the positive x direction, (b) the negative x direction, or (c) the negative y direction? 2 Answer According to the right hand, which has your thumb pointing the propagation direction, the direction of magnetic field should be along positive x. 18.2 The electromagnetic spectrum 18.2.1 Radio waves The lowest-frequency electromagnetic waves of practical importance are the radio and television waves. The frequency ranges from 106 Hz to 109 Hz. 18.2.2 Microwaves Electromagnetic radiation with frequencies from 109 Hz to about 1012 Hz is referred to as microwaves. Waves in this frequency range are used to carry long distance telephone conversations, as well as to cook our food. Microwaves, with wavelengths of about 1 mm to 30 mm, are the highest-frequency electromagnetic waves that can be produced by electronic circuitry. 18.2.3 Infrared waves Electromagnetic waves with frequencies just below that of light – roughly 1012 Hz to 4.3×1014 Hz – are known as infrared rays. Most remote controls operate on a beam of infrared light with a wavelength of about 1000 nm. This infrared light is so close to the visible spectrum and so low in intensity that it cannot be felt as heat. 3 18.2.4 Visible light The full range of colors in a rainbow as seen by human eyes is visible light. The frequency range is 4.3×1014 Hz to 7.5×1014 Hz. 18.2.5 Ultraviolet light When electromagnetic waves have frequencies just above that of violet light – from about 7.5×1014 Hz to 1017 Hz – they are called ultraviolet or UV rays. More prolonged or intense exposure to UV rays can have harmful consequences, including an increased probability of developing a skin cancer. 18.2.6 X-ray The frequency range of X-ray is about 1017 Hz to 1020 Hz. Typically, the X-rays used in medicine are generated by the rapid deceleration of high-speed electrons projected against a metal target. These energetic rays, which are only weakly absorbed by the skin and soft tissues, pass through our bodies rather freely, except when they encounter bones, teeth, to other relatively dense material. This property makes X-rays most valuable for medical diagnosis, research, and treatment. 18.2.7 Gamma rays Electromagnetic waves with frequencies above about 1020 Hz are generally referred to as gamma (γ) rays. These rays, which are even more energetic than X-rays, are often produced as neutrons and protons rearrange themselves within a nucleus, or when a particle collides with its antiparticle, and the two annihilate each other. Gamma rays are highly penetrating and destructive to living cells. It is for this reason that they are used to kill cancer cells and, more recently, microorganisms in food. 4 18.3 Polarization To understand what is meant by the polarization of light, or any other electromagnetic wave, consider the electromagnetic waves pictured in the figure. Each of these waves has an electric field that points along a single line. For example, the electric field that points in either the positive or negative z direction. We say, then, that this wave is linearly polarized in the z direction. A wave of this sort might be produced by a straight-wire antenna oriented along the z axis. In general, the polarization of an electromagnetic wave refers to the direction of its electric field. Polarization is an evidence for electromagnetic waves being a transverse wave. A beam of unpolarized light can be polarized in a number of ways, e.g. by passing it through a polarizer. To be specific, a polarizer is a material this is composed of long, thin, electrically conductive molecules oriented in a specific direction. When a beam of light strikes a polarizer, it is readily absorbed if its electric field is parallel to the molecules; light whose electric field is perpendicular to the molecules passes through the material 5 with little absorption, As a result, the light that passes through a polarizer is preferentially polarized along a specific direction. A simple mechanical analog of a polarizer is shown in the following figure. Remarks: Polaroid is made from tiny crystals of quinine iodosulphate all lined up in the same direction in a sheet of nitrocellulose. Crystals which transmit light vibrations, i.e. electric field variations, in one particular plane and absorb those in a mutually perpendicular plane are said to be dichroic. We now consider what happens when a beam of light polarized in one direction encounters a polarizer oriented in a different direction. In the left figure in next page, we see light with a vertical polarization and intensity I 0 passes through a polarizer with its preferred direction – its transmission axis – at θ an angle to the vertical. The component of E along the transmission axis is E cosθ . Recalling that the intensity of light is proportional to the electric field squared, we see that the intensity, I, of the transmitted beam is reduced to I = I 0 cos 2 θ . When an unpolarized beam of intensity I 0 passes through a polarizer (the right figure in next page), the transmitted beam has an intensity of 1 I 0 and is polarized in the 2 direction of the polarizer. 