University:
California State University, Northridge
Course:
PHYS 100B | General Physics II
Academic year:

2008
Views:

294

Pages:

18
Author:

Shaylee Vance

Chapter 25 Electromagnetic Waves Outline 25-1 The production of Electromagnetic Waves 25-3 The Electromagnetic Spectrum 25-4 Energy gy and momentum in Electromagnetic g Waves 25-5 Polarization 25-1 The production of Electromagnetic Waves Electromagnetic waves: Visible light (440 ~ 700nm) is a kind of electromagnetic waves Electromagnetic waves are transverse waves: the vibration of each point is vertical to the propagation direction. An Electromagnetic wave consists ofG two vertical G components: E (electric filed) and B (Magnetic field) field). G G The electric field vector E , magnetic g field B , and the propagation p p g are all vertical each other ! The Electromagnetic waves travel through vacuum (air) at the same speed: C = 3.00 × 108 m/s, the speed of light ! Figure 25-3 An Electromagnetic Wave at an instant moment t. Direction of Propagation for Electromagnetic Waves G Point the fingers g of yyour g hand in the direction of the E , curl G right your fingers toward B , and your thumb will point in the direction of the propagation. 25-3 Electromagnetic Spectrum When white light passes through a prism, it spread out into a rainbow of different colors. This indicates that the electromagnetic waves have different wavelengths, since different colors have different wavelength. The wavelength and frequency is associated each other as, c = fλ (25 − 4) Where c is the speed of the electromagnetic wavelength in vacuum and is a constant. C is in m/s. λ Is in m f is in Hz. Wavelength λ, frequency f, and propagation direction. Example a p e 25-3 53 Find the frequency of red light with a wavelength of 700.0 nm. Solution Since c=fλ, we have, 3.00 × 108 m / s = f (700.0 × 10−9 m) 3.00 3 00 ×108 m / s 14 f = = 4.29 × 10 Hz −9 700.0 × 10 m Example 25-3 The spectrum of white light. Figure 25-8 The Electromagnetic Spectrum 25-4 Energy and momentum in Electromagnetic Waves Since energy density (energy per volume) of electric field (Section 206) is, 1 uE = ε 0 E 2 2 And since energy density of magnetic field (Section 23-9) is, uB = 1 2μ0 B2 As the electromagnetic wave is composed of electric and magnetic fields, the total energy density of the electromagnetic wave is 1 1 2 2 u = uB + uE = ε 0 E + B 2 2μ0 (25 − 5) It can be shown that electric and magnetic energy densities are equal. l So, S one h has 1 1 2 1 2 2 2 u = ε0E + B = ε0E = B 2 2μ0 μ0 (25 − 6) Since E and B are sinusoidal functions of the time, we must use RMS value to calculate the “average” average value, value 1 1 2 1 2 2 2 uav = ε 0 Erms + Brms = ε 0 Erms = Brms 2 2μ0 μ0 (25 − 7) Where, Emax 2 B = max 2 Erms = Brms (25 − 8) The relationship of E and B Since, We have ε0E2 = E= Pemittivity of Permeability of 1 μ0 B2 , 1 ε 0 μ0 from (25 − 6) B free space ε 0 = 8.85 × 10−12 C 2 /( N ⋅ m 2 ) free space μ0 = 4π × 10−7 T ⋅ m / A c = 1/ ε 0 μ0 = 3.00 × 108 m / s so we eventually have E = cB ( Erms = cBrms ) (25 − 9) (19 − 12) (22 − 8) Exercise 25-3 At a given instant of time, the electric field in a beam of sunlight has a magnitude of 510 N/C. What is the magnitude of the magnetic field at this instant? Solution: E = Bc 510 N / C = B(3.00 × 108 m / s ) E 510 N / C −6 B= = = 1.7 × 10 T 8 c 3.00 ×10 m / s Intensity I The amount of the energy gy a wave delivers to a unit area in a unit time is called Intensity I. Also I = P/A In figure 25-9, the intensity I is I= ΔU u ( AcΔt ) = = uc AΔt AΔt Substitute (25 (25-6), 6) we have c 2 1 1 2 2 2 I = cε 0 E + cB = cε 0 E = B 2 2μ0 μ0 Figure 25-9 The Energy in a Beam of Light (25 − 10) To calculate the average intensity, we only need to replace the E and B with their rms value in the above formula. Example 25-4 A garage is illuminated by a single light bulb. If the bulb radiate light uniformly in all directions, and consumes an average electrical power of 50.0 W. Assume that 5.00% of the electric power consumed by the bulb is converted to light. What are (a) The average intensity of the bulb at a distance of 1.00 m from the bulb, and (b) the rms values of E and B at a distance of 1.00 m from the bulb? Active Example 25-1 A smallll llaser emits it a cylindrical li d i l b beam off lilight ht 1 1.00 00 mm iin di diameter t with average power of 5.00 mW. Find the maximum values of E and B in the beam. Summary 1) Electromagnetic waves are transverse waves and it consists of two vertical components: electric filed and magnetic field field. 2)) Energy gy in Electromagnetic g Waves