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 EM 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
/λ
v
. Thus,
λ
w
= 2
.
25
/
3
.
00
λ
v
= 442
nm.
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 Summer '08
 Lee
 Applied Physics, Light, total power

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