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899 Chapter 31 Properties of Light Conceptual Problems 1 • [SSM] A ray of light reflects from a plane mirror. The angle between the incoming ray and the reflected ray is 70°. What is the angle of reflection? (a) 70°, (b) 140°, (c) 35°, (d) Not enough information is given to determine the reflection angle. Determine the Concept Because the angles of incidence and reflection are equal, their sum is 70° and the angle of reflection is 35°. ( )c is correct. 2 • A ray of light passes in air is incident on the surface of a piece of glass. The angle between the normal to the surface and the incident ray is 40°, and the angle between the normal and the refracted ray 28°. What is the angle between the incident ray and the refracted ray? (a) 12°, (b) 28°, (c) 40° ,(d) 68° Determine the Concept The angle between the incident ray and the refracted ray is the difference between the angle of incidence and the angle of refraction. ( )a is correct. 3 • During a physics experiment, you are measuring refractive indices of different transparent materials using a red helium-neon laser beam. For a given angle of incidence, the beam has an angle of refraction equal to 28° in material A and an angle of refraction equal to 26° in material B. Which material has the larger index of refraction? (a) A, (b) B, (c) The indices of refraction are the same. (d) You cannot determine the relative magnitudes of the indices of refraction from the data given. Determine the Concept The refractive index is a measure of the extent to which a material refracts light that passes through it. Because the angles of incidence are the same for materials A and B and the angle of refraction is smaller (more bending of the light) for material B, its index of refraction is larger than that of A. ( )b is correct. 4 • A ray of light passes from air into water, striking the surface of the water at an angle of incidence of 45º. Which, if any, of the following four quantities change as the light enters the water: (a) wavelength, (b) frequency, (c) speed of propagation, (d) direction of propagation, (e) none of the above? Chapter 31 900 Determine the Concept When light passes from air into water its wavelength changes ( waterairwater nλλ = ), its speed changes ( waterwater ncv = ), and the direction of its propagation changes in accordance with Snell’s law. Its frequency does not change, so ( )a , ( )c and ( )d change. 5 • Earth’s atmosphere decreases in density as the altitude increases. As a consequence, the index of refraction of the atmosphere also decreases as altitude increases. Explain how one can see the Sun when it is below the horizon. (The horizon is the extension of a plane that is tangent to Earth’s surface.) Why does the setting Sun appear flattened? Determine the Concept The decrease in the index of refraction n of the atmosphere with altitude results in refraction of the light from the Sun, bending it toward the normal to the surface of Earth. Consequently, the Sun can be seen even after it is just below the horizon. The setting Sun appears flattened because refraction is greater for light from the lower part of the Sun than for light from the upper part. Earth Atmosphere 6 • A physics student playing pocket billiards wants to strike her cue ball so that it hits a cushion and then hits the eight ball squarely. She chooses several points on the cushion and then measures the distances from each point to the cue ball and to the eight ball. She aims at the point for which the sum of these distances is least. (a) Will her cue ball hit the eight ball? (b) How is her method related to Fermat’s principle? Neglect any effects due to ball rotation. Determine the Concept (a) Yes. (b) Her procedure is based on Fermat’s principle of least time. The ball presumably bounces off the cushion with an angle of reflection equal to the angle of incidence, just as a light ray would do if the cushion were a mirror. The least time would also be the shortest distance of travel for the light ray. 7 • [SSM] A swimmer at point S in Figure 31-53 develops a leg cramp while swimming near the shore of a calm lake and calls for help. A lifeguard at point L hears the call. The lifeguard can run 9.0 m/s and swim 3.0 m/s. She knows physics and chooses a path that will take the least time to reach the swimmer. Which of the paths shown in the figure does the lifeguard take? Properties of Light 901 Determine the Concept The path through point D is the path of least time. In analogy to the refraction of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction equals the ratio of the speeds of the lifeguard in each medium. Careful measurements from the figure show that path LDS is the path that best satisfies this criterion. 8 • Material A has a higher index of refraction than material B. Which material has the larger critical angle for total internal reflection when the material is in air? (a) A, (b) B, (c) The angles are the same. (d) You cannot compare the angles based on the data given. Determine the Concept Because the product of the index of refraction on the incident side of the interface and the sine of the critical angle is equal to one (the index of refraction of air), the material with the smaller index of refraction will have the larger critical angle. ( )b is correct. 9 • [SSM] A human eye perceives color using a structure which is called a cone that is is located on the retina. Three types of molecules compose these cones and each type of molecule absorbs either red, green, or blue light by resonance absorption. Use this fact to explain why the color of an object that appears blue in air appears blue underwater, in spite of the fact that the wavelength of the light is shortened in accordance with Equation 31-6. Determine the Concept In resonance absorption, the molecules respond to the frequency of the light through the Einstein photon relation E = hf. Neither the wavelength nor the frequency of the light within the eyeball depend on the index of refraction of the medium outside the eyeball. Thus, the color appears to be the same in spite of the fact that the wavelength has changed. 10 • Let θ be the angle between the transmission axes of two polarizing sheets. Unpolarized light of intensity I is incident upon the first sheet. What is the intensity of the light transmitted through both sheets? (a) I cos2 θ, (b) (I cos2 θ)/2, (c) (I cos2 θ)/4, (d) I cos θ, (e) (I cos θ)/4, (f) None of the above Picture the Problem The intensity of the light transmitted by the second polarizer is given by ,II θ20trans cos= where .210 II = Therefore, θ221trans cosII = and )(b is correct. Chapter 31 902 11 •• [SSM] Draw a diagram to explain how Polaroid sunglasses reduce glare from sunlight reflected from a smooth horizontal surface, such as the surface found on a pool of water. Your diagram should clearly indicate the direction of polarization of the light as it propagates from the Sun to the reflecting surface and then through the sunglasses into the eye. Determine the Concept The following diagram shows unpolarized light from the sun incident on the smooth surface at the polarizing angle for that particular surface. The reflected light is polarized perpendicular to the plane of incidence, i.e., in the horizontal direction. The sunglasses are shown in the correct orientation to pass vertically polarized light and block the reflected sunlight. Smooth surface Light from the sun Polaroid sunglasses θ θ θr PP 12 • Why is it far less dangerous to stand in front of an intense beam of red light than in front of a verylow-intensity beam of gamma rays? Determine the Concept The red light photons contain considerably less energy than the gamma photons so, even though there are likely to be fewer photons in the low-intensity gamma beam, each one is potentially dangerous. 13 • Three energy states of an atom are A, B and C. State B is 2.0 eV above state A and state C is 3.00 eV above state B. Which atomic transition results in the emission of the shortest wavelength of light? (a) B → A, (b) C → B, (c) C → A, (d) A→ C Determine the Concept Because the wavelength of the light emitted in an atomic transition is inversely proportional to the energy difference between the energy levels, the highest energy difference produces the shortest wavelength light. ( )c is correct. 14 • In Problem 13, if the atom is initially in state A, which transition results in the emission of the longest wavelength light? (a) A → B, (b) B → C, (c) A → C, (d) B→ A Properties of Light 903 Determine the Concept Because the energy required to induce an atomic transition varies inversely with the wavelength of the light that must be absorbed to induce the transition and the transition from A to B is the lowest energy transition, the transition B→ A results in the longest wavelength light. ( )d is correct. 