5/6/2014Week 13: EM Waves and Propagation of Light Week 13: EM Waves and Propagation of Light Due: 6:00am on Monday, April 21, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Triangle Electromagnetic Wave Learning Goal: To show how a propagating triangle electromagnetic wave can satisfy Maxwell's equations if the wave travels at speed c. Light, radiant heat (infrared radiation), X rays, and radio waves are all examples of traveling electromagnetic waves. Electromagnetic waves consist of mutually compatible combinations of electric and magnetic fields ("mutually compatible" in the sense that changes in the electric field generate the magnetic field, and vice versa). The simplest form for a traveling electromagnetic wave is a plane wave. One particularly simple form for a plane wave is known as a "triangle wave," in which the electric and magnetic fields are linear in position and time (rather than sinusoidal). In this problem we will investigate a triangle wave traveling in the x direction whose electric field is in the y direction. This wave is linearly polarized along the y axis; in other words, the electric field is always directed along the y axis. Its electric and magnetic fields are given by the following expressions: E y (x, t) = E0 (x − vt)/a and Bz (x, t) = B0 (x − vt)/a, where E 0 , B0 , and a are constants. The constant a, which has dimensions of length, is introduced so that the constants E 0 and B0 have dimensions of electric and magnetic field respectively. This wave is pictured in the figure at time t = 0. Note that we have only drawn the field vectors along the x axis. In fact, this idealized wave fills all space, but the field vectors only vary in the x direction. We expect this wave to satisfy Maxwell's equations. For it to do so, we will find that the following must be true: 1. The amplitude of the electric field must be directly proportional to the amplitude of the magnetic field. 2. The wave must travel at a particular velocity (namely, the speed of light). Part A → What is the propagation velocity v of the electromagnetic wave whose electric and magnetic fields are given by the expressions in the introduction? Express → v ^ ^ in terms of v and the unit vectors ^ i , j , and k. The answer will not involve c; we have not yet shown that this wave travels at the speed of light. Hint 1. Phase velocity Typesetting All points math: along 98% the wave will propagate with the same velocity. You may find it easiest to concentrate on the point where Ey (x, t) = Bz (x, t) = 0. At t = 0, this point occurs at x = 0. Where is this point when t is http://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 1/45 5/6/2014 Week 13: EM Waves and Propagation of Light some later time tf ? Express the location xf , where the field amplitude is zero at time tf , in terms of tf and any necessary quantities from the problem introduction. ANSWER: xf = vtf ANSWER: → v = ^ vi Correct In the next few parts, we will use Faraday's law of induction to find a relationship between E 0 and B0 . Faraday's law relates the line integral of the electric field around a closed loop to the rate of change in magnetic flux through this loop: ∮ C → → → E ( r , t) ⋅ d l = − d dt ∫ S → → → B ( r , t) ⋅ d A . Part B To use Faraday's law for this problem, you will need to constuct a suitable loop, around which you will integrate the electric field. In which plane should the loop lie to get a nonzero electric field line integral and a nonzero magnetic flux? ANSWER: the xy plane the yz plane the zx plane http://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 2/45 5/6/2014 Week 13: EM Waves and Propagation of Light Correct Part C Consider the loop C1 shown in the figure. It is a square loop with sides of length L, with one corner at the origin and the opposite corner at the coordinates x = L , y = L . Recall that → → ^ E ( r , t) = [E0 (x − vt)/a]j . What is the value of the line integral of the electric field around loop C1 at arbitrary time t? Express the line integral in terms of E 0 , L , a, v, and/or t. Hint 1. Integrating along segments 1 and 2 Note that the loop C1 is divided into four segments, labeled 1 through 4. First, consider the line integral → along segment 1. Along this segment, → E since Ex = 0 → ⋅d l d l ^ = dx i . When we dot this into the electric field vector, we find ^ ^ ^ ^ = (Ex i + E y j + Ez k ) ⋅ (dx i ) = E x dx = 0 , . Only the y component of the electric field is nonzero. Therefore, the line integral along segment 1 is equal to zero. Now consider the line integral along segment 2. In this case, → d l ^ = dy j , and → E → ⋅d l = Ey dy , which leads to a nonvanishing integral. Find the value of the line integral along segment 2 at arbitrary time t. Express line integral along segment 2 in terms of E0 , L , v, a, and t. Hint 1. Evaluating the integral Note that ∫ L 0 x = L along segment 2, so that the integral reduces to E y (x = L, t) dy = ∫ L 0 [E0 (L − vt)/a] dy = [E 0 (L − vt)/a] ∫ L 0 dy . ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 3/45 5/6/2014 Week 13: EM Waves and Propagation of Light ∫ L 0 Ey (x = L, t) dy = E 0 (L−vt)L a Hint 2. Integrating along segments 3 and 4 → → → Along segment 3, d l = −dx ^ i , and we find that E ⋅ d l = 0 , just as it did along segment 1. The line integral along segment 4, however, will be nonzero at general time t. Along segment 4, → d l ^ ^ = dy (−j ) = −dy j , and → → E ⋅d l = −Ey dy . (If we include the minus sign in d remember to integrate dy from 0 to L, rather than from L → l , we must to 0.) Find the value of the line integral along segment 4 at arbitrary time t. Express the line integral along segment 4 in terms of E0 , L , v, a, and t. Hint 1. Evaluating the integral Note that ∫ L 0 x = 0 along segment 4, so that −E y (x = 0, t) dy = − ∫ L 0 [E0 (0 − vt)/a] dy = [E 0 vt/a] ∫ L 0 dy . ANSWER: ∫ L 0 − Ey (x = 0, t) dy = E 0 vtL a Hint 3. Integrating around the entire loop The value of the line integral around the loop C1 will be equal to the sum of the integrals along each of the segments. You should have found that the only nonzero contributions to the total line integral come from the integrations along segments 2 and 4. When you add these contributions together, you should find that the time t cancels out of the final result. The value of the line integral of the electric field along loop C1 is therefore independent of time. ANSWER: ∮ C1 → → → E ( r , t) ⋅ d l = E0 L 2 a Correct Part D Recall that → → ^ B ( r , t) = [B0 (x − vt)/a]k . Find the value of the magnetic flux through the surface S 1 in the xy plane that is bounded by the loop C1 , at arbitrary time t. B0 L http://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 4/45 you find that → → dA ^ = dx dy k → → B ( r . When you take the . The surface S1 is in the xy plane and is bounded by the curve C1 . t) ⋅ d l .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 ( . the unit vector that is normal (i. and in Part D 5/45 . Evaluating the integral Using the result from the previous hint. v. Because the curve C1 is oriented in a counterclockwise direction. Simplifying the integrand The quantity appearing in the integrand is → → → B ( r . you found an expression for the left-hand side of this equation.masteringphysics. dot product of → B → and d A . t) ⋅ d A . t) dx dy . ∮ C1 → → http://session. perpendicular) to the surface S1 is ^ k. t) dx dy = B0 ∫ L 0 [(x − vt)/a] dx ∫ L 0 dy . Hint 1. t) ⋅ d A = Bz (x. . L Hint 1. t) ⋅ d A = ∫ S1 What is the value of the integral ∫ L 0 Bz (x. we find that the integral reduces to ∫ S1 → → → B ( r . ANSWER: ∫ L 0 (x − vt) dx L = 2 2 − vtL ANSWER: ∫ S1 → → → B ( r . Therefore. Using Faraday's law By using Faraday's law. v.. and t. t) ⋅ d A . and/or t. t) ⋅ d → → → E ( r .e.. (x − vt) dx? Express your answer in terms of L. t) ⋅ d A = B 0L 2 ( 1 2 L−vt) a Correct Part E Now use Faraday's law to establish a relationship between E 0 and B0 . Express E 0 in terms of B0 and other quantities given in the introduction. Hint 2. we know that ∮ C1 → → → E ( r . t) ⋅ d l = − d dt ∫ S1 → → → B ( r . In Part C. a.5/6/2014 Week 13: EM Waves and Propagation of Light Express the magnetic flux in terms of B0 . → you found an expression for ∫S Use the result of Part D to find an expression for the 1 right-hand side of Faraday's law. ANSWER: − d dt ∫ S1 → → → B ( r . In this problem. In which plane should the loop lie to get a nonzero magnetic field line integral and hence nonzero electric flux? ANSWER: the xy plane the yz plane the zx plane Correct http://session. The Ampère-Maxwell law relates the line integral of the magnetic field around a closed loop to the rate of change in electric flux through this loop: ∮ C → → → B ( r . We assume that the electromagnetic wave is propagating through a vacuum. In these last few parts (again. but this time you will integrate the magnetic field around the loop. v.) Part F To use the Ampère-Maxwell law you will once again need to construct a suitable loop. t) ⋅ d A . t) ⋅ d A in terms of B0 . (For I to be nonzero.5/6/2014 Week 13: EM Waves and Propagation of Light → → B ( r .