Name St.No. - Date(YY/MM/DD) / / Section UNIT 102-6: ELECTROMAGNETIC WAVES AND POLARIZATION Approximate Time Three 100-minute Sessions Hey diddle diddle, what kind of riddle Is this nature of light? Sometimes it s a wave, Other times particle... But which answer will be marked right? Jon Scieszka OBJECTIVES 1. To understand electromagnetic waves and how they propagate. 2. To understand how to determine the energy density and the intensity of EM waves. 3. To investigate linear polarization of light. 4. To explore polarization by reflection.
Page 2 Physics for Life Sciences II Activity Guide SFU OVERVIEW Electricity and Magnetism Maxwell's Equations James Clerk Maxwell, a Scottish physicist, was in his prime when Faraday retired from active teaching and research. He had more of a mathematical bent than Faraday and pulled many of the basic equations describing electric and magnetic effects into a set of four very famous equations. These equations are shown below in simplified integral form for situations in which no dielectric or magnetic materials are present. I ~E d ~A = Q 0 Gauss Law in Electricity\oint I\vec{E}\cdot d\vec{a} = \frac{q}{\epsilon_0} ~B d ~A = 0 Gauss Law in Magnetism I I ~E d~s = d m dt! d B d s = µ 0 I + 0 µ 0 dt \oint \vec{b}\cdot d\vec{s} = \mu_0 I + \epsilon_0 \mu_0\frac{d \Phi_e}{dt} Faraday s Law If we add the Lorentz force F = q E + q v B e! Ampère-Maxwell Law to Maxwell's equations, then we can derive a complete description of all classical electromagnetic interactions from this set of equations. Perhaps the most exciting intellectual outcome of Maxwell's equations is their prediction of electromagnetic waves and our eventual understanding of the self-propagating nature of these waves. This picture of electromagnetic wave propagation was not fully appreciated until scientists abandoned the idea that all waves had to propagate through an elastic medium and accepted Einstein's theory of special relativity; these changes occurred in the early part of the twentieth century. To the vast majority of the world's population the practical consequences of Maxwell's formulation assume much more importance than its purely intellectual joys. The famous physicist Richard Feynmann wrote the following: Now we realize that the phenomena of chemical interaction and ultimately of life itself are to be understood in terms of electromagnetism.... The electrical forces, enormous as they are, can also be very tiny, and we can control them and use them in many ways... From a long view of the history of mankind seen from, say, ten thousand years from now there can be little doubt that the most significant event of the nineteenth century will be judged as Maxwell's discovery of the laws of electrodynamics. The American civil
Unit 102-6 Electromagnetic Waves and Polarization Page 3 Author: Sarah Johnson war will pale into provincial insignificance in comparison with this important scientific event of the same decade. In this unit you will learn about the behaviour of electromagnetic waves. Maxwell predicted the existence of electromagnetic waves in 1864, a long time before anyone was able to create or detect them. It was Maxwell s four famous equations that led him to this prediction which was finally confirmed experimentally by Hertz in 1887. Maxwell argued that if a changing magnetic field could create a changing electric field then the changing electric field would in turn create another changing magnetic field. He predicted that these changing fields would continuously generate each other and so propagate and carry energy with them. The mathematics involved further indicated that these fields would propagate as waves. When one solves for the wave speed one gets: wave speed: c = 1 ε 0 µ 0 (6.1) Where ε 0 and µ 0 are the electrical constants we have already been introduced to. The magnitude of this speed led Maxwell to further hypothesize that light is an electromagnetic wave. Once you have learned a few things about the nature of electromagnetic waves, you will go on to investigate the phenomenon of polarization.