15 • [SSM] What role does the helium play in a helium–neon laser? Determine the Concept The population inversion between the state E2,Ne and the state 1.96 eV below it (see Figure 31-51) is achieved by inelastic collisions between neon atoms and helium atoms excited to the state E2,He. 16 • When a beam of visible white light that passes through a gas of atomic hydrogen at room temperature is viewed with a spectroscope, dark lines are observed at the wavelengths of the hydrogen atom emission series. The atoms that participate in the resonance absorption then emit light of the same wavelength as they return to the ground state. Explain why the observed spectrum nevertheless exhibits pronounced dark lines. Determine the Concept Although the excited atoms emit light of the same frequency on returning to the ground state, the light is emitted in a random direction, not exclusively in the direction of the incident beam. Consequently, the beam intensity is greatly diminished at this frequency. 17 • [SSM] Which of the following types of light would have the highest energy photons? (a) red (b) infrared (c) blue (d) ultraviolet Determine the Concept The energy of a photon is directly proportional to the frequency of the light and inversely proportional to its wavelength. Of the portions of the electromagnetic spectrum include in the list of answers, ultraviolet light has the highest frequency. ( )d is correct. Estimation and Approximation 18 • Estimate the time required for light to make the round trip during Galileo’s experiment to measure the speed of light. Compare the time of the round trip to typical human response times. How accurate do you think this experiment is? Picture the Problem We can use the distance, rate, and time relationship to estimate the time required to travel 6 km. Chapter 31 904 Express the distance d the light in Galileo’s experiment traveled in terms of its speed c and the elapsed time Δt: tcd Δ= ⇒ c dt =Δ Substitute numerical values and evaluate Δt: s 102m/s10998.2 km6 Δ 58 −×=×=t Because human reaction is approximately 0.3 s: 4 5 reaction 102 s102 s 3.0 Δ Δ ×≈×= −t t or ( ) tt Δ102Δ 4reaction ×≈ Because human reaction time is so much longer than the travel time for the light, there was no way that Galileo’s experiment could demonstrate that the speed of light was not infinite. 19 • Estimate the time delay in receiving a light on your retina when you are wearing eyeglasses compared to when you are not wearing your eyeglasses. Determine the Concept We’ll assume that the source of the photon is a distance L from your retina and express the difference in the photon’s travel time when you are wearing your glasses. Let the thickness of your glasses by 2 mm and the index of refraction of the material from which they are constructed by 1.5. When you are not wearing your glasses, the time required for a light photon, originating a distance L away, to reach your retina is given by: c Lt =0Δ If glass of thickness d and index of refraction n is inserted in the path of the photon, its travel time becomes: ( ) ( ) c dnt c dnL nc d c dLttt 1 Δ 1 Δ 0 glassin airin −+=−+= +−=+= The time delay is the difference between Δt and Δt0: ( ) c dnttt 1ΔΔ 0delay −=−= Substitute numerical values and evaluate tdelay: ( )( ) ps 3 m/s10998.2 mm 215.1 8delay ≈× −=t Properties of Light 905 20 •• Estimate the number of photons that enter your eye if you look for a tenth of a second at the Sun. What energy is absorbed by your eye during that time, assuming that all the photons are absorbed? The total power output of the Sun is 4.2 × 1026 W. Picture the Problem The rate at which photons enter your eye is the ratio of power incident on your pupil to the energy per photon. We’ll assume that the electromagnetic radiation from the Sun is at 550 nm and that, therefore, its photons have energy (given by λhchfE == ) of 2.25 eV. The rate at which photons enter your eye is the ratio of the rate at which energy is incident on your pupil to the energy carried by each photon: photonper pupilon incident E P dt dN = (1) The intensity of the radiation from the Sun is given by: Sun thefrom distance sEarth'at sphere Sun pupil pupilon incident Sun A P A P I == Solving for pupilon incident P yields: Sun Sun thefrom distance sEarth'at sphere pupil pupilon incident PA A P = (2) Substituting in equation (1) yields: photonper Sun Sun thefrom distance sEarth'at sphere pupil E P A A dt dN = Substitute for the two areas and simplify to obtain: photonper Sun 2 -SunEarth pupil photonper Sun 2 -SunEarth 2 pupil4 1 4 4 E P R d E P R d dt dN ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= = π π Substitute numerical values and evaluate dN/dt: ( ) 11519 262 11 s 10237.3 eV J 101.602eV 25.2 W102.4 m 1050.14 mm 1 − − ×=×× ×⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×=dt dN or, separating variables and integrating this expression, ( )tN 115 s 10237.3 −×= Chapter 31 906 Evaluating N for t = 0.1 s yields: ( ) ( )( ) photons 103 s 1.0s 10237.3s 1.0 14 115 ×≈ ×= −N The energy deposited, assuming all the photons are absorbed, is the product of the rate at which energy is incident on the pupil and the time during which it is delivered: tPE pupilon incident = Substituting for pupilon incident P from equation (2) yields: tP A A E Sun Sun thefrom distance sEarth'at sphere pupil= Substitute for the two areas and simplify to obtain: tP R d tP R d E Sun 2 -SunEarth pupil Sun2 -SunEarth 2 pupil4 1 4 4 ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= = π π Substitute numerical values and evaluate E for t = 0.1 s: ( ) ( ) ( )( ) mJ 0.1mJ 1167.0s 1.0 W102.4m 1050.14 mm 1s 1.0 26 2 11 ≈=×⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×=E 21 •• Römer was observing the eclipses of Jupiter’s moon Io with the hope that they would serve as a highly accurate clock that would be independent of longitude. (Prior to GPS, such a clock was needed for accurate navigation.) Io eclipses (enters the umbraof Jupiter’s shadow) every 42.5 h. Assuming an eclipse of Io is observed on Earth on June 1 at midnight when Earth is at location A (as shown in Figure 31-54), predict the expected time of observation of an eclipse one-quarter of a year later when Earth is at location B, assuming (a) the speed of light is infinite and (b) the speed of light is 2.998 × 108 m/s. Picture the Problem We can use the period of Io’s motion and the position of Earth at B to find the number of eclipses of Io during Earth’s movement and then use this information to find the number of days before a night-time eclipse. During the 42.5 h between eclipses of Jupiter’s moon, Earth moves from A to B, increasing the distance from Jupiter by approximately the distance from Earth to the Sun, making the path for the light longer and introducing a delay in the onset of the eclipse. Properties of Light 907 (a) Find the time it takes Earth to travel from point A to point B: h4.2191 d h24 4 d24.365 4 earth = ×==→ Tt BA Because there are 42.5 h between eclipses of Io, the number of eclipses N occurring in the time it takes for Earth to move from A to B is: 56.51 h42.5 h4.2191 Io === → T tN BA Hence, in one-fourth of a year, there will be 51.56 eclipses. Because we want to find the next occurrence that happens in the evening hours, we’ll use 52 as the number of eclipses. We’ll also assume that Jupiter is visible so that the eclipse of Io can be observed at the time we determine. Relate the time t(N) at which the Nth eclipse occurs to N and the period TIo of Io: ( ) IoNTNt = Evaluate t(52) to obtain: ( ) ( ) d083.92 h24 d1h5.425252 = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×=t Subtract the number of whole days to find the clock time t: ( ) a.m.00:2 h992.1 d h24 d0.083 d92d92.083d9252 ≈ =×= −=−= tt Because June, July, and August have 30, 31, and 31 d, respectively, the date is: 1September (b) Express the time delay Δt in the arrival of light from Io due to Earth’s location at B: c rt sun-earth=Δ Substitute numerical values and evaluate Δt: min34.8 s 60 min 1s500 m/s10998.2 m105.1 8 11 = ×=× ×=Δt Hence, the eclipse will actually occur at 2:08 a.m., September 1 Chapter 31 908 22 •• If the angle of incidence is small enough, the small angle approximation sin θ ≈ θ may be used to simplify Snell’s law of refraction. Determine the maximum value of the angle that would make the value for the angle differ by no more than one percent from the value for the sine of the angle. (This approximation will be used in connection with image formation by spherical surfaces in Chapter 32.) Picture the Problem We can express the relative error in using the small angle approximation and then either use 1) trial-and-error methods, 2) a spreadsheet program, or 3) the Solver capability of a scientific calculator to solve the transcendental equation that results from setting the error function equal to 0.01. Express the relative error δ in using the small angle approximation: ( ) 1 sinsin sin −=−= θ θ θ θθθδ A spreadsheet program was used to plot the following graph of δ (θ ). 