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 6/45 .masteringphysics. t) ⋅ d A . a. and/or t. L . In this problem. the current I is zero. t) ⋅ d A 2 = B0 L v a ANSWER: E0 = B0 v Correct If the electric and magnetic fields given in the introduction are to be self-consistent. we would need charged particles moving around. t) ⋅ d l = μ I + μ ϵ0 0 0 d dt ∫ S → → → E ( r . they must obey all of Maxwell's equations. there are no charged particles present. including the Ampère-Maxwell law. Express − d dt ∫ S1 → → → B ( r . most of which are hidden) we will use the Ampère-Maxwell law to show that self-consistency requires the electromagnetic wave described in the introduction to propagate at the speed of light. → → → B ( r . If you have finished Part C. t) ⋅ d A . with only minor modifications. ANSWER: http://session. t) ⋅ d A . ϵ0 . and/or t. v L .masteringphysics. t) ⋅ d l in Part C. How to approach the problem By using the Ampère-Maxwell law. . and find an expression for Combine these results to find a relationship between E0 and B0 . Express the line integral in terms of B0 . Knowing that → → ^ B ( r . we know that ∮ C2 → → → B ( r . a. t) ⋅ d l you can take a very similar approach to that used to find 2 → → → E ( r . t) = [B0 (x − vt)/a]k . You can follow the same steps for this part. Express E 0 in terms of B0 . ∫ S2 → → → E ( r . Hint 1. t) ⋅ d l . μ 0 . z = L . find the value of the line integral of the magnetic field around loop C2 at arbitrary time t. d dt ∮ ∫ C2 S2 → → → E ( r .5/6/2014 Week 13: EM Waves and Propagation of Light Part G Use the Ampère-Maxwell law to find a new relationship between E 0 and B0 . and other quantities given in the introduction. you can click on "Show complete part" next to Part C's answer to review the steps that led to an expression for the line integral of the electric field. Hint 1. a square loop with sides of length L with one corner at the origin and the opposite corner at the coordinates x = L . Using the methods of previous parts To find ∮ C ∮ C1 → → → B ( r . t) ⋅ d l = μ0 ϵ0 Find an expression for the left-hand side of this equation.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 7/45 . Hint 2. Find an expression for the left-hand side of the equation Consider the new loop C2 . One thing to notice is the direction of the unit vector normal to the surface S 2 . . v.the surface in the xz plane bounded by the loop C2 . μ0 . you can click on "Show complete part" next to its answer to review the steps that led to the final result. t) ⋅ d l = B0 L 2 a Hint 3.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 8/45 . Express the electric flux in terms of E0 . your thumb then points in the direction of the unit vector that is normal to the surface S 2 ) you should find that the unit vector normal to S 2 is ^ −j . a. find an expression for the electric flux through S2 .5/6/2014 Week 13: EM Waves and Propagation of Light ∮ C2 → → → B ( r . t) ⋅ d A . and/or t. it is nearly identical 1 to what you need to do to find the electric flux through the new loop C2 . a. Use the Ampère-Maxwell law Use the result of the previous subpart to find an expression for the right-hand side of the Ampère-Maxwell law. http://session. ϵ0 . t) = E 0 [(x − vt)/a]^ j . L . at arbitrary time t. Find an expression for the right-hand side of the equation Knowing that the electric field is given by → → E ( r . t) ⋅ d A in terms of E0 . This will affect the sign of your final result. Hint 1. ANSWER: ∫ S2 → → → E ( r . Using the methods of previous parts Try looking over the method you used in Part D to find ∫S → → → B ( r . v L . Express μ0 ϵ0 d dt ∫ S2 → → → E ( r .masteringphysics. If you have finished Part D. Using the righthand rule (with your fingers curled in the direction of the loop C2 . and/or t. t) ⋅ d A 2 = −E 0 L ( L 2 −vt) a Hint 4. Using one of Maxwell's equations. Using another of Maxwell's equations.5/6/2014 Week 13: EM Waves and Propagation of Light ANSWER: d μ0 ϵ0 dt ∫ S2 → → → E ( r . ANSWER: v = 1 ϵ 0μ √ 0 http://session. You derived this in Part E. You derived this in Part I. Faraday's law. t) ⋅ d A 2 = μ0 ϵ 0 E0 L v a ANSWER: E0 B0 = ϵ 0μ v 0 Correct Part H Finally we are ready to show that the electric and magnetic fields given in the introduction describe an electromagnetic wave propagating at the speed of light. we found what appears to be a different relationship between E 0 and B0 . we found a certain relationship between E 0 and B0 . the AmpèreMaxwell law. what does this imply that the speed of propagation v must be? Express v in terms of only μ0 and ϵ0 . If the electric and magnetic fields are to be self-consistent.masteringphysics. If the results of Parts E and I are to agree.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 9/45 . they must obey all of Maxwell's equations. 5/6/2014 Week 13: EM Waves and Propagation of Light Correct You have just worked through the details of one of the great triumphs of physics: Maxwell's equations predict a form of traveling wave consisting of a matched pair of electric and magnetic fields moving at a very high velocity − − − − v = 1/√μ 0 ϵ0 . and these experimentally determined values lead to a speed of v 8 = 2. After thousands of years of speculation about the nature of light.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 10/45 .masteringphysics. Maxwell had developed a plausible and quantitatively testable theory about it.10V/m . as demonstrated by the Faraday effect. 2. with two possible polarizations (which agreed with an already known characteristic of light). Faraday had a hunch that light and magnetism were related. This theory showed that lower frequency waves could be created and detected by their interactions with currents in wires (later called antennas) and paved the way to the creation and detection of radio waves. what is the magnitude of the magnetic field of the wave at this same point in space and instant in time? ANSWER: B = 1. It was transverse. It had an extraordinarily high velocity (relative to waves in air or on strings) that agreed with the experimentally determined value for the speed of light. Any doubt that light waves were in fact electromagnetic waves vanished as various optical phenomena (such as the behavior of electromagnetic waves at glass surfaces) were predicted and found to agree with the behavior of light. will rotate the plane of polarization of light that passes through it. the speed of light c. We can measure μ 0 and ϵ0 independently in the laboratory.03×10−8 T Correct Part B What is the direction of the magnetic field? ANSWER: http://session.3 A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction. put in a large magnetic field. Part A If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 3.998 × 10 m/s . Exercise 32.) Now Maxwell had predicted an electromagnetic wave with the following properties: 1. (Glass. . B0 .5/6/2014 Week 13: EM Waves and Propagation of Light .direction . use these variables ( E .