622 18 Interference, Diffraction and Polarization of Light Therefore, Eq 18.29 can be used to measure λ if the grating spacing d and θm are known For monochromatic light, Fig 18.14 shows a sharp intensity distribution for the maxima and a broad distribution in the dark areas I 2 λ /d λ /d m sin θ λ /d λ /d Fig 18.14 A sketch of the intensity versus sin θ for a diffraction grating The zeroth-, first-, and secondorder maxima are shown The sharpness of the maxima and the broadness of the minima are shown We can use this technique to distinguish and identify light of several unknown wavelengths We cannot that with the double-slit arrangement of Sect 18.2, even though the same equation and wavelength dependencies apply there In a double-slit interference, the bright fringes due to different wavelengths overlap too much to be distinguished Resolving Power of the Diffraction Gratings Diffraction gratings are useful tools for accurately measuring wavelengths To resolve two similar light sources with nearly equal wavelengths λ1 and λ2 (near a wavelength λ), the diffraction grating should have a high resolving power R, defined in terms of the average wavelength λav and the wavelength difference R= λav λ λav = λ1 + λ2 λ, λ as: λ = λ2 − λ (18.30) If N is the number of illuminated slits in the grating, then it can be shown that the resolving power in the mth-order diffraction is: R=N m (18.31) Thus, R increases as N and m increase When m = 0, we know that all the wavelengths are indistinguishable and hence R = as expected 18.5 Diffraction Gratings 623 Example 18.9 A diffraction grating has 7,000 lines per centimeter When the grating is illuminated normally with a monochromatic light, the second order spectral line is found at 62◦ (a) What is wavelength of the light? (b) Where can we observe the third order maximum? Solution: (a) First, we must find the slit separation d as follows: d= cm = 1.429 × 10−4 cm = 1,429 nm 7,000 Then, we use d sin θm = m λ with m = to find the wavelength: λ= (1,429 nm) sin 62◦ d sin θ2 = = 631 nm 2 (b) For m = 3, we calculate sin θ3 as follows: sin θ3 = 3λ × 631 nm = = 1.33 d 1,429 nm Since sin θ3 cannot exceed unity, then this order cannot be observed Example 18.10 A diffraction grating cm wide has N = 1,000 equally spaced slits across its width The diffraction grating is illuminated at normal incidence by a sodiumvapor yellow lamp The yellow light (known as the sodium doublet) contains two colors, one with wavelengths λ1 = 589.0 nm and the other with wavelength λ2 = 589.6 nm (a) What is the separation between the slits of the grating? (b) How many bright fringes are seen for both colors? (c) What must the resolving power of the grating be if the two colors are to be resolved (distinguished)? (d) How many slits of this grating must be illuminated in order to resolve these two colors in the fourth-order? Solution: (a) The ruling separation distance d is: d= cm = 10−3 cm = 104 nm 1,000 (b) Maxima occur at sin θm = m λ/d (m = 0, 1, 2, ) This condition is accepted only if m λ/d < 1, or m < d/λ We select the larger wavelength λ2 = 589.6 nm to find the possible values of m as follows: 624 18 Interference, Diffraction and Polarization of Light m< d 104 nm = 16.96 = λ2 589.6 nm Thus, the orders m = 0, 1, 2, , 16 are seen in the diffraction pattern (c) With λav = (λ1 + λ2 )/2 = 589.3 nm and calculate the resolving power as follows: R= λ = 0.6 nm, we use Eq 18.30 to λav 589.3 nm = = 982 λ 0.6 nm (d) Using Eq 18.31 and R = 982, we find that: N = 982 R = = 246 m Thus, in order to resolve the yellow sodium doublet up to a 4th -order maximum, we must illuminate at least N = 246 slits 18.6 Polarization of Light Waves Light propagates in vacuum with a speed c = 2.9979 × 108 m/s, see Chap 27 As shown in Fig 18.15a, light waves have the properties of transverse electromag→ → netic waves, with an electric field vector E and a magnetic field vector B vibrating in → → two planes perpendicular to each other In addition, E and B propagate with velocity → c in the direction of the light wave, which is perpendicular to both → The direction of vibration of E for an individual wave is defined as the direction of polarization of the wave However, an ordinary beam of light contains a large number of electromagnetic waves emitted by atoms having random vibrational orientations, → and hence the direction of the electric field vector E in the beam is random In this case the beam of light is unpolarized → An electromagnetic wave is said to be polarized if the electric field vector E at a given position