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# Chapter-33 - 1281 Chapter 33 1(a From Fig 33-2 we find the...

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Unformatted text preview: 1281 Chapter 33 1. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm. (b) Similarly, the larger wavelength is approximately 610 nm. (c) From Fig. 33-2 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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Chapter-33 - 1281 Chapter 33 1(a From Fig 33-2 we find the...

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