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 B (t) 11/45 . You will also need the permittivity of x Note: To indicate the square of a trigonometric function in your answer.z . and ω) in your answers. we explore the properties of a plane electromagnetic wave traveling at the speed of light axis through vacuum. Energy density in an electric field Recall that the energy density uE in an electric field E is given by uE = ϵ0 E 2 2 . and also makes a useful connection between the energy density of a plane electromagnetic wave and the Poynting vector. use the notation sin(x)^2NOT sin^2(x). Its electric and magnetic field vectors are as follows: → E c along the x ^ = E0 sin (kx − ωt)j .masteringphysics. free space ϵ0 and the permeability of free space μ0 .direction + y . respectively. Part A What is the instantaneous energy density uE (t) in the electric field of the wave? Give your answer in terms of some or all of the variables in E = E 0 sin (kx − ωt) . k.direction Correct Energy in Electromagnetic Waves Electromagnetic waves transport energy. B.y . Hint 1. . This problem shows you which parts of the energy are stored in the electric and magnetic fields. E 0 . ANSWER: 2 uE (t) = ϵ 0 (E 0 sin(kx−ωt)) 2 Correct Part B http://session. → ^ B = B0 sin (kx − ωt)k Throughout. In this problem. 5/6/2014 Week 13: EM Waves and Propagation of Light What is the instantaneous energy density uB (t) in the magnetic field of the wave? Give your answer in terms of some or all of the variables in B = B0 sin (kx − ωt) .masteringphysics. ANSWER: ⟨u ⟩ = 2 E0 ϵ 0 E 4 Correct Part D What is the average energy density ⟨uB ⟩ in the magnetic field of the wave? Give your answer in terms of μ0 and 2 Hint 1. Average value of sin2 (α) The average of sin2 (α) over a full period is equal to 1/2. (α) http://session. Hint 1. ANSWER: 2 uB (t) = (B sin(kx−ωt)) 0 2μ 0 Correct Part C What is the average energy density ⟨uE ⟩ in the electric field of the wave? Give your answer in terms of ϵ0 and E0 . Hint 1. Energy density in a magnetic field Recall that the energy density uB in a magnetic field B is given by uB = B 2 2μ0 . Average value of sin 2 B0 .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 (α) 12/45 . Relationship among ⟨u⟩.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 ⟨ ⟩ 13/45 . for electromagnetic waves in vacuum In vacuum c 2 = 1 μ0 ϵ 0 . Relationship among c. It is defined by the relation → S = → → E ×B μ . Hint 2. the average energy density in the whole wave. ⟨uE ⟩ .5/6/2014 Week 13: EM Waves and Propagation of Light The average of sin2 (α) over a full period is equal to 1/2. Relationship between E and B for electromagnetic waves in vacuum In vacuum E0 = cB0 Hint 3. ANSWER: B0 ⟨u ⟩ = B 2 μ0 4 Correct Part E From the previous results. ϵ0 and μ 0 . and ⟨uB ⟩ The average energy density is simply the sum of the contributions from the electric and magnetic fields: ⟨u⟩ = ⟨uE ⟩ + ⟨uB ⟩ . derive an expression for ⟨u⟩. Express the average energy density in terms of E 0 and only. ϵ0 Hint 1. 0 → http://session. ANSWER: ⟨u⟩ = ϵ 0E 0 2 2 Correct Part F The Poynting vector → S gives the energy flux per unit area of electromagnetic waves.masteringphysics. Relationship among c.5/6/2014 Week 13: EM Waves and Propagation of Light → Calculate the time-averaged Poynting vector ⟨ S ⟩ of the wave considered in this problem. Thus. Do not use E and B for electromagnetic waves in vacuum Give your answer in terms of E 0 . or Poynting vector → S . We say that "electrical energy is dissipated" by the resistor. and resistance R with a steady current http://session. Hint 1. which is a logical result. where did it come from? Did it come from the voltage source through the wires? This problem will show you an alternative way to think about the flow of energy and will introduce a picture in which the energy flows in many unexpected places--but not through the wires! We will calculate the Poynting flux. ANSWER: → ⟨S ⟩ = ϵ 0E 0 2 2 c ^ i Correct If you compare this expression for the time-averaged Poynting flux to the one obtained for the overall energy density. Consider a cylindrical resistor of radius the cylinder. that is. you find the simple relation ∣ →∣ ∣S ∣ = c u ∣ ∣ ⟨ ⟩. converted into heat. j . the energy density of the electromagnetic field times the speed at which it moves gives the energy flux. Relationship between μ0 or B0 . where μ0 is the permeability of free space.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 i flowing along the axis of 14/45 . ϵ0 and μ 0 . and/or k. In vacuum E0 = cB0 Hint 2. r0 . length l. But if energy is dissipated. the flow of electromagnetic energy. Poynting Flux and Power Dissipation in a Resistor When a steady current flows through a resistor.masteringphysics. c and ϵ0 ^ ^ and unit vectors ^ i . The Poynting flux. for electromagnetic waves in vacuum In vacuum c 2 = 1 μ0 ϵ 0 . the resistor heats up. has units of energy per unit area per unit time and is related to the electric field vector and the magnetic field vector → B → E by the equation → μ0 S → → = E × B . across the surface of the resistor. The magnetic field vector is everywhere tangential to circles centered on the axis of the cylinder. Hint 1. The magnetic field vector points along the axis of the cylinder in the direction of the current. and other given variables. r0 ∣ . you need Ampère's law. where r is the distance from the ∣ axis of the cylinder. in terms of i . You will also need π and μ0 .masteringphysics.5/6/2014 Week 13: EM Waves and Propagation of Light Part A Which of the following is the most accurate qualitative description of the the magnetic field vector cylindrical resistor? → B inside the ANSWER: The magnetic field vector points radially away from the axis of the cylinder. Correct Part B ∣ → ∣ Find the magnitude of the magnetic field ∣ B (r)∣ inside the cylindrical resistor.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 15/45 . l. r. Hint 2. Ampère's law To calculate the magnetic field. The magnetic field vector points inward toward the axis of the cylinder. which relates the integral of the magnetic field vector around a loop to the flow of current through it: → → ∮ B ⋅ dl = μ i encl 0 . How to set up the integral http://session. Ignore fringing effects at the ends of the cylinder. Hint 3. However. Amount of current through a loop The amount of current that passes through a loop of radius r centered on the axis is simply total current×area of the loop total cross section of the resistor . the better way to proceed is to exploit the symmetry of the situation. The electric field vector is zero inside the resistor and on its surface.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 i through it. The electric field vector is confined to the surface of the resistor and points in the ^ k direction. The electric field vector is everywhere tangential to circles centered on the axis of the resistor that lie in the plane perpendicular to the current direction. ANSWER: ∣→ ∣ ∣ B (r)∣ ∣ ∣ = μ i 0 2 2πr0 r Correct Part C What can you say about the electric field vector → E inside the resistor? ANSWER: The electric field vector points along the axis of the resistor in the direction of the current. i . The electric field vector points radially outward--away from the axis of the cylinder. choosing a loop in a plane perpendicular to the axis of the cylinder. and it might be interesting for you to take different loops inside the resistor. The magnetic field always reflects the symmetry of the problem. Hint 1.masteringphysics. Correct Part D ∣ →∣ What is the magnitude of the electric field vector ∣ E ∣? ∣ ∣ Give the magnitude of the electric field vector in terms of R. making it circular. and centering it on the axis of the resistor.5/6/2014 Week 13: EM Waves and Propagation of Light What kind of loop do you need? You could take any loop. 16/45 . Use Ohm's law Use Ohm's law to express the potential drop V across the resistor in terms of the current R i http://session. and other parameters of the problem. ^ ^ ^ ×θ = k r .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 17/45 . ^ ^ ^ k × θ = −r . just like the electric field inside a parallel plate capacitor.5/6/2014 Week 13: EM Waves and Propagation of Light Give you answer in term of resistance R and current i ANSWER: = V Ri Hint 2. ANSWER: ^ i The Poynting vector is zero inside the resistor including its surface.masteringphysics. Or. Relationship between → E and V Inside the resistor. therefore the field is defined by the relationship ∣ →∣ ∣E ∣ = ∣ ∣ V l . the electric field is uniform. rearranging this. the magnitude of the Poynting vector at the surface of the resistor (not at the circular ends of the ∣ ∣ http://session. ANSWER: → ∣ ∣ ∣E ∣ ∣ ∣ = iR l Correct Part E → In what direction does the Poynting vector S point? Hint 1. ^ −r ^ θ Correct Part F ∣ → ∣ Calculate ∣ S ∣ . Cross products in cylindrical coordinates In cylindrical coordinates. similar to the way wind propels a sailboat. thus maximizing the momentum transfer from the incident radiation. rather than absorbs http://session. i . the force of the radiation pressure would have to be be greater than the gravitational attraction from the star emitting the photons. Part A Consider a perfectly reflecting mirror oriented so that solar radiation of intensity I is incident upon.masteringphysics. If such a spacecraft were to be simply pushed away from a star by the incident photons. Solar Sail A solar sail allows a spacecraft to use radiation pressure for propulsion. 