vibrates in the same direction at all times The plane of polarization → is defined as the plane containing E and the direction of propagation → c Fig 18.15a displays a schematic diagram for a light wave polarized along the y axis Fig 18.15b represents an unpolarized light beam and Fig 18.15c represents a polarized beam, both viewed along the direction of propagation A bean of unpolarized light can be polarized by reflection, refraction, scattering, or absorption 18.6 Polarization of Light Waves 625 Representation of unpolarized light y E Representation of polarized light E E B x c z c ⊥ to the page c ⊥ to the page (b) (c) (a) Fig 18.15 (a) An electromagnetic wave propagating in the x direction with velocity → c , where → c is → → perpendicular to both E and B (b) A sketch representing an unpolarized light beam (c) A sketch representing a polarized light beam In 1938, E H Land invented a polarizing sheet called Polaroid This sheet trans→ mits waves whose electric fields E vibrate in a certain direction (called the transmission axis) and absorbs waves whose electric fields vibrate in a perpendicular direction Figure 18.16 represents an unpolarized light beam incident first on a polarizing sheet, called the polarizer Because the transmission axis is vertical in the figure, the light transmitted through this sheet is polarized vertically with an electric field vector → denoted by E ◦ A second polarizing sheet, called the analyzer, with transmission axis making an angle θ with the polarizer, intercepts the beam The only component that → is allowed through by the analyzer is E cos θ, while other components are absorbed Polarizer E ° Analyzer θ Unpolarized light Transmission axis E cos θ ° Polarized light Fig 18.16 Two polarizing sheets whose transmission axes make an angle θ with each other Only a fraction of the polarized light incident on the analyzer is transmitted through it Since the intensity of the transmitted beam varies as the square of the electric field, then the intensity of the polarized beam transmitted through the analyzer varies with θ, and is given by Malus’s law as: 626 18 Interference, Diffraction and Polarization of Light I = I◦ cos2 θ (18.32) where I◦ is the intensity of the polarized beam incident on the analyzer This expression applies to any two polarizing materials where their transmission axes are placed at an angle θ to each other Example 18.11 → A plane-polarized light wave E◦ sin ωt of intensity I◦ makes an angle θ with the transmission axis of a Polaroid sheet What fraction of the original light is transmitted through the Polaroid? Solution: The incident polarized wave is equivalent to two mutually perpendicular → → components One component is parallel to the transmission axis E = (E◦ cos θ ) → → → sin ω t and the other is perpendicular to it E⊥ = (E◦ sin θ ) sin ωt Since E is the only transmitted component through the Polaroid with an intensity I proportional to the square of its amplitude, see Fig 18.17, then: → → I (E ◦ cos θ ) • (E◦ cos θ ) = = cos2 θ → → I◦ E◦ • E◦ Or: I = I◦ cos2 θ Which is Malus’s law Fig 18.17 E° Polaroid I° θ E° cos θ I Transmission axis Example 18.12 Unpolarized light of intensity I◦ is incident on a Polaroid sheet with a vertical transmission axis, see Fig 18.18 What is the intensity I of the transmitted beam? Solution: We recall that the incident wave consists of a multitude of randomly oriented electric fields Then, Eq 18.32 applies to each electric field, but with 18.6 Polarization of Light Waves 627 angle θ ranging from 0◦ to 360◦ Because the orientation is random, all values of θ will occur equally As a result, Eq 18.32 will give us the transmitted intensity if we use the average value of cos2 θ Thus, I = I◦ cos2 θ , where cos2 θ = 21 Then: I = 21 I◦ (as indicated in Fig 18.18) Fig 18.18 Unpolarized light I° Polarized light I = 12 I° Transmission axis 18.7 Exercises Sections 18.1 and 18.2 Interference of Light Waves and Young’s Double-Slit Experiment (1) Two identical narrow slits are separated by a distance d = 0.3 mm The slits are illuminated by a monochromatic red light of wavelength λ = 630 nm An interference pattern is observed on a screen at a distance D = 1.2 m from the plane of the slits Find the separation between adjacent bright fringes (2) Two narrow slits are separated by a distance d = 0.05 mm and are m away from a screen When the slits are illuminated by a monochromatic light of unknown wavelength λ, we obtain a second-order bright fringe cm from the central line Find the wave length of the light (3) When white light is