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332
CHAPTER 24
From the figure, note that
()
2
22
2 2
2
Rr R
t rR R
t
t
2
=
+−=+− +
which reduces to
. Since
t
will be very small in comparison to either
r
or
R
, we may neglect the
term
, leaving
2
2
rR
t
=−
2
t
2
t
2
≈
t
.
For a dark fringe,
2
film
m
n
t
λ
=
so the radii of the dark rings will be
2
2
film
m
n
≈
=
film
mR
n
for
0, 1, 2,
m
=
…
24.65
If the signal from the antenna to the
receiver station is to be completely
polarized by reflection from the water,
the angle of incidence where it strikes
the water must equal the polarizing
angle from Brewster’s law. This is
given by
11
water
air
tan
tan
1.33
53.1
p
n
n
θ
−−
==
() ()
=
°
From the triangle RST is the sketch, the horizontal distance from the point of refection,
T, to shore is given by
( )
90.0 m tan
90.0 m
1.33
1
p
x
20 m
(
=
and from triangle ABT, the horizontal distance from the antenna to this point is
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This document was uploaded on 10/20/2011.
 Spring '09
 Physics, Work

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