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.00 0.05 0.10 0.15 0.20 0.25 0.30 theta (radians) de lta (th et a) From the graph, we can see that δ (θ ) < 1% for θ ≤ 0.24 radians. In degree measure, °≤ 14θ Remarks: Using the Solver program on a TI-85 gave θ = 0.244 radians. Properties of Light 909 The Speed of Light 23 • Mission Control sends a brief wake-up call to astronauts in a spaceship that is far from Earth. 5.0 s after the call is sent, Mission Control can hear the groans of the astronauts. How far from Earth is the spaceship? (a) 7.5 × 108 m, (b) 15 × 108 m, (c) 30 × 108 m, (d) 45 × 108 m, (e) The spaceship is on the moon. Picture the Problem We can use the distance, rate, and time relationship to find the distance to the spaceship. Relate the distance d to the spaceship to the speed of electromagnetic radiation in a vacuum and to the time for the message to reach the astronauts: tcd Δ= Noting that the time for the message to reach the astronauts is half the time for Mission Control to hear their response, substitute numerical values and evaluate d: ( )( ) m105.7 s5.2m/s10998.2 8 8 ×= ×=d and )(a is correct. 24 • The distance from a point on the surface of Earth to a point on the surface of the moon is measured by aiming a laser light beam at a reflector on the surface of the moon and measuring the time required for the light to make a round trip. The uncertainty in the measured distance Δx is related to the uncertainty in the measured time Δt by Δx = 12 cΔt. If the time intervals can be measured to ±1.00 ns, (a) find the uncertainty of the distance. (b) Estimate the percentage uncertainty in the distance. Picture the Problem We can use the given information that the uncertainty in the measured distance Δx is related to the uncertainty in the time Δt by Δx = cΔt to evaluate Δx. (a) The uncertainty in the distance is: tcx ΔΔ 21= Substitute numerical values and evaluate Δx: ( )( ) cm0.15 ns00.1m/s10998.2Δ 821 ±= ±×=x Chapter 31 910 (b)The percent uncertainty in the distance to the Moon is: %10 m 103.84 cm 0.15Δ 8 8 Moon Earth to Moon Earth to −≈ ×=x x 25 •• [SSM] Ole Römer discovered the finiteness of the speed of light by observing Jupiter’s moons. Approximately how sensitive would the timing apparatus need to be in order to detect a shift in the predicted time of the moon’s eclipses that occur when the moon happens to be at perigee (3.63×105 km ) and those that occur when the moon is at apogee ( 4.06 ×105 km )? Assume that an instrument should be able to measure to at least one-tenth the magnitude of the effect it is to measure. Picture the Problem His timing apparatus would need to be sensitive enough to measure the difference in times for light to travel to Earth when the moon is at perigee and at apogee. The sensitivity of the timing apparatus would need to be one-tenth of the difference in time for light to reach Earth from the two positions of the moon of Jupiter: Sensitivity = tΔ101 where Δt is the time required for light to travel between the two positions of the moon. The time required for light to travel between the two positions of the moon is given by: c dd t perigeeat moon apogeeat moon Δ − = Substituting for Δt yields: c dd 10 y Sensitivit perigeeat moon apogeeat moon − = Substitute numerical values and evaluate the required sensitivity: ( )( ) ms 14 m/s 10998.210 km 1063.3km 1006.4 Δ 8 55 = × ×−×=t Remarks: Instruments with this sensitivity did not exist in the 17th century. Reflection and Refraction 26 • Calculate the fraction of light energy reflected from an air–water interface at normal incidence. Picture the Problem Use the equation relating the intensity of reflected light at normal incidence to the intensity of the incident light and the indices of refraction of the media on either side of the interface. Properties of Light 911 Express the intensity I of the light reflected from an air-water interface at normal incidence in terms of the indices of refraction and the intensity I0 of the incident light: 0 2 waterair waterair I nn nnI ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −= Solve for the ratio I/I0: 2 waterairwaterair 0 ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −= nn nn I I Substitute numerical values and evaluate I/I0: %0.233.100.1 33.100.1 2 0 =⎟⎠ ⎞⎜⎝ ⎛ + −= I I 27 • A ray of light is incident on one of a pair of mirrors set at right angles to each A ray of light is incident on one of two mirrors that are set at right angles to each other. The plane of incidence is perpendicular to both mirrors. Show that after reflecting from each mirror, the ray will emerge traveling in the direction opposite to the incident direction, regardless of the angle of incidence. Picture the Problem The diagram shows ray 1 incident on the vertical surface at an angle θ1, reflected as ray 2, and incident on the horizontal surface at an angle of incidence θ3. We’ll prove that rays 1 and 3 are parallel by showing that θ1 = θ4, i.e., by showing that they make equal angles with the horizontal. Note that the law of reflection has been used in identifying equal angles of incidence and reflection. 1 2 3 θ θ θ θ θ θ 1 1 2 33 4 We know that the angles of the right triangle formed by ray 2 and the two mirror surfaces add up to 180°: ( ) °=−°+°+ 1809090 12 θθ or 21 θθ = The sum of θ2 and θ3 is 90°: 23 90 θθ −°= Because 21 θθ = : 13 90 θθ −°= The sum of θ4 and θ3 is 90°: °=+ 9043 θθ Chapter 31 912 Substitute for θ3 to obtain: ( ) °=+−° 9090 41 θθ ⇒ 41 θθ = 28 •• (a) A ray of light in air is incident on an air–water interface. Using a spreadsheet or graphing program, plot the angle of refraction as a function of the angle of incidence from 0º to 90º. (b) Repeat Part (a), but for a ray of light in water that is incident on a water–air interface. [For Part (b), there is no reflected ray for angles of incidence that are greater than the critical angle.] Picture the Problem Diagrams showing the light rays for the two cases are shown below. In (a) the light travels from air into water and in (b) it travels from water into air. (a) Air Water θ θ 1 2 1n 2n (b) Air Water 1n 2n θ θ 1 2 (a) Apply Snell’s law to the air- water interface to obtain: 2211 sinsin θθ nn = where the angles of incidence and refraction are θ1 and θ2, respectively. Solving for θ2 yields: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 1 2 11 2 sinsin θθ n n Properties of Light 913 A spreadsheet program to graph θ2 as a function of θ1 is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Content/Formula Algebraic Form B1 1 n1 B2 1.33333 n2 A6 0 θ1 (deg) A7 A6 + 5 θ1 + Δθ B6 A6*PI()/180 1801 πθ × C6 ASIN(($B$1/$B$2)*SIN(B6)) ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛− 1 2 11 sinsin θ n n D6 C6*180/PI() πθ 180 2 × A B C D 1 n1= 1 2 n2= 1.33333 3 4 θ1 θ1 θ2 θ2 5 (deg) (rad) (rad) (deg) 6 0 0.00 0.000 0.00 7 1 0.02 0.013 0.75 8 2 0.03 0.026 1.50 9 3 0.05 0.039 2.25 21 87 1.52 0.847 48.50 22 88 1.54 0.847 48.55 23 89 1.55 0.848 48.58 24 90 1.57 0.848 48.59 Chapter 31 914 A graph of θ2 as a function of θ1 follows: 0 5 10 15 20 25 30 35 40 45 50 0 10 20 30 40 50 60 70 80 90 Angle of incidence, deg A ng le o f r ef ra ct io n, d eg (b) Change the contents of cell B1 to 1.33333 and the contents of cell B2 to 1 to obtain the following graph: 0 10 20 30 40 50 60 70 80 90 0 10 20 30 40 50 Angle of incidence, deg A ng le o f r ef ra ct io n, d eg Note that as the angle of incidence approaches the critical angle for a water-air interface (48.6°), the angle of refraction approaches 90°. 