used instead of the monochromatic light in Exercise 2, the first-order fringe of the observed interference pattern resembles a rainbow of violet and red light at the fringe border The approximate locations of the violet and red light on the screen are y1V = 16 mm and y1R = 28 mm from the central line Estimate the wavelengths of the violet and red light (4) In a Young’s double-slit experiment, monochromatic light is diffracted from two narrow slits 0.4 mm apart Near the central line, successive bright and dark fringes that are mm apart are both viewed on a screen m away Find the wavelength and frequency of the light 628 18 Interference, Diffraction and Polarization of Light (5) In a double-slit experiment, the fifth-order bright fringe produced by light of wavelength 450 nm is observed at an angle of 30◦ from the central line How far apart are the two slits? (6) What are the expected angles of all dark fringes preceding the fifth-order bright fringe of Exercise 5? (7) A blue light of wavelength λB = 475 nm and yellow light of wavelength λY = 570 nm pass through a Young’s double-slit apparatus Blue and yellow patterns of fringes are formed on the screen of the experiment At the central bright fringe, where mV = mB = 0, both the blue and yellow light mix and form a green fringe What is the next order of the blue and yellow fringes that overlap on the screen to form a green fringe? (8) Two narrow slits that are 0.015 mm apart in a Young’s double-slit experiment are illuminated by a green laser beam of wavelength λG = 510 nm (a) What will be the total number of bright fringes that will be formed on both sides of a very large distant screen? (b) What angle does the most distant bright fringe from the central fringe make with respect to the original direction of the laser beam? (9) Two very narrow slits are 1.5 µm apart and are 25 cm away from a screen Light of wavelength 450 nm passes through the double slits, forming an interference pattern that does not satisfy the condition λ d, i.e., tan θ = sin θ What is the distance between the first and second dark fringes of the interference pattern on the screen? (10) A double-slit experiment is designed for easy viewing in a classroom such that the distance between the central maximum and the first maximum is 30 cm The wavelength of the He-Ne red laser light used in the experiment is 634 nm, and the viewing screen is m away from the double slits What slit separation is required for such an interference pattern? (11) Light of wavelength 600 nm passes through two slits that are 0.5 mm apart What is the phase difference between two parallel diffracted light rays making an angle 15◦ with the central line? (12) The peak intensity of a two-slit interference pattern is denoted by I◦ and the variation of the intensity as a function of the phase difference is given by Eq 18.12 Assume a point in the pattern where the phase difference between waves from the two slits is π/3 rad (a) What is the intensity of the pattern at this point? (b) What is the path difference between waves at this point when light of wavelength λ = 567 nm is used? 18.7 Exercises 629 (13) Two narrow slits 0.15 mm apart are illuminated by 634 nm light and their interference pattern is viewed on a screen 1.2 m meters away from the slits Find the intensity (relative to the central maximum) of the interference pattern 3.5 cm above the central line (14) Show that the relation 2d sin θ = (m + 21 )λ, m = 0, 1, 2, gives the angle θ at which the double-slit intensity is one-half of the peak value, i.e., when I = 21 I◦ (15) Using the relation obtained in Exercise 14 for half the intensity at the peak, show that the angular displacement from the half-intensity position on one side of the central maximum to the half-intensity position on the other side is given by the relation θ = λ/2d (Hint: sin θ θ when θ is small) Section 18.3 Thin Films—Change of Phase due to Reflection (16) An oil film has an index of refraction n = 1.5 Monochromatic light of wavelength λ = 600 nm is incident normally on the film (a) What is the smallest thickness of the film that will give a maximum interference in the reflected light view? (b) Would tripling the thickness calculated in part (a) produce maximum interference in the reflected light view? (17) A soap bubble has an index of refraction n = 1.32 and is 130 nm thick White light is incident normally on the outer surface of the bubble What is the wavelength of the color that reflected from the bubble’s outer surface? (18) Two glass plates are in contact at one edge and