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英语代写-V00910618
时间：2021-03-04
2021/2/3 Xinyu Wang
V00910618
Experiment 6
Interferometry
Abstract: Michelson Interferometer is important in model physics. It is designed for
Michelson-Morley Experiment to verify the principle of constant speed of light. The
Interferometer is used to generate interference fringes. By studying the change of
interference fringes, we could do some research on refractive index of different objects.
Introduction and Theory:
Interferometers produce interference
patterns by the division and recombination of
light. A sketch of the essential parts of a
Michelson interferometer is shown in Figure
1.
Light from source S is divided into two
paths at the beam splitter A by the partially
silvered front surface mirror. One beam
passes through A to fixed mirror M2 while
the other reflects from A, through the
compensator plate C, to mirror M1. Referring
to Figure 1, on reflection from mirrors M1
and M2 the beams retrace their paths to A
where they recombine to form interference
fringes.
When the compensator plate is used, the
light has the same path length in glass in each
arm of the interferometer (since the plates A
and C are identical in thickness and are
parallel). The difference in the path of the
two beams is then said to be entirely “in-air”.
If the Michelson interferometer is used in
conjunction with a monochromatic source,
such as a sodium lamp or the green line of a
mercury lamp, interference fringes are found
to exist over a path length difference of a few
centimeters. As the path length difference is
gradually increased, the contrast of the
fringes diminishes and finally the fringes
disappear. The fringes eventually vanish
when the difference in path length exceeds
the predominant coherence path length of the
source. Thus, for fringes to be visible, there
must not be an independent change in phase
of the light wave train in either path. The
laser, on the other hand, is said to have a
temporal coherence or coherence path length
orders of magnitude longer than any other
source.
Referring to Figure 1, to have a change of one
fringe requires a change in path length
between the rays to M1 and M2 of one
wavelength. For a change of n fringes,
2 = (1)
Where 2 is the path difference for a
change of fringes and is the
wavelength in air.
Apparatus:
Fig 1
Apparatus(1)
Apparatus(2)
The length of vacuum cell
Procedure:
Use the equation (1) and we know that
wavelength of the He-Ne laser is 632.8nm.
So the value of 2 for a count of 100 fringes
is
2 = 100 × 632.8 = 632.8 × 10−7
The path difference for a count of 100
fringes is
62.1 × 5.09 × 10−7 = 316.089 × 10−7
The reading of the gear ratio is 62.3. So
2 determined from the gear ratio is
2 = 632.178 × 10−7
I think the value above which is 632.8 ×
10−7 is more accurate. eecause the
refractive index of air is not 1, though we
often take it as 1. The refractive index of
vacuum is 1. We should use equation (3)
added the refractive index of air to calculate
the path difference,
2 = (3)
Where = index of refraction of air.
To determine the refractive index of air,
use the equation (2),
2( − 1) = (2)
where,
= wavelength of the laser light in vacuum
= number of fringes observed
= length of path through the cell (to be
measured with travelling microscope).
The length of path through the cell is measure,
= 51.28
The laser light in vacuum is He-Ne laser. The
wavelength of the laser light is,
= 632.8
In experiment, we could see 44 fringes. In
this case = 44.
So,
=
2
+ 1
=
44 × 632.8
2 × 51.28
+ 1 ≈ 1.00027
Use equation (3),
2 = (3)
Where,
2 = 2 × 1.00027 × 316.089 × 10−7
= 632.35 × 10−7
The error could be represented by,
= (632.8 − 632.35) ÷ 632.8 × 100%
= 0.07%
eecause reading the gear ratio has an error,
= 0.1 ÷ 62.3 × 100% = 0.16%
The calculated error is less than the reading
error. We could consider the previous
calculation result to be correct.
Discussion and Question:
Derive equation (2)
2( − 1) = (2)
Where,
= index of refraction of air
= wavelength of the laser light in vacuum
= number of fringes observed
= length of path through the cell (to be
measured with travelling microscope).
The refractive index of vacuum could see
as 1. eecause the path difference is caused by
the difference of the refractive index of air
and vacuum, which could be represented by,
2( − 1)
It could be seemed as 2. Use equation (1).
We have equation (2),
2( − 1) = (2)
The different path has two values,
theoretical value and measurements. The
theoretical value is more accurate. eecause
the value we measured contains a lot of errors
and something should be corrected. In the
initial calculation, we use the equation (1)
which ignores the refractive index of air .
In fact, the thickness of mirrors should be
considered also. If we use the same mirrors,
the influence of the thickness of mirrors
could be ignored. eecause the rays to M1 and
M2 travel the same length in mirrors.
The errors could be concluded as three
points. First, the scale value of the instrument
will cause error. It just depends on the
accuracy of the instrument. Second, what we
read from gear ratio has an error about ±0.5.
If we use the length as the unit to express the
error, it could be ±0.5 × 5.09 × 10−7 .
The influence of this error could be reduced
by increasing the number of fringes we count.
In the above calculation, we only count 100
fringes. The relative error could be smaller, if
we count 200 fringes or more fringes. Third,
the reading of the gear ratio has an accidental
error. We should read the gear ratio more
times and more different 100 fringes to
reduce the error.
Conclusion:
This lab uses interference equation to
calculate the refractive index of air. The
result of calculation of path difference
in theory is different from the path
difference which we get from reading the
gear ratio. We could infer that there is
something wrong. Because the refractive
index of air is ignored. We could use
equation (2) to calculate the refractive
index of air. The result is very close to
the theoretical value. However, the
accuracy of experiment is not high enough.
The result has an error, which is
inevitable, compared with the theoretical
value.