29 •• The red light from a helium-neon laser has a wavelength of 632.8 nm in air. Find the (a) speed, (b) wavelength, and (c) frequency of helium-neon laser light in air, water, and glass. (The glass has an index of refraction equal to 1.50.) Picture the Problem We can use the definition of the index of refraction to find the speed of light in the three media. The wavelength of the light in each medium is its wavelength in air divided by the index of refraction of the medium. The frequency of the helium-neon laser light is the same in all media and is equal to its value in air. The wavelength of helium-neon laser light in air is 632.8 nm. Properties of Light 915 The speed of light in a medium whose index of refraction is n is given by: n cv = (1) The wavelength of light in a medium whose index of refraction is n is given by: n air n λλ = (2) The frequency of the light is equal to its frequency in air independently of the medium in which the light is propagating: Hz 1074.4 nm 632.8 m/s 10998.2 14 8 ×= ×== λ cf Substitute numerical values in equation (1) and evaluate vwater: m/s1025.2 33.1 m/s10998.2 8 8 water ×= ×=v Substitute numerical values in equation (2) and evaluate λwater: nm 47633.1 nm 8.632 water ==λ The other speeds and wavelengths are found similarly and are summarized in the following table: (a) speed (m/s) (b) wavelength (nm) (c) frequency (Hz) Air 81000.3 × 633 141074.4 × Water 81025.2 × 476 141074.4 × glass 81000.2 × 422 141074.4 × 30 •• The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength in air equal to 400 nm and 1.61 for red light that has a wavelength in air equal to 700 nm. A ray of 700-nm–wavelength red light and a ray of 400-nm-wavelength violet light both have angles of refraction equal to 30º upon entering the glass from air. (a) Which is greater, the angle of incidence of the ray of red light or the angle of incidence of the ray of violet light? Explain your answer. (b) What is the difference between the angles of incidence of the two rays? Picture the Problem Let the subscript 1 refer to the air and the subscript 2 to the silicate glass and apply Snell’s law to the air-glass interface. Chapter 31 916 (a) Because the index of refraction for violet light is larger than that of red light, for a given incident angle violet light would refract more than red light. Thus to exhibit the same refraction angle, violet light would require an angle of incidence larger than that of red light. (b) Express the difference in their angles of incidence: red1,violet1,Δ θθθ −= (1) Apply Snell’s law to the air-glass interface to obtain: 2211 sinsin θθ nn = Solving for θ1 yields: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 2 1 21 1 sinsin θθ n n Substitute for violet1,θ and red1,θ in equation (1) to obtain: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛−⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= −− red ,2 air red1 violet,2 air violet1 sinsinsinsinΔ θθθ n n n n For °== 30red 2, violet2, θθ : ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛−⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ °−⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ °= −−−− air red1 air violet1 air red1 air violet1 2 sin 2 sin30sinsin30sinsinΔ n n n n n n n nθ Substitute numerical values and evaluate Δθ: ( ) ( ) °=°−°=⎟⎟⎠ ⎞⎜⎜⎝ ⎛−⎟⎟⎠ ⎞⎜⎜⎝ ⎛= −− 49.261.5310.56 00.12 61.1sin 00.12 66.1sinΔ 11θ Remarks: Note that Δθ is positive. This means that the angle for violet light is greater than that for red light and confirms our answer in Part (a). 31 •• [SSM] A slab of glass that has an index of refraction of 1.50 is submerged in water that has an index of refractionof 1.33. Light in the water is incident on the glass. Find the angle of refraction if the angle of incidence is (a) 60º, (b) 45º, and (c) 30º. Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to the glass and apply Snell’s law to the water-glass interface. Properties of Light 917 Apply Snell’s law to the water- glass interface to obtain: 2211 sinsin θθ nn = Solving for θ2 yields: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 1 2 11 2 sinsin θθ n n (a) Evaluate θ2 for θ1 = 60°: °=⎟⎠ ⎞⎜⎝ ⎛ °= − 5060sin 50.1 33.1sin 12θ (b) Evaluate θ2 for θ1 = 45°: °=⎟⎠ ⎞⎜⎝ ⎛ °= − 3945sin 50.1 33.1sin 12θ (c) Evaluate θ2 for θ1 = 30°: °=⎟⎠ ⎞⎜⎝ ⎛ °= − 2630sin 50.1 33.1sin 12θ 32 •• Repeat Problem 31 for a beam of light initially in the glass that is incident on the glass–water interface at the same angles. Picture the Problem Let the subscript 1 refer to the glass and the subscript 2 to the water and apply Snell’s law to the glass-water interface. Apply Snell’s law to the water- glass interface to obtain: 2211 sinsin θθ nn = Solving for θ2 yields: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 1 2 11 2 sinsin θθ n n (a) Evaluate θ2 for θ1 = 60°: °=⎟⎠ ⎞⎜⎝ ⎛ °= − 7860sin 33.1 50.1sin 12θ (b) Evaluate θ2 for θ1 = 45°: °=⎟⎠ ⎞⎜⎝ ⎛ °= − 5345sin 33.1 50.1sin 12θ (c) Evaluate θ2 for θ1 = 30°: °=⎟⎠ ⎞⎜⎝ ⎛ °= − 3430sin 33.1 50.1sin 12θ 33 •• A beam of light in air strikes a glass slab at normal incidence. The glass slab has an index of refraction of 1.50. (a) Approximately what percentage of the incident light intensity is transmitted through the slab (in one side and out the other)? (b) Repeat Part (a) if the glass slab is immersed in water. Chapter 31 918 Picture the Problem Let the subscript 1 refer to the medium to the left (air) of the first interface, the subscript 2 to glass, and the subscript 3 to the medium (air) to the right of the second interface. Apply the equation relating the intensity of reflected light at normal incidence to the intensity of the incident light and the indices of refraction of the media on either side of the interface to both interfaces. We’ll neglect multiple reflections at glass-air interfaces. 1I 1r,I 2r,I 2I 3I 00.11 =n 50.12 =n 00.13 =n (a) Express the intensity of the transmitted light in the second medium: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −−= ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −−=−= 2 21 21 1 1 2 21 21 1r,112 1 nn nnI I nn nnIIII Express the intensity of the transmitted light in the third medium: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −−= ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −−=−= 2 32 32 2 2 2 32 32 2r,223 1 nn nnI I nn nnIIII Substitute for I2 to obtain: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −−⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −−= 2 32 32 2 21 21 13 11 nn nn nn nnII Solve for the ratio I3/I1 to obtain: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −−⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + −−= 2 32 32 2 21 21 1 3 11 nn nn nn nn I I Substitute numerical values and evaluate I3/I1: %92 00.150.1 00.150.11 50.100.1 50.100.11 22 1 3 = ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛ + −− ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛ + −−= I I Properties of Light 919 (b) With n1 = n3 = 1.33: %99 33.150.1 33.150.11 50.133.1 50.133.11 22 1 3 = ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛ + −− ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛ + −−= I I 34 •• This problem is a refraction analogy. A band is marching down a football field with a constant speed v1. About midfield, the band comes to a section of muddy ground that has a sharp boundary making an angle of 30º with the 50-yd line, as shown in Figure 31-55. In the mud, each marcher moves at a speed equal to 1 2 v1 in a direction perpendicular to the row of markers they are in. (a) Diagram how each line of marchers is bent as it encounters the muddy section of the field so that the band is eventually marching in a different direction. Indicate the original direction by a ray and the final direction by a second ray. (b) Find the angles between these rays and the line normal to the boundary. Is their direction of motion ″bent″ toward the normal or away from it? Explain your answer in terms of refraction. Picture the Problem We can apply Snell’s law to find the angle of refraction of the line of marchers as they enter the muddy section of the field (a) °30 θ θ 1 2 (b) Apply Snell’s law at the interface to obtain: 2211 sinsin θθ nn = Solving for θ2 yields: ⎥⎦ ⎤⎢⎣ ⎡= − 1 2 11 2 sinsin θθ n n The ratio of the indices of refraction is the reciprocal of the ratio of the speeds of the marchers in the two media: 2 1 1 12 1 1 2 2 1 2 1 ==== v v v v v v v v n n Chapter 31 920 Because the left and right sides of the 30° angle and θ1 are mutually perpendicular, θ1 = 30°. Substitute numerical values and evaluate θ2: [ ] °=°= − 1430sinsin 2112θ As the line enters the muddy field, its speed is reduced by half and the direction of the forward motion of the line is changed. In this case, the forward motion in the muddy field makes an angle θ2 with respect to the normal to the boundary line. Note that the separation between successive lines in the muddy field is half that in the dry field. 35 •• [SSM] In Figure 31-56, light is initially in a medium that has an index of refraction n1. It is incident at angle θ1 on the surface of a liquid that has an index of refraction n2. The light passes through the layer of liquid and enters glass that has an index of refraction n3. If θ3 is the angle of refraction in the glass, show that n1 sin θ1 = n3 sin θ3. That is, show that the second medium can be neglected when finding the angle of refraction in the third medium. Picture the Problem We can apply Snell’s law consecutively, first to the n1-n2 interface and then to the n2-n3 interface. Apply Snell’s law to the n1-n2 interface: 2211 sinsin θθ nn = Apply Snell’s law to the n2-n3 interface: 3322 sinsin θθ nn = Equate the two expressions for 22 sinθn to obtain: 3311 sinsin θθ nn = 36 •• On a safari, you are spear fishing while wading in a river. You observe a fish gliding by you. If your line of sight to the fish is 64.0o degrees below the horizontal in air, and assuming the spear follows a straight-line path through the air and water after it is released, determine the angle below the horizontal that you should aim your spear gun in order to catch dinner. Assume the spear gun barrel is 1.50 m above the water surface, the fish is 1.20 m below the surface, and the spear travels in a straight line all the way to the fish. Picture the Problem The following pictorial representation summarizes the information given in the problem statement. We can use the geometry of the diagram and apply Snell’s law at the air-water interface to find the aiming angle α. Properties of Light 921 θ θ 1 2 α°0.64m 50.1= h m 20.1= d lL Air Water 1n 2n Use the pictorial representation to express the aiming angle α: ⎥⎦ ⎤⎢⎣ ⎡ + += − lL dh1tanα Referring to the diagram, note that: 1tanθhL = and 2tanθd=l Substituting for L and l yields: ⎥⎦ ⎤⎢⎣ ⎡ + += − 21 1 tantan tan θθα dh dh Apply Snell’s law at the air-water interfaceto obtain: 2211 sinsin θθ nn = Solving for θ2 yields: ⎥⎦ ⎤⎢⎣ ⎡= − 1 2 11 2 sinsin θθ n n Substitute for θ2 to obtain: ⎥⎥ ⎥⎥ ⎥ ⎦ ⎤ ⎢⎢ ⎢⎢ ⎢ ⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡+ += − − 1 2 11 1 1 sinsintantan tan θθ α n ndh dh Chapter 31 922 Noting that θ1 is the complement of 64.0°, substitute numerical values and evaluate α: ( ) ( ) °= ⎥⎥ ⎥⎥ ⎦ ⎤ ⎢⎢ ⎢⎢ ⎣ ⎡ ⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ °+° += − − 9.66 0.26sin 33.1 00.1sintanm 20.10.26tanm 50.1 m 20.1m 50.1tan 1 1α That is, you should aim 66.9° below the horizontal. 