separated by a very fine separator at the other end to form a wedge as shown in Fig 18.19 The wedge is illuminated normally from above by light of wavelength λ = 589 nm When the wedge is viewed vertically, six dark fringes are observed between the left edge and the last dark fringe located at the separator, see Fig 18.19 Find the height of the separator Fig 18.19 See Exercise(18) λ Almost normal Air film (Exaggerated) Dark fringes d D D D D D D Separator 630 18 Interference, Diffraction and Polarization of Light (19) Assume that the separator in Fig 18.19 has a height d = 6,000 nm (a) How many dark and bright bands will be seen in the wedge area? (b) Assume that the glass plates are 20.5 cm long and the dark and bright bands are equal in thickness How far apart are the bright bands? (20) A soap film has an index of refraction n = 1.32 The film is illuminated normally by light of wavelength λ = 445 nm (a) What is the smallest thickness of the film that it will appear black? (b) Why would the film also appear black if the film thickness is much less than the wavelength? (21) A lens of index of refraction nL = 1.52 appears green (λ = 510 nm, i.e., reflects most of the green) when white light is shined on its surface One solution to avoid this effect is to coat the surface of the lens with a film of material that has an index of refraction nC = 1.25 What is the minimum thickness of coating required such that the coating interferes constructively with the green light? (22) Figure 18.20 shows a very thin film of oil (no = 1.5) with variable thickness The oil is floating on water (nw = 1.33) The film is illuminated from above by white light, which causes a sequence of highly distinguished colors to appear as shown in the figure Take blue to have λ = 445 nm, yellow to have λ = 570 nm, and red to have λ = 650 nm (a) Find the minimum and maximum thickness of the variable thickness of the oil film (b) Explain the existence of the dark Oil Water Red Yellow Blue Red Yellow Blue Dark Blue Yellow Red region in the oil film Air no = 1.5 nw = 1.33 Fig 18.20 See Exercise (22) (23) The radius of curvature of the convex surface of a plano-convex thin lens is R = m The convex surface is placed down on a plane glass plate and illuminated from above by light of wavelength λ = 450 nm, see Fig 18.21 (a) What is the change in thickness of the air film between the third d3 and the sixth d6 bright fringes in the reflected light view? (b) What is the radius r4 of the fourth bright fringe? 18.7 Exercises R r 631 R r r λ r4 56 Air film (Exaggerated) d P Almost normal O Fig 18.21 See Exercise(23) (24) When viewing the Newton’s rings from above, use the geometry of Fig 18.21 to show that the radius rm of the mth dark ring when rm R is given by √ rm = mλR/nfilm , where nfilm is refractive index of the film [Hint: Use the binomial expansion (1 − x)n − nx, when x 1.] (25) Use the result of Exercise 24 to show that the distance r between adjacent √ dark Newton’s rings of order m and m + is given by r = λR/4mnfilm , when m (26) The maximum ring radius in Fig 18.21 is r = 1.5 cm and corresponds to the 32th dark ring (a) What is the radius of curvature of the plano-convex lens if light of wavelength λ = 570 nm is used? (b) What is the focal length of the lens if its refractive index is n = 1.52? Sections 18.4 and 18.5 Diffraction of Light Waves and Diffraction Gratings (27) A beam of red light from a helium-neon laser is diffracted by a slit of width a = 0.5 mm A diffraction pattern is formed on a screen at a distance D = 1.9 m from the slit The distance between the zero intensities on either side of the central peak is 4.81 mm (a) Find the wavelength of the laser light (b) Calculate the ratio of the intensities of the third maximum to the central maximum (28) Monochromatic light of wavelength λ = 480 nm is diffracted by a single slit of width a = 4.5 × 10−3 mm A diffraction pattern is formed on a screen at 632 18 Interference, Diffraction and Polarization of Light a distance D = m away from the slit Assuming that the angle of the first maximum is equal to the average of the angles of the first and second minima, estimate how far the first maximum is from the central maximum (29) A single slit diffracts light of wavelength λ = 650 nm When the screen is at a distance D = m away from the slit, the central maximum is cm wide What is the width of the slit? (30) (a) What will be the minimum value of a single slit of width a that will not produce diffraction minima for a given wavelength λ? (b) What will be the minimum value of the