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Unformatted text preview: 1281 Chapter 33 1. (a) From Fig. 332 we find the smaller wavelength in question to be about 515 nm. (b) Similarly, the larger wavelength is approximately 610 nm. (c) From Fig. 332 the wavelength at which the eye is most sensitive is about 555 nm. (d) Using the result in (c), we have 8 14 3.00 10 m/s 5.41 10 Hz 555 nm c f × = = = × λ . (e) The period is T = 1/ f = (5.41 × 10 14 Hz) –1 = 1.85 × 10 –15 s. 2. In air, light travels at roughly c = 3.0 × 10 8 m/s. Therefore, for t = 1.0 ns, we have a distance of d ct = = × × = ( . . 30 10 0 30 8 9 m / s) (1.0 10 s) m. 3. Since ∆λ << λ , we find ∆ f is equal to ∆ ∆λ c c λ λ F H G I K J ≈ = × × × = × 2 8 9 9 9 30 10 632 8 10 7 49 10 ( . ( . . m / s)(0.0100 10 m) m) Hz. 2 4. (a) The frequency of the radiation is f c = = × × × = × λ 30 10 10 10 6 4 10 4 7 10 8 5 6 3 . ( . )( . . m / s m) Hz. (b) The period of the radiation is T f = = × = = 1 1 4 7 10 212 3 32 3 . min Hz s s. 5. If f is the frequency and λ is the wavelength of an electromagnetic wave, then f λ = c . The frequency is the same as the frequency of oscillation of the current in the LC circuit of the generator. That is, f LC = 1 2 / π , where C is the capacitance and L is the inductance. Thus CHAPTER 33 1282 λ π 2 LC c = . The solution for L is L Cc = = × × × = × λ π π 2 2 2 9 2 2 12 8 2 21 4 550 10 4 17 10 2 998 10 500 10 m F m / s H. c h c hc h . . This is exceedingly small. 6. The emitted wavelength is ( 29 ( 29( 29 8 6 12 2 2 2.998 10 m/s 0.253 10 H 25.0 10 F 4.74m. c c LC f λ = = π = π × × × = 7. If P is the power and ∆ t is the time interval of one pulse, then the energy in a pulse is E P t = = × × = × ∆ 100 10 10 10 10 10 12 9 5 W s J. c hc h . . 8. The amplitude of the magnetic field in the wave is B E c m m = = × × = × 320 10 2 998 10 107 10 4 8 12 . . . V / m m / s T. 9. (a) The amplitude of the magnetic field is 9 9 8 2.0V/m 6.67 10 T 6.7 10 T. 2.998 10 m/s m m E B c = = = × ≈ × × (b) Since the wave E r oscillates in the z direction and travels in the x direction, we have B x = B z = 0. So, the oscillation of the magnetic field is parallel to the y axis. (c) The direction (+ x ) of the electromagnetic wave propagation is determined by E B × r r . If the electric field points in + z , then the magnetic field must point in the – y direction. With SI units understood, we may write ( 29 ( 29 15 15 8 9 15 2.0cos 10 / cos 10 3.0 10 6.7 10 cos 10 y m t x c x B B t c x t c π = π× = × = × π 1283 10. (a) The amplitude of the magnetic field in the wave is B E c m m = = × = × 500 2 998 10 167 10 8 8 ....
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 Summer '10
 Prof.Yang
 Physics

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