Modern Physics Homework: Interferometry and Binomial Approximation, Assignments of Physics

A modern physics homework problem about interferometry and the binomial approximation. The problem involves calculating the expected fringe shift for a given interferometer setup and discussing the significance of observing or not observing this small shift. The document also mentions the work of michelson and morley and their improvements to the interferometer design.

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Pre 2010

Uploaded on 02/13/2009

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Homework 1
Physics 2130
Modern Physics
Due: Wednesday September 4, 2002
1. Taylor and Zaratos, problem 1.8.
ABC
D
v
u
u
u
A
D
C
Figure 1: Geometry for problem #1.
We're given that the wind speed
v
= 30 m/s and that sound propagates in the air with a sp eed
of
u
= 330 m/s:
u
A
=
u
,
v
= 300 m/s (1)
u
B
=
u
+
v
= 360 m/s (2)
u
D
=
p
u
2
,
v
2
=328
:
6m/s (3)
2. Taylor and Zaratos:
Exploration of the binomial approximation: (1
,
x
)
n
1
,
nx
for
j
x
j
<<
1.
(a) Problem 1.10.
x
(1
,
x
)
,
1
=
2
1+
x=
2 % dierence
0.001 1.0005004 1.0005000 0.00004%
0.01 1.00504 1.00500 0.004%
0.1 1.054 1.050 0.4%
0.5 1.414 1.250 12%
(b) Problem 1.12.
Using the binomial approximation: (1
,
10
,
20
)
,
1
,
1
1+10
,
20
,
1=
10
,
20
.
3. Taylor and Zaratos, problem 1.15.
The expected fringe shift is (eqn. 1.10 of T&Z):
N
=
2
`
2
=2
0
:
5
m
590
10
,
9
m
3
10
4
m=s
3
10
8
m=s
!
2
=0
:
017 radians
;
(4)
where
=
v=c
. This is about 1/60 of a complete fringe. The negative result of not observing this
small fringe shift was not taken seriously. Michelson and Morley later lengthened the arms of the
interferometer by a factor of about 20 to 11 m, however, and increased this value to about 1/3 of a
complete fringe which was easily observable.
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Homework 1

Physics 2130

Mo dern Physics

Due: Wednesday Septemb er 4, 2002

  1. Taylor and Za ratos, problem 1.8.

A B C

D v

u

u

A u

D C

Figure 1: Geometry for problem #1.

We're given that the wind sp eed v = 30 m/s and that sound propagates in the air with a sp eed of u = 330 m/s:

uA = u v = 300 m/s (1)

uB = u + v = 360 m/s (2) uD =

p

u^2 v 2 = 328 : 6 m/s (3)

2. Taylor and Za ratos: Exploration of the binomial approximation: (1 x)n^  1 nx for

jxj << 1.

(a) Problem 1.10.

x (1 x)^1 =^2 1 + x= 2 % di erence

(b) Problem 1.12. Using the binomial approximation: (1 10 ^20 )^1 1  1 + 10 ^20 1 =

10 ^20.

  1. Taylor and Za ratos, problem 1.15. The exp ected fringe shift is (eqn. 1.10 of T&Z):

N =

2 ` 2

0 : 5 m

590  10 ^9 m

3  104 m=s

3  108 m=s

= 0 : 017 radians; (4)

where = v =c. This is ab out 1/60 of a complete fringe. The negative result of not observing this small fringe shift was not taken seriously. Michelson and Morley later lengthened the arms of the interferometer by a factor of ab out 20 to 11 m, however, and increased this value to ab out 1/3 of a complete fringe which was easily observable.

  1. Taylor and Za ratos, problem 1.16. If one of the arms lengthens by amount , the interferometer will have two arms of length and + . To nd the resulting phase shift, N , caused by the increased one-way length =T , we must compare this extra length traveled by light along the lengthened arm. Light will travel an extra distance 2= 2T when you include its transit down and back the lengthened arm. The phase shift will dep end on the size of this distance relative to the wavelength, , of light:

N =

2 `T

2  (0: 5  C ^1 )  (0: 5 m)  (10^2  C ) 590  10 ^9 m

= 0 : 2 radians: (5)

Note: There is the p otential for notational confusion here. In this case, N is the phase shift caused by the di erence in arm lengths of the interferometer pro duced by temp erature variations across the interferometer. In the text, the phase shift caused by the earth's motion relative to the aether frame was referred to as N and N was the phase shift up on rotation of the interferometer.