Page 4 Physics for Life Sciences II Activity Guide SFU SESSION ONE: ELECTROMAGNETIC WAVES Figure 6.1: An electromagnetic wave. An electromagnetic wave that is travelling in the positive z-direction with its electric field oscillating parallel to the x-axis and its magnetic field oscillating parallel to the y-axis (as shown in Figure 6.1) can be represented mathematically using two sinusoidal functions of position (z) and time (t): E(x, t) = E 0 sin(kz t) (6.2a) B(x, t) = B 0 sin(kz t) (6.2b) where ~ E 0 and ~ B 0 are the amplitudes of the fields, ω is the angular frequency which is equal to 2πf, where f is the frequency of the wave, and k is the wave number, which is equal to 2π/λ, where λ is the wavelength. It is important to note that Figure 6.1 is an abstract representation of an electromagnetic wave that represents the magnitude and direction of the electric and magnetic fields at points along the z-axis. In a real electromagnetic wave travelling through space, for each line parallel to the z-axis there is a similar picture. These arrows, which represent field vectors, do not indicate a sideways displacement of anything. Also, as we ll soon see, there can be many axes along which the fields oscillate in one electromagnetic wave.
Unit 102-6 Electromagnetic Waves and Polarization Page 5 Author: Sarah Johnson Activity 6-1: Wave Speed (a) Plug in the values for ε 0, the permittivity of free space and µ 0, the permeability of free space, into the following equation and verify that the wave speed of electromagnetic waves that you get is consistent with the speed of light in a vacuum, c = 299,792,458 m/s: wave speed: c = 1 ε 0 µ 0 (6.1) The speed of a wave is equal to a product of its wavelength λ and its frequency f. So. for an electromagnetic wave we get: c = λf (6.3) This means there is an inverse relationship between wavelength and frequency for electromagnetic waves, i.e. the longer the wavelength of a wave, the lower the frequency. Electromagnetic (EM) waves are often sorted into what s know as the Electromagnetic Spectrum, where the types of electromagnetic waves are sorted in order of their wavelength or frequency. One example of an electromagnetic spectrum where the waves are sorted by wavelength is shown below in Figure 6.2. As you know, modern technology takes advantage of the existence of many of these electromagnetic waves, which all travel in a vacuum at the speed of light, c.
Page 6 Physics for Life Sciences II Activity Guide SFU Figure 6.2: The Electromagnetic Spectrum. Activity 6-2: The Electromagnetic Spectrum (a) Using Figure 6.2 estimate the average wavelength and frequency of the following EM waves: 1) Gamma rays 2) Ultraviolet waves 3) Microwaves 4) Radio waves
Unit 102-6 Electromagnetic Waves and Polarization Page 7 Author: Sarah Johnson (b) On the grid below, graph log(frequency) vs. log(wavelength) for these 4 EM waves. (c) Does your graph show the relationship between frequency and wavelength that you expect? What is this relationship? Why do you think it was necessary to take logarithms of both quantities?
Page 8 Physics for Life Sciences II Activity Guide SFU The Wave Equation Using the definitions of ω and k, ω = 2πf and k = 2π/ λ, we can rewrite 6.3 as: c = λf = 2π k ω 2π = ω k (6.4) Using this result we can rewrite our two wave equations as: apple z E(z, t) = E 0 sin t (6.5a) c \vec{e}(x,t)= \vec{e}_0 \sin\left[\omega\left(\frac{z}{c}- t\right)\right apple z B(z, t) = B 0 sin t c (6.5b)) One can use Faraday s Law to determine the ratio of the electric and magnetic fields in an EM wave. This ratio is: E 0 B 0 = E B = c (6.6) Thus if you know the magnitude of one of the two fields you can always easily determine the magnitude of the other in an electromagnetic wave. Activity 6-3: The Wave Equation Suppose you are given the following wave equation for an EM wave: apple E x = (4.0 V/m) sin t, E y = 0, E z = 0 z 5 1015 s 1 c E_x = (4.0 {\rm~v/m})\sin \left[ \left(\frac{\pi}{5} \times 10^{15}{\rm~s}^{-1}\right)\left(\frac{z}{c} -t\right)\right], ~E_y=0, ~E_z=0 Determine the following quantities for this electromagnetic wave. (a) ω =? (b) f =?
