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we find
12
0
2
cos( / 2)sin(
/ 2)
EE E
t
φ
ωφ
+=
+
where
E
o
= 2.00 μV/m,
ω
= 1.26
×
10
15
rad/s, and
=
39.6 rad.
This shows that the
electric field amplitude of the resultant wave is
0
2
cos( / 2)
2(2.00 V/m)cos(19.2 rad)
2.33 V/m
EE
µµ
==
=
.
(b) Eq. 3522 leads to
2
00
4
cos ( / 2) 1.35
I
II
at point
P
,
and
2
center
0
0
4c
o
s
(
0
)4
I
at the center . Thus,
center
/
1.35/ 4
0.338
.
(c) The phase difference
φ
(in wavelengths) is gotten from
φ
in radians by dividing by 2
π.
Thus,
φ = 39.6/2π = 6.3
wavelengths.
Thus, point
P
is between the sixth side maximum
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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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