6 An unpolarized beam of intensity I 0 is polarized in the vertical direction by a polarizer with a vertical transmission axis. Next, it passes through another polarizer, the analyzer, whose transmission axis is at an angle θ relative to the transmission axis of the polarizer. The final intensity of the beam is I = 1 I 0 cos 2 θ . 2 Example Vertically polarized light with an intensity of 515 W/m2 passes through a polarizer oriented at an angle θ to the vertical. Find the transmitted intensity of the light for (a) θ = 10.0o , (b) θ = 45.0o , and (c) θ = 90.0o . Answer We apply the formula I = I 0 cos 2 θ . (a) = I (515 = W / m 2 ) cos 2 10.0o 499 W / m 2 (b) = I (515 = W / m 2 ) cos 2 45.0o 258W / m 2 (c) = I (515 = W / m 2 ) cos 2 90.0o 0 W / m 2 7 Example In a polarization experiment, the final intensity of the beam is 0.200 I 0 . What is the angle θ between the transmission axes of the analyzer and polarizer? Answer The intensity of the light is first reduced to I = 1 I 0 and then to 0.200 I 0 . That is, 2 1 I 0 cos 2 θ 0.200 I 0 . We therefore have = 2 1 cos 2 θ = 0.200 . 2 Solving for θ, we obtain θ = 50.8o . Since the analyzer absorbs part of the light as the beam passes through, it also absorbs energy. Thus, in principle, the analyzer would experience a slight heating. The applications of crossed polarizers include photoelastic stress analysis and LCD. The operation of a liquid-crystal display (LCD) is simply “ON” and “OFF”. The structure of LCD is shown in the figure. (a) Off (b) On There are basically three essential elements to an LCD – two crossed polarizers and a thin cell that holds a fluid formed of long, thin molecules known as a liquid crystal. 8 The liquid crystal is selected for its ability to rotate the direction of polarization, and the thickness of the liquid-crystal cell is adjusted to give a rotation of 90o. Thus, in its “OFF“ state, the liquid crystal rotates the direction of polarization, and light passes through the crossed polarizers. In this state, the LCD is transparent. When a voltage is applied to the liquid crystal it no longer rotates the direction of polarization, and light is no longer transmitted. Thus, it is in the “ON” state. The advantage of LCD is its energy efficient. A very little energy is required to give the voltage necessary to turn a liquid crystal cell “ON”. The light that the LCD uses is already present in the environment. 18.4 Polarization by reflection Unpolarized light can be partially polarized by reflection. When unpolarized light strikes a reflecting surface between two optical materials, preferential reflection occurs for those waves in which the electric-field vector is parallel to the reflecting surface. At one particular angle of incidence, called the polarization angle θ p , only the light for which the E vector is perpendicular to the plane of incidence is reflected. The reflected light is therefore linearly polarized perpendicular to the plane of incidence. θp θp Sir David Brewster noticed that when the angle of incidence is equal to the polarizing angle θ p , the reflected ray and the refracted ray are perpendicular to each other, so r = 90 − θ p or sin r = cosθ p . From the law of refraction, n1 sin θ p = n2 sin r , which gives n1 sin θ p = n2 cosθ p , and thus the Brewster’s law: tan θ p = n2 . n1 9 Remark: When light is incident at the polarizing angle, none of the E field component parallel to the plane of incidence is reflected; this component is transmitted 100% in the refracted beam. So the reflected light is completely polarized. The refracted light is a mixture of the component parallel to the plane of incidence, all of which is refracted, and the remainder of the perpendicular component; it is therefore partially polarized. Example Sunlight reflects off the smooth surface of an unoccupied swimming pool. (a) At what angle of reflection is the light completely polarized? (b) What is the corresponding angle of refraction for the light that is transmitted (refracted) into the water? (c) At night an underwater floodlight is turned on in the pool. Repeat parts (a) and (b) for rays from the floodlight that strike the smooth surface from below. Answer (a),(b) The light is first in the air, then in the water. Take n a = 1.0 (air) and n b = 1.33 (water). The Brewster’s law tan θ p = nb gives na −1 nb −1 1.33 o = = θ p tan tan= 53.1 . 1.0 na The reflected and refracted rays are perpendicular, so θ p + r = 90o and thus r = 90o − θ p = 90o − 53.1o = 36.9o . (c) Now the light is first in the water, then in the air. Take n a = 1.33 (water) and n b = 1.00 (air). The Brewster’s law tan θ p = nb gives na 10 −1 nb −1 1.0 o = θ p tan = tan= 36.9 . n 1.33 a The reflected and refracted rays are perpendicular, so θ p + r = 90o and thus r = 90o − θ p = 90o − 36.9o = 53.1o . 18.5 Polarization by scattering 11