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 8143.0010 m/s5.41 10Hz555 nmcf×===×λ. (e) The period is T = 1/f= (5.41 ×1014Hz)–1= 1.85 ×10–15s.
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