37 ••• You are standing on the edge of a swimming pool and looking directly across at the opposite side. You notice that the bottom edge of the opposite side of the pool appears to be at an angle of 28° degrees below the horizontal. However, when you sit on the pool edge, the bottom edge of the opposite side of the pool appears to be at an angle of only 14o below the horizontal. Use these observations to determine the width and depth of the pool. Hint: You will need to estimate the height of your eyes above the surface of the water when standing and sitting. Picture the Problem The following diagrams represent the situations when you are standing on the edge of the pool (the diagram to the left) and when you are sitting on the edge of the pool (the diagram to the right). We can use Snell’s law and the geometry of the pool to determine the width and depth of the pool. θ θ 1 2 α h L l d Standing Water Air 2n 1n standingh θ θ 1 2 α h L l d Sitting ' ' ' ' ' Air Water 1n 2n sittingh Use the fact that angles α and θ1 are complementary, as are α′ and θ1′ to determine θ1 and θ1′: °=°−°=−°= 622890901 αθ and °=°−°=−°= 761490'90'1 αθ Express the distances L and L′ in terms of θ1 and θ1′: 1standing tanθhL = and 'tan' 1sitting θhL = Properties of Light 923 Assuming that your eyes are 1.7 m above the level of the water when you are standing and 0.7 m above the water when you are sitting, evaluate L and L′: ( ) m 197.362tanm 7.1 =°=L and ( ) m 808.276tanm 7.0' =°=L Referring to the pictorial representations, note that: h Ld h −== l2tanθ (1) and h Ld h '''tan 2 −== lθ Divide the first of these equations by the second to obtain: ''tan tan 2 2 Ld Ld − −=θ θ Solving for d yields: 'tantan 'tantan' 22 22 θθ θθ − −= LLd (2) Apply Snell’s law to the air-water interface when you are standing: ⎥⎦ ⎤⎢⎣ ⎡= − 1 2 11 2 sinsin θθ n n Substitute numerical values and evaluate 2θ : °=⎥⎦ ⎤⎢⎣ ⎡ °= − 60.4162sin 33.1 00.1sin 12θ Apply Snell’s law to the air-water interface when you are sitting: ⎥⎦ ⎤⎢⎣ ⎡= − 'sinsin' 1 2 11 2 θθ n n Substitute numerical values and evaluate '2θ : °=⎥⎦ ⎤⎢⎣ ⎡ °= − 85.4676sin 33.1 00.1sin' 12θ Substitute numerical values in equation (2) and evaluate d: ( ) ( ) widem 5.1m 130.5 85.46tan60.41tan 85.46tanm 197.360.41tanm 808.2 ==°−° °−°=d Solving equation (1) for h yields: 2tanθ Ldh −= Substitute numerical values and evaluate h: deep m 2.2 60.41tan m 197.3m 130.5 =° −=h Chapter 31 924 38 ••• Figure 31-57 shows a beam of light incident on a glass plate of thickness d and index of refraction n. (a) Find the angle of incidence so that the separation b between the ray reflected from the top surface and the ray reflected from the bottom surface and exiting the top surface is a maximum. (b) What is this angle of incidence if the index of refraction of the glass is 1.60? (c) What is the separation of the two beams if the thickness of the glass plate is 4.0 cm? Picture the Problem Let x be the perpendicular separation between the two rays and let l be the separation between the points of emergence of the two rays on the glass surface. We can use the geometry of the refracted and reflected rays to express x as a function of l, d, θr, and θi. Setting the derivative of the resulting equation equal to zero will yield the value of θi that maximizes x. Air Glass Air θ θ θ θ θ r r i i i l d x (a) Express l in terms of d and the angle of refraction θr: rtan2 θd=l Express x as a function of l, d, θr, and θi: ir costan2 θθdx = Differentiate x with respect toθi: ( ) ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ +−== i r ir 2 irir ii cossecsintan2costan2 θ θθθθθθθθθ d dd d dd d dx (1) Apply Snell’s law to the air-glass interface: r2i1 sinsin θθ nn = (2) or, because n1 = 1 and n2 = n, ri sinsin θθ n= Differentiate implicitly with respect toθI to obtain: rrii coscos θθθθ dnd = or r i i r cos cos1 θ θ θ θ nd d = Properties of Light 925 Substitute in equation (1) to obtain: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ +−= r ir r 3 i 2 r i r 2 i i r r i cos sinsin cos cos12 cos cos cos cos1sin cos sin2 θ θθ θ θ θ θ θ θθθ θ θ ndndd dx Substitute i 2sin1 θ− for i2cos θ and isin 1 θ n for rsinθ to obtain: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −−= r i 2 r 3 i 2 i cos sin cos sin1 2 θ θ θ θ θ nndd dx Multiply the second term in parentheses by r 2 r 2 coscos θθ and simplify to obtain: ( )r2i2i2 r 3 r 3 r 2 i 2 r 3 i 2 i cossinsin1 cos 2 cos cossin cos sin1 2 θθθθθ θθ θ θ θ −−=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −−= n d nn d d dx Substitute r 2sin1 θ− for r2cos θ : ( )[ ]r2i2i2 r 3 i sin1sinsin1 cos 2 θθθθθ −−−= n d d dx Substitute isin 1 θ n for rsinθ to obtain: ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛ −−−= i22i2i2 r 3 i sin11sinsin1 cos 2 θθθθθ nn d d dx Factor out 1/n2, simplify, and set equal to zero to obtain: [ ] extremafor 0sin2sin cos 2 2 i 22 i 4 r 33 i =+−= nn n d d dx θθθθ If dx/dθ1 = 0, then it must be true that: 0sin2sin 2i 22 i 4 =+− nn θθ Solve this quartic equation for θi to obtain: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −−= − 2 1 i 1 11sin n nθ Chapter 31 926 (b) Evaluate θI for n = 1.60: ( ) °= ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −−= − 5.48 60.1 1116.1sin 2 1 iθ (c) In (a) we showed that: ir costan2 θθdx = Solve equation (2) for θr: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − i 2 11 r sinsin θθ n n Substitute numerical values and evaluate θr: °=⎟⎠ ⎞⎜⎝ ⎛ °= − 9.275.48sin 60.1 1sin 1rθ Substitute numerical values and evaluate x: ( ) cm8.2 5.48cos9.27tancm0.42 = °°=x Total Internal Reflection 39 • [SSM] What is the critical angle for light traveling in water that is incident on a water–air interface? Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to the air and use Snell’s law under total internal reflection conditions. Use Snell’s law to obtain: 2211 sinsin θθ nn = When there is total internal reflection: c1 θθ = and °= 902θ Substitute to obtain: 22c1 90sinsin nnn =°=θ Solving for θc yields: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 1 21 c sin n nθ Substitute numerical values and evaluate θc: °=⎟⎠ ⎞⎜⎝ ⎛= − 8.48 33.1 00.1sin 1cθ 40 • A glass surface (n = 1.50) hasa layer of water (n = 1.33) on it. Light in the glass is incident on the glass–water interface. Find the critical angle for total internal reflection. Properties of Light 927 Picture the Problem Let the index of refraction of glass be represented by n1 and the index of refraction of water by n2 and apply Snell’s law to the glass- water interface under total internal reflection conditions. Glass Water 50.11 =n 33.12 =n θ θ 1 2 Apply Snell’s law to the glass-water interface: 2211 sinsin θθ nn = At the critical angle, θ1 = θc and θ2 = 90°: °= 90sinsin 2c1 nn θ Solve for θc: ⎥⎦ ⎤⎢⎣ ⎡ °= − 90sinsin 1 21 c n nθ Substitute numerical values and evaluate θc: °=⎥⎦ ⎤⎢⎣ ⎡ °= − 5.6290sin 50.1 33.1sin 1cθ 41 • A point source of light is located 5.0 m below the surface of a large pool of water. Find the area of the largest circle on the pool’s surface through which light coming directly from the source can emerge. Picture the Problem We can apply Snell’s law to the water-air interface to express the critical angle θc in terms of the indices of refraction of water (n1) and air (n2) and then relate the radius of the circle to the depth d of the point source and θc. 90º Air Water 33.11 =n 00.12 =n m 0.5=d r θ θ c c Express the area of the circle whose radius is r: 2rA π= Relate the radius of the circle to the depth d of the point source and the critical angle θc: ctanθdr = Apply Snell’s law to the water-air interface to obtain: 22c1 90sinsin nnn =°=θ Chapter 31 928 Solving for θc yields: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 1 21 c sin n nθ Substitute for r and θc to obtain: [ ] 2 1 21 2 c sintan tan ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎭⎬ ⎫ ⎩⎨ ⎧= = − n nd dA π θπ Substitute numerical values and evaluate A: ( ) 22 2 1 m100.1 33.1 1sintanm0.5 ×= ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎭⎬ ⎫ ⎩⎨ ⎧= −πA 42 •• Light traveling in air strikes the largest face of an isosceles-right- triangle prism at normal incidence. What is the speed of light in this prism if the prism is just barely able to produce total internal reflection? Picture the Problem We can use the definition of the index of refraction to express the speed of light in the prism in terms of the index of refraction n1 of the prism. The application of Snell’s law at the prism-air interface will allow us to relate the index of refraction of the prism to the critical angle for total internal reflection. Finally, we can use the geometry of the isosceles-right- triangle prism to conclude that θc = 45°. 