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27. (a) To get to the detector, the wave from
S
1
travels a distance
x
and the wave from
S
2
travels a distance
√
d
2
+
x
2
. The phase diFerence (in terms of wavelengths) between the two waves is
p
d
2
+
x
2
−
x
=
mλ
m
=0
,
1
,
2
,...
where we are requiring constructive interference. The solution is
x
=
d
2
−
m
2
λ
2
2
mλ
.
The largest value of
m
that produces a positive value for
x
is
m
= 3. This corresponds to the
maximum that is nearest
S
1
,at
x
=
(4
.
00 m)
2
−
9(1
.
00 m)
2
(2)(3)(1
.
00 m)
=1
.
17 m
.
±or the next maximum,
m
=2and
x
=3
.
00 m. ±or the third maximum,
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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