学霸联盟
V00910618
Experiment 6
Interferometry
Abstract: Michelson Interferometer is important in model physics. It is designed for
Michelson-Morley Experiment to verify the principle of constant speed of light. The
Interferometer is used to generate interference fringes. By studying the change of
interference fringes, we could do some research on refractive index of different objects.
Introduction and Theory:
Interferometers produce interference
patterns by the division and recombination of
light. A sketch of the essential parts of a
Michelson interferometer is shown in Figure
1.
Light from source S is divided into two
paths at the beam splitter A by the partially
silvered front surface mirror. One beam
passes through A to fixed mirror M2 while
the other reflects from A, through the
compensator plate C, to mirror M1. Referring
to Figure 1, on reflection from mirrors M1
and M2 the beams retrace their paths to A
where they recombine to form interference
fringes.
When the compensator plate is used, the
light has the same path length in glass in each
arm of the interferometer (since the plates A
and C are identical in thickness and are
parallel). The difference in the path of the
two beams is then said to be entirely “in-air”.
If the Michelson interferometer is used in
conjunction with a monochromatic source,
such as a sodium lamp or the green line of a
mercury lamp, interference fringes are found
to exist over a path length difference of a few
centimeters. As the path length difference is
gradually increased, the contrast of the
fringes diminishes and finally the fringes
disappear. The fringes eventually vanish
when the difference in path length exceeds
the predominant coherence path length of the
source. Thus, for fringes to be visible, there
must not be an independent change in phase
of the light wave train in either path. The
laser, on the other hand, is said to have a
temporal coherence or coherence path length
orders of magnitude longer than any other
source.
Referring to Figure 1, to have a change of one
fringe requires a change in path length
between the rays to M1 and M2 of one
wavelength. For a change of n fringes,
2 = (1)
Where 2 is the path difference for a
change of fringes and is the
wavelength in air.
Apparatus:
Fig 1
Apparatus(1)
Apparatus(2)
The length of vacuum cell
Procedure:
Use the equation (1) and we know that
wavelength of the He-Ne laser is 632.8nm.
So the value of 2 for a count of 100 fringes
is
2 = 100 × 632.8 = 632.8 × 10−7
The path difference for a count of 100
fringes is
62.1 × 5.09 × 10−7 = 316.089 × 10−7
The reading of the gear ratio is 62.3. So
2 determined from the gear ratio is
2 = 632.178 × 10−7
I think the value above which is 632.8 ×
10−7 is more accurate. eecause the
refractive index of air is not 1, though we
often take it as 1. The refractive index of
vacuum is 1. We should use equation (3)
added the refractive index of air to calculate
the path difference,
2 = (3)
Where = index of refraction of air.
To determine the refractive index of air,
use the equation (2),
2( − 1) = (2)
where,
= wavelength of the laser light in vacuum
= number of fringes observed
= length of path through the cell (to be
measured with travelling microscope).
The length of path through the cell is measure,
= 51.28
The laser light in vacuum is He-Ne laser. The
wavelength of the laser light is,
= 632.8
In experiment, we could see 44 fringes. In
this case = 44.
So,
=
2
+ 1
=
44 × 632.8
2 × 51.28
+ 1 ≈ 1.00027
Use equation (3),
2 = (3)
Where,
2 = 2 × 1.00027 × 316.089 × 10−7
= 632.35 × 10−7
The error could be represented by,
= (632.8 − 632.35) ÷ 632.8 × 100%
= 0.07%
eecause reading the gear ratio has an error,
= 0.1 ÷ 62.3 × 100% = 0.16%
The calculated error is less than the reading
error. We could consider the previous
calculation result to be correct.
Discussion and Question:
Derive equation (2)
2( − 1) = (2)
Where,
= index of refraction of air
= wavelength of the laser light in vacuum
= number of fringes observed
= length of path through the cell (to be
measured with travelling microscope).
The refractive index of vacuum could see
as 1. eecause the path difference is caused by
the difference of the refractive index of air
and vacuum, which could be represented by,
2( − 1)
It could be seemed as 2. Use equation (1).
We have equation (2),
2( − 1) = (2)
The different path has two values,
theoretical value and measurements. The
theoretical value is more accurate. eecause
the value we measured contains a lot of errors
and something should be corrected. In the
initial calculation, we use the equation (1)
which ignores the refractive index of air .
In fact, the thickness of mirrors should be
considered also. If we use the same mirrors,
the influence of the thickness of mirrors
could be ignored. eecause the rays to M1 and
M2 travel the same length in mirrors.
The errors could be concluded as three
points. First, the scale value of the instrument
will cause error. It just depends on the
accuracy of the instrument. Second, what we
read from gear ratio has an error about ±0.5.
If we use the length as the unit to express the
error, it could be ±0.5 × 5.09 × 10−7 .
The influence of this error could be reduced
by increasing the number of fringes we count.
In the above calculation, we only count 100
fringes. The relative error could be smaller, if
we count 200 fringes or more fringes. Third,
the reading of the gear ratio has an accidental
error. We should read the gear ratio more
times and more different 100 fringes to
reduce the error.
Conclusion:
This lab uses interference equation to
calculate the refractive index of air. The
result of calculation of path difference
in theory is different from the path
difference which we get from reading the
gear ratio. We could infer that there is
something wrong. Because the refractive
index of air is ignored. We could use
equation (2) to calculate the refractive
index of air. The result is very close to
the theoretical value. However, the
accuracy of experiment is not high enough.
The result has an error, which is
inevitable, compared with the theoretical
value.
学霸联盟