45º 45º θ 00.12 =n 1n c Express the speed of light v in the prism in terms of its index of refraction n1: 1n cv = Apply Snell’s law to the prism-air interface to obtain: 190sinsin 2c1 =°= nn θ Solving for n1 yields: c 1 sin 1 θ=n Substitute for n1 and simplify to obtain: csinθcv = Properties of Light 929 Substitute numerical values and evaluate v: ( ) m/s101.2 45sinm/s10998.2 8 8 ×= °×=v 43 •• A point source of light is located at the bottom of a steel tank, and an opaque circular card of radius 6.00 cm is placed horizontally over it. A transparent fluid is gently added to the tank so that the card floats on the fluid surface with its center directly above the light source. No light is seen by an observer above the surface until the fluid is 5.00 cm deep. What is the index of refraction of the fluid? Picture the Problem The observer above the surface of the fluid will not see any light until the angle of incidence of the light at the fluid-air interface is less than or equal to the critical angle for the two media. We can use Snell’s law to express the index of refraction of the fluid in terms of the critical angle and use the geometry of card and light source to express the critical angle. θ θ c c θ 2 r d n n 1 2 Apply Snell’s law to the fluid-air interface to obtain: 2211 sinsin θθ nn = Light is seen by the observer when θ1 = θc and θ2 = 90°: 22c1 90sinsin nnn =°=θ Because the medium above the interface is air, n2 = 1. Solve for n1 to obtain: c 1 sin 1 θ=n From the geometry of the diagram: d r=ctanθ ⇒ ⎟⎠ ⎞⎜⎝ ⎛= − d r1 c tanθ Substitute for cθ to obtain: ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛= − d r n 1 1 tansin 1 Chapter 31 930 Substitute numerical values and evaluate n1: 30.1 cm00.5 cm00.6 tansin 1 1 1 = ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − n 44 •• An optical fiber allows rays of light to propagate long distances by using total internal reflection. Optical fibers are used extensively in medicine and in digital communications. As shown in Figure 31-58 the fiber consists of a core material that has an index of refraction n2 and radius b surrounded by a cladding material that has an index of refraction n3 < n2. The numerical aperture of the fiber is defined as sinθ1, where θ1 is the angle of incidence of a ray of light that impinges on the center of the end of the fiber and then reflects off the core- cladding interface just at the critical angle. Using the figure as a guide, show that the numerical aperture is given by sinθ1 = n22 − n32 assuming the ray is initially in air. Hint: Use of the Pythagorean theorem may be required. Picture the Problem We can use the geometry of the figure, the law of refraction at the air-n1 interface, and the condition for total internal reflection at the n1-n2 interface to show that the numerical aperture is given by 23 2 21sin nn −=θ . Referring to the figure, note that: c a n n == 2 3 csinθ and c b=2sinθ Apply the Pythagorean theorem to the right triangle to obtain: 222 cba =+ or 12 2 2 2 =+ c b c a Solving for c b yields: 2 2 1 c a c b −= Substitute for c a and c b to obtain: 2 2 2 3 2 1sin n n−=θ Use the law of refraction to relate θ1 and θ2: 2211 sinsin θθ nn = Substitute for sinθ2, let n1 = 1 (air), and simplify to obtain: 2 3 2 22 2 2 3 21 1sin nnn nn −=−=θ 45 •• [SSM] Find the maximum angle of incidence θ1 of a ray that would propagate through an optical fiber that has a core index of refraction of 1.492, a core radius of 50.00 μm, and a cladding index of 1.489. See Problem 44. Properties of Light 931 Picture the Problem We can use the result of Problem 44 to find the maximum angle of incidence under the given conditions. From Problem 44: 2 3 2 21sin nn −=θ Solve for θ1 to obtain: ( )232211 sin nn −= −θ Substitute numerical values and evaluate θ1: ( ) ( ) °= ⎟⎠ ⎞⎜⎝ ⎛ −= − 5 489.1492.1sin 2211θ 46 •• Calculate the difference in time needed for two pulses of light to travel down 15.0 km of the fiber that is described in Problem 44. Assume that one pulse enters the fiber at normal incidence, and the second pulse enters the fiber at the maximum angle of incidence calculated in Problem 45. In fiber optics, this effect is known as modal dispersion. Picture the Problem We can derive an expression for the difference in the travel times by expressing the travel time for a pulse that enters the fiber at the maximum angle of incidence and a pulse that enters the fiber at normal incidence. Examination of Figure 31-58 reveals that, if the length of the tube is L, the distance traveled by the pulse that entersat an angle θ1 is the ratio of c to a multiplied by L. The difference in the travel times Δt is given by: incidence normal1 ttt −=Δ θ (1) The travel time for the pulse that enters the fiber at the maximum angle of incidence is: 2 medium in the speed traveleddistance 1 n c a cL t ==θ The travel time for the pulse that enters the fiber normally is: 2 incidence normal n c Lt = Substitute for 1θt and incidence normalt in equation (1) to obtain: 22 n c L n c a cL t −=Δ Chapter 31 932 Simplify to obtain: ⎟⎠ ⎞⎜⎝ ⎛ −=Δ 12 a c c Lnt (2) Referring to the figure, note that: c a=csinθ From Snell’s law, the sine of the critical angle is also given by: 2 3 csin n n=θ ⇒ 2 3 n n c a = Substitute for c/a in equation (2) to obtain: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −=Δ 1 3 22 n n c Lnt Substitute numerical values and evaluate Δt: ( )( ) ns150 1 489.1 492.1 m/s10998.2 km15492.1 Δ 8 = ⎟⎠ ⎞⎜⎝ ⎛ −×=t 47 ••• Investigate how a thin film of water on a glass surface affects the critical angle for total reflection. Use n = 1.50 for glass and n = 1.33 for water. (a) What is the critical angle for total internal reflection at the glass–water interface? (b) Does a range of incident angles exist such that the angles are greater than θc for glass-to-air refraction and for which the light rays will leave the glass, travel through the water and then pass into the air? Picture the Problem Let the index of refraction of glass be represented by n1, the index of refraction of water by n2, and the index of refraction of air by n3. We can apply Snell’s law to the glass-water interface under total internal reflection conditions to find the critical angle for total internal reflection. The application of Snell’s law to glass-air and glass-water interfaces will allow us to decide whether there are angles of incidence greater than θc for glass-to-air refraction for which light rays will leave the glass and the water and pass into the air. 00.13 =n 33.12 =n 50.11 =n Air Water Glass θ θ θ θ1 2 2 3 Properties of Light 933 (a) Apply Snell’s law to the glass- water interface: 2211 sinsin θθ nn = At the critical angle, θ1 = θc and θ2 = 90°: °= 90sinsin 2c1 nn θ Solving for θc yields: ⎥⎦ ⎤⎢⎣ ⎡ °= − 90sinsin 1 21 c n nθ Substitute numerical values and evaluate θc: °=⎥⎦ ⎤⎢⎣ ⎡ °= − 5.6290sin 50.1 33.1sin 1cθ (b) Apply Snell’s law to a water-air interface at the critical angle for a water-air interface: °= 90sinsin 3c2 nn θ Solving for cθ yields: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 2 31 c sin n nθ Substitute numerical values and evaluate cθ : °=⎟⎠ ⎞⎜⎝ ⎛= − 8.48 33.1 00.1sin 1cθ Apply Snell’s law to a ray incident at the glass-water interface: 2211 sinsin θθ nn = and ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 2 1 21 1 sinsin θθ n n For c2 θθ = : ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛⎥⎦ ⎤⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= −− 1 31 2 3 1 21 1 sinsin n n n n n nθ Substitute numerical values and evaluate θ1: °=⎟⎠ ⎞⎜⎝ ⎛= − 8.41 50.1 00.1sin 11θ Yes, if θ ≥ 41.8°, where θ is the angle of incidence for the rays in glass that are incident on the glass-water boundary, the rays will leave the glass through the water and pass into the air. 48 ••• A laser beam is incident on a plate of glass that is 3.0-cm thick (Figure 31-57). The glass has an index of refraction of 1.5 and the angle of incidence is 40º. The top and bottom surfaces of the glass are parallel. What