width a that will not produce diffraction minima for the whole range of visible light (with the approximate range from 400 nm to 700 nm)? (31) A single slit 1.5 µm wide is illuminated by 634 nm light and its diffraction pattern is viewed on a screen 53.6 cm away from the slit (a) What is the height of the first minimum above the central maximum? (b) As a fraction of the central maximum’s intensity I◦ , determine the light intensity 10 cm above the central maximum (32) The secondary maxima in Eq 18.26 not occur precisely at the maximum of the sine function This is because the denominator of the intensity function causes the intensity to decrease more rapidly than the sine function causes it to increase Consequently, the intensity reaches a maximum slightly before the sine function reaches its maximum By differentiating Eq 18.26 with respect to δ/2, show that the secondary maxima occur when δ/2 satisfies the condition tan δ/2 = δ/2 (33) A diffraction grating has 8,000 grooves per centimeter The first order of the spectral line is observed to be diffracted at an angle of 30◦ What is the wavelength of the light used? (34) A diffraction grating has 5,000 lines per centimeter The grating is illuminated normally with the green light of mercury, which has a wavelength λ = 546.1 nm What is the angular separation between the first and second-order green lines? (35) A diffraction grating has N = 4,000 equally spaced slits per centimeter The diffraction grating is illuminated at normal incidence by the doublet colors of wavelengths λ1 = 732 nm and λ2 = 733 nm, emitted by a singly-ionized Oxygen atom (a) What is the separation between the slits of the grating? (b) How many bright fringes are seen for both colors? (c) What should the 18.7 Exercises 633 resolving power of the grating be if these two colors are to be distinguished? (d) How many slits of this grating must be illuminated in order to resolve these two colors in the second-order? (36) A diffraction grating has 4,500 lines per centimeter The grating is illuminated normally by white light (wavelengths ranging from 400 nm to 700 nm) How many spectral orders can be observed? (37) Use the data of Exercise 36 to find the width of the first-order spectrum on a screen 0.5 m away from the grating (38) Find the range of wavelengths of the second-order spectra of a diffraction grating of white light that overlap with the range of wavelengths of the thirdorder spectra (39) A diffraction grating has 6,200 lines per centimeter and is illuminated by light with λ = 525 nm The light falls normally on its surface (a) What is the maximum possible order for this grating? (b) What order gives the best resolution? (c) How close can two wavelengths be if they are to be resolved in the maximum order? Section 18.6 Polarization of Light Waves (40) A beam of polarized light has one-fifth of its initial intensity after passing through an analyzer What is the angle between the axis of the analyzer and the initial polarization direction of the beam? (41) The transmission axes of two polarizers are oriented at 60◦ to one another Unpolarized light of intensity I◦ falls on them What fraction of light is transmitted through them? (42) If the light in Exercise 41 was polarized and the transmitted intensity from the two polarizers is 0.125 I◦ , what is the angle between the axis of the first polarizer and the initial polarization direction of the beam? (43) Two polarizers are oriented 60◦ to one another Light of intensity I◦ gets polarized as it passes through these polarizers at half the orientation angle between them What fraction of light intensity is transmitted through both of them? (44) Determine the angle in Exercise 43 that will make the two polarizers transmit only half of the incident light intensity (45) An ordinary light of intensity I◦ is incident on one Polaroid sheet and then falls on a second Polaroid sheet whose transmission axis makes an angle θ = 30◦ 634 18 Interference, Diffraction and Polarization of Light with the first, see Fig 18.22 (a) Find the intensity fractions I1 /I◦ , I2 /I1 , and I2 /I◦ (b) If the second Polaroid is rotated until the transmitted intensity is 10% of the incident intensity I◦ , what is the new angle? Unpolarized light Polaroid θ I ° Transmission axis Fig 18.22 See Exercise(45) I1 I2 Polaroid Part V Electricity Electric Force 19 In this chapter, we study one of the fundamental forces of nature, the electric force Electrical forces play an important role in the structure of atoms, molecules, and nuclei We will discuss the following: (1) The existence of electric charges and electric forces (2) The basic properties of electrostatic forces (3) Coulomb’s law, which is the fundamental law governing electric forces between charged particles (4) The application of Coulomb’s law to simple charge distributions 19.1 Electric Charge Many simple experiments indicate the existence of electric forces and charges It is possible to impart an electric charge to any solid material by rubbing it with another material The rubbed solid material is said to be electrified, or electrically charged For example, a comb becomes electrified when it is used to brush dry hair This is justified by observing that the comb will attract bits of paper Many experiments conducted by Benjamin Franklin reveal that there are two types of electric charges: positive and negative A glass rod that has been rubbed with silk is commonly used as an example for identifying positive and negative charges Another common example is a hard rubber rod that has been rubbed with fur Using Franklin’s convention, positive charges are formed on a glass rod that has been rubbed with silk, and negative charges are formed on a rubber rod that has been rubbed with fur H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_19, © Springer-Verlag Berlin Heidelberg 2013 637 638 19 Electric Force When a positively charged glass rod is brought close to a suspended negatively charged rubber rod, the two rods attract each other, see Fig 19.1a Conversely, if two positively charged glass rods (or two negatively charged rubber rods) are brought close to each other, the two rods repel each other, see Fig 19.1b F Rubber Glass F F Glass (a) F Glass (b) Fig 19.1 (a) A negatively charged rubber rod attracting a positively charged glass rod (b) A positively charged glass rod repelling another positively charged glass rod Based on these observations, we conclude that there are two kinds of charges in nature; one is positive and the other is negative, and they obey the following properties: Spotlight Like charges repel each other and unlike charges attract each other Additionally, it was found that when one object is rubbed with another, charge is transferred between them, i.e charge is not created in the rubbing process That is: Spotlight The total charge in any isolated system is conserved In 1909, Robert Millikan discovered that an electric charge always occurs in integral multiples of a fundamental charge e In a modern view, the electric charge q is said to be quantized and we can write q = ne, where n is an integer (n = ±1, ±2, ) That is: 19.1 Electric Charge 639 Spotlight Charge is quantized In today’s modern scientific views, an electric charge is considered to be a basic property of atoms As we all know, an atom is the fundamental entity of which all matter is formed Atoms themselves are composed of three types of particles— protons, electrons, and neutrons A proton carries one unit of positive charge +e, an electron carries one unit of negative charge −e, and a neutron carries no charge; it is electrically neutral Based on the charge conservation and the atomic structure, we find that when a glass rod is rubbed with silk, electrons are transferred from the glass to the silk giving the silk a net negative charge and consequently leaving a net positive charge of the same magnitude on the glass, see Fig 19.2 Similarly, when a rubber rod is rubbed with fur, electrons are transferred from the fur to the rod, giving the rod a net negative charge, leaving the fur with a net positive charge Fig 19.2 When a neutral Neutral Glass Glass glass rod is rubbed with a neutral silk cloth, electrons are transferred from the glass to Silk Neutral Silk the silk, leaving the glass positively charged Before rubbing 19.2 After rubbing Charging Conductors and Insulators Materials can be classified according to their ability to conduct electrical charge In some materials, such as metals (copper, aluminum, etc), tap water, and the human body, some of the negative charges (electrons) can move rather freely We call such materials conductors Spotlight Conductors are materials containing some electrons that can move freely 640 19 Electric Force In contrast, charges cannot move freely in some other materials such as glass, rubber, and plastic We call such materials nonconductors or insulators Spotlight Insulators are materials that contain electrons that are bound to their atoms and cannot move freely through the material Semiconductors are materials that lie somewhere between conductors and insulators, such as silicon