Unit 102-6 Electromagnetic Waves and Polarization Page 9 Author: Sarah Johnson (c) λ =? (d) k =? (e) Write expressions for the three components (x,y,z) of the magnetic field of this wave below. (f) What type of electromagnetic wave do these equations describe? (radio wave?, x-ray? etc...) Energy Transport by Electromagnetic Waves We know that the energy per unit volume or energy density stored in an electric field of magnitude E is: u elec = 1 2 ε 0E 2 (6.7)
Page 10 Physics for Life Sciences II Activity Guide SFU We also know that the energy per unit volume stored in a magnetic field of magnitude B is: u mag = B2 2µ 0 (6.8) Using the fact that E = cb for an electromagnetic wave, it is easy to show that these two energy densities are equal for EM waves: u elec = 1 2 ε 0E 2 = 1 2 ε 0(cB) 2 = 1 2 ε 0c 2 B 2 = 1 2 ε 0c 2 2µ 0 u mag = c 2 1 u mag = u mag c 2 where we have used Equation 6.1 to replace ε 0 µ 0 with 1/c2. This equality of the two energy densities is true everywhere along an electromagnetic wave. The total energy density for an electromagnetic wave is the sum of the two energy densities: u total = u elec + u mag = 1 2 ε 0 E 2 + B2 2µ 0 (6.9) Because the two densities are equal, one can also write: u total = 2u elec = ε 0 E 2 (6.10) u total = 2u mag = B2 µ 0 (6.11) These two equations are useful if you only know one of the two field strengths. One can use these expressions to determine the rate of energy transport per unit area for an EM wave. This instantaneous power per area also known as the instantaneous intensity I is given as: I = 1 cµ 0 E 2 = " 0 ce 2 (6.12)
Unit 102-6 Electromagnetic Waves and Polarization Page 11 Author: Sarah Johnson or in terms of the magnetic field strength as: I = c µ 0 B 2 = " 0 c 3 B 2 (6.13) Activity 6-4: Electromagnetic Wave Average Intensity (a) Using equation 6.12, show that for an EM wave travelling in the positive z-direction the average intensity Ī is: Ī = 1 cµ 0 E 2 0sin 2 (kz!t) (b), Use the fact that the time-average of sin 2 (kz-ωt) in this case equals 1/2 to remove the sine function from your expression for average intensity. (c) Now rewrite the average intensity in terms of the root-mean square value of the magnitude of the electric field Erms = E 0 / 2.
Page 12 Physics for Life Sciences II Activity Guide SFU (d) Instead write the average intensity in terms of the root-mean square of the magnetic field: Brms = B 0 / 2. (e) Finally write the average intensity in terms of both Erms and Brms. In the last activity you developed some expressions for the average intensity, or average power per area for an EM wave. In the next activity you can use one of these expressions to determine the rms field strengths of a laser beam. It might be worth checking your results for Activity 6-4 with those in your textbook before proceeding. Problem Activity 6-5: Intensity of a Laser (a) A 13.5-mW laser puts out a narrow beam with a circular crosssection of radius 0.85 mm. What is the average intensity of this laser beam? (Hint: Remember that intensity is power per area.)
Unit 102-6 Electromagnetic Waves and Polarization Page 13 Author: Sarah Johnson (b) Find Erms and Brms for this laser beam.
Page 14 Physics for Life Sciences II Activity Guide SFU SESSION TWO: POLARIZATION Theory of Polarization To completely describe light as an electromagnetic wave one must specify its frequency, its direction of propagation and its state of polarization. Our interest in this session is with polarization, so let us assume that we have monochromatic light propagating along the +z direction of a right-handed co-ordinate system. Light is a transverse electromagnetic wave the electric field is always perpendicular to the direction of propagation. Because the direction of propagation is along the +z axis, the electric field vector E must lie in the plane formed by the x and y axes. This can be expressed mathematically as follows: ~E(x, y, z, t) =E x (z,t)î + E y (z,t)ĵ (6.14) The components of the electric field Ex and Ey do not depend on x and y, only z, because we the wave is a plane wave propagating along the +z direction. Light is linearly polarized if the electric field vector E is always parallel to the same line which is perpendicular to the direction of propagation. In general, mathematically you could write the following expression for the EM wave: ~E(x, y, z, t) =E 0x sin(kz!t)î + E 0y sin(kz!t)ĵ (6.15) The amplitudes E0x and E0y are constants. As we saw in the last session, the intensity of the light is proportional to the square of the amplitude. Figure 6.3 shows the electric field vectors at a fixed time along a line in the direction of propagation. Figure 6.3a illustrates light polarized along the x direction (E0x 0, E0y = 0) and 6.3b shows polarization along the y direction (E0x = 0, E0y 0). As time goes on the entire pattern moves in the +z direction. Light polarized in an arbitrary plane is a superposition of these two independent possibilities (E0x 0, E0y 0). The plane of polarization is determined by the relative magnitudes of E0x and E0y.
