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Moduson table
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223 008
The results are summarised in the table below:
The angles are given in algebraic notation relative to the graduation zero.
θ minimum
minimum difference
θ maximum
minimum difference
We observe that for angles in the range of -25 deg. to +25 deg., the variation of the angle θ
from one peak to the adjacent peak (and from a minimum to the adjacent minimum) is
constant and equal to:
As for the number of peaks and of minima, we have:
The differences between the peaks and the minima is identical and there are regularly
alternated.
Note: The results are similar with other values. For example with a=8 cm and D=30 cm, the mean
difference is ∆θ ≈ 6°
3.2 Step 3: Exploitation of the results
Theoretical study of the interference phenomenon
The wavelength is λ.
In the above setup schematic, the course difference is expressed as δ=d
d
and d
2
The following relationships can be expressed:
2
d
=hM
2
2
d
=hM
1
The difference is written as follows:
2
d
2
or δ=2 aDsin(θ)/(d
or else:
ENGLISH
Not measurable
Not measurable
∆θ ≈ 12°
•
8 peaks (the 9th one could not be read)
•
7 minima (the 8th one could not be read)
The superposition model of the waves from the two synchronous sources is used.
•
At an arbitrary point M, the waves from the sources S
difference δ=d
-d
.
2
1
•
If the two waves arrive in phase at point M, they have constructive a effect and their
amplitudes are added: this yields a maximum amplitude point called internode.
This condition is rendered by
•
If the two waves arrive in opposition of phase at point M, they have a destructive
effect and their amplitudes are subtracted: this yields a point with zero amplitude (if
the amplitudes are equal) or a low amplitude (equal to the amplitude difference),
called node.
This condition is rendered by
Calculation of the course difference
are the hypotenuses of two rectangular triangles (MhS
1
2
2
2
+hS
=Ov
+ (Oh+a/2)
2
2
2
2
+hS
=Ov
+ (Oh- a/2)
1
)= δ(d
2
- d
=(d
-d
)(d
+d
+d
1
2
1
2
1
2
+d
)
2
1
δ
=
2
a
+
+
2
D
4
-32
-18
-6
14
12
-38
-24
-12
14
12
12
δ=(2k+1). λ/2.
2
2
2
=D
cos
(θ)+(Dsin(θ)+a/2)
2
2
2
=D
cos
(θ)+(Dsin(θ)- a/2)
)=2 aDsin(θ)
1
θ
2
aD
sin(
)
2
a
θ
+
+
−
2
aD
sin(
)
D
4
48
6
18
32
12
12
14
15
0
11.5
24
38
11.5
12.5
14
and S
exhibit a course
1
2
δ=k. λ.
-d
2
1.
and MhS
).
1
2
2
2
2
=D
+a
/4+aDsin(θ)
2
2
2
=D
+a
/4- aDsin(θ)
θ
aD
sin(
)
47
54
16
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