and germanium The electrical properties of semiconductors can be changed drastically by adding specific amounts of certain atoms (impurities) Generally, the conductivity of semiconductors increases with increasing temperature, in contrast to metallic conductors The microelectronic revolution that has changed our lives is due to devices constructed of semiconductors Spotlight Semiconductors are materials that have electrical properties that lie somewhere between conductors and insulators, such as silicon and germanium Charging a Conductor by Rubbing When a person rubs a copper rod with wool while holding it in his hand, he will not be able to charge the rod The reason is that both the rod and his body are conductors The rubbing will cause a charge imbalance on the rod, but the excess charge will immediately flow from the rod through his body to the Earth, and the rod will be neutralized immediately Conversely, if the experiment is repeated while the rod is held by an insulating handle, we would eliminate the conducting path to Earth, and the rod can then be charged Charging a Conductor by Induction Another way of charging a conductor is shown in Fig 19.3 Figure 19.3a shows a negatively charged plastic rod and an isolated neutral copper rod that is suspended by an insulated twistable wire When the plastic rod is brought into the vicinity of the copper rod, many of the conduction electrons in the closer end of the copper rod 19.2 Charging Conductors and Insulators 641 are repelled by the negative charge on the plastic rod They move to the far end of the copper rod, leaving the near end depleted in electrons Such repulsion of negative charges from the near end leaves that end positively charged This positive charge is attracted to the negative charge in the plastic rod as shown in Fig 19.3b Although as a whole, the copper rod is still neutral, it is said to have an induced charge At this state, if we ground the copper rod, as shown in Fig 19.3c, some of the negative charges move out of the rod through the wire into the Earth Earth can accept or provide electrons freely with negligible effect on its electrical characteristics When the wire to the ground is removed, the conducting copper rod remains in an induced positive charge state, see Fig 19.3d When the plastic rod is removed from the vicinity of the copper rod, this induced positive charge remains and is distributed on the rod, see Fig 19.3e This process can be repeated with a positively charged glass rod to obtain a negatively charged copper rod Induced charge Neutral copper F F F F F F Charged copper Charged plastic (a) (b) (c) (d) (e) Fig 19.3 (a) A negatively charged plastic rod is kept far away from a neutral copper rod (b) The electrons on the copper rod are redistributed when the charged plastic rod is brought into the vicinity of the copper rod (c) When the copper rod is grounded, some of the electrons move into the Earth (d) Removing the ground connection (e) Removing the plastic rod to obtain a positively charged conductor Charging an Insulator by Induction In most of the neutral molecules of insulators, the center of positive charge coincides with the center of negative charge In the presence of a charged object, the charged centers of each molecule in the insulator may shift slightly The molecule is then said to be electrically polarized This produces a layer of induced charge on the insulator surface as shown in Fig 19.4a Consequently, the charged object and the insulator will attract each other, see Fig 19.4b 642 19 Electric Force Polarized molecule Induced charge Charged comb Charged object Neutral insulator Neutral bits of paper (a) (b) Fig 19.4 (a) A negatively charged object produces an induced charge on the surface of an insulator because charges in the molecules of the insulator are electrically polarized (b) A charged comb attracts small bits of dry paper due to the effect of molecular polarization 19.3 Coulomb’s Law In an experiment to measure the magnitude of the electrical force F between two charged particles separated by a distance r and having charges q1 and q2 , Charles Coulomb was able to find that: F=k |q1 | |q2 | r2 (Coulomb’s law) (19.1) This formula is known as Coulomb’s Law, and k is a constant called the Coulomb constant Coulomb found that each charged particle (also called a point charge) exerts a force of this magnitude on the other particle, and the two forces form an action–reaction pair It was found that charges of the same sign repel each other, while charges of opposite signs attract each other, see Fig 19.5 The SI unit of a charge is the coulomb (abbreviated by C) and is derived from the SI unit of electric current, the ampere (abbreviated by A) which will be defined in Chap 24 The form given by Eq 19.1 resembles Newton’s force law that describes the universal gravitation between two objects of masses m1 and m2 that are separated by a distance r, see Fig 19.6 That is: F =G m1 m2 r2 (Newton’s gravitational law) where G = 6.67 × 10−11 N.m2 /kg2 is the gravitational constant (19.2) [...]