Unit 102-6 Electromagnetic Waves and Polarization Page 15 Author: Sarah Johnson Linearly Polarized Light x E (a) x-direction x (b) y-direction y z y E Figure 6.3: The electric field vectors of linearly polarized light. Most light sources do not produce polarized light. Individual atoms in the source radiate independently. Although at any instant in time light received from a radiating atom in the source has a definite state of polarization the state of polarization changes rapidly with time. You may think of unpolarized light as having two polarization components (x/y) which are radiated independently and randomly. Between polarized light which has a fixed relation in time between the amplitude and phase of the two polarization components and unpolarized light which has a random relation in time between the two polarization components, is partially polarized light. Partially polarized light is a mixture of polarized and unpolarized light. Polarization Filters A linear polarizer produces polarized light by selectively absorbing light polarized perpendicular to the polarization axis and transmitting light parallel to the transmission axis. Our HN22 linear polarizers absorb more than 99.99% of the perpendicular component and transmit 44% of the parallel component. Thus the linear polarizer transmits 22% of incident unpolarized light. A perfect polarizer would transmit 50% of incident unpolarized light. Consider unpolarized light falling on a polarizer which only transmits the components parallel to the y-axis. The filtered light has amplitude E 0 and intensity I = (E 0 ) 2. Let this filtered light fall on a second polarizer whose polarization axis y is at an angle θ with respect to the y-axis. (See Figure 6.4 below where the y axis is depicted as vertical.) The incident light has component E 0 cos θ along the y axis. If both polarizers are perfect, the intensity after the second polarizer would be (E 0 cos θ) 2 but for the non-ideal polarizers the transmitted intensity is 0.44 (E 0 cos θ) 2.
Page 16 Physics for Life Sciences II Activity Guide SFU Figure 6.4: A vertical polaroid transmits only the vertical component of an EM wave incident upon it. For the following experiments, mount the light source and filters on the optical bench as shown in Figure 6.4. Now you will do some experiments to examine linear polarization. For the next few activities you will need the following equipment: From the Optics Kit: Optical Bench Light Source All component holders 3 Linear Polarization Filters Viewing Screen For the following experiments, mount the light source and filters on the optical bench as shown in Figure 6.5. Polarizing Filters or Retardation Plates Light Source Figure 6.5: Experimental set-up for exploring polarization.
Unit 102-6 Electromagnetic Waves and Polarization Page 17 Author: Sarah Johnson Activity 6-6: Linear Polarization (a) Look at the light source through a linear polarizer while rotating the polarizer through 360. Try to determine if the light source is polarized. Describe your findings below. (b) Now produce linearly polarized light by placing a linear polarizer in front of your light source. Look at this light through a second linear polarizer, called the analyzer. For the angle θ between the two polarizers axes given in the table below describe how the intensity of the output light compares to the intensity at θ = 0. (i.e. brighter, a lot brighter, dimmer, a lot dimmer, about the same) θ (degrees) Intensity comparison 45 90 135 180 225 270 315
Page 18 Physics for Life Sciences II Activity Guide SFU (c) Try to explain in words in the results you see in (b). Were any of the results surprising to you? Malus Law As mentioned in the last activity, when we deal with linear polarizers we call the first filter the unpolarized light falls upon the polarizer and the second polarizer the light encounters the analyzer. Malus Law states that the intensity of the light exiting the analyzer I depends on θ, the angle between the polarization axes of the polarizer and the analyzer, according to the following formula: I = I 0 cos 2 θ (6.16) where I 0 is the intensity of the linearly polarized light entering the analyser. As shown in the introduction to this session, the cosine indicates that only the component of the field parallel to the polarization axis is allowed to pass through. The cosine-squared dependence arises because the intensity is proportional to the electric field strength squared. See Figure 6.6 below.
