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34. (a) From
f
=
c/λ
and Eq. 3830, we get
λ
0
=
λ
s
1
−
β
1+
β
=(
λ
0
+∆
λ
)
s
1
−
β
1+
β
.
Dividing both sides by
λ
0
leads to
1=(1+
z
)
s
1
−
β
1+
β
.
We solve for
β
:
β
=
(1 +
z
)
2
−
1
(1 +
z
)
2
+1
=
z
2
+2
z
z
2
+2
z
+2
.
(b) Now
z
=4
.
43, so
β
=
(4
.
43)
2
+2(4
.
43)
(4
.
43)
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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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