... Coulomb’s Law In an experiment to measure the magnitude of the electrical force F between two charged particles separated by a distance r and having charges q1 and q2 , Charles Coulomb was able to find that: F=k |q1 | |q2 | r2 (Coulomb’s law) (19.1) This formula is known as Coulomb’s Law, and k is a constant called the Coulomb constant Coulomb found that each charged particle (also called a point charge)... materials conductors Spotlight Conductors are materials containing some electrons that can move freely 640 19 Electric Force In contrast, charges cannot move freely in some other materials such as glass, rubber, and plastic We call such materials nonconductors or insulators Spotlight Insulators are materials that contain electrons that are bound to their atoms and cannot move freely through the material... Electric Force Polarized molecule Induced charge Charged comb Charged object Neutral insulator Neutral bits of paper (a) (b) Fig 19.4 (a) A negatively charged object produces an induced charge on the surface of an insulator because charges in the molecules of the insulator are electrically polarized (b) A charged comb attracts small bits of dry paper due to the effect of molecular polarization 19.3 Coulomb’s... possible order for this grating? (b) What order gives the best resolution? (c) How close can two wavelengths be if they are to be resolved in the maximum order? Section 18.6 Polarization of Light Waves (40) A beam of polarized light has one-fifth of its initial intensity after passing through an analyzer What is the angle between the axis of the analyzer and the initial polarization direction of the beam?... example for identifying positive and negative charges Another common example is a hard rubber rod that has been rubbed with fur Using Franklin’s convention, positive charges are formed on a glass rod that has been rubbed with silk, and negative charges are formed on a rubber rod that has been rubbed with fur H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI:... exerts a force of this magnitude on the other particle, and the two forces form an action–reaction pair It was found that charges of the same sign repel each other, while charges of opposite signs attract each other, see Fig 19.5 The SI unit of a charge is the coulomb (abbreviated by C) and is derived from the SI unit of electric current, the ampere (abbreviated by A) which will be defined in Chap 24 The... The transmission axes of two polarizers are oriented at 60◦ to one another Unpolarized light of intensity I◦ falls on them What fraction of light is transmitted through them? ( 42) If the light in Exercise 41 was polarized and the transmitted intensity from the two polarizers is 0. 125 I◦ , what is the angle between the axis of the first polarizer and the initial polarization direction of the beam? (43)... has changed our lives is due to devices constructed of semiconductors Spotlight Semiconductors are materials that have electrical properties that lie somewhere between conductors and insulators, such as silicon and germanium Charging a Conductor by Rubbing When a person rubs a copper rod with wool while holding it in his hand, he will not be able to charge the rod The reason is that both the rod and. .. Conductor by Induction Another way of charging a conductor is shown in Fig 19.3 Figure 19. 3a shows a negatively charged plastic rod and an isolated neutral copper rod that is suspended by an insulated twistable wire When the plastic rod is brought into the vicinity of the copper rod, many of the conduction electrons in the closer end of the copper rod 19 .2 Charging Conductors and Insulators 641 are repelled... material Semiconductors are materials that lie somewhere between conductors and insulators, such as silicon and germanium The electrical properties of semiconductors can be changed drastically by adding specific amounts of certain atoms (impurities) Generally, the conductivity of semiconductors increases with increasing temperature, in contrast to metallic conductors The microelectronic revolution that