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24.
(a) We note that the cross section area of the beam is
πd
2
/
4, where
d
is the diameter of the spot
(
d
=2
.
00
λ
). The beam intensity is
I
=
P
2
/
4
=
5
.
00
×
10
−
3
W
π
[(2
.
00)(633
×
10
−
9
m)]
2
/
4
=3
.
97
×
10
9
W
/
m
2
.
(b) The radiation pressure is
p
r
=
I
c
=
3
.
97
×
10
9
W
/
m
2
2
.
998
×
10
8
m
/
s
=13
.
2Pa
.
(c) In computing the corresponding force, we can use the power and intensity to eliminate the area
(mentioned in part (a)). We obtain
F
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 Fall '08
 SPRUNGER
 Physics, Radiation

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