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Physics 9BA
Assignment #6
Cole
[36Q10]
Using Rayleigh's criterion, the angle between two points that are at the limit of being
resolved is
θ
λ
=
122
.
D
, so the shorter wavelengths will give a smaller angle because the diffraction
caused by the apperture of the telescope will be smaller.
The shorter wavelengths will provided
more detail so ultraviolet is better.
[36P19]
The intensity for the interference looks like:
4
2
0
2
4
0.2
0.4
0.6
0.8
1
Intensity for
d = 4a & d/¬ = 8
Note the
m = 4
fringe is missing.
a)
If there are only 5 fringes the
m = 3
fringe will be missing.
The first diffraction minimum for the single slit light (∫ = 2π) must be happening where the third
two source interference maximum is occuring ƒ = 3(2π).
Taking the ratio of the phases:
φ
π
=
2
d
sin
and
β
=
2
a
sin
õ
==
()
=
d
a
32
2
3
b)
The central maximum has a range
of 4π and contains two fringes on either side of the central fringe. The first
diffraction max to the
side of the central runs from ∫ = 2π  4π, so it must contain 2 fringes.
Look at the first drawing,
where it contains 3 fringes just as there are 3 fringes to either side of the central fringe.
Example a system where
d/a
= 5
2
2
4
6
8
10
12
0.2
0.4
0.6
0.8
1
Set #6, 11/12/04
Page 1 of 5
UC Davis
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View Full Document [
36.24]
The diffraction minima occur at
∫ =
n(2π)
and the interference max's at ƒ =
m
(2π).
As in
38.17
take the ratio
φ
β
π
==
()
=
d
a
m
n
2
2
3
õ
m
= 3
n
where
n = 1, 2, 3.
...
[38.31]
The maxima are given by
λ
θπ
2
2
dm
sin
õ
sin
θ
=
m
d
where
m
d
≤
1
.
There are two values of
m.
=
×
×
−
−
−
sin
(.
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This note was uploaded on 05/19/2008 for the course PHY 009B taught by Professor Cole during the Fall '07 term at UC Davis.
 Fall '07
 Cole
 Diffraction

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