page 161
35.5.0.1.3 - Interpreting Interference Patterns
• Lets consider a lapped surface that is examined under an optical flat, with a helium light (with a wavelength of 23.2 µ in.).
Case 1: Parallel equally spaced lines indicate that the surface is flat.
Question: why are the lines
in a diagonal pattern instead of vertical?
Case 2: Curves lines in a regular pattern
This end is in contact with the work.
a
b
This surface curves up in the centre, but in a uniform way. Therefore the part might look something like the exaggerated view below,
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a |
× |
λ |
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b |
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page 162
Case 3: There is a sudden change in the pattern
n2 = 9
This end is in contact 
n1 = 3 with the work.
The slope of the surface changes part way along the optical flat. The fact that in both of the two regions the lines are parallel and uniform, means the work surface is flat.
y
x
x = n1 × λ--2 =
y = n2 × λ--2 =
Case 4: A radial circular pattern is observed
Practice: Draw the Section this would represent.
page 163
35.5.0.1.4 - Types of Interferometers
•There are three basic types of interferometers for surface texture measurement
-Interference microscopes
-flatness type interferometer
-gauge block interferometer
•Interference Microscopes - used for analysis of surface texture analysis. The example below shows the instrument being used to examine a gauge block.
b |
S |
microscope
T
optical flat
lapping
texture
scratch depth |
= |
S |
× |
λ |
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-- |
-- |
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b |
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2 |
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texture height |
= |
T |
× |
λ |
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-- |
-- |
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b |
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2 |
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T |
× |
λ |
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Ra(estimated) |
= |
----- |
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4b |
2 |
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gauge block
This assumes a triangular shape of the surface as discussed earlier.
Note:
-The lay on gauge blocks is along the length
-The Mercury wavelength is approximately 20 µ in.
•Flatness type interferometer - this type of interferometer is used to compare the parallelism of the top and bottom faces of the gauge blocks. This microscope has a base plate with a lapped
page 164
finish. The gauge blocks are wrung to the surface, and then the patterns on the top and bottom are compared.
microscope
optical flat
base plate fringe pattern
gauge block
gauge block fringe pattern
Because the fringe patterns for the gauge block, and the base plate have the same spacing, the two sides of the gauge block are parallel.
page 165
Practice Problem: Draw sections for the views below, and estimate errors in
flatness and parallelism. (These have been taken on a flatness type interferometer, using 20 in. wavelength light)
A
Front
C
A |
B |
B |
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Front |
D |
D |
C |
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Front |
Front |
• Gauge Block Interferometer - This instrument is used for height measurements. using four different frequencies of light. (red, green, blue, violet). The example below shows and example of its use.
page 166
view
b
proof plane
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a |
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h |
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front |
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H |
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G
front
The gauge block is a 0.3” specimen, and it will be measured with the red, green, and violet light components.
G = H – h = n × |
λ |
a |
× |
λ |
= |
λ |
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a |
-- + -- |
-- |
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n + -- |
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2 |
b |
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2 |
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2 |
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b |
where, |
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λ |
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a |
λ |
n = whole number of --2 |
intervals, and |
b-- |
= the fraction of --2 |
For each frequency of light used, there will be a different ‘n’ value. In addition, the a/b fractions will also differ.
f1 = |
ared |
f2 |
= |
agreen |
f3 |
aviolet |
b---------red |
b-------------green |
= ------------- |
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bviolet |
Next, we will relate the three readings,
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λ 1 |
( n1 |
+ f1) = |
λ 2 |
( n2 |
+ f2) = |
λ 3 |
( n3 |
+ f3) |
G = |
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----- |
----- |
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2 |
2 |
2 |
page 167
Next, use the ideal value of the gauge block, and then combine it in with the previous equations,
G |
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= |
λ 1 |
( N |
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+ F ) = |
λ |
2 |
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+ F ) |
= |
λ 3 |
( N |
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+ F ) |
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1 |
-----( N |
2 |
----- |
3 |
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N |
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2 |
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1 |
2 |
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where, |
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GN |
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F |
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GN |
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N1 |
= int |
---------- |
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= ---------- – N |
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λ 1 |
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λ 1 |
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----- |
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N2, F2, N3, F3 is similar to the above equations
Assuming G > GN, we can combine the equations,
G – G |
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λ 1 |
[ ( n |
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– N ) + ( f |
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– F ) ] |
N |
----- |
1 |
1 |
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1 |
1 |
The equation above is similar for the other two colours
The calculations continue to find values for (n1-N1), (n2-N2), (n3-N3), these will be small whole numbers.
• To consider an example of measuring the height of a gauge block using the interferometer,
page 168
Given that a gauge block has a height of 0.3”, and we want to find the height in more detail. We are using the gauge block interferometer, with red, green, and violet
light.
The basic observations, and known values to start are,
red |
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f |
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a |
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0.35 |
λ |
1 |
= |
25.348478µ |
in. |
1 |
-- |
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b |
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green |
λ |
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= |
20.023055µ |
in. |
f2 |
= |
a |
= |
0.35 |
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2 |
-- |
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f |
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a |
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0.95 |
violet |
λ |
3 |
= |
18.418037µ |
in. |
3 |
-- |
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b |
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The following equations must be solved, |
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λ |
1 |
[ ( n1 – N1) + ( f1 |
– F1) ] |
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G – GN = |
----- |
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2 |
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λ |
2 |
[ ( n2 – N2) + ( f2 |
– F2) ] |
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G – GN = |
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2 |
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λ |
3 |
[ ( n3 – N3) + ( f3 |
– F3) ] |
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G – GN = |
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The solution to these equations is not direct, but a table can help find the values,
colour |
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f |
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N |
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F |
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f-F |
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coincide |
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mean error |
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at |
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in gauge |
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X10-6” |
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fraction |
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block error |
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a/b |
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red |
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25.35 |
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0.35 |
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23670.06 |
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0.06 |
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0.29 |
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2.29 |
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28.7 µ in. |
green |
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20.02 |
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0.35 |
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29965.46 |
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0.46 |
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0.89** |
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2.84 |
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blue |
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18.42 |
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0.95 |
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32576.76 |
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0.76 |
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0.19 |
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3.13 |
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** this value is arrived at by 1.35 - 0.46, not 0.35 - 0.46