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Unformatted text preview: 60. (a) In a reference frame ﬁxed on Earth, the ether travels leftward with speed v . Thus, the speed of the
light beam in this reference frame is c − v as the beam travels rightward from M to M1 , and c + v
as it travels back from M1 to M . The total time for the round trip is therefore given by
t1 = d1
d1
2cd1
.
+
=2
c−v c+v
c − v2 (b) In a reference frame ﬁxed on the ether, the mirrors travel rightward with speed v , while the speed
of the light beam remains c. Thus, in this reference frame, the total distance the beam has to travel
is given by
d2 = 2 d2 + v
2 2 t2
2 [see Fig. 3637(h)(j)]. Thus,
t2 = d2
2
=
c
c which we solve for t2 :
t2 = √ 2 t2
2 d2 + v
2 , 2d2
.
c2 − v 2 (c) We use the binomial expansion (Appendix E)
(1 + x)n = 1 + nx + · · · ≈ 1 + nx
In our case let x = v/c (x 1) . 1, then L1 = 2c2 d1
v
= 2d1 1 −
2 − v2
c
c and
L2 = √ 2 −1 ≈ 2d1 1 + ≈ 2d2 1 + 1
2 2 − 1/ 2 2cd2
v
= 2d2 1 −
2 − v2
c
c 2 v
c , v
c 2 . Thus, if d1 = d2 = d then
∆L = L1 − L2 ≈ 2d 1 + v
c 2 − 2d 1 + 1
2 v
c 2 = dv 2
.
c2 (d) In terms of the wavelength, the phase diﬀerence is given by
dv 2
∆L
= 2.
λ
λc
(e) We now must reverse the indices 1 and 2, so the new phase diﬀerence is
−∆L
dv 2
=− 2 .
λ
λc
The shift in phase diﬀerence between these two cases is
shift = ∆L
λ −− ∆L
λ = 2dv 2
.
λc2 (f) Assume that v is about the same as the orbital speed of the Earth, so that v ≈ 29.8 km/s (see
Appendix C). Thus,
shift = 2dv 2
2(10 m)(29.8 × 103 m/s)2
=
= 0.40 .
λc2
(500 × 10−9 m)(2.998 × 108 m/s)2 ...
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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, Light

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