0 where → E and → B are the electric field and magnetic fields. respectively. and the speed of light c. Thus. Hint 1. and other parameters of the problem. we recover the familiar expression for the power P dissipated in a resistor R through which a current i flows: P = Ri 2 . what is Frad .5/6/2014 Week 13: EM Waves and Propagation of Light cylinder). The sails of such spacecraft are made out of enormous reflecting panels. The critical parameter is the area density (mass per unit area) of the sail. Hint 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 18/45 . How to approach the problem Radiation pressure arises from the photon momentum transfer as the photons strike the mirror. you can determine the average force exerted on the mirror. and perpendicular to. The area of the panels is maximized to catch the largest number of incident photons. the mirror's surface area A . To answer this you need to take r = r0 . If the mirror has surface area A. Notice that when writing an expression for the momentum transfer you'll need to take into account the fact that the mirror reflects the photons. Definition of the Poynting vector The Poynting vector → S is defined as → S = 1 μ → → E × B . ANSWER: ∣ →∣ ∣S ∣ ∣ ∣ = i 2 2πr0 R l Correct Multiplying this value of the Poynting flux by the surface area of the resistor (which in this case is equivalent to integrating the Poynting vector over the surface of the resistor). the reflective surface of the mirror. if you find an expression for the total momentum transferred to the mirror by the photons that strike it. Give your answer in terms of R. the magnitude of the average force due to the radiation pressure of the sunlight on the mirror? Express your answer in terms of the intensity I . Reflection vs. Find the total momentum transfer What is the total momentum Δp transferred to the mirror by the photons in a time interval Express your answer in terms of the time interval and the speed of light c. absorption When an object absorbs a photon of energy U . A Hint 1. ANSWER: Δp = 2I AΔt c Hint 3. This ratio also holds for the total momentum and energy of the photons striking the mirror. the total momentum transfer for photon reflection is twice as much as in the case of photon absorption. where c is the speed of light in vacuum. the mirror's surface area . Hint 3.masteringphysics. Energy of the photons and their momentum The momentum p of a photon can be expressed in terms of the photon energy p= U c U as . ANSWER: Frad = 2AI c http://session. When an object reflects a photon of energy U .5/6/2014 Week 13: EM Waves and Propagation of Light them. Hint 2. the intensity I . Then the magnitude of the average force exerted on the mirror is F rad = Δp Δt .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 19/45 . the object must not only stop the photon (as is the case when the photon is absorbed) but also send it back in the opposite direction. Δt Δt ? . where I is the intensity of the radiation and A is the surface area of the mirror. Thus. Radiation intensity and energy The total energy of the photons striking the mirror during a time interval U = I AΔt Δt is given by . Hint 2. Force and change in momentum Let Δp interval be the total momentum transferred to the mirror by the photons that strike the mirror during a time Δt . it receives momentum equal to U /c . you should be able to solve for the area density of the mirror.masteringphysics. Express your answer symbolically in terms of the gravitational constant G . so the critical value of the area density turns out to be independent of R. the mass of the sun. the magnitude of the gravitational force due to the sun that acts on the mirror. in units of kilograms per meter squared. Part B Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. . the mass of the mirror. ANSWER: Fgrav = GM M sun R 2 Hint 2.60×10−3 2 kg/m http://session. R. M sun = 2. R 25 I sun (R) = and the gravitational constant −11 G = 6. Hint 1. given in the problem. Find the force due to gravity Suppose the mirror has mass M . What is the critical value of area density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun? Express your answer numerically.0 × 10 30 kg . Find a general expression for Fgrav . ANSWER: mass/area = 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 20/45 . Note that the expression for 2 the intensity. . and the mirror's distance from the sun. Solving for area density By equating the force due to the sun's radiation (Frad found in Part A) and the force due to the sun's gravitational pull. M . the intensity of sunlight as a function of the distance. M sun .67 × 10 3 3.5/6/2014 Week 13: EM Waves and Propagation of Light Correct To solve the second part of this problem you will need to know the following: the mass of the sun.2×10 R 2 m /(kg ⋅ s ) 2 W . has a factor of 1/R . just like the expression for the gravitational force. to two significant figures. from the sun. damage to the retina can occur. Given a representative thickness of the sail of 1 μm.45mm in diameter can have so that it can be considered safe to view head-on? ANSWER: Pmax = 0. roughly one-fifth that of iron. strength.60 3 g/cm . area density. These have a density of 1. Part A What is the largest average power (in mW) that a laser beam 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 21/45 .13×10−7 T Correct http://session. and reflectivity are the principal concerns.5/6/2014 Week 13: EM Waves and Propagation of Light Correct In selecting the material for a solar sail. Exercise 32.165 mW Correct Part B What is the maximum value of the electric field for the beam in part A? ANSWER: Emax = 274 V/m Correct Part C What is the maximum value of the magnetic field for the beam in part A? ANSWER: Bmax = 9.27 If the eye receives an average intensity greater than 100W/m2 .masteringphysics. This quantity is called the damage threshold of the retina. one of the few currently existing materials with a sufficiently low density and high strength can be made from carbon fibers. 97 × 10 7 .44 The electric field of a sinusoidal electromagnetic wave obeys the equation 15 E = −(375V/m) sin[(5.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 22/45 .99 × 10 rad/m)x] Part A What is the amplitude of the electric field of this wave? ANSWER: E = 375 V/m Correct Part B What is the amplitude of the magnetic field of this wave? ANSWER: B = 1.masteringphysics.165 U mJ Correct Part E Express the damage threshold in W/cm2 .25 μT Correct http://session. rad/s)t + (1.00×10−2 2 W/cm Correct Problem 32. ANSWER: I = 1.5/6/2014 Week 13: EM Waves and Propagation of Light Part D How much energy would the beam in part A deliver per second to the retina? ANSWER: = 0. this wavelength is too short to be visible.5/6/2014 Week 13: EM Waves and Propagation of Light Part C What is the frequency of the wave? ANSWER: f = 9. Correct Part G What is the speed of the wave? http://session. this light is visible to humans.50×1014 Hz Correct Part D What is the wavelength of the wave? ANSWER: λ = 316 nm Correct Part E What is the period of the wave? ANSWER: T = 1.masteringphysics. Yes.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 23/45 .05×10−15 s Correct Part F Is this light visible to humans? ANSWER: No. and n . the speed of the light in the glass? Express your answer in terms of f . http://session. How to approach the problem You'll need to find the speed of light in the glass in terms of n and f .00×108 m/s Correct Wavelength. Hint 1. the wavelength of the light in the glass? Express your answer in terms of n and L . the constant c should not appear in any of your answers. In this problem. and Speed of Light in Different Media A beam of light from a monochromatic laser shines into a piece of glass. The wavelength of the laser light in vacuum is L/10 and its frequency is f . Find the wavelength in glass What is λglass . The speed of light The speed of light in a given medium is the product of the frequency and wavelength in that medium. Find the frequency in glass What is f glass . the frequency of the light in the glass? Express your answer in terms of L. Frequency. Hint 1. Find the speed of light in the glass What is v glass .