is the distance b Chapter 31 934 between the beam formed by reflection off the top surface of the glass and the beam reflected off the bottom surface of the glass. Picture the Problem The situation is shown in the adjacent figure. We can use the geometry of the diagram and trigonometric relationships to derive an expression for d in terms of the angles of incidence and refraction. Applying Snell’s law will yield θr. Glass Air θ θ θ θ i i r r θi x 5.1=n t b Express the distance x in terms of t and θr: rtan2 θtx = The separation of the reflected rays is: icosθxb = Substitute for x to obtain: ir costan2 θθtb = (1) Apply Snell’s law at the air-glass interface to obtain: ri sinsin θθ n= ⇒ ⎟⎠ ⎞⎜⎝ ⎛= − n i1 r sinsin θθ Substitute for rθ in equation (1) to obtain: i i1 cossinsintan2 θθ ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛= − n tb Substitute numerical values and evaluate b: ( ) cm2.2 40cos 5.1 40sinsintancm0.32 1 = °⎥⎦ ⎤⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛ °= −b Dispersion 49 • A beam of light strikes the plane surface of silicate flint glass at an angle of incidence of 45º. The index of refraction of the glass varies with wavelength (see Figure 31-59). How much smaller is the angle of refraction for violet light of wavelength 400 nm than the angle of refraction for red light of wavelength 700 nm? Picture the Problem We can apply Snell’s law of refraction to express the angles of refraction for red and violet light in silicate flint glass. Properties of Light 935 Express the difference between the angle of refraction for violet light and for red light: violetr,redr, θθθ −=Δ (1) Apply Snell’s law of refraction to the interface to obtain: rsin45sin θn=° ⇒ ⎟⎠ ⎞⎜⎝ ⎛= − n2 1sin 1rθ Substituting for rθ in equation (1) yields: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛−⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛=Δ −− violet 1 red 1 2 1sin 2 1sin nn θ Substitute numerical values and evaluate Δθ : ( ) ( ) °=°−°= ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛− ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − − 0.12.252.26 66.12 1sin 60.12 1sinΔ 1 1θ 50 •• In many transparent materials, dispersion causes different colors (wavelengths) of light to travel at different speeds. This can cause problems in fiber-optic communications systems where pulses of light must travel very long distances in glass. Assuming a fiber is made of silicate crown glass (see Figure 31-59), calculate the difference in travel times that two short pulses of light take to travel 15.0 km in this fiber if the first pulse has a wavelength of 700 nm and the second pulse has a wavelength of 500 nm. Picture the Problem The transit times will be different because the speed with which light of various wavelengths propagates in silicate crown glass is dependent on the index of refraction. We can use Figure 31-19 to estimate the indices of refraction for pulses of wavelengths 500 and 700 nm. Express the difference in time needed for two short pulses of light to travel a distance L in the fiber: 700500 v L v Lt −=Δ Substitute for L, v500, and v700 and simplify to obtain: ( )700500700500 nnc L c Ln c Lnt −=−=Δ Use Figure 31-59 to find the indices of refraction of silicate crown glass for the two wavelengths: 51.1500 ≈n and 50.1700 ≈n Chapter 31 936 Substitute numerical values and evaluate Δt: ( ) s5.0 50.151.1 m/s10998.2 km0.15 Δ 8 μ≈ −×=t Polarization 51 • [SSM] What is the polarizing angle for light in air that is incident on (a) water (n = 1.33), and (b) glass (n = 1.50)? Picture the Problem The polarizing angle is given by Brewster’s law: 12ptan nn=θ where n1 and n2 are the indices of refraction on the near and far sides of the interface, respectively.Use Brewster’s law to obtain: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 1 21 p tan n nθ (a) For n1 = 1 and n2 = 1.33: °=⎟⎠ ⎞⎜⎝ ⎛= − 1.53 00.1 33.1tan 1pθ (b) For n1 = 1 and n2 = 1.50: °=⎟⎠ ⎞⎜⎝ ⎛= − 3.56 00.1 50.1tan 1pθ 52 • Light that is horizontally polarized is incident on a polarizing sheet. It is observed that only 15 percent of the intensity of the incident light is transmitted through the sheet. What angle does the transmission axis of the sheet make with the horizontal? Picture the Problem The intensity of the transmitted light I is related to the intensity of the incident light I0 and the angle the transmission axis makes with the horizontal θ according to .cos20 θII = Express the intensity of the transmitted light in terms of the intensity of the incident light and the angle the transmission axis makes with the horizontal: θ20 cosII = ⇒ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − 0 1cos I Iθ Substitute numerical values and evaluate θ : ( ) °== − 6715.0cos 1θ Properties of Light 937 53 • Two polarizing sheets have their transmission axes crossed so that no light gets through. A third sheet is inserted between the first two so that its transmission axis makes an angle θ with the transmission axis of the first sheet. Unpolarized light of intensity I0 is incident on the first sheet. Find the intensity of the light transmitted through all three sheets if (a) θ = 45º and (b) θ = 30º. Picture the Problem Let In be the intensity after the nth polarizing sheet and use θ20 cosII = to find the intensity of the light transmitted through all three sheets for θ = 45° and θ = 30°. (a) The intensity of the light between the first and second sheets is: 02 1 1 II = The intensity of the light between the second and third sheets is: 04 12 02 1 2,1 2 12 45coscos IIII =°== θ The intensity of the light that has passed through the third sheet is: 08 12 04 1 3,2 2 23 45coscos IIII =°== θ (b) The intensity of the light between the first and second sheets is: 02 1 1 II = The intensity of the light between the second and third sheets is: 08 32 02 1 2,1 2 12 30coscos IIII =°== θ The intensity of the light that has passed through the third sheet is: 032 32 08 3 3,2 2 23 60coscos IIII =°== θ 54 • A horizontal 5.0 mW laser beam that is vertically polarized is incident on a sheet that is oriented with its transmission axis vertical. Behind the first sheet is a second sheet that is oriented so that its transmission axis makes an angle of 27º with respect to the vertical. What is the power of the beam transmitted through the second sheet? Picture the Problem Because the light is polarized in the vertical direction and the first polarizer is also vertically polarized, no loss of intensity results from the first transmission. We can use Malus’s law to find the intensity of the light after it has passed through the second polarizer. The intensity of the beam is the ratio of its power to cross-sectional area: A PI = Chapter 31 938 Express the intensity of the light between the first and second polarizers: 01 II = and 01 PP = Express the law of Malus in terms of the power of the beam: θ20 cos A P A P = ⇒ θ20 cosPP = Express the power of the beam after the second transmission: 12 2 02,1 2 12 coscos θθ PPP == Substitute numerical values and evaluate P2: ( ) mW0.427cosmW0.5 22 =°=P 55 •• [SSM] The polarizing angle for light in air that is incident on a certain substance is 60º. (a) What is the angle of refraction of light incident at this angle? (b) What is the index of refraction of this substance? Picture the Problem Assume that light is incident in air (n1 = 1.00). We can use the relationship between the polarizing angle and the angle of refraction to determine the latter and Brewster’s law to find the index of refraction of the substance. (a) At the polarizing angle, the sum of the angles of polarization and refraction is 90°: °=+ 90rp θθ ⇒ pr 90 θθ −°= Substitute for θp to obtain: °=°−°= 306090rθ (b) From Brewster’s law we have: 1 2 ptan n n=θ or, because n1 = 1.00, p2 tanθ=n Substitute for θp and evaluate n2: 7.160tan2 =°=n 56 •• Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that its transmission axis makes an angle θ with the transmission axis of the first sheet. (a) Derive an expression for the intensity of the transmitted light as a function of θ. (b) Show that the intensity transmitted through all three sheets is maximum when θ = 45º. Properties of Light 939 Picture the Problem Let In be the intensity after the nth polarizing sheet and use θ20 cosII = to find the intensity of the light transmitted through the three sheets. (a) The intensity of the light between the first and second sheets is: 02 1 1 II = The intensity of the light between the second and third sheets is: θθ 20212,1212 coscos III == Express the intensity of the light that has passed through the third sheet and simplify to obtain: ( ) ( ) θ θθ θθ θθ θ 2sin sincos2 sincos 90coscos cos 2 08 1 2 08 1 22 02 1 22 02 1 3,2 2 23 I I I I II = = = −°= = (b) Because the sine function is a maximum when its argument is 90°, the maximum value of I3 occurs when θ = 45°. 