Unit 102-6 Electromagnetic Waves and Polarization Page 19 Author: Sarah Johnson Figure 6.6: Unpolarized light travelling through two linear polarizers. Activity 6-7: Malus Law (a) Fill in the table below with the calculated value of the ratio of the outgoing intensity to the incident intensity: I/I 0 for the given angles. θ (degrees) I/I 0 0 45 90 135 180 225 270 315
Page 20 Physics for Life Sciences II Activity Guide SFU (b) Compare the numerical values for the intensity ratios in the table above with the qualitative intensity comparisons you did in the previous activity. Do the two generally agree with each other for all of the angles? (Note: Remember that our polarizers are not ideal.) Activity 6-8: Linear Polarization Puzzler (a) Orient the first polarizer and the analyzer so that a minimum amount of intensity exits the analyzer. (This should mean that the polarization axes are oriented at a 90 degree angle with respect to each other.) Take a third linear polarizer and place it in between the first polarizer and the analyzer. Start this middle polarizer oriented with its axis aligned parallel to the first polarizer. Now rotate it through 360 degrees and record the relative intensity of the light emitted by the analyzer (as compared to the intensity seen at 0 degrees) in the table below. θ (degrees) Intensity comparison 45 90 135 180 225 270 315
Unit 102-6 Electromagnetic Waves and Polarization Page 21 Author: Sarah Johnson (b) Can you think of an explanation for what you see. Why does adding a third linear polarizer between the other two allow more light to pass when very little light is emitted when there are only two? (Hint: Think about Malus Law and what happens to the angles of orientation when you add the third polarizer. Look at Figure 6.6 again.) (c) If we call the intensity leaving the first polarizer I 0, the intensity leaving the second I and the intensity leaving the third I, show that: I ʹ = I 0 cos 2 θ cos 2 ( θ ʹ θ) where θ is the angle between the axes of the first and second polarizers and θ is the angle between the axes of the first and third polarizers.
Page 22 Physics for Life Sciences II Activity Guide SFU (d) If θ = 90 degrees, calculate the ratio of I to I 0 for θ = 45, 90, 135 and 180 degrees. (e) Are your results in (d) consistent with your observations in (a)? Explain. Polarization by Reflection When unpolarized light reflects from a nonmetallic surface, it is at least partially polarized after reflection. The reflected beam of light will be polarized preferentially in the plane parallel to the surface. The amount of polarization depends on the angle of incidence, varying from no polarization at normal incidence to 100% at an angle known as Brewster s Angle. In the following activity you will do an experiment to determine Brewster s Angle for the Cylindrical Lens in the Optics Kit. For the next activity you will need the following equipment:
Unit 102-6 Electromagnetic Waves and Polarization Page 23 Author: Sarah Johnson From the Optics Kit: Optical Bench Light Source 1 Linear Polarization Filter 1 Slit Plate and Holder Viewing Screen and holder for Ray Table Ray Table Cylindrical Lens Polarizing Filter Light source Screen Figure 6.7: Set-up to measure Brewster s Angle. Activity 6-9: Brewster s Angle (a) Reflect a ray off the flat surface of the Cylindrical Lens and observe the reflected ray through a linear polarizer. (You may look directly at the reflected light or project it onto a screen.) Do you observe any polarization dependence of the reflected ray? Do the same observation on the refracted ray emerging from the curved surface of the Cylindrical Lens.
Page 24 Physics for Life Sciences II Activity Guide SFU (b) Now polarize the incident ray by setting the linear polarizer at 90 and placing it between the light source and the Cylindrical Lens. Project the reflected ray onto a screen. Rotate the Ray Table and Cylindrical Lens and observe the intensity of the reflected ray as a function of angle. You should see a minimum in the reflected intensity of the polarized ray, note the angle at which this occurs. Also record the angles of the reflected and refracted rays at this position. (c) The incident angle at which you see minimum reflection in the experiment in (b) is Brewster s angle for our set-up. According to the theory the sum of Brewster s angle and the angle of refraction of the light ray should equal 90 0. Is this the case for the angles you measured in (b)? How far off were you?