Chapter33 - 1. (a) From Fig. 33-2 we find the smaller...

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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.
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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 dc t == × × = (. . 30 10 0 30 89 m / s)(1.0 10 s) m.
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3. Since ∆λ << λ , we find f is equal to ∆λ cc λλ F H G I K J ≈= ×× × 2 89 9 9 30 10 632 8 10 749 10 (. . m / s)(0.0100 10 m) m) Hz. 2
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4. (a) The frequency of the radiation is f c == × ×× λ 30 10 10 10 6 4 10 47 10 8 56 3 . (. ) . m/s m) Hz. (b) The period of the radiation is T f × 11 212 3 32 3 . min Hz s s .
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λ π 2 LC c = . The solution for L is L Cc == × ×× λ π π 2 22 9 2 21 2 8 2 21 4 550 10 4 17 10 2 998 10 500 10 m Fm / s H. c h ch c h . . This is exceedingly small. 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, fL C = 12 / π , where C is the capacitance and L is the inductance. Thus
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6. The emitted wavelength is () ( ) ( ) 86 1 2 2 2 2.998 10 m/s 0.253 10 H 25.0 10 F 4.74m. c cL C f −− λ= =π × × × =
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7. If P is the power and t is the time interval of one pulse, then the energy in a pulse is EP t ==× × = × 100 10 10 10 10 10 12 9 5 Ws J . c hc h ..
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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.
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9. (a) The amplitude of the magnetic field is 99 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 G 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 EB × GG . If the electric field points in + z , then the magnetic field must point in the – y direction. With SI units understood, we may write ( ) () 15 15 8 91 5 2.0cos 10 / cos 10 3.0 10 6.7 10 cos 10 ym txc x BB t c x t c π− ⎡⎤ ⎛⎞ × = ⎜⎟ ⎢⎥ × ⎝⎠ ⎣⎦ π
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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 . . . V/m m/s T. (b) The intensity is the average of the Poynting vector: IS E c m = ×⋅ × avg 2 Tm /A m /s W/m 2 0 2 78 2 2 2 4 10 2 998 10 331 10 µ . . .. bg c hc h π
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11. The intensity is the average of the Poynting vector: IS cB m == = ×× × avg 2 m/s T H/m W/m 2 0 84 2 6 2 6 2 30 10 10 10 2 126 10 12 10 µ .. . c hc h ch
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12. The intensity of the signal at Proxima Centauri is I P r == × × 4 10 10 4 4 3 9 46 10 48 10 2 6 15 2 29 π π . .. W ly m / ly W/m 2 b gc h
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13. (a) The magnetic field amplitude of the wave is B E c m m == × 20 2 998 10 67 10 8 9 .
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter33 - 1. (a) From Fig. 33-2 we find the smaller...

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