57 •• If the middle polarizing sheet in Problem 56 is rotating at an angular speed ω about an axis parallel with the light beam, find an expression for the intensity transmitted through all three sheets as a function of time. Assume that θ = 0 at time t = 0. Picture the Problem Let In be the intensity after the nth polarizing sheet, use θ20 cosII = to find the intensity of the light transmitted through each sheet, and replace θ with ωt. The intensity of the light between the first and second sheets is: 02 1 1 II = The intensity of the light between the second and third sheets is: tIII ωθ 20212,1212 coscos == Express the intensity of the light that has passed through the third sheet and simplify to obtain: ( ) ( ) tI ttI ttI ttI II ω ωω ωω ωω θ 2sin sincos2 sincos 90coscos cos 2 08 1 2 08 1 22 02 1 22 02 1 3,2 2 23 = = = −°= = Chapter 31 940 58 •• A stack of N + 1 ideal polarizing sheets is arranged so that each sheet is rotated by an angle of π/(2N) rad with respect to the preceding sheet. A linearly polarized light wave of intensity I0 is incident normally on the stack. The incident light is polarized along the transmission axis of the first sheet and is therefore perpendicular to the transmission axis of the last sheet in the stack. (a) Show that the intensity of the light transmitted through the entire stack is given by I0 cos 2N π 2N( )⎡⎣ ⎤⎦ . (b) Using a spreadsheet or graphing program, plot the transmitted intensity as a function of N for values of N from 2 to 100. (c) What is the direction of polarization of the transmitted beam in each case? Picture the Problem Let In be the intensity after the nth polarizing sheet and use θ20 cosII = to find the ratio of In+1 to In. (a) Find the ratio of In+1 to In: NI I n n 2 cos21 π=+ Because there are N such reductions of intensity: ⎟⎠ ⎞⎜⎝ ⎛== ++ NI I I I NNN 2 cos2 0 1 1 1 π and ⎟⎠ ⎞⎜⎝ ⎛=+ NII N N 2 cos201 π (b)A spreadsheet program to graph IN+1/I0 as a function of N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Content/Formula Algebraic Form A2 2 N A3 A2 + 1 N + 1 B2 (cos(PI()/(2*A2))^(2*A2) ⎟⎠ ⎞⎜⎝ ⎛ N N 2 cos2 π A B 1 N I/I0 2 2 0.250 3 3 0.422 4 4 0.531 5 5 0.605 95 95 0.974 96 96 0.975 97 97 0.975 98 98 0.975 99 99 0.975 Properties of Light 941 100 100 0.976 A graph of I/I0 as a function of N follows. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 10 20 30 40 50 60 70 80 90 100 N I/ I 0 (c) The transmitted light, if any, is polarized parallel to the transmission axis of the last sheet. (For N = 2 there is no transmitted light.) Remarks: The correct values for the intensity of the transmitted light would be significantly smaller than predicted above because of reflection losses. 59 •• [SSM] The device described in Problem 58 could serve as a polarization rotator, which changes the linear plane of polarization from one direction to another. The efficiency of such a device is measured by taking the ratio of the output intensity at the desired polarization to the input intensity. The result of Problem 58 suggests that the highest efficiency is achieved by using a large value for the number N. A small amount of intensity is lost regardless of the input polarization when using a real polarizer. For each polarizer, assume the transmitted intensity is 98 percent of the amount predicted by the law of Malus and use a spreadsheet or graphing program to determine the optimum number of sheets you should use to rotate the polarization 90º. Picture the Problem Let In be the intensity after the nth polarizing sheet and use θ20 cosII = to find the ratio of In+1 to In. Because each sheet introduces a 2% loss of intensity, the net transmission after N sheets (0.98)N. Find the ratio of In+1 to In: ( ) NI I n n 2 cos98.0 21 π=+ Chapter 31 942 Because there are N such reductions of intensity: ( ) ⎟⎠ ⎞⎜⎝ ⎛=+ NI I NNN 2 cos98.0 2 0 1 π A spreadsheet program to graph IN+1/I0 for an ideal polarizer as a function of N, the percent transmission, and IN+1/I0 for a real polarizer as a function of N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Content/Formula Algebraic Form A3 1 N B2 (cos(PI()/(2*A2))^(2*A2) ⎟⎠ ⎞⎜⎝ ⎛ N N 2 cos2 π C3 (0.98)^A3 ( )N98.0 D4 B3*C3 ( ) ⎟⎠ ⎞⎜⎝ ⎛ N NN 2 cos98.0 2 π A B C D 1 Ideal Percent Real 2 N Polarizer Transmission Polarizer 3 1 0.000 0.980 0.000 4 2 0.250 0.960 0.240 5 3 0.422 0.941 0.397 6 4 0.531 0.922 0.490 7 5 0.605 0.904 0.547 8 6 0.660 0.886 0.584 9 7 0.701 0.868 0.608 10 8 0.733 0.851 0.624 11 9 0.759 0.834 0.633 12 10 0.781 0.817 0.638 13 11 0.798 0.801 0.639 14 12 0.814 0.785 0.638 15 13 0.827 0.769 0.636 16 14 0.838 0.754 0.632 17 15 0.848 0.739 0.626 18 16 0.857 0.724 0.620 19 17 0.865 0.709 0.613 20 18 0.872 0.695 0.606 21 19 0.878 0.681 0.598 22 20 0.884 0.668 0.590 Properties of Light 943 A graph of I/I0 as a function of N for the quantities described above follows: 0.0 0.2 0.4 0.6 0.8 1.0 0 5 10 15 20 Number of sheets (N ) I /I 0 Ideal Polarizer Percent Transmission Real Polarizer Inspection of the table, as well as of the graph, tells us that the optimum number of sheets is .11 Remarks: The correct values for the intensity of the transmitted light would be significantly smaller than predicted above because of reflection losses. 60 •• Show mathematically that a linearly polarized wave can be thought of as a superposition of a right and a left circularly polarized wave. Picture the Problem A circularly polarized wave is said to be right circularly polarized if the electric and magnetic fields rotate clockwise when viewed along the direction of propagation and left circularly polarized if the fields rotate counterclockwise. For a circularly polarized wave, the x and y components of the electric field are given by: tEEx ωcos0= and tEEy ωsin0= or tEEy ωsin0−= for left and right circular polarization, respectively. For a wave polarized along the x axis: i iiEE ˆcos2 ˆcosˆcos 0 00leftright tE tEtE ω ωω = +=+ rr 61 •• Suppose that the middle sheet in Problem 53 is replaced by two polarizing sheets. If the angle between the transmission axes in the second, third, and fourth sheets in the stack make angles of 30º, 60º and 90º, respectively, with the transmission axis of the first sheet, (a) what is the intensity of the transmitted Chapter 31 944 light? (b) How does this intensity compare with the intensity obtained in Part (a) of Problem 53? Picture the Problem Let In be the intensity after the nth polarizing sheet and use θ20 cosII = to find the intensity of the light transmitted by the four sheets. (a) The intensity of the light between the first and second sheets is: 02 1 1 II = The intensity of the light between the second and third sheets is: 08 32 02 1 2,1 2 12 30coscos IIII =°== θ The intensity of the light between the third and fourth sheets is: 032 92 08 3 3,2 2 23 30coscos IIII =°== θ The intensity of the light to the right of the fourth sheet is: 0 0128 272 032 9 4,3 2 34 211.0 30coscos I IIII = =°== θ (b) The intensity with four sheets at angles of 0°, 30°, 30° and 90° is greater the intensity of three sheets at angles of 0°, 45° and 90° by a factor of 1.69. Remarks: We could also apply the result obtained in Problem 58(a) to solve this problem. 62 •• Show that the electric field of a circularly polarized wave propagating parallel with the x axis can be expressed by ( ) ( )kjE ˆcosˆsin 00 tkxEtkxE ωω +++=r . Picture the Problem A circularly polarized wave may be resolved into two linearly polarized waves, of equal amplitude, 90 degrees apart and with their planes of polarization at right angles to each other. The components of the electric field E r of a circularly polarized wave propagating parallel to the x axis are given by tEEx ωsin0= and tEEy ωcos0= . The electric field E r is the vector sum of its components: kjE ˆˆ yx EE += r For an electric field propagating parallel with the x axis and in the −x direction: ( )tkxEEx ω+= sin0 and ( )tkxEEy ω+= cos0 Properties of Light 945 Substituting for Ex and Ey yields: ( ) ( )kjE ˆcosˆsin 00 tkxEtkxE ωω +++=r 63 •• [SSM] A circularly polarized wave is said to be right circularly polarized if the electric and magnetic fields rotate clockwise when viewed along the direction of propagation and left circularly polarized if the fields rotate counterclockwise. (a) What is the sense of the circular polarization for the wave described by the expression in Problem 62? (b) What would be the expression for the electric field of a circularly polarized wave traveling in the same direction as the wave in Problem 60, but with the fields rotating in the opposite direction? Picture the Problem We can apply the given definitions of right and left circular polarization to the electric field and magnetic fields of the wave. (a) The electric field of the wave in Problem 62 is: ( ) ( )kjE ˆcosˆsin 00 tkxEtkxE ωω +++=r The corresponding magnetic field is: ( ) ( ) jkB ˆcosˆsin 00 tkxBtkxB ωω +−+=r Because these fields rotate clockwise when viewed along the direction
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