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 24/45 .masteringphysics. The glass has a thickness L and an index of refraction n . ANSWER: f glass = f Hint 2. Hint 2. Hint 3.5/6/2014 Week 13: EM Waves and Propagation of Light ANSWER: v = 3. L . and n . f . Part A How long does it take for a short pulse of the light to travel from one end of the glass to the other? Express the time t in terms of f and n . Then use the fact that distance is the product of velocity and time. the speed of the light wave changes (since it is the product of the frequency and wavelength).com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 25/45 . you are seeing light reflected from the object that you are looking at. ANSWER: http://session. In this problem. As a result. but the frequency remains the same. Part A Consider a spotlight shining onto a horizontal mirror . This is reflection of a different sort: diffuse reflection. If the light from the spotlight strikes the mirror at an angle θ a to the normal. the wavelength changes.masteringphysics. However. what angle θ r to the normal would you expect for the reflected rays? Express your answer in terms of θ a . you will see how diffuse reflection actually arises from the same law of reflection that you are accustomed to for reflections from mirrors. Diffuse Reflection The law of reflection is quite useful for mirrors and other flat.5/6/2014 Week 13: EM Waves and Propagation of Light ANSWER: λglass = L 10n ANSWER: v glass = fL 10n ANSWER: t = 10n f Correct When light travels from air into another medium. (This sort of reflection is called specular reflection). you've likely been told that when you look at something. shiny surfaces. masteringphysics. Consider the same spotlight but now reflecting from the surface of a table . Express the angle between the reflected ray and the vertical in terms of α and θa . Throughout this problem. then add the incoming ray.5/6/2014 Week 13: EM Waves and Propagation of Light θr = θa Correct This simple rule of reflection no longer seems to hold for diffuse reflection. The two blue dashed lines represent horizontal and vertical. inclined downward from the horizontal by an angle α . it's best to redraw the figure with only the minimum information you need. Find the angle between the normal line and vertical Find the angle between the normal line and vertical. and finally add in the reflected ray. The angle between the incoming ray and the vertical is θ a . Once you have found the angle between the normal and the vertical. Add new information as it becomes necessary. The red line represents the surface and the red dotted line indicates the normal to this surface (the normal line). and vertical. You can find the angle by considering triangles and what the sum of http://session. you can determine the correct sign rules to generalize your results later. Hint 1. If it didn't. being sure to label all right angles. horizontal. then you'd only be able to see tables when you were at a specific angle to the lights above you! To understand why the light reflects in all directions. It will be helpful to redraw the figure without the incoming ray. Unlike the light reflected from the mirror. Consider a flat surface. normal line.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 26/45 . you must first look at a slightly simpler problem.) Part B Find the angle θ r between the reflected ray and the vertical. How to approach the problem For this problem. assume that θ a is larger than α but smaller than 2α. the light reflected from the table seems to go in all directions. Begin with only the physical setup: surface. Hint 2. (If you wish. Express your answer in terms of θ a and . you can find the angle between the incoming ray and the normal line by subtraction (since you already have the angle between the incoming ray and vertical). if it is not already there. You will need to add the incoming ray to your drawing. This angle is also the complement of the angle that you are looking for (i. Find the angle between the normal line and the reflected ray Find the angle between the normal line and the reflected ray.e. α ANSWER: Angle between the incoming ray and the normal line = θa − α Hint 4. α ANSWER: Angle between the normal line and the reflected ray = θa − α ANSWER: θr = 2α − θ a Correct http://session. Once you know the angle between the normal line and vertical. Consider the right triangle Look at the right triangle that contains α and has the surface and the dashed vertical and horizontal lines as its three sides. Hint 1. the third angle (between the surface and the vertical) must be 90 − α degrees. the angle between the vertical and the normal). ANSWER: Angle between the normal line and vertical = α Hint 3. Now. the vertical and horizontal lines intersect in a right angle). you will need to add the reflected ray to your drawing. Find the angle between the incoming ray and the normal line Find the angle between the incoming ray and the normal line. Express your answer in terms of α. Use the law of reflection to find the angle.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 27/45 .. Express your answer in terms of θ a and .masteringphysics. Use this information to find the value of the angle you are looking for. Because this is a right triangle (by definition.5/6/2014 Week 13: EM Waves and Propagation of Light their angles must be. na . just substitute β for α in the formula from the previous part. and θw . http://session. Part A If the light strikes the plastic (from the water) at an angle θ w . and α β . when light from a spotlight (or any other source) strikes a seemingly flat surface. Thus. To find the angle between the reflected ray and the vertical for the surface tilted at an angle β . However. Underwater Optics Your eye is designed to work in air. light reflects in all directions. Instead. In this way. Consider a flat piece of plastic (index of refraction n p ) with water (index of refraction n w ) on one side and air (index of refraction n a ) on the other.5/6/2014 Week 13: EM Waves and Propagation of Light Part C Suppose that the spotlight shines so that different parts of the beam reflect off of different two surfaces.masteringphysics. Hint 1. Remember that the inverse sine of a number x should be entered as asin(x)in the answer box. What would the angular separation Δθ be between the rays reflected from the two surfaces? Assume that the light comes at an angle θ a to the vertical. one inclined at an angle α (from the horizontal) and one inclined at an angle β . Finding the angular separation You already know the angle between the reflected ray and the vertical for the surface tilted at an angle α. it will be refracted twice: once at the water/plastic interface and once at the plastic/air interface. frequently sending rays that began very close to one another in very different directions. Consequently. Express your answer in terms of some or all of the angles θ a .