hw_ch25 - Ch. 25 HW Solutions 2.) Given the wavelength, ,...

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Ch. 25 HW Solutions 2.) Given the wavelength, λ , the frequency is found using c = λf or f = c/λ = 3 × 10 8 / 633 × 10 - 9 = 4 . 739 × 10 14 Hz. 3.) Given the wavelength in vacuum and knowing the speed of light in vacuum, one can calculate the frequency in vacuum, f = c/λ = 5 . 09 × 10 14 Hz. The frequency of a wave in water will be the same as the frequency in vacuum. (Imagine a wave incident on the water. The E-M wave at the point in space where the water begins will have the frequency of the wave in vacuum, and thus the same frequency will propagate into the water. The wave- lengths in the different media will change as a result of the speed of light being different in different media; this will be discussed in greater detail in later chapters.) Since the frequency is the same in both vacuum and water, we can write c v = λ v f and c w = λ w f where v and w represent vacuum and water respectively. Diving the second equation by the first yields c w /c v = λ w v . Thus, λ w
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This note was uploaded on 07/16/2011 for the course PHY 2005 taught by Professor Lee during the Summer '08 term at University of Florida.

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hw_ch25 - Ch. 25 HW Solutions 2.) Given the wavelength, ,...

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