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(EM) Waves Are Forms of Energy That Have Magnetic and Electric Components Physics 202-Section 2G Worksheet 9-Electromagnetic Radiation and Polarizers Formulas and Concepts Electromagnetic (EM) waves are forms of energy that have magnetic and electric components. EM waves carry energy, not matter. EM waves all travel at the speed of light, which is about 3*108 m/s. The speed of light is often represented by the letter c. Only small part of the EM spectrum is visible to us (colors). Waves are defined by their frequency (measured in hertz) and wavelength (measured in meters). These quantities are related to velocity of waves according to the formula: 풗 = 흀풇 and for EM waves: 풄 = 흀풇 Intensity is a property of EM waves. Intensity is defined as power per area. 푷 푺 = , where P is average power 푨 o Intensity is related the magnetic and electric fields associated with an EM wave. It can be calculated using the magnitude of the magnetic field or the electric field. o Additionally, both the average magnetic and average electric fields can be calculated from the intensity. ퟐ ퟐ 푩 푺 = 풄휺ퟎ푬 = 풄 흁ퟎ Polarizability is a property of EM waves. o Unpolarized waves can oscillate in more than one orientation. Polarizing the wave (often light) decreases the intensity of the wave/light. o When unpolarized light goes through a polarizer, the intensity decreases by 50%. ퟏ 푺 = 푺 ퟏ ퟐ ퟎ o When light that is polarized in one direction travels through another polarizer, the intensity decreases again, according to the angle between the orientations of the two polarizers. ퟐ 푺ퟐ = 푺ퟏ풄풐풔 휽 this is called Malus’ Law. 1. A radio wave has a frequency of 1605 kHz. What is the wavelength of the radio wave? 186.92 m. Radio waves are electromagnetic waves, and therefore travel at the speed of light. Use the formula 풄 = 흀풇 plugging in 3*108 for c and 1605000 Hz for f (converting kilohertz to hertz). Divide to get 186.92. Wavelength is measured in meters. 2. The magnetic field emitted by a laser is 0.003 T. a) What is the average intensity of the waves emitted by the laser? ퟐ 2.15 *109 use the formula 푺 = 풄 푩 . Plug in 0.003 T for B, c is 3e8, and mu is the magnetic 흁ퟎ constant: 4(pi)e-7. b) What is the magnitude of the average electric field emitted by the laser? 900,000 N/C. Take the intensity from the last question and use the formula that includes electric ퟐ field. 푺 = 풄휺ퟎ푬 . Substitute in 3e8 for c, 2.15e9 for S, and the permittivity of free space (a constant) for epsilon (8.85e-12). Solve for E. 3. Unpolarized light of intensity 2000 W/m2 passes through three polarizers (A, B, and C). a) Polarizer A is oriented to the vertical. What is the intensity of the light after it passes through Polarizer A? 1000 W/m2. No matter the orientation, if UNPOLARIZED light travels through ANY polarizer, it’s intensity will be cut in half. b) The light from Polarizer A proceeds to pass through Polarizer B. Polarizer B is oriented 18 degrees from the vertical. What is the intensity of the light exiting Polarizer B? 904.5 W/m2. Use Malus’ Law. S2 (the new intensity) = S1 (old intensity) * cos2(angle between 2 polarizers). S2 = 1000cos2(18°) = 904.5. c) The light from Polarizer B proceeds to pass through Polarizer C. Polarizer C is oriented 44 degrees from the vertical. What is the intensity of the light exiting Polarizer C? 730.7 W/m2. Use Malus’ Law again. S1 is now the answer to the previous question (904.5), and the angle is the angle between the 44-degree polarizer and the 18-degree polarizer, which is 26 degrees. 904.5cos2(26) = 730.7. 4. Horizontally polarized light enters a polarizer and exits with an intensity of 12.8 W/m2. If the polarizer was oriented at 38 degrees from the horizontal, what was the intensity of the light incident on the polarizer? 20.6 W/m2. Rearrange Malus’ Law. We know the exiting intensity (S2 = 12.8) and we know the angle between the two polarizers (38 degrees). Plug in and solve for S1, the incident intensity. 5. If you stand 4 feet away from a plane mirror, how far away from your image in the mirror are you? Is your image real or virtual? 8 feet away, virtual. When you stand 4 feet away from a mirror, your image will appear to be behind the mirror (virtual) and 4 feet away from the mirror as well. That would put you and your image 8 feet apart. Plane mirrors always for virtual images. .
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