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 28/45 . If light is to move from the water into the air. on a microscopic scale. Surrounding it with water impairs its ability to form images. In most surfaces. at what angle θ a does it emerge from the plastic (into the air)? Express your answer in terms of n w . flat surfaces inclined at different angles to the horizontal. scuba divers wear masks to allow them to form images properly underwater. then the angular separation between the reflected rays would be 180∘ . the roughness can be represented as a large number of small. You can see that if α = 45∘ and β = −45∘ (the negative angle indicates that the second surface is inclined above the horizontal). they are covered in ridges and pits. as you will calculate. ANSWER: Δθ = 2α − 2β Correct Most surfaces are not smooth like the surface of a mirror. this does affect the perception of distance. np . the light is reflected in all directions. This microscopic roughness is why most surfaces do not form images as mirrors do. The difference between these two angles will be the angular separation between the two reflected angles. . the angle inside the plastic (i. Angles inside the plastic There are two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface). Remember that the inverse sine of a number x should be entered as asin(x)in the answer box. respectively. http://session. Important theorem from geometry If two parallel lines are cut by a transversal (a third line not parallel to the first two). To find out how they relate. then alternate interior angles are congruent. of refraction for the medium from which light is incident. Your brain can then use that information to determine the angle θ w as shown in the figure . where θ 1 is the angle of incidence. is D. and thus their normals are parallel lines. is performed unconsciously. Once you have labeled both angles. draw a picture with the path the light follows in the plastic and the normals to both surfaces. Hint 1.masteringphysics. and np . In the figure. keep in mind that the surfaces are parallel. The most important method for estimating distance. the angle to the normal immediately after the first refraction at the water/plastic interface). Hint 2. Hint 3.5/6/2014 Week 13: EM Waves and Propagation of Light Hint 1. θ2 n1 is the index is the angle of refraction. ANSWER: θp = asin( nw np sin(θw )) ANSWER: θa = asin( nw na sin(θw )) Correct Notice that np does not appear in this equation.. triangulation. An important theorem from geometry will give you the relationship between the angles. Humans estimate distance based on several different factors. point O. The distance between your eyes is 2l. nw .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 29/45 . such as shadows and relative positions. Snell's law Snell's law states that n 1 sin(θ 1 ) = n 2 sin(θ 2 ) . and n 2 is the index of refraction for the medium into which light emerges. Express your answer in terms of θ w . Find the angle in the plastic Using Snell's law.e. Triangulation is based on the fact that light from distant objects strikes each eye at a slightly different angle. and the distance to the object. find θ p . points L and R represent your left and right eyes. instead of the correct geometric angle θ w . unless the light rays are bent before they reach your eyes. Underwater. Your eyes will calculate an apparent distance d using the angle θ a that reaches your eyes. This is the same θ a that you calculated in Part A. the situation changes.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 30/45 . ANSWER: D = l tan(θ w ) Correct Part C If the distance to the object is more than about 0. ANSWER: D = l θw Correct Your eyes determine θ w by assuming that θ a and θ w (in the figure) are equal.5/6/2014 Week 13: EM Waves and Propagation of Light Part B What is the distance to the object in terms of θ w and l? Express your answer in terms of θ w and l.masteringphysics. then you can use the small-angle approximation tan(θ) ≈ θ . as they are if you're wearing a scuba mask underwater. What is the formula for the distance D to the object. if you make use of this approximation? Express your answer in terms of θ w and l. as shown in the figure . because the refraction takes place so close to your eyes. This is true.4 m. If the problem discussed someone looking out of http://session. Note that there are no important geometric considerations arising from the refraction except the substitution of θ a for θ w . the geometry would become more complicated. Find d Because of the refraction your eyes use θ a instead of θ w . and na . you may use the approximations sin(x) ≈ x and . Apply both small-angle approximations to this equation to get a simpler expression for θ a .masteringphysics.5/6/2014 Week 13: EM Waves and Propagation of Light the porthole of a submarine. you have the expression nw θa = asin( n a sin(θw )) . Since you are putting θa into the equation instead of θ w . and θw . Once you've used the small-angle approximations. Since we are dealing with small angles. na . plug your equation for θ a into the equation from Part C D = l/θ w . Part D Find the ratio d/D. asin(x) ≈ x Express your answer in terms of n w and na . θw . Express your answer in terms of l. Express your answer in terms of n w . ANSWER: θa nw = na θw Hint 2. nw .com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 31/45 . ANSWER: d = lna nw θ w ANSWER: http://session. this gives the apparent distance d. Use the small-angle approximations From Part A. Hint 1. 5/6/2014 Week 13: EM Waves and Propagation of Light d D = na nw Correct Part E Given that n w = 1.6)/1] × 100%). For example. Since the index of refraction depends on the light's wavelength.33 and n a = 1. The surrounding air has n air = 1.410 for red light.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 32/45 .000 . the different colors that comprise white light will spread out as they pass through the diamond. by what percent do objects underwater appear closer than they actually are? Express your answer to two significant figures.450 for blue light. and n blue = 2. Hint 1. then divide by the true distance. You can then convert this fraction to a percent to get the final answer.6 m away then it is 40% closer ([(1 − 0.00 . ANSWER: 25 % Correct A Sparkling Diamond A beam of white light is incident on the surface of a diamond at an angle θ a . Note that the angles in the figure are not to scale. http://session. The indices of refraction in diamond are n red = 2. if an object is 1 m away and it appears to be 0. simply find the difference between the apparent distance and the true distance. How to approach the problem To find the fraction by which objects appear closer. and θa . ANSWER: v blue = 1. θa http://session. Hint 1.5/6/2014 Week 13: EM Waves and Propagation of Light Part A Calculate vred . To four significant figures.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 air = 1 33/45 . Express your answer in terms of θ a and n blue . where θ red is the angle of the red refracted ray. These angles are relative to the normal of the interface between the two materials. Express the angle in terms of n red . Type an expression for sin θ blue .224×108 m/s Correct Part C Derive a formula for δ . which passes through material b. To four significant figures. the speed of blue light in the diamond. ANSWER: v red = 1.998 × 10 m/s . where θ blue is the angle of the blue refracted ray. where θ a is the angle of the incident ray as it passes through and θ b is the angle of the refracted ray. Express your answer in meters per second to four significant digits. Apply Snell's law for blue light Snell's law states that material a n a sin θ a = n b sin θ b . Use n air = 1 . ANSWER: sin θ blue = 1 nblue sin(θa ) Hint 2. Use n air = 1 .masteringphysics. Express your answer in meters per second to four significant digits. Remember that the proper way to enter the inverse sine of x in this case is asin(x). the angle between the red and blue refracted rays in the diamond. 8 c = 2.998 × 10 8 m/s . c = 2. n blue .244×108 m/s Correct Part B Calculate vblue . Apply Snell's law for red light Write an expression for sin θ red . the speed of red light in the diamond. 287 ∘ Correct The red and blue light rays are split by almost a third of a degree as they pass through the diamond. Part E θc http://session. Use n air = 1 . ANSWER: δ = 0. ANSWER: sin θ red = 1 nred sin(θa ) Hint 3.masteringphysics. Compare the angles Which refracted angle is larger? ANSWER: θ red θ blue ANSWER: δ = −asin( 1 nblue sin(θa )) + asin( 1 nred sin(θa )) Correct Part D Calculate δ numerically for θ a ∘ = 45 . each color will shine out of a different facet on the surface of the diamond. This explains why diamonds are cut to have faceted surfaces--if the rays are spread out enough. Express your answer in degrees to three significant figures.