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(j) When
1.500 (
3/ 2)
Lm
λ
∆=
=
, the interference is fully destructive.
(k) Using the formula obtained in part (a), we have
x
= 2.29
µ
m for
3/2
m
=
.
(l) When
2.00 (
2)
=
, the interference is fully constructive.
(m) Using the formula obtained in part (a), we have
1.50 m
x
=
for
m
= 2.
(n) When
2.500 (
5/ 2)
=
, the interference is fully destructive.
(o) Using the formula obtained in part (a), we have
x
= 0.975
m for
5/2
m
=
.
122. (a) To get to the detector, the wave from
S
1
travels a distance
x
and the wave from
S
2
travels a distance
dx
22
+
. The phase difference (in terms of wavelengths) between the
two waves is
0,1, 2,
dxx
m
m
+−
=
λ
=
…
where we are requiring constructive interference. The solution is
x
dm
m
=
−
2
2
λ
λ
.
We see that setting
m
= 0 in this expression produces
x
=
∞
; hence, the phase difference
between the waves when
P
is very far away is 0.
(b) The result of part (a) implies that the waves constructively interfere at
P
.
(c) As is particularly evident from our results in part (d), the phase difference increases as
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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