Phys 181b Problem Set 9 Solution

Phys 181b Problem Set 9 Solution - 2. (a) From Fig. 33-2 we...

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2. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm, (b) and the larger wavelength to be 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 (5.41 × 10 14 Hz) –1 = 1.85 × 10 –15 s.
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4. 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) Hz. 2
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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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10. 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) We use I = 2 m E /2 µ 0 c to calculate E m : EI mc == × × × 2 2 4 10 140 10 2 998 10 103 10 0 73 8 3 π Tm /A W /m m /s V/m. 2 c hc hc h .. . (b) The magnetic field amplitude is therefore B m × × E c m 2 998 10 343 10 4 8 6 . . . V/m m/s T.
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37. As the polarized beam of intensity I 0 passes the first polarizer, its intensity is reduced to
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Phys 181b Problem Set 9 Solution - 2. (a) From Fig. 33-2 we...

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