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 θc 34/45 . Hint 1. producing a brilliant sparkle. How to approach the problem Plug the given numerical values into the formula you derived from Part C.5/6/2014 Week 13: EM Waves and Propagation of Light Express your answer in terms of θ a and n red . the light ray is reflected back into the diamond. If θ c . ANSWER: θd = 90 ∘ Hint 2. at θ c = θ crit . it will be refracted into the air . If θ c ≤ θ crit = θ crit . If θ c is larger than the critical angle θ crit . Express your answer in terms of n blue and θd . what is θd ? Express your answer in degrees. the light will not be refracted out into the air. Remember that the proper way to enter the inverse sine of x in this case is asin(x). ANSWER: θc = asin( n 1 blue sin(θd )) http://session.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 35/45 . This means that. Hint 1.5/6/2014 Week 13: EM Waves and Propagation of Light Now consider θ c . the light ray must go exactly along the surface of the diamond. What's going on at the critical angle? If θ c < θ crit the air. Use n air = 1 . where θ d is the angle at which the light ray is refracted into the air. Hint 1. Find the refracted angle when θc = θcrit If the light ray hits the bottom surface of the diamond at an angle θ c with an angle of θ d (with respect to the normal of the surface). Apply Snell's law Use Snell's law to express θc in terms of θ d . the light ray is refracted into > θ crit . but instead it will be totally internally reflected back into the diamond. Express your answer in degrees to four significant figures.masteringphysics. the angle at which the blue refracted ray hits the bottom surface of the diamond. Find θ crit . Express your answer in terms of \texttip{\theta _{\rm c}}{theta_c}. What is this third angle? Express your answer in terms of some or all of the angles α. find the largest possible value of the incident angle θ a such that the blue light is totally internally reflected off the bottom surface. For α = 45∘ .5/6/2014 Week 13: EM Waves and Propagation of Light ANSWER: θ crit = 24.09 ∘ Correct Part F A diamond is cut such that the angle between its top surface and its bottom surface is α. and θ red . θ c .masteringphysics. Hint 1. \texttip{\alpha }{alpha}. ANSWER: \texttip{\theta _{\rm a}}{theta_a} = {\asin}\left(n_{\rm{blue}} {\sin}\left({\alpha}-{\theta}_{c}\right)\right) http://session. Use the normal to find the unknown angle You can use the normal to the bottom surface and the normal to the top surface to define a triangle that is made out of the angles θ blue . Express your answer in degrees to four significant figures. Hint 1. Determine θblue Find an expression for θ blue (defined in the figure) in terms of θ c and α. θc . ANSWER: θ inside = 180 − α ANSWER: θ blue = α − θc Hint 2.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 36/45 . Remember that the proper way to enter the inverse sine of \texttip{x}{x} in this case is asin(x). Apply Snell's law Use Snell's law to find an expression for θ a . and \texttip{n_{\rm blue}}{n_blue}. and a third angle that we will call θ inside . The creation of Cerenkov radiation occurs in much the same way that a sonic boom is created when a plane is moving faster then the speed of sound in the air.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 37/45 .7519 \texttip{c}{c} Correct Part B What is the threshold velocity \texttip{v_{\rm threshold}}{v_threshold} for creating Cerenkov light of a charged particle http://session. Cerenkov Radiation Electromagnetic radiation is emitted when a charged particle moves through a medium faster then the local speed of light (which is always lower then the speed of light in vacuum). See your answer for part E. This radiation is known as Cerenkov radiation.97 ^\circ Correct The angle at which a diamond is cut plays an important role in how brightly it sparkles.masteringphysics. When is \texttip{\theta _{\rm a}}{theta_a} maximized? The maximum value of \texttip{\theta _{\rm a}}{theta_a} that can give total internal reflection will occur when the blue light ray hits the bottom surface of the diamond at exactly the critical angle: \theta_{\rm c}=\theta_{\rm crit} (with respect to the normal of the bottom surface). Part A What is the threshold velocity \texttip{v_{\rm threshold}}{v_threshold} for creating Cerenkov light of a charged particle as it travels through water (which has an index of refraction n=1. maximizing the diamond's "fire.33)? Express your answer as a multiple of \texttip{c}{c} to three significant figures. For more information on the physics of diamonds.folds.net/diamond/. The various wavefronts that propagate in the material add coherently to create an effective shock wave. ANSWER: \texttip{v_{\rm threshold}}{v_threshold} = 0. check out http://www. It should be stressed that the particle is never going faster then the speed of light in vacuum (or \texttip{c}{c}). ANSWER: \texttip{\theta _{\rm a}}{theta_a} = 60. just faster then the speed of light in the material (which is always less then \texttip{c}{c}). In this problem you will become familiar with this type of radiation and learn how to use its properties to get information about the particles that created it.5/6/2014 Week 13: EM Waves and Propagation of Light Hint 3. The proper choice for \texttip{\alpha }{alpha} will ensure that a large fraction of the light gets totally internally reflected back toward your eyes." Generally \texttip{\alpha }{alpha} is chosen to be between 39^\circ and 42^\circ. Cerenkov radiation is found in many interesting places such as particle detectors and nuclear reactors and can even be seen by astronauts when cosmic rays traverse their eyes. this interaction can then release radiation (in this case in the form of Cerenkov light). Hint 1.masteringphysics. it will emit the Cerenkov radiation in a cone. ANSWER: \texttip{v_{\rm threshold}}{v_threshold} = 0. Remember that the local speed of light is lower than that of the particle. Correct When a charged particle passes through a material it interacts with the atom's electric and magnetic fields.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 38/45 . Geometry of the problem Consider that the particle is emitting light along every point along its path through the medium. Neutral particles don't interact with the electromagnetic fields of the medium. \texttip{c}{c}. Charged particles hit the nuclei of the material more often and thus create more radiation. Neutral particles can't move above the threshold velocity for Cerenkov radiation. Part D If a particle is traveling straight through a material with index of refraction \texttip{n}{n} at a speed \texttip{v}{v}.36)? Express your answer as a multiple of \texttip{c}{c} to three significant figures. Next we will calculate how the cone angle is correlated to the speed of the particle. When a charged particle passes straight through a medium faster than the local speed of light. Pick one point along the particle's path and look at how far the light from this point would have traveled in a certain amount of time \texttip{t}{t} compared to how far the particle itself traveled. and \texttip{n}{n}.5/6/2014 Week 13: EM Waves and Propagation of Light as it travels through ethanol (with index of refraction n=1. what is the angle \texttip{\Theta }{Theta} that the cone of light makes with the particle's trajectory? In other words what is the angle between the vector of the propagating Cerenkov radiation and the vector in the direction of the propagating particle? Express your answer in terms of \texttip{v}{v}.7353 \texttip{c}{c} Correct Part C Which of the following best explains why neutrally charged particles can't give off Cerenkov radiation? ANSWER: Neutral particles are too heavy to emit radiation. Use this information to determine the angle of the light cone. http://session. This gives us a right triangle from which one can derive the angle of the propagating light cone. Using the geometry One can draw a circle about a point along the particle's path that represents the distance that light travels in time \texttip{t}{t} from when the particle passed that point and then draw a line tangent to the circle and intersecting the point at which the particle is at time \texttip{t}{t}.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 39/45 . Let us look at what decisions need to go into designing a Cerenkov detector. You can use a ring-imaging Cerenkov detector consisting of a thin slab of material separated from an array of photomultiplier tubes (devices used to detect weak light signals) by an arbitrary open space. Use these constraints and the equation for \texttip{\Theta }{Theta} from Part D to determine which of the following substrate materials is best suited to giving you the greatest precision in determining particle velocity. Since this line is tangent to the circle of propagating light it is perpendicular to a line from the center of the circle.masteringphysics. http://session.5/6/2014 Week 13: EM Waves and Propagation of Light Hint 2. Your photomultiplier tubes. ANSWER: \texttip{\Theta }{Theta} = \large{{\acos}\left(\frac{c}{v n}\right)} Answer Requested Ring-imaging Cerenkov detectors are devices that can accurately measure the velocity of charged particles as they pass through them. They are very useful as subdetectors in large particle detector systems. can only resolve a finite change in the angle of the ring created by the Cerenkov radiation. The detector works on the principle that the Cerenkov radiation emitted in the thin slab will be a cone of light that can be measured with the array of photomultiplier tubes. having a finite width. Part E Suppose you wish to accurately measure the speed of high energy particles with velocities greater then 98% the speed of light in vacuum. com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 40/45 .8 degrees but can't tell the difference between Cerenkov light emitted at 15 degrees and that emitted at 13.2\.03) for the Cerenkov material and the photomultiplier tube array had a resolution of \Delta\Theta = 1. What is the highest velocity \texttip{v_{\rm max}}{v_max} at which a charged particle can be accurately measured to be below the speed of light in vacuum ( \texttip{c}{c})? Express your answer as a multiple of \texttip{c}{c} to three significant figures.52) ice (n = 1. Determine the angle for a particle going the speed of light What is the Cerenkov angle \texttip{\Theta }{Theta} for a charged particle passing through the aerogel at the http://session.5/6/2014 Week 13: EM Waves and Propagation of Light Hint 1.417) crown glass (n = 1. This will allow you to determine which of the materials provides the best angular resolution for the velocities. ANSWER: diamond (n = 2.99c and v_2 = 0.9 degrees. Compare this difference using each of the different materials. such as v_1 = 0. Using example velocities Look at the difference in angle between two different particle velocities.masteringphysics. Hint 1. This means the detector can distinguish between Cerenkov light emitted at an angle of 15 degrees and that emitted ar 13. {\rm degrees}.3) aerogel (n = 1.03) vacuum (n = 1) Correct Part F Suppose our detector used aerogel (n= 1.98c. \texttip{n}{n} is the index of refraction of the material the particle is passing through. and \texttip{v}{v} is the actual speed of the particle (for which \texttip{c}{c} is an upper limit). ANSWER: \texttip{\Theta }{Theta} = 12. \texttip{c}{c} is the speed of light in vacuum. it must measure an angle \Theta_{\rm max}-\Delta\Theta. and \texttip{v}{v} is the actual speed of the particle (always a bit below \texttip{c}{c}).9 degrees give the resolution of 1.2 degrees in the detector? Express your answer in degrees.masteringphysics. Hint 1. http://session. ANSWER: \texttip{\Theta }{Theta} = 13. Obtaining the Cerenkov angle from the speed of a particle Recall that the speed of the charged particle can be determined from the angle of the emitted Cerenkov radiation using the formula \large{\cos(\theta) = \frac{c}{n v}}. Determine the resolved angle What is the next lower angle that can be resolved as being different from 13.5/6/2014 Week 13: EM Waves and Propagation of Light speed of light in vacuum \texttip{c}{c}? Express your answer in degrees to three significant figures. \texttip{c}{c} is the speed of light in vacuum.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 41/45 . Hint 1. Obtaining the speed of a particle from the angle Recall that the speed of the charged particle can be determined from the angle of the emitted Cerenkov radiation using the formula \large{\cos(\theta) = \frac{c}{n v}}. where \Theta_{\rm max} is the Cerenkov angle from a particle going the speed of light and \Delta\Theta is the angular resolution. Angular resolution For the detector to measure that a particle travels slower than the speed of light \texttip{c}{c}. This new angle will allow one to calculate the highest velocity that can be distinguished from the speed of light.7 \rm degrees Hint 3. where \texttip{\theta }{theta} is the angle between the Cerenkov light cone and the line the propagating particle follows. where \texttip{\theta }{theta} is the angle between the Cerenkov light cone and the line the propagating particle follows. \texttip{n}{n} is the index of refraction of the material the particle is passing through.9 \rm degrees Hint 2. Use the index of refraction of the material to calculate this speed. Requirement for Cerenkov radiation To generate Cerenkov radiation.97 \texttip{c}{c} Correct Exercise 33. the particle must be traveling faster than the local speed of light.8 A laser beam shines along the surface of a block of transparent material. (See the figure . while the other half travels through the block and then hits the detector. Part A http://session. Hint 1. ANSWER: \texttip{v_{\rm min}}{v_min} = 0.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 42/45 .) Half of the beam goes straight to a detector.995 \texttip{c}{c} Correct Part G What is the lowest velocity \texttip{v_{\rm min}}{v_min} that a charged particle can have and still emit Cerenkov radiation in the aerogel? Express your answer as a multiple of \texttip{c}{c} to two significant figures. The time delay between the arrival of the two light beams at the detector is 6.50{\rm {\rm ns}} .5/6/2014 Week 13: EM Waves and Propagation of Light ANSWER: \texttip{v_{\rm max}}{v_max} = 0. as measured in air.44 A light ray in air strikes the right-angle prism shown in the figure (\angle B=29. n_2 = 1. has an index of refraction of 1.5/6/2014 Week 13: EM Waves and Propagation of Light What is the index of refraction of this material? ANSWER: n = 1.masteringphysics. When it emerges at face AB.78 Correct Problem 33.34. a transparent.13 Correct Exercise 33. ANSWER: n_1. This ray consists of two different wavelengths.50^\circ . it has been split into two different rays that diverge from each other by 8.1. gelatinous fluid that fills most of the eyeball.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 43/45 .09.2 The vitreous humor. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. Part A http://session. Visible light ranges in wavelength from 400{\rm {\rm nm}} (violet) to 700{\rm {\rm nm}} (red). Enter your answers in ascending order separated by a comma.0{\rm ^\circ}). Part A Find the index of refraction of the prism for each of the two wavelengths. \. Separate your answers with a comma.4 Light with a frequency of 6.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 44/45 .29×1014.522 {\rm nm} Correct Part B What are the ranges of the frequency of the light just as it approaches the retina within the vitreous humor? Answer in the order indicated. ANSWER: \lambda_{\rm min}.\.7.5/6/2014 Week 13: EM Waves and Propagation of Light What are the ranges of the wavelength of the light just as it approaches the retina within the vitreous humor? Answer in the order indicated.\lambda_{\rm max} = 299.masteringphysics. ANSWER: f_{\rm min}.50×1014 {\rm Hz} Correct Part C What is the speed of the light just as it approaches the retina within the vitreous humor? ANSWER: v = 2.00×108{\rm m/s} for the speed of light in a vacuum.f_{\rm max} = 4. Separate your answers with a comma.30×1014{\rm Hz} travels in a block of glass that has an index of refraction of 1.53.76×10−7 \rm m http://session. ANSWER: \lambda = 4.24×108 {\rm m/s} Correct Exercise 33. Part A What is the wavelength of the light in vacuum? Use 3. 72 .9 ^\circ Correct Score Summary: Your score on this assignment is 99. ANSWER: \theta = 67.60.11×10−7 \rm m Correct Exercise 33.87 out of a possible total of 17 points.5{\rm ^\circ}. http://session. and the ray originates in the glass with n= 1.72 and 1. You received 16.15 A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=2778958 45/45 .2%.masteringphysics.5/6/2014 Week 13: EM Waves and Propagation of Light Correct Part B What is the wavelength of the light in the glass? ANSWER: \lambda = 3. The angle